Symbolic Logic

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Dec 10, 2013 ... An Accessible Introduction to Serious Mathematical Logic ... in mathematical logic with metalogical components often cast the barest glance at.
Symbolic Logic An Accessible Introduction to Serious Mathematical Logic Tony Roy version 7.5 November 19, 2017

Preface There is, I think, a gap between what many students learn in their first course in formal logic, and what they are expected to know for their second. While courses in mathematical logic with metalogical components often cast only the barest glance at mathematical induction or even the very idea of reasoning from definitions, a first course may also leave these untreated, and fail explicitly to lay down the definitions upon which the second course is based. The aim of this text is to integrate material from these courses and, in particular, to make serious mathematical logic accessible to students I teach. The first parts introduce classical symbolic logic as appropriate for beginning students; the last parts build to Gödel’s adequacy and incompleteness results. A distinctive feature of the last section is a complete development of Gödel’s second incompleteness theorem. Accessibility, in this case, includes components which serve to locate this text among others: First, assumptions about background knowledge are minimal. I do not assume particular content about computer science, or about mathematics much beyond high school algebra. Officially, everything is introduced from the ground up. No doubt, the material requires a certain sophistication — which one might acquire from other courses in critical reasoning, mathematics or computer science. But the requirement does not extend to particular contents from any of these areas. Second, I aim to build skills, and to keep conceptual distance for different applications of ‘so’ relatively short. Authors of books that are completely correct and precise may assume skills and require readers to recognize connections not fully explicit. It may be that this accounts for some of the reputed difficulty of the material. The results are often elegant. But this can exclude a class of students capable of grasping and benefiting from the material, if only it is adequately explained. Thus I attempt explanations and examples to put the student at every stage in a position to understand the next. In some cases, I attempt this by introducing relatively concrete methods for reasoning. The methods are, no doubt, tedious or unnecessary for the experienced logician. However, I have found that they are valued by students, insofar as students i

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are presented with an occasion for success. These methods are not meant to wash over or substitute for understanding details, but rather to expose and clarify them. Clarity, beauty and power come, I think, by getting at details, rather than burying or ignoring them. Third, the discussion is ruthlessly directed at core results. Results may be rendered inaccessible to students, who have many constraints on their time and schedules, simply because the results would come up in, say, a second course rather than a first. My idea is to exclude side topics and problems, and to go directly after (what I see as) the core. One manifestation is the way definitions and results from earlier sections feed into ones that follow. Thus simple integration is a benefit. Another is the way predicate logic with identity is introduced as a whole in Part I. Though it is possible to isolate sentential logic from the first parts of chapter 2 through chapter 7, and so to use the text for separate treatments of sentential and predicate logic, the guiding idea is to avoid repetition that would be associated with independent treatments for sentential logic, or perhaps monadic predicate logic, the full predicate logic, and predicate logic with identity. Also (though it may suggest I am not so ruthless about extraneous material as I would like to think), I try to offer some perspective about what is accomplished along the way. In addition, this text may be of particular interest to those who have, or desire, an exposure to natural deduction in formal logic. In this case, accessibility arises from the nature of the system, and association with what has come before. In the first part, I introduce both axiomatic and natural derivation systems; and in Part III, show how they are related. There are different ways to organize a course around this text. For students who are likely to complete the whole, the ideal is to proceed sequentially through the text from beginning to end (but postponing chapter 3 until after chapter 6). Taken as wholes, Part II depends on Part I; Parts III and IV on Parts I and II. Part IV is mostly independent of Part III. I am currently working within a sequence that isolates sentential logic from quantificational logic, treating them in separate quarters, together covering all of chapters 1 - 7 (except 3). A third course picks up leftover chapters from the first two parts (3 and 8) with Part III; and a fourth the leftover chapters from the first parts with Part IV. Perhaps not the most efficient arrangement, but the best I have been able to do with shifting student populations. Other organizations are possible! A remark about chapter 7 especially for the instructor: By a formal system for reasoning with semantic definitions, chapter 7 aims to leverage derivation skills from earlier chapters to informal reasoning with definitions. I have had a difficult time convincing instructors to try this material — and even been told flatly that these skills “cannot be taught.” In my experience, this is false (and when I have been

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able to convince others to try the chapter, they have quickly seen its value). Perhaps the difficulty is just that the strategy is unfamiliar. Of course, if one is presented with students whose mathematical sophistication is sufficient for advanced work, it is not necessary. But if, as is often the case especially for students in philosophy, one obtains one’s mathematical sophistication from courses in logic, this chapter is an important part of the bridge from earlier material to later. Additionally, the chapter is an important “take-away” even for students who will not continue to later material. The chapter closes an open question from chapter 4 — how it is possible to demonstrate quantificational validity. But further, the ability to reason closely with definitions is a skill from which students in (sentential or) predicate logic, even though they never go on to formalize another sentence or do another derivation, will benefit both in philosophy and more generally. Another remark about the (long) sections 13.3, 13.4 and 13.5. These develop in PA the “derivability conditions” for Gödel’s second theorem. They are perhaps for enthusiasts. Still, in my experience many students are enthusiasts and, especially from an introduction, benefit by seeing how the conditions are derived. There are different ways to treat the sections. One might work through them in some detail. However, even if you decide to pass them by, there is an advantage having a panorama at which to wave and say “thus it is accomplished!” Naturally, results in this book are not innovative. If there is anything original, it is in presentation. Even here, I am greatly indebted to others, especially perhaps Bergmann, Moor and Nelson, The Logic Book, Mendelson, Introduction to Mathematical Logic, and Smith, An Introduction to Gödel’s Theorems. I thank my first logic teacher, G.J. Mattey, who communicated to me his love for the material. And I thank especially my colleagues John Mumma and Darcy Otto for many helpful comments. Hannah Baehr and Catlin Andrade made comments and produced answers to exercises for certain parts. In addition I have received helpful feedback from Steve Johnson, along with students in different logic classes at CSUSB. I welcome comments, and expect that your sufferings will make it better still. This text evolved over a number of years starting modestly from notes originally provided as a supplement to other texts. It is now long (!) and perhaps best conceived in separate volumes for Parts I and II and then Parts III and IV. With the addition of Part IV it now complete. (But chapter 11, which I rarely get to in teaching, remains a stub that could be developed in different directions.) Most of the text is reasonably stable, though I shall be surprised if I have not introduced errors in the last part both substantive and otherwise. I think this is fascinating material, and consider it great reward when students respond “cool!” as they sometimes do. I hope you will have that response more than

PREFACE once along the way. T.R. Fall 2017

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Contents Preface

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Contents

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Named Definitions

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Quick Reference Guides

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I The Elements: Four Notions of Validity

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Logical Validity and Soundness 1.1 Consistent Stories . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 The Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Some Consequences . . . . . . . . . . . . . . . . . . . . . . . . .

5 6 12 24

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Formal Languages 2.1 Introductory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Sentential Languages . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Quantificational Languages . . . . . . . . . . . . . . . . . . . . . .

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Axiomatic Deduction 3.1 General . . . . . 3.2 Sentential . . . . 3.3 Quantificational . 3.4 Application: PA .

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Semantics 4.1 Sentential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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CONTENTS 4.2

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Quantificational . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

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Translation 140 5.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 5.2 Sentential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 5.3 Quantificational . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

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Natural Deduction 6.1 General . . . . . . . . 6.2 Sentential . . . . . . . 6.3 Quantificational . . . . 6.4 Applications: Q and PA 6.5 The System ND+ . . .

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II Transition: Reasoning About Logic

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Direct Semantic Reasoning 7.1 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Sentential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Quantificational . . . . . . . . . . . . . . . . . . . . . . . . . . . .

335 336 339 355

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Mathematical Induction 8.1 General Characterization . . . . . 8.2 Preliminary Examples . . . . . . . 8.3 Further Examples (for Part III) . . 8.4 Additional Examples (for Part IV)

382 382 388 402 412

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III Classical Metalogic: Soundness and Adequacy 9

Preliminary Results 9.1 Semantic Validity Implies Logical Validity 9.2 Validity in AD Implies Validity in ND . . 9.3 Validity in ND Implies Validity in AD . . 9.4 Extending to ND+ . . . . . . . . . . . .

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430 430 435 442 463

10 Main Results 468 10.1 Soundness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469 10.2 Sentential Adequacy . . . . . . . . . . . . . . . . . . . . . . . . . 476

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10.3 Quantificational Adequacy: Basic Version . . . . . . . . . . . . . . 487 10.4 Quantificational Adequacy: Full Version . . . . . . . . . . . . . . . 502 11 More Main Results 11.1 Expressive Completeness . . . . . 11.2 Unique Readability . . . . . . . . . 11.3 Independence . . . . . . . . . . . . 11.4 Isomorphic Models . . . . . . . . . 11.5 Compactness and Isomorphism . . . 11.6 Submodels and Löwenheim-Skolem

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IV Logic and Arithmetic: Incompleteness and Computability 12 Recursive Functions and Q 12.1 Recursive Functions . . . . . . . 12.2 Expressing Recursive Functions 12.3 Capturing Recursive Functions . 12.4 More Recursive Functions . . . 12.5 Essential Results . . . . . . . .

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13 Gödel’s Theorems 13.1 Gödel’s First Theorem . . . . . . . . . . . 13.2 Gödel’s Second Theorem: Overview . . . . 13.3 The Derivability Conditions: Background . 13.4 The Second Condition: .P ! Q/ ! .P 13.5 The Third Condition: P ! P . . . . 13.6 Reflections on the Theorem . . . . . . . . .

518 518 523 526 530 540 541

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554 556 564 575 592 613

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14 Logic and Computability 743 14.1 Turing Computable Functions . . . . . . . . . . . . . . . . . . . . 743 14.2 Essential Results . . . . . . . . . . . . . . . . . . . . . . . . . . . 757 14.3 Church’s Thesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 763 Concluding Remarks

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Answers to Selected Exercises 786 Chapter One . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 787 Chapter Two . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 792

CONTENTS Chapter Three . . Chapter Four . . Chapter Five . . . Chapter Six . . . Chapter Seven . . Chapter Eight . . Chapter Nine . . Chapter Ten . . . Chapter Eleven . Chapter Twelve . Chapter Thirteen Chapter Fourteen

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801 808 820 834 857 866 875 888 893 895 903 1010

Bibliography

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Index

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Named Definitions AR LV LS IT VT

chapter 1 Argument . . . . . . Logical Validity . . . Logical Soundness . Invalidity Test . . . . Validity Test . . . . .

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VC TR FR AB FR0

chapter 2 Vocabulary (sentential) . . . . . . Formulas (sentential) . . . . . . . Subformulas . . . . . . . . . . . . Immediate Subformula . . . . . . Atomic Subformula . . . . . . . . Main Operator (formal) . . . . . . Abbreviation (sentential) . . . . . Abbreviated Formulas (sentential) Vocabulary . . . . . . . . . . . . Terms . . . . . . . . . . . . . . . Formulas . . . . . . . . . . . . . Abbreviation . . . . . . . . . . . Abbreviated Formulas . . . . . .

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MP AV AS AU AE AD

chapter 3 Modus Ponens . . . . . Axiomatic Consequence AD Sentential . . . . . . AD Quantificational . . . AD Equality . . . . . . . AD Axioms (summary) .

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VC FR SB IS AS MO AB FR 0

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Named Definitions PA

x Peano Axioms . . . . . . . . . . . . . . . . . . . . .

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QI TA SF B(s) B(r) B() B(!) B(8) TI QV B(^) B(_) B($) B(9) SF0

chapter 4 Satisfaction as Truth (Sentential) . . Characteristic Table () . . . . . . Characteristic Table (!) . . . . . . Sentential Validity . . . . . . . . . Characteristic Table (_) . . . . . . . Characteristic Table (^) . . . . . . . Characteristic Table ($) . . . . . . Truth for Abbreviations (Sentential) Quantificational Interpretations . . . Term Assignment . . . . . . . . . . Satisfaction . . . . . . . . . . . . . Branch Condition (s) . . . . . . . . Branch Condition (r) . . . . . . . . Branch Condition () . . . . . . . . Branch Condition (!) . . . . . . . Branch Condition (8) . . . . . . . . Truth on an Interpretation . . . . . . Quantificational Validity . . . . . . Branch Condition (^) . . . . . . . . Branch Condition (_) . . . . . . . . Branch Condition ($) . . . . . . . Branch Condition (9) . . . . . . . . Satisfaction for Abbreviations . . .

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100 100 100 107 112 112 112 115 116 121 123 125 125 125 125 125 129 131 135 135 135 136 138

CG DC SO CS MO TF TP

chapter 5 Criterion of Goodness for Translation Declarative Sentences . . . . . . . . . Sentential Operator . . . . . . . . . . Compound and Simple . . . . . . . . Main Operator (informal) . . . . . . . Truth Functional Operator . . . . . . Translation Procedure . . . . . . . . .

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141 147 147 147 147 147 149

N1 SD FA

chapter 6 Natural Derivation One . . . . . . . . . . . . . . . . 211 Subderivation . . . . . . . . . . . . . . . . . . . . . 219 Accessible Formula . . . . . . . . . . . . . . . . . . 219

ST

T() T(!) SV T(_) T(^) T($) ST 0

Named Definitions SA R !E !I ^E ^I I E ?I _I _E $E $I SG SC

8E 9I 8I 9E SG SC DI DE (8E) (9I) (8I) (9E) Q IN PA7 ?E MT NB DS HS DN Com Assoc

xi Accessible Subderivation . . . . . . . . . ND Reiteration . . . . . . . . . . . . . . ND ! Exploitation . . . . . . . . . . . . ND ! Introduction . . . . . . . . . . . . ND ^ Exploitation . . . . . . . . . . . . ND ^ Introduction . . . . . . . . . . . . ND  Introduction . . . . . . . . . . . . ND  Exploitation . . . . . . . . . . . . ND ? Introduction . . . . . . . . . . . . ND _ Introduction . . . . . . . . . . . . ND _ Exploitation . . . . . . . . . . . . ND $ Exploitation . . . . . . . . . . . . ND $ Introduction . . . . . . . . . . . . Strategies for a Goal (Sentential) . . . . . Strategies for a Contradiction (Sentential) ND 8 Exploitation . . . . . . . . . . . . ND 9 Introduction . . . . . . . . . . . . . ND 8 Introduction . . . . . . . . . . . . ND 9 Exploitation . . . . . . . . . . . . . Strategies for a Goal . . . . . . . . . . . Strategies for a Contradiction . . . . . . . ND D Introduction . . . . . . . . . . . . ND D Exploitation . . . . . . . . . . . . (8) Exploitation . . . . . . . . . . . . . . (9) Introduction . . . . . . . . . . . . . . (8) Introduction . . . . . . . . . . . . . . (9) Exploitation . . . . . . . . . . . . . . Robinson Arithmetic Axioms . . . . . . . Mathematical Induction . . . . . . . . . . Peano Induction Axiom . . . . . . . . . . ND ? Exploitation . . . . . . . . . . . . ND+ Modus Tollens . . . . . . . . . . . ND+ Negated Biconditional . . . . . . . ND+ Disjunctive Syllogism . . . . . . . ND+ Hypothetical Syllogism . . . . . . . ND+ Double Negation . . . . . . . . . . ND+ Commutation . . . . . . . . . . . . ND+ Association . . . . . . . . . . . . .

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Named Definitions Idem Impl Trans DeM Exp Equiv Dist QN BQN ST (Í) T(Í) ST

com idm dem cnj dsj neg bot SV exs ins cnd bcnd abv ST 0

dst SF SF0 TI QV unv qn TA eq SF( R )

xii ND+ Idempotence . . . . . . . . . ND+ Implication . . . . . . . . . . ND+ Transposition . . . . . . . . . ND+ DeMorgan . . . . . . . . . . . ND+ Exportation . . . . . . . . . . ND+ Equivalence . . . . . . . . . . ND+ Distribution . . . . . . . . . . ND+ Quantifier Negation . . . . . . ND+ Bounded Quantifier Negation .

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chapter 7 Satisfaction for Stroke . . . . . . . . . . . . . Characteristic Table (Í) . . . . . . . . . . . . . Sentential Truth (formalized) . . . . . . . . . . Commutation (metalinguistic) . . . . . . . . . Idempotence (metalinguistic) . . . . . . . . . . DeMorgan (metalinguistic) . . . . . . . . . . . Conjunctive rules (metalinguistic) . . . . . . . Disjunctive rules (metalinguistic) . . . . . . . . Negation Rules (metalinguistic) . . . . . . . . Bottom Introduction (metalinguistic) . . . . . . Sentential Validity (formalized) . . . . . . . . Existential rules (metalinguistic) . . . . . . . . Inspection . . . . . . . . . . . . . . . . . . . . Conditional rules (metalinguistic) . . . . . . . Biconditional rules (metalinguistic) . . . . . . Abbreviation (metalinguistic) . . . . . . . . . . Abbreviations for Sentential Truth (formalized) Distribution (metalinguistic) . . . . . . . . . . Satisfaction (formalized) . . . . . . . . . . . . Abbreviations for Satisfaction (formalized) . . Truth on an Interpretation (formalized) . . . . . Quantificational Validity (formalized) . . . . . Universal rules (metalinguistic) . . . . . . . . . Quantifier negation (metalinguistic) . . . . . . Term Assignment (formalized) . . . . . . . . . Equality rules (metalinguistic) . . . . . . . . . Satisfaction for relation symbols (formalized) .

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338 339 340 340 340 340 340 340 340 340 343 343 343 348 348 348 348 350 355 355 357 357 357 358 360 360 361

Named Definitions

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SF(8) SF0 .9/ def

Satisfaction for 8 (formalized) . . . . . . . . . . . . 365 Satisfaction for 9 (formalized) . . . . . . . . . . . . 365 Definition (metalinguistic) . . . . . . . . . . . . . . 371

AI

chapter 8 Term Assignment on an Interpretation . . . . . . . . 415 chapter 9

Con ./ Max .?/ Scgt .??/

chapter 10 Consistency . . . . . . . . . . Core Thesis (sentential) . . . . Maximality . . . . . . . . . . Core Thesis (preliminary) . . Scapegoat Set . . . . . . . . . Core Thesis (quantificational)

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477 478 482 490 494 502

A1() A1(!) A2() A2(!) IS EE ST SM CT ES

chapter 11 Table for Independence () . Table for Independence (!) Table for Independence () . Table for Independence (!) Isomorphism . . . . . . . . Elementary Equivalence . . Satisfiability . . . . . . . . . Submodel . . . . . . . . . . Cantor’s Theorem . . . . . . Elementary Submodel . . .

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527 527 530 530 530 532 540 543 544 545

chapter 12 zero . . . . . . . . . . identity . . . . . . . . successor . . . . . . . Composition . . . . . n-place zero . . . . . . constant n . . . . . . . Recursion . . . . . . . plus . . . . . . . . . . times . . . . . . . . . factorial . . . . . . . . Recursion Theorem . .

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557 557 557 558 558 558 558 559 559 560 561

zero./ j idntk suc.x/

CM zeron .x1 : : : xn / y n

RC plus.x; y/ times.x; y/ fact.y/

RT

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Named Definitions RM RF PR power.x; y/

EXr EXf CP 0 pred.y/ subc.x; y/ absval.x - y/ sg.y/ csg.y/

Cf EQ.x; y/ LEQ.x; y/ LESS.x; y/ NEG.P.E x// DSJ.P.E x/; Q.Ey// IMP.P.E x/; Q.Ey// .9y  z/P.Ex; y/ .9y < z/P.Ex; y/ .8y  z/P.Ex; y/ .8y < z/P.Ex; y/ .y  z/P.Ex; y/ f.Ex/ FCTR.m; n/ PRIME.n/ pi.n/ exp.n; i/ len.n/ rm.m; n/ qt.m; n/ cncat.m; n/ VAR.n/ TERMSEQ.m; n/ TERM.n/ ATOMIC.n/

xiv regular minimization . . . Recursive Functions . . . primitive recursive . . . . power . . . . . . . . . . . Expressionr . . . . . . . . Expressionf . . . . . . . . Capture . . . . . . . . . . Delta Formulas . . . . . . predecessor . . . . . . . . subtraction with cutoff . . absolute value . . . . . . . sign . . . . . . . . . . . . converse sign . . . . . . . characteristic function . . equality . . . . . . . . . . less than or equal . . . . . less than . . . . . . . . . . negation . . . . . . . . . . disjunction . . . . . . . . implication . . . . . . . . exists less than or equal to exists less than . . . . . . all less than or equal to . . all less than . . . . . . . . bounded minimization . . definition by cases . . . . factor . . . . . . . . . . . prime . . . . . . . . . . . prime sequence . . . . . . prime exponent . . . . . . prime length . . . . . . . . remainder . . . . . . . . . quotient . . . . . . . . . . concatenation . . . . . . . variable . . . . . . . . . . term sequence . . . . . . . term . . . . . . . . . . . . atomic formula . . . . . .

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562 563 563 563 565 565 576 577 593 593 593 593 594 594 595 595 595 596 596 596 596 596 596 596 597 597 598 598 599 599 599 601 601 603 604 604 605 605

Named Definitions WFF.n/

xv

well-formed formula . . . . . . . formula sequence . . . . . . . . . PRFADS.m; n/ sentential proof . . . . . . . . . . AXIOMADS.n/ sentential axiom . . . . . . . . . . cnd.n; o/ conditional . . . . . . . . . . . . neg.n/ negation . . . . . . . . . . . . . . unv.v; n/ universal . . . . . . . . . . . . . . MP.m; n; o/ recursive modus ponens . . . . . . TSUBSEQ.m; n; t; v; s; u/ substitution sequence for terms . . TERMSUB.t; v; s; u/ substitution in terms . . . . . . . ATOMSUB.p; v; s; q/ substitution in atomics . . . . . . FSUBSEQ.m; n; p; v; s; q/ substitution sequence for formulas FORMSUB.p; v; s; q/ substitution in formulas . . . . . . formusb.p; v; s/ formsub (function) . . . . . . . . FREEt.t; v/ free in term . . . . . . . . . . . . FREEf.p; v/ free in formula . . . . . . . . . . SENT.n/ sentence . . . . . . . . . . . . . . FREEFOR.s; v; u/ free for . . . . . . . . . . . . . . FFSEQ.m; s; v; u/ free for sequence . . . . . . . . . AXIOMAD4.n/ axiom 4 . . . . . . . . . . . . . . AXIOMAD5.n/ axiom 5 . . . . . . . . . . . . . . GEN.m; n/ gen rule . . . . . . . . . . . . . . AXIOMAD6.n/ axiom 6 . . . . . . . . . . . . . . AXIOMAD7.n/ axiom 7 . . . . . . . . . . . . . . ICON.m; n; o/ immediate consequence . . . . . . AXIOMPA7.n/ PA axiom 7 . . . . . . . . . . . . AXIOMAD.n/ axiom of AD . . . . . . . . . . . PRFAD.m; n/ proof in AD . . . . . . . . . . . . AXIOMQ.n/ axiom of Q . . . . . . . . . . . . PRFQ.m; n/ proof in Q . . . . . . . . . . . . . AXIOMPA.n/ axiom of PA . . . . . . . . . . . . PRFPA.m; n/ proof in PA . . . . . . . . . . . . num.n/ number of numeral for value of n . FORMSEQ.m; n/

f .x; E y; E zE/ vQ.x; E v/ .y  z/Q.x; E y/

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605 605 606 606 606 606 606 606 608 608 609 609 609 609 610 610 610 610 610 611 611 611 611 612 612 612 613 613 613 613 613 613 615

chapter 13 composition . . . . . . . . . . . . . . . . . . . . . . 649 minimization . . . . . . . . . . . . . . . . . . . . . 650 bounded minimization . . . . . . . . . . . . . . . . 651

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xvi

rm qt ˇ suc.x/ zero./ j idntk .x1 : : : xj / -: j Pr Rp G d lcm plm maxp maxs h.i/ CF CO CR pred sg csg ex exc.m; n; i / val.n; i / val .m; n; i / gvar.n/ numseq.n/ †? LR

remainder . . . . . . . . quotient . . . . . . . . . beta function . . . . . . defined successor . . . . defined zero . . . . . . . defined identity function dot minus . . . . . . . . Factor . . . . . . . . . . Prime . . . . . . . . . . Rprime . . . . . . . . . Good . . . . . . . . . . least good . . . . . . . . lcm . . . . . . . . . . . plm . . . . . . . . . . . maxp . . . . . . . . . . maxs . . . . . . . . . . . h(i) . . . . . . . . . . . coordinate functions . . coordinate operator . . . coordinate relations . . . pred . . . . . . . . . . . sg . . . . . . . . . . . . csg . . . . . . . . . . . . ex . . . . . . . . . . . . exc . . . . . . . . . . . . val . . . . . . . . . . . . val* . . . . . . . . . . . number of variable n . . numseq . . . . . . . . . Sigma Star Formulas . . Löb Rule . . . . . . . .

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654 654 654 655 655 655 660 663 664 664 664 664 666 667 668 668 670 680 681 681 683 683 683 691 693 693 693 709 712 716 737

CT RL IR AC MKU

chapter 14 Church’s thesis . . . . . . . . . . . Language for recursion . . . . . . . Interpretation of recursive language Algorithmic computability . . . . . MKU Machine . . . . . . . . . . .

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Quick Reference Guides Negation and Quantity . . . . . . . . . . . . . . Countability . . . . . . . . . . . . . . . . . . . . Parts of a Formula . . . . . . . . . . . . . . . . . More on Countability . . . . . . . . . . . . . . . Grammar Quick Reference . . . . . . . . . . . . AD Quick Reference . . . . . . . . . . . . . . . Peano Arithmetic (AD) . . . . . . . . . . . . . . Semantics Quick Reference (Sentential) . . . . . Basic Notions of Set Theory . . . . . . . . . . . Semantics Quick Reference (quantificational) . . Definitions for Translation . . . . . . . . . . . . Cause and Conditional . . . . . . . . . . . . . . Definitions for Auxiliary Assumptions . . . . . . ND Quick Reference (Sentential) . . . . . . . . . ND Quick Reference (Quantificational) . . . . . . LNT reference . . . . . . . . . . . . . . . . . . . Robinson and Peano Arithmetic (ND) . . . . . . ND+ Quick Reference . . . . . . . . . . . . . . Metalinguistic Quick Reference (sentential) . . . Metalinguistic Quick Reference (quantificational) Theorems of Chapter 7 . . . . . . . . . . . . . . Induction Schemes . . . . . . . . . . . . . . . . First Theorems of Chapter 8 . . . . . . . . . . . Final Theorems of Chapter 8 . . . . . . . . . . . Theorems of Chapter 9 . . . . . . . . . . . . . . Some Arithmetic Relevant to Gödel Numbering . More Arithmetic Relevant to Gödel Numbering . Theorems of Chapter 10 . . . . . . . . . . . . . . xvii

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22 36 40 50 64 89 97 115 117 138 147 167 219 242 303 308 320 327 351 373 379 389 411 426 467 481 493 517

Quick Reference Guides Cantor’s Theorem . . . . . . . . The Recursion Theorem . . . . . Arithmetic for the Beta Function First Results of Chapter 12 . . . Final Results of Chapter 12 . . . Additional Theorems of PA . . . First theorems of chapter 13 . . Font conventions . . . . . . . . Second theorems of chapter 13 . Substitution Vectors . . . . . . . Final theorems of chapter 13 . . Simple Time Dilation . . . . . . Theorems of chapter 14 . . . . .

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Part I

The Elements: Four Notions of Validity

1

Introductory Symbolic logic is a tool for argument evaluation. In this part of the text we introduce the basic elements of that tool. Those parts are represented in the following diagram.

 Ordinary Arguments

-

Symbolic Language

Truth and Validity

@ @ @ R Metalogical @  Consideration

@ @ @ R @

Proof and Validity

The starting point is ordinary arguments. Such arguments come in various forms and contexts — from politics and ordinary living, to mathematics and philosophy. Here is a classic, simple case.

(A)

All men are mortal. Socrates is a man. Socrates is mortal.

This argument has premises listed above a line, with a conclusion listed below. The premises are supposed to demonstrate the conclusion. Here is another case which may seem less simple.

(B)

If the maid did it, then it was done with a revolver only if it was done in the parlor. But if the butler is innocent, then the maid did it unless it was done in the parlor. The maid did it only if it was done with a revolver, while the butler is guilty if it did happen in the parlor. So the butler is guilty.

It is fun to think about this; from the given evidence, it follows that the butler did it! Here is an argument that is both controversial and significant. 2

PART I.

(C)

THE ELEMENTS

3

There is evil. If god is good, then there is no evil unless god has morally sufficient reasons for allowing it. If god is both omnipotent and omniscient, then god does not have morally sufficient reasons for allowing evil. So god is not good, omnipotent and omniscient.

A being is omnipotent if it is all-powerful, and omniscient if all-knowing. This is a version of the famous “problem of evil” for traditional theism. It matters whether the conclusion is true! Roughly, an argument is good if it does what it is supposed to do, if its premises demonstrate the conclusion, and bad if they do not. So a theist (someone who believes in god) may hold that (C) is a bad argument, but an atheist (someone who does not believe in god) that it is good. We begin in chapter 1 with an account of success for ordinary arguments (the leftmost box). So we say what it is for an argument to be good or bad. This introduces us to the fundamental notions of logical validity and logical soundness. These will be our core concepts for argument evaluation. But just as it is one thing to know what integrity is, and another to know whether someone has it, so it is one thing to know what logical validity and logical soundness are, and another to know whether arguments have them. In some cases, it may be obvious. But others are not so clear — as, for example, cases (B) or (C) above, along with complex arguments in mathematics. Thus symbolic logic is introduced as a sort of machine or tool to identify validity and soundness. This machine begins with certain symbolic representations of ordinary arguments (the box second from the left). That is why it is symbolic logic. We introduce these representations in chapter 2, and translate from ordinary arguments to the symbolic representations in chapter 5. Once arguments have this symbolic representation, there are different methods of operating upon them. An account of truth and validity is developed for the symbolic representations in chapter 4 and chapter 7 (the upper box). On this account, truth and validity are associated with clearly defined criteria for their evaluation. And validity from this upper box implies logical validity for the ordinary arguments that are symbolically represented. Thus we obtain clearly defined criteria to identify the logical validity of arguments we care about. Evaluation of validity for the butler and evil cases is entirely routine given the methods from chapter 2, chapter 4 and chapter 5 — though the soundness of (C) will remain controversial! Accounts for proof and validity are developed for the symbolic representations in chapter 3 and chapter 6 (the lower box). Again, on this account, proof and validity are associated with clearly defined criteria for their evaluation. And validity from the lower box implies logical validity for the ordinary arguments that are symbolically represented. The result is another well-defined approach to the identification of

PART I.

THE ELEMENTS

4

logical validity. Evaluation of validity for the butler and evil cases is entirely routine given the methods from, say, chapter 2, chapter 3 and chapter 5, or alternatively, chapter 2, chapter 5 and chapter 6 — though, again, the soundness of (C) will remain controversial. These, then, are the elements of our logical “machine” — we start with the fundamental notion of logical validity; then there are symbolic representations of ordinary reasonings, along with approaches to evaluation from truth and validity, and from proof and validity. These elements are developed in this part. In later parts we turn to thinking about how these parts work together (the right-hand box). In particular, we begin thinking how to reason about logic (Part II), demonstrate that the same arguments come out valid by the truth method and by the proof method (Part III), and establish limits on application of logic and computing to arithmetic (Part IV). But first we have to say what the elements are. And that is the task we set ourselves in this part.

Chapter 1

Logical Validity and Soundness Symbolic logic is a tool or machine for the identification of argument goodness. It makes sense to begin, however, not with the machine, but by saying something about this argument goodness that the machinery is supposed to identify. That is the task of this chapter. But first, we need to say what an argument is. An argument is made up of sentences one of which is taken to be supported by the others. AR An argument is some sentences, one of which (the conclusion) is taken to be supported by the remaining sentences (the premises). (Important definitions are often offset and given a short name as above; then there may be appeal to the definition by its name, in this case, ‘AR’.) So an argument has premises which are taken to support a conclusion. Such support is often indicated by words or phrases of the sort, ‘so’, ‘it follows’, ‘therefore’, or the like. We will typically indicate the division by a simple line between premises and conclusion. Roughly, an argument is good if the premises do what they are taken to do, if they actually support the conclusion. An argument is bad if they do not accomplish what they are taken to do, if they do not actually support the conclusion. Logical validity and soundness correspond to different ways an argument can go wrong. Consider the following two arguments:

(A)

Only citizens can vote Hannah is a citizen

(B)

Hannah can vote

All citizens can vote Hannah is a citizen Hannah can vote

5

CHAPTER 1. LOGICAL VALIDITY AND SOUNDNESS

6

The line divides premises from conclusion, indicating that the premises are supposed to support the conclusion. Thus these are arguments. But these arguments go wrong in different ways. The premises of argument (A) are true; as a matter of fact, only citizens can vote, and Hannah (my daughter) is a citizen. But she cannot vote; she is not old enough. So the conclusion is false. Thus, in argument (A), the relation between the premises and the conclusion is defective. Even though the premises are true, there is no guarantee that the conclusion is true as well. We will say that this argument is logically invalid. In contrast, argument (B) is logically valid. If its premises were true, the conclusion would be true as well. So the relation between the premises and conclusion is not defective. The problem with this argument is that the premises are not true — not all citizens can vote. So argument (B) is defective, but in a different way. We will say that it is logically unsound. The task of this chapter is to define and explain these notions of logical validity and soundness. I begin with some preliminary notions, then turn to official definitions of logical validity and soundness, and finally to some consequences of the definitions.

1.1

Consistent Stories

Given a certain notion of a possible or consistent story, it is easy to state definitions for logical validity and soundness. So I begin by identifying the kind of stories that matter. Then we will be in a position to state the definitions, and apply them in some simple cases. Let us begin with the observation that there are different sorts of possibility. Consider, say, “Hannah could make it in the WNBA.” This seems true. She is reasonably athletic, and if she were to devote herself to basketball over the next few years, she might very well make it in the WNBA. But wait! Hannah is only a kid — she rarely gets the ball even to the rim from the top of the key — so there is no way she could make it in the WNBA. So she both could and could not make it. But this cannot be right! What is going on? Here is a plausible explanation: Different sorts of possibility are involved. When we hold fixed current abilities, we are inclined to say there is no way she could make it. When we hold fixed only general physical characteristics, and allow for development, it is natural to say that she might. The scope of what is possible varies with whatever constraints are in play. The weaker the constraints, the broader the range of what is possible. The sort of possibility we are interested in is very broad, and constraints are correspondingly weak. We will allow that a story is possible or consistent so long as it involves no internal contradiction. A story is impossible when it collapses from

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within. For this it may help to think about the way you respond to ordinary fiction. Consider, say, J.K. Rowling’s Harry Potter and the Prisoner of Azkaban (much loved by my youngest daughter). Harry and his friend Hermione are at wizarding school. Hermione acquires a “time turner” which allows time travel, and uses it in order to take classes that are offered at the same time. Such devices are no part of the actual world, but they fit into the wizarding world of Harry Potter. So far, then, the story does not contradict itself. So you go along. At one stage, though, Harry is at a lakeshore under attack by a bunch of fearsome “dementors.” His attempts to save himself appear to have failed when a figure across the lake drives the dementors away. But the figure who saves Harry is Harry himself who has come back from the future. Somehow, then, as often happens in these stories, the past depends on the future, at the same time as the future depends on the past: Harry is saved only insofar as he comes back from the future, but he comes back from the future only insofar as he is saved. This, rather than the time travel itself, generates an internal conflict. The story makes it the case that you cannot have Harry’s rescue apart from his return, and cannot have Harry’s return apart from his rescue. This might make sense if time were always repeating in an eternal loop. But, according to the story, there were times before the rescue and after the return. So the story faces internal collapse. Notice: the objection does not have anything to do with the way things actually are — with existence of time turners or the like; it has rather to do with the way the story hangs together internally.1 Similarly, we want to ask whether stories hold together internally. If a story holds together internally, it counts for our purposes as consistent and possible. If a story does not hold together, it is not consistent or possible. In some cases, stories may be consistent with things we know are true in the real world. Thus Hannah could grow up to play in the WNBA. There is nothing about our world that rules this out. But stories may remain consistent though they do not fit with what we know to be true in the real world. Here are cases of time travel and the like. Stories become inconsistent when they collapse internally — as when a story says that some time both can and cannot happen apart from another. As with a movie or novel, we can say that different things are true or false in our 1 In

more consistent cases of time travel (in fiction) time seems to move on different paths so that after yesterday and today, there is another yesterday and another today. So time does not return to the very point at which it first turns back. In the trouble cases, time seems to move in a sort of “loop” so that a point on the path to today (this very day) goes through tomorrow. With this in mind, it is interesting to think about say, the Terminator and Back to the Future movies along with, maybe more consistent, Hermione’s “academic” travel or Groundhog Day. Even if I am wrong, and the Potter story is internally consistent, the overall point should be clear. And it should be clear that I am not saying anything serious about time travel.

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stories. In Harry Potter it is true that Harry and Hermione travel through time with a timer turner, but false that they go through time in a DeLorean (as in the Back to the Future films). In the real world, of course, it is false that there are time turners, and false that DeLoreans go through time. Officially, a complete story is always maximal in the sense that any sentence is either true or false in it. A story is inconsistent when it makes some sentence both true and false. Since, ordinarily, we do not describe every detail of what is true and what is false when we tell a story, what we tell is only part of a maximal story. In practice, however, it will be sufficient for us merely to give or fill in whatever details are relevant in a particular context. But there are a couple of cases where we cannot say when sentences are true or false in a story. The first is when stories we tell do not fill in relevant details. In The Wizard of Oz, it is true that Dorothy wears red shoes. But neither the movie nor the book have anything to say about whether she likes Twinkies. By themselves, then, neither the book nor the movie give us enough information to tell whether “Dorothy likes Twinkies” is true or false in the story. Similarly, there is a problem when stories are inconsistent. Suppose according to some story, (a) All dogs can fly (b) Fido is a dog (c) Fido cannot fly Given (a), all dogs fly; but from (b) and (c), it seems that not all dogs fly. Given (b), Fido is a dog; but from (a) and (c) it seems that Fido is not a dog. Given (c), Fido cannot fly; but from (a) and (b) it seems that Fido can fly. The problem is not that inconsistent stories say too little, but rather that they say too much. When a story is inconsistent, we will refuse to say that it makes any sentence (simply) true or false.2 It will be be helpful to consider some examples of consistent and inconsistent stories: (a) The real story, “Everything is as it actually is.” Since no contradiction is actually true, this story involves no contradiction; so it is internally consistent and possible. (b) “All dogs can fly: over the years, dogs have developed extraordinarily large and muscular ears; with these ears, dogs can fly.” It is bizarre, but not obviously inconsistent. If we allow the consistency of stories according to which monkeys fly, 2 The intuitive picture developed above should be sufficient for our purposes. However, we are on the verge of vexed issues. For further discussion, you may want to check out the vast literature on “possible worlds.” Contributions of my own include the introductory article, “Modality,” in The Continuum Companion to Metaphysics.

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as in The Wizard of Oz, or elephants fly, as in Dumbo, then we should allow that this story is consistent as well. (c) “All dogs can fly, but my dog Fido cannot; Fido’s ear was injured while he was chasing a helicopter, and he cannot fly.” This is not internally consistent. If all dogs can fly and Fido is a dog, then Fido can fly. You might think that Fido retains a sort of flying nature — just because Fido remains a dog. In evaluating internal consistency, however, we require that meanings remain the same. F 

(C)

able to fly



All dogs can fly Q T

Q Q Q  

Fido is a dog

 Q Q Q TQ

Fido cannot fly

flying nature

 F

If “can fly” means “is able to fly” then in the story it is true that Fido cannot fly, but not true that all dogs can fly (since Fido cannot). If “can fly” means “has a flying nature” then in the story it is true that all dogs can fly, but not true that Fido cannot (because he remains a dog). The only way to keep both ‘all dogs fly’ and ‘Fido cannot fly’ true is to switch the sense of “can fly” from one use to another. So long as “can fly” means the same in each use, the story is sure to fall apart insofar as it says both that Fido is and is not that sort of thing. (d) “Germany won WWII; the United States never entered the war; after a long and gallant struggle, England and the rest of Europe surrendered.” It did not happen; but the story does not contradict itself. For our purposes, then it counts as possible. (e) “1 1 D 3; the numerals ‘2’ and ‘3’ are switched (the numerals are ‘1’, ‘3’, ‘2’, ‘4’, ‘5’, ‘6’, ‘7’. . . ); so that one and one are three.” This story does not hang together. Of course numerals can be switched — so that people would correctly say, ‘1 1 D 3’. But this does not make it the case that one and one are three! We tell stories in our own language (imagine that you are describing a foreign-language film in English). Take a language like English except that ‘fly’ means ‘bark’; and consider a movie where dogs are ordinary, so that people in the movie correctly assert, in their language, ‘dogs fly’. But changing the words people use to describe a situation does not change the situation. It would be a mistake to tell a friend, in English, that you saw a movie in which there were flying dogs. Similarly, according to our story, people correctly assert, in their language, ‘1 1 D 3’. But it is a mistake to say in English (as our story does), that this makes one and one equal to three.

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Some authors prefer talk of “possible worlds,” “possible situations” or the like to that of consistent stories. It is conceptually simpler to stick with stories, as I have, than to have situations and distinct descriptions of them. However, it is worth recognizing that our consistent stories are or describe possible situations, so that the one notion matches up directly with the others. As you approach the following exercises, note that answers to problems indicated by star are provided in the back of the book. It is essential to success that you work a significant body of exercises successfully and independently. So do not neglect exercises! E1.1. Say whether each of the following stories is internally consistent or inconsistent. In either case, explain why. *a. Smoking cigarettes greatly increases the risk of lung cancer, although most people who smoke cigarettes do not get lung cancer. b. Joe is taller than Mary, but Mary is taller than Joe. *c. Abortion is always morally wrong, though abortion is morally right in order to save a woman’s life. d. Mildred is Dr. Saunders’s daughter, although Dr. Saunders is not Mildred’s father. *e. No rabbits are nearsighted, though some rabbits wear glasses. f. Ray got an ‘A’ on the final exam in both Phil 200 and Phil 192. But he got a ‘C’ on the final exam in Phil 192. *g. Barack Obama was never president of the United States, although Michelle is president right now. h. Egypt, with about 100 million people is the most populous country in Africa, and Africa contains the most populous country in the world. But the United States has over 200 million people. *i. The death star is a weapon more powerful than that in any galaxy, though there is, in a galaxy far, far away, a weapon more powerful than it. j. Luke and the Rebellion valiantly battled the evil Empire, only to be defeated. The story ends there.

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E1.2. For each of the following sentences, (i) say whether it is true or false in the real world and then (ii) say, if you can, whether it is true or false according to the accompanying story. In each case, explain your answers. Do not forget about contexts where we refuse to say whether sentences are true or false. The first problem is worked as an example. a. Sentence: Aaron Burr was never a president of the United States. Story: Aaron Burr was the first president of the United States, however he turned traitor and was impeached and then executed. (i) It is true in the real world that Aaron Burr was never a president of the United States. (ii) But the story makes the sentence false, since the story says Burr was the first president. b. Sentence: In 2006, there were still buffalo. Story: A thundering herd of buffalo overran Phoenix, Arizona in early 2006. The city no longer exists. *c. Sentence: After overrunning Phoenix in early 2006, a herd of buffalo overran Newark, New Jersey. Story: A thundering herd of buffalo overran Phoenix, Arizona in early 2006. The city no longer exists. d. Sentence: There has been an all-out nuclear war. Story: After the all-out nuclear war, John Connor organized the Resistance against the machines — who had taken over the world for themselves. *e. Sentence: Jack Nicholson has swum the Atlantic. Story: No human being has swum the Atlantic. Jack Nicholson and Bill Clinton and you are all human beings, and at least one of you swam all the way across! f. Sentence: Some people have died as a result of nuclear explosions. Story: As a result of a nuclear blast that wiped out most of this continent, you have been dead for over a year. *g. Sentence: Your instructor is not a human being. Story: No beings from other planets have ever made it to this country. However, your instructor made it to this country from another planet.

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h. Sentence: Lassie is both a television and movie star. Story: Dogs have super-big ears and have learned to fly. Indeed, all dogs can fly. Among the many dogs are Lassie and Rin Tin Tin. *i. Sentence: The Yugo is the most expensive car in the world. Story: Jaguar and Rolls Royce are expensive cars. But the Yugo is more expensive than either of them. j. Sentence: Lassie is a bird who has learned to fly. Story: Dogs have super-big ears and have learned to fly. Indeed, all dogs can fly. Among the many dogs are Lassie and Rin Tin Tin.

1.2

The Definitions

The definition of logical validity depends on what is true and false in consistent stories. The definition of soundness builds directly on the definition of validity. Note: in offering these definitions, I stipulate the way the terms are to be used; there is no attempt to say how they are used in ordinary conversation; rather, we say what they will mean for us in this context. LV An argument is logically valid if and only if (iff) there is no consistent story in which all the premises are true and the conclusion is false. LS An argument is logically sound iff it is logically valid and all of its premises are true in the real world. Observe that logical validity has entirely to do with what is true and false in consistent stories. Only with logical soundness is validity combined with premises true in the real world. Logical (deductive) validity and soundness are to be distinguished from inductive validity and soundness or success. For the inductive case, it is natural to focus on the plausibility or the probability of stories — where an argument is relatively strong when stories that make the premises true and conclusion false are relatively implausible. Logical (deductive) validity and soundness are thus a sort of limiting case, where stories that make premises true and conclusion false are not merely implausible, but impossible. In a deductive argument, conclusions are supposed to be guaranteed; in an inductive argument, conclusions are merely supposed to be made probable or plausible. For mathematical logic, we set the inductive case to the side, and focus on the deductive.

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Also, do not confuse truth with validity and soundness. A sentence is true in the real world when it correctly represents how things are in the real world, and true in a story when it correctly represents how things are in the story. An argument is valid when there is no consistent story that makes the premises true and conclusion false, and sound when it is valid and all its premises are true in the real world. The definitions for validity and soundness depend on truth and falsity for the premises and conclusion in stories and then in the real world. But, just as it would be a mistake to say that the number three weighs eleven pounds, so truth and falsity do not even apply to arguments themselves, which may be valid or sound.3

1.2.1

Invalidity

It will be easiest to begin thinking about invalidity. From the definition, if an argument is logically valid, there is no consistent story that makes the premises true and conclusion false. So to show that an argument is invalid, it is enough to produce even one consistent story that makes premises true and conclusion false. Perhaps there are stories that result in other combinations of true and false for the premises and conclusion; this does not matter for the definition. However, if there is even one story that makes premises true and conclusion false then, by definition, the argument is not logically valid — and if it is not valid, by definition, it is not logically sound. We can work through this reasoning by means of a simple invalidity test. Given an argument, this test has the following four stages. IT

a. List the premises and negation of the conclusion. b. Produce a consistent story in which the statements from (a) are all true. c. Apply the definition of validity. d. Apply the definition of soundness.

We begin by considering what needs to be done to show invalidity. Then we do it. Finally we apply the definitions to get the results. For a simple example, consider the following argument,

(D)

Eating brussels sprouts results in good health Ophelia has good health Ophelia has been eating brussels sprouts

3 From an introduction to philosophy of language, one might wonder (with good reason) whether the proper bearers of truth are sentences rather than, say, propositions. This question is not relevant to the simple point made above.

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The definition of validity has to do with whether there are consistent stories in which the premises are true and the conclusion false. Thus, in the first stage, we simply write down what would be the case in a story of this sort. a. List premises and negation of conclusion.

In any story with the premises true and conclusion false, 1. Eating brussels sprouts results in good health 2. Ophelia has good health 3. Ophelia has not been eating brussels sprouts

Observe that the conclusion is reversed! At this stage we are not giving an argument. Rather we merely list what is the case when the premises are true and conclusion false. Thus there is no line between premises and the last sentence, insofar as there is no suggestion of support. It is easy enough to repeat the premises for (1) and (2). Then for (3) we say what is required for the conclusion to be false. Thus, “Ophelia has been eating brussels sprouts” is false if Ophelia has not been eating brussels sprouts. I return to this point below, but that is enough for now. An argument is invalid if there is even one consistent story that makes the premises true and the conclusion false. Thus, to show invalidity, it is enough to produce a consistent story that makes the premises true and conclusion false. b. Produce a consistent story in which the statements from (a) are all true.

Story: Eating brussels sprouts results in good health, but eating spinach does so as well; Ophelia is in good health but has been eating spinach, not brussels sprouts.

For the statements listed in (a): the story satisfies (1) insofar as eating brussels sprouts results in good health; (2) is satisfied since Ophelia is in good health; and (3) is satisfied since Ophelia has not been eating brussels sprouts. The story explains how she manages to maintain her health without eating brussels sprouts, and so the consistency of (1) - (3) together. The story does not have to be true — and, of course, many different stories will do. All that matters is that there is a consistent story in which the premises of the original argument are true, and the conclusion is false. Producing a story that makes the premises true and conclusion false is the creative part. What remains is to apply the definitions of validity and soundness. By LV, an argument is logically valid only if there is no consistent story in which the premises are true and the conclusion is false. So if, as we have demonstrated, there is such a story, the argument cannot be logically valid.

CHAPTER 1. LOGICAL VALIDITY AND SOUNDNESS c. Apply the definition of validity.

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This is a consistent story that makes the premises true and the conclusion false; thus, by definition, the argument is not logically valid.

By LS, for an argument to be sound, it must have its premises true in the real world and be logically valid. Thus if an argument fails to be logically valid, it automatically fails to be logically sound. d. Apply the definition of soundness.

Since the argument is not logically valid, by definition, it is not logically sound.

Given an argument, the definition of validity depends on stories that make the premises true and the conclusion false. Thus, in step (a) we simply list claims required of any such story. To show invalidity, in step (b), we produce a consistent story that satisfies each of those claims. Then in steps (c) and (d) we apply the definitions to get the final results; for invalidity, these last steps are the same in every case. It may be helpful to think of stories as a sort of “wedge” to pry the premises of an argument off its conclusion. We pry the premises off the conclusion if there is a consistent way to make the premises true and the conclusion not. If it is possible to insert such a wedge between the premises and conclusion, then a defect is exposed in the way premises are connected to the conclusion. Observe that the flexibility we allow in consistent stories (with flying dogs and the like) corresponds directly to the strength of the required connection between premises and conclusion. If the connection is sufficient to resist all such attempts to wedge the premises off the conclusion, then it is significant indeed. Observe also that our method reflects what we did with argument (A) at the beginning of the chapter: Faced with the premises that only citizens can vote and Hannah is a citizen, it was natural to worry that she might be under-age and so cannot vote. But this is precisely to produce a story that makes the premises true and conclusion false. Thus our method is not “strange” or “foreign”! Rather, it makes rigorous what has seemed natural from the start. Here is another example of our method. Though the argument may seem on its face not to be a very good one, we can expose its failure by our methods — in fact, again, our method may formalize or make rigorous a way you very naturally think about cases of this sort. Here is the argument, I shall run for president (E) I shall be one of the most powerful men on earth To show that the argument is invalid, we turn to our standard procedure.

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a. In any story with the premise true and conclusion false, 1. I shall run for president 2. I shall not be one of the most powerful men on earth b. Story: I do run for president, but get no financing and gain no votes; I lose the election. In the process, I lose my job as a professor and end up begging for scraps outside a Domino’s Pizza restaurant. I fail to become one of the most powerful men on earth. c. This is a consistent story that makes the premise true and the conclusion false; thus, by definition, the argument is not logically valid. d. Since the argument is not logically valid, by definition, it is not logically sound. This story forces a wedge between the premise and the conclusion. Thus we use the definition of validity to explain why the conclusion does not properly follow from the premises. It is, perhaps, obvious that running for president is not enough to make me one of the most powerful men on earth. Our method forces us to be very explicit about why: running for president leaves open the option of losing, so that the premise does not force the conclusion. Once you get used to it, then, our method may appear as a natural approach to arguments. If you follow this method for showing invalidity, the place where you are most likely to go wrong is stage (b), telling stories where the premises are true and the conclusion false. Be sure that your story is consistent, and that it verifies each of the claims from stage (a). If you do this, you will be fine. E1.3. Use our invalidity test to show that each of the following arguments is not logically valid, and so not logically sound. Understand terms in their most natural sense. *a. If Joe works hard, then he will get an ‘A’ Joe will get an ‘A’ Joe works hard b. Harry had his heart ripped out by a government agent Harry is dead c. Everyone who loves logic is happy Jane does not love logic Jane is not happy

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d. Our car will not run unless it has gasoline Our car has gasoline Our car will run e. Only citizens can vote Hannah is a citizen Hannah can vote

1.2.2

Validity

Suppose I assert that no student at California State University San Bernardino is from Beverly Hills, and attempt to prove it by standing in front of the library and buttonholing students to ask if they are from Beverly Hills — I do this for a week and never find anyone from Beverly Hills. Is the claim that no CSUSB student is from Beverly Hills thereby proved? Of course not – for there may be students I never meet. Similarly, failure to find a story to make the premises true and conclusion false does not show that there is not one — for all we know, there might be some story we have not thought of yet. So, to show validity, we need another approach. If we could show that every story which makes the premises true and conclusion false is inconsistent, then we could be sure that no consistent story makes the premises true and conclusion false — and so, from the definition of validity, we could conclude that the argument is valid. Again, we can work through this by means of a procedure, this time a validity test. VT

a. List the premises and negation of the conclusion. b. Expose the inconsistency of such a story. c. Apply the definition of validity. d. Apply the definition of soundness.

In this case, we begin in just the same way. The key difference arises at stage (b). For an example, consider this argument.

(F)

No car is a person My mother is a person My mother is not a car

Since LV has to do with stories where the premises are true and the conclusion false, as before, we begin by listing the premises together with the negation of the conclusion.

CHAPTER 1. LOGICAL VALIDITY AND SOUNDNESS a. List premises and negation of conclusion.

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In any story with the premises true and conclusion false, 1. No car is a person 2. My mother is a person 3. My mother is a car

Any story where “My mother is not a car” is false, is one where my mother is a car (perhaps along the lines of the 1965 TV series, My Mother the Car). For invalidity, we would produce a consistent story in which (1) - (3) are all true. In this case, to show that the argument is valid, we show that this cannot be done. That is, we show that no story that makes each of (1) - (3) true is a consistent story. b. Expose the inconsistency of such a story.

In any such story, Given (1) and (3), 4. My mother is not a person Given (2) and (4), 5. My mother is and is not a person

The reasoning should be clear if you focus just on the specified lines. Given (1) and (3), if no car is a person and my mother is a car, then my mother is not a person. But then my mother is a person from (2) and not a person from (4). So we have our goal: any story with (1) - (3) as members contradicts itself and therefore is not consistent. Observe that we could have reached this result in other ways. For example, we might have reasoned from (1) and (2) that (40 ), my mother is not a car; and then from (3) and (40 ) to the result that (50 ) my mother is and is not a car. Either way, an inconsistency is exposed. Thus, as before, there are different options for this creative part. Now we are ready to apply the definitions of logical validity and soundness. First, c. Apply the definition of validity.

So no consistent story makes the premises true and conclusion false; so by definition, the argument is logically valid.

For the invalidity test, we produce a consistent story that “hits the target” from stage (a), to show that the argument is invalid. For the validity test, we show that any attempt to hit the target from stage (a) must collapse into inconsistency: no consistent story includes each of the elements from stage (a) so that there is no consistent story in which the premises are true and the conclusion false. So by application of LV the argument is logically valid. Given that the argument is logically valid, LS makes logical soundness depend on whether the premises are true in the real world. Suppose we think the premises of our argument are in fact true. Then,

CHAPTER 1. LOGICAL VALIDITY AND SOUNDNESS d. Apply the definition of soundness.

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In the real world no car is a person and my mother is a person, so all the premises are true; so since the argument is also logically valid, by definition, it is logically sound.

Observe that LS requires for logical soundness that an argument is logically valid and that its premises are true in the real world. Thus we are no longer thinking about merely possible stories! Soundness depends on the way things are in the real world. And we do not say anything at this stage about claims other than the premises of the original argument! Thus we do not make any claim about the truth or falsity of the conclusion, “my mother is not a car.” Rather, the observations have entirely to do with the two premises, “no car is a person” and “my mother is a person.” When an argument is valid and the premises are true in the real world, by LS, it is logically sound. But it will not always be the case that a valid argument has true premises. Say My Mother the Car is (surprisingly) a documentary about a person reincarnated as a car (the premise of the show) and therefore a true account of some car that is a person. Then some cars are persons and the first premise is false; so you would have to respond as follows, d0 . Since in the real world some cars are persons, the first premise is not true. So, though the argument is logically valid, by definition it is not logically sound. Another option is that you are in doubt about reincarnation into cars, and in particular about whether some cars are persons. In this case you might respond as follows, d00 . Although in the real world my mother is a person, I cannot say whether no car is a person; so I cannot say whether the first premise is true. So though the argument is logically valid, I cannot say whether it is logically sound. So once we decide that an argument is valid, for soundness there are three options: (i) You are in a position to identify all of the premises as true in the real world. In this case, you should do so, and apply the definition for the result that the argument is logically sound. (ii) You are in a position to say that at least one of the premises is false in the real world. In this case, you should do so, and apply the definition for the result that the argument is not logically sound.

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(iii) You cannot identify any premise as false, but neither can you identify them all as true. In this case, you should explain the situation and apply the definition for the result that you are not in a position to say whether the argument is logically sound. So given a valid argument, there remains a substantive questions about soundness. In some cases, as for example (C) on p. 3, this can be the most controversial part. Again, given an argument, we say in step (a) what would be the case in any story that makes the premises true and the conclusion false. Then, at step (b), instead of finding a consistent story in which the premises are true and conclusion false, we show that there is no such thing. Steps (c) and (d) apply the definitions for the final results. Observe that only one method can be correctly applied in a given case! If we can produce a consistent story according to which the premises are true and the conclusion is false, then it is not the case that no consistent story makes the premises true and the conclusion false. Similarly, if no consistent story makes the premises true and the conclusion false, then we will not be able to produce a consistent story that makes the premises true and the conclusion false. For showing validity, the most difficult steps are (a) and (b), where we say what happens in every story where the premises true and the conclusion false. For an example, consider the following argument.

(G)

All collies can fly All collies are dogs All dogs can fly

It is invalid. We can easily tell a story that makes the premises true and the conclusion false — say one where collies fly but dachshunds do not. Suppose, however, that we proceed with the validity test as follows, a. In any story with the premises true and conclusion false, 1. All collies can fly 2. All collies are dogs 3. No dogs can fly b. In any such story, Given (1) and (2), 4. Some dogs can fly Given (3) and (4), 5. Some dogs can and cannot fly

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c. So no consistent story makes the premises true and conclusion false; so by definition, the argument is logically valid. d. Since in the real world collies cannot fly, the first premise is not true. So, though the argument is logically valid, by definition it is not logically sound. The reasoning at (b), (c) and (d) is correct. Any story with (1) - (3) is inconsistent. But something is wrong. (Can you see what?) There is a mistake at (a): It is not the case that every story that makes the premises true and conclusion false includes (3). The negation of “All dogs can fly” is not “No dogs can fly,” but rather, “Not all dogs can fly” (or “Some dogs cannot fly”). All it takes to falsify the claim that all dogs fly is some dog that does not. Thus, for example, all it takes to falsify the claim that everyone will get an ‘A’ is one person who does not (on this, see the extended discussion on p. 22). So for argument (G) we have indeed shown that every story of a certain sort is inconsistent, but have not shown that every story which makes the premises true and conclusion false is inconsistent. In fact, as we have seen, there are consistent stories that make the premises true and conclusion false. Similarly, in step (b) it is easy to get confused if you consider too much information at once. Ordinarily, if you focus on sentences singly or in pairs, it will be clear what must be the case in every story including those sentences. It does not matter which sentences you consider in what order, so long as in the end, you reach a contradiction according to which something is and is not so. So far, we have seen our procedures applied in contexts where it is given ahead of time whether an argument is valid or invalid. But not all situations are so simple. In the ordinary case, it is not given whether an argument is valid or invalid. In this case, there is no magic way to say ahead of time which of our two tests, IT or VT applies. The only thing to do is to try one way — if it works, fine. If it does not, try the other. It is perhaps most natural to begin by looking for stories to pry the premises off the conclusion. If you can find a consistent story to make the premises true and conclusion false, the argument is invalid. If you cannot find any such story, you may begin to suspect that the argument is valid. This suspicion does not itself amount to a demonstration of validity! But you might try to turn your suspicion into such a demonstration by attempting the validity method. Again, if one procedure works, the other better not! E1.4. Use our validity procedure to show that each of the following is logically valid, and decide (if you can) whether it is logically sound.

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Negation and Quantity In general you want to be careful about negations. To negate any claim P it is always correct to write simply, it is not the case that P . You may choose to do this for conclusions in the first step of our procedures. At some stage, however, you will need to understand what the negation comes to. We have chosen to offer interpreted versions in the text. It is easy enough to see that, My mother is a car

and

My mother is not a car

negate one another. However, there are cases where caution is required. This is particularly the case with terms involving quantities. Say the conclusion of your argument is, ‘there are at least ten apples in the basket’. Clearly a story according to which there are, say, three apples in the basket makes this conclusion false. However, there are other ways to make the conclusion false — as if there are two apples or seven. Any of these are fine for showing invalidity. But when you show that an argument is valid, you must show that any story that makes the premises true and conclusion false is inconsistent. So it is not sufficient to show that stories with (the premises true and) three apples in the basket contradict. Rather, you need to show that any story that includes the premises and less than ten apples fails. Thus in step (a) of our procedure we always say what is so in every story that makes the premises true and conclusion false. So, in (a) you would have the premises and, ‘there are less than ten apples in the basket’. If a statement is included in some range of consistent stories, then its negation says what is so in all the others — all the ones where it is not so. not-P

all consistent stories

P

That is why the negation of ‘there are at least ten’ is ‘there are less than ten’. The same point applies with other quantities. Consider some grade examples: First, if a professor says, “everyone will not get an ‘A’,” she says something disastrous — nobody in your class will get an ‘A’. In order too deny it, to show that she is wrong, all you need is at least one person that gets an ‘A’. In contrast, if she says, “someone will not get an ‘A’,” she says only what you expect from the start — that not everyone will get an ‘A’. To deny this, you would need that everyone gets an ‘A’. Thus the following pairs negate one another. Everyone will not get an ‘A’

and

Someone will get an ‘A’

Someone will not get an ‘A’

and

Everyone will get an ‘A’

It is difficult to give rules to cover all the cases. The best is just to think about what you are saying, perhaps with reference to examples like these.

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*a. If Bill is president, then Hillary is first lady Hillary is not first lady Bill is not president b. Only fools find love Elvis was no fool Elvis did not find love c. If there is a good and omnipotent god, then there is no evil There is evil There is no good and omnipotent god d. All sparrows are birds All birds fly All sparrows fly e. All citizens can vote Hannah is a citizen Hannah can vote E1.5. Use our procedures to say whether the following are logically valid or invalid, and sound or unsound. Hint: You may have to do some experimenting to decide whether the arguments are logically valid or invalid — and so decide which procedure applies. a. If Bill is president, then Hillary is first lady Bill is president Hillary is first lady b. Most professors are insane TR is a professor TR is insane *c. Some dogs have red hair Some dogs have long hair Some dogs have long, red hair

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d. If you do not strike the match, then it does not light The match lights You strike the match e. Shaq is taller than Kobe Kobe is at least as tall as TR Kobe is taller than TR

1.3

Some Consequences

We now know what logical validity and soundness are, and should be able to identify them in simple cases. Still, it is one thing to know what validity and soundness are, and another to know why they matter. So in this section I turn to some consequences of the definitions.

1.3.1

Soundness and Truth

First, a consequence we want: The conclusion of every sound argument is true in the real world. Observe that this is not part of what we require to show that an argument is sound. LS requires just that an argument is valid and that its premises are true. However it is a consequence of validity plus true premises that the conclusion is true as well. sound



valid true premises true conclusion

To see this, consider a two-premise argument. Say the real story describes the real world; so the sentences of the real story are all true in the real world. Then in the real story, the premises and conclusion of our argument must fall into one of the following combinations of true and false: 1 T T T

2 T T F

3 T F T

4 F T T

5 T F F

6 F T F

7 F F T

8 F F F

combinations for the real story

These are all the combinations of T and F. Say the argument is logically valid; then no consistent story makes the premises true and the conclusion false. But the real story is a consistent story. So we can be sure that the real story does not result in combination (2). So far, the real story might result in any of the other combinations. Thus the

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conclusion of a valid argument may or may not be true in the real world. Now say the argument is sound; then it is valid and all its premises are true in the real world. Again, since it is valid, the real story does not result in combination (2). And since the premises of a sound argument are true in the real world, we can be sure that the premises do not fall into any of the combinations (3) - (8). (1) is the only combination left: in the real story, and so in the real world, the conclusion of a sound argument is true. And not only in this case but in general, if an argument is sound its conclusion is true in the real world: Since a sound argument is valid, there is no consistent story where its premises are true and the conclusion is false, and since the premises really are true, the conclusion has to be true as well. Put another way, if an argument is sound, its premises are true in the real story; but then if the conclusion is false, the real story has the premises true and conclusion false — and the argument is not valid. So if an argument is sound, if it is valid and its premises are true, its conclusion must be true. Note again: we do not need that the conclusion is true in the real world in order to decide that an argument is sound; saying that the conclusion is true is no part of our procedure for validity or soundness! Rather, by discovering that an argument is logically valid and that its premises are true, we establish that it is sound; this gives us the result that its conclusion therefore is true. And that is just what we want.

1.3.2

Validity and Form

It is worth observing a connection between what we have done and argument form. Some of the arguments we have seen so far are of the same general form. Thus both of the arguments on the left have the form on the right.

(H)

If Joe works hard, then he will get an ‘A’ Joe works hard

If Hannah is a citizen then she can vote Hannah is a citizen

Joe will get an ‘A’

Hannah can vote

If P then Q P Q

As it turns out, all arguments of this form are valid. In contrast, the following arguments with the indicated form are not.

(I)

If Joe works hard then he will get an ‘A’ Joe will get an ‘A’

If Hannah can vote, then she is a citizen Hannah is a citizen

Joe works hard

Hannah can vote

If P then Q Q P

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There are stories where, say, Joe cheats for the ‘A’, or Hannah is a citizen but not old enough to vote. In these cases, there is some way to obtain condition Q other than by having P — this is what the stories bring out. And, generally, it is often possible to characterize arguments by their forms, where a form is valid iff every instance of it is logically valid. Thus the first form listed above is valid, and the second not. In chapters to come, we take advantage of certain very general formal or structural features of arguments to identify ones that are valid and ones that are invalid. For now, though, it is worth noting that some presentations of critical reasoning (which you may or may not have encountered), take advantage of patterns like those above, listing typical ones that are valid, and typical ones that are not (for example, Cederblom and Paulsen, Critical Reasoning). A student may then identify valid and invalid arguments insofar as they match the listed forms. This approach has the advantage of simplicity — and one may go quickly to applications of the logical notions for concrete cases. But the approach is limited to application of listed forms, and so to a very narrow range of arguments. In contrast, our approach based on definition LV has application to arbitrary arguments. Further, a mere listing of valid forms does not explain their relation to truth, where the definition is directly connected. Finally, for our logical machine, within a certain range we shall develop develop an account of validity for quite arbitrary forms. So we are pursuing a general account or theory of validity that goes well beyond the mere lists of these other more traditional approaches.4

1.3.3

Relevance

Another consequence seems less welcome. Consider the following argument.

(J)

Snow is white Snow is not white All dogs can fly

It is natural to think that the premises are not connected to the conclusion in the right way — for the premises have nothing to do with the conclusion — and that this argument therefore should not be logically valid. But if it is not valid, by definition, there is a consistent story that makes the premises true and the conclusion false. And, in this case, there is no such story, for no consistent story makes the premises true. 4 Some authors introduce a notion of formal validity (maybe in the place of logical validity as above) such that an argument is formally valid iff it has some valid form. As above, formal validity is parasitic on logical validity, together with a to-be-specified notion of form. But if an argument is formally valid, it is logically valid. So if our logical machine is adequate to identify formal validity, it identifies logical validity as well.

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Thus, by definition, this argument is logically valid. The procedure applies in a straightforward way. Thus, a. In any story that makes the premises true and conclusion false, 1. Snow is white 2. Snow is not white 3. Some dogs cannot fly b. In any such story, Given (1) and (2), 4. Snow is and is not white c. So no consistent story makes the premises true and conclusion false; so by definition, the argument is logically valid. d. Since in the real world snow is white, the second premise is not true. So, though the argument is logically valid, by definition it is not logically sound. This seems bad! Intuitively, there is something wrong with the argument. But, on our official definition, it is logically valid. One might rest content with the observation that, even though the argument is logically valid, it is not logically sound. But this does not remove the general worry. For this argument, There are fish in the sea (K) Nothing is round and not round has all the problems of the other and is logically sound as well. (Why?) One might, on the basis of examples of this sort, decide to reject the (classical) account of validity with which we have been working. Some do just this.5 But, for now, let us see what can be said in defense of the classical approach. (And the classical approach is, no doubt, the approach you have seen or will see in any standard course on critical thinking or logic.) As a first line of defense, one might observe that the conclusion of every sound argument is true and ask, “What more do you want?” We use arguments to demonstrate the truth of conclusions. And nothing we have said suggests that sound arguments 5 Especially the so-called “relevance” logicians. For an introduction, see Graham Priest, NonClassical Logics. But his text presumes mastery of material corresponding to Part I and Part II (or at least Part I with chapter 7) of this one. So the non-classical approaches develop or build on the classical one developed here.

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do not have true conclusions: An argument whose premises are inconsistent is sure to be unsound. And an argument whose conclusion cannot be false is sure to have a true conclusion. So soundness may seem sufficient for our purposes. Even though we accept that there remains something about argument goodness that soundness leaves behind, we can insist that soundness is useful as an intellectual tool. Whenever it is the truth or falsity of a conclusion that matters, we can profitably employ the classical notions. But one might go further, and dispute even the suggestion that there is something about argument goodness that soundness leaves behind. Consider the following two argument forms. (ds)

P or Q, not-P

(add)

Q

P P or Q

According to ds (disjunctive syllogism), if you are given that P or Q and that not-P , you can conclude that Q. If you have cake or ice cream, and you do not have cake, you have ice cream; if you are in California or New York, and you are not in California, you are in New York; and so forth. Thus ds seems hard to deny. And similarly for add (addition). Where ‘or’ means ‘one or the other or both’, when you are given that P , you can be sure that P or anything. Say you have cake, then you have cake or ice cream, cake or brussels sprouts, and so forth; if grass is green, then grass is green or pigs have wings, grass is green or dogs fly, and so forth. Return now to our problematic argument. As we have seen, it is valid according to the classical definition LV. We get a similar result when we apply the ds and add principles. 1. 2. 3. 4.

Snow is white Snow is not white Snow is white or all dogs can fly All dogs can fly

premise premise from 1 and add from 2 and 3 and ds

If snow is white, then snow is white or anything. So snow is white or dogs fly. So we use line 1 with add to get line 3. But if snow is white or dogs fly, and snow is not white, then dogs fly. So we use lines 2 and 3 with ds to reach the final result. So our principles ds and add go hand-in-hand with the classical definition of validity. The argument is valid on the classical account; and with these principles, we can move from the premises to the conclusion. If we want to reject the validity of this argument, we will have to reject not only the classical notion of validity, but also one of our principles ds or add. And it is not obvious that one of the principles should go. If we decide to retain both ds and add then, seemingly, the classical definition of validity

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should stay as well. If we have intuitions according to which ds and add should stay, and also that the definition of validity should go, we have conflicting intuitions. Thus our intuitions might, at least, be sensibly resolved in the classical direction. These issues are complex, and a subject for further discussion. For now, it is enough for us to treat the classical approach as a useful tool: It is useful in contexts where what we care about is whether conclusions are true. And alternate approaches to validity typically develop or modify the classical approach. So it is natural to begin where we are, with the classical account. At any rate, this discussion constitutes a sort of acid test: If you understand the validity of the “snow is white” and “fish in the sea” arguments (J) and (K), you are doing well — you understand how the definition of validity works, with its results that may or may not now seem controversial. If you do not see what is going on in those cases, then you have not yet understood how the definitions work and should return to section 1.2 with these cases in mind. E1.6. Use our procedures to say whether the following are logically valid or invalid, and sound or unsound. Hint: You may have to do some experimenting to decide whether the arguments are logically valid or invalid — and so decide which procedure applies. a. Bob is over six feet tall Bob is under six feet tall Bob is disfigured b. Marilyn is not over six feet tall Marilyn is not under six feet tall Marilyn is not in the WNBA c. There are fish in the sea Nothing is round and not round *d. Cheerios are square Chex are round There is no round square e. All dogs can fly Fido is a dog Fido cannot fly I am blessed

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E1.7. Respond to each of the following. a. Create another argument of the same form as the first set of examples (H) from section 1.3.2, and then use our regular procedures to decide whether it is logically valid and sound. Is the result what you expect? Explain. b. Create another argument of the same form as the second set of examples (I) from section 1.3.2, and then use our regular procedures to decide whether it is logically valid and sound. Is the result what you expect? Explain. E1.8. Which of the following are true, and which are false? In each case, explain your answers, with reference to the relevant definitions. The first is worked as an example. a. A logically valid argument is always logically sound. False. An argument is sound iff it is logically valid and all of its premises are true in the real world. Thus an argument might be valid but fail to be sound if one or more of its premises is false in the real world. b. A logically sound argument is always logically valid. *c. If the conclusion of an argument is true in the real world, then the argument must be logically valid. d. If the premises and conclusion of an argument are true in the real world, then the argument must be logically sound. *e. If a premise of an argument is false in the real world, then the argument cannot be logically valid. f. If an argument is logically valid, then its conclusion is true in the real world. *g. If an argument is logically sound, then its conclusion is true in the real world. h. If an argument has contradictory premises (its premises are true in no consistent story), then it cannot be logically valid. *i. If the conclusion of an argument cannot be false (is false in no consistent story), then the argument is logically valid. j. The premises of every logically valid argument are relevant to its conclusion.

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E1.9. For each of the following concepts, explain in an essay of about two pages, so that (high-school age) Hannah could understand. In your essay, you should (i) identify the objects to which the concept applies, (ii) give and explain the definition, and give and explicate examples of your own construction (iii) where the concept applies, and (iv) where it does not. Your essay should exhibit an understanding of methods from the text. a. Logical validity b. Logical soundness

E1.10. Do you think we should accept the classical account of validity? In an essay of about two pages, explain your position, with special reference to difficulties raised in section 1.3.3.

Chapter 2

Formal Languages Having said in chapter 1 what validity and soundness are, we now turn to our logical machine. As depicted in the picture of elements for symbolic logic on p. 2, this machine begins with symbolic representations of ordinary reasoning. In this chapter we introduce the formal languages by introducing their grammar. After some brief introductory remarks, the chapter divides into sections that introduce grammar for a sentential language Ls (section 2.2), and then the grammar for an extended quantificational language Lq (section 2.3).

2.1

Introductory

There are different ways to introduce a formal language. It is natural to introduce expressions of a new language in relation to expressions of one that is already familiar. Thus, a standard course in a foreign language is likely to present vocabulary lists of the sort, cabbage small

chou: petit: :: :

But the terms of a foreign language are not originally defined by such lists. Rather French, in this case, has conventions of its own such that sometimes ‘chou’ corresponds to ‘cabbage’ and sometimes it does not. It is not a legitimate criticism of a Frenchman who refers to his sweetheart as mon petit chou to observe that she is no cabbage! Though it is possible to use such correlations to introduce the conventions of a new language, it is also possible to introduce a language “as itself” — the way a native speaker learns it. In this case, one avoids the danger of importing conventions 32

CHAPTER 2. FORMAL LANGUAGES

33

and patterns from one language onto the other. Similarly, the expressions of a formal language might be introduced in correlation with expressions of, say, English. But this runs the risk of obscuring just what the official definitions accomplish. Since we will be concerned extensively with what follows from the definitions, it is best to introduce our languages in their “pure” forms. In this chapter, we develop the grammar of our formal languages. Consider the following algebraic expressions, a

b=c

a

= c

Until we know what numbers are assigned to the terms (as a = 1; b = 2; c = 3/, we cannot evaluate the first for truth or falsity. Still, we can confidently say that it is grammatical where the other is not. We shall be able to evaluate the grammar of formal languages in a similar way. Though, eventually, our goal is to represent ordinary reasonings in a formal language, we do not have to know what the language represents in order to decide if a sentence is grammatically correct. Again, just as a computer can check the spelling and grammar of English without reference to meaning, so we can introduce the vocabulary and grammar of our formal languages without reference to what their expressions mean or what makes them true. The grammar, taken alone, is completely straightforward. Taken this way, we work directly from the definitions, without “pollution” from associations with English or whatever. So we want the definitions. Even so, it may be helpful to offer some hints that foreshadow how things will go. Do not take these as defining anything! Still, it is nice to have a sense of how it fits together. Consider some simple sentences of an ordinary language, say, ‘The butler is guilty’ and ‘The maid is guilty’. It will be convenient to introduce capital letters corresponding to these, say, B and M . Such sentences may combine to form ones that are more complex as, ‘It is not the case that the butler is guilty’ or ‘If the butler is guilty, then the maid is guilty’. We shall find it convenient to express these, ‘the butler is guilty’ and ‘The butler is guilty ! the maid is guilty’, with operators  and !. Putting these together we get, B and B ! M . Operators may be combined in obvious ways so that B ! M says that if the butler is guilty then the maid is not. And so forth. We shall see that incredibly complex expressions of this sort are possible! In this case, simple sentences, ‘The butler is guilty’ and ‘The maid is guilty’ are “atoms” and complex sentences are built out of them. This is characteristic of the sentential languages to be considered in section 2.2. For the quantificational languages of section 2.3, certain sentence parts are taken as atoms. So quantificational languages expose structure beyond that for the sentential case. Perhaps, though, this

CHAPTER 2. FORMAL LANGUAGES

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will be enough to give you a glimpse of the overall strategy and aims for the formal languages of which we are about to introduce the grammar.

2.2

Sentential Languages

Just as algebra or English have their own vocabulary or symbols, and then grammatical rules for the way the vocabulary is combined, so our formal language has its own vocabulary, and then grammatical rules for the way the vocabulary is combined. In this section, we introduce the vocabulary for a sentential language, introduce the grammatical rules, and conclude with some discussion of abbreviations for official expressions.

2.2.1

Vocabulary

We begin, then, with the vocabulary. In this section, we say which symbols are included in the language, and introduce some conventions for talking about the symbols. In the sentential case, vocabulary includes, VC

(p) Punctuation symbols: . / (o) Operator symbols:  ! (s) A non-empty countable collection of sentence letters

And that is all.  is tilde and ! is arrow.1 In order to fully specify the vocabulary of any particular sentential language, we need to identify its sentence letters — so far as definition VC goes, different languages may differ in their collections of sentence letters. The only constraint on such specifications is that the collections of sentence letters be non-empty and countable. A collection is non-empty iff it has at least one member. So any sentential language has at least one sentence letter. A collection is countable iff its members can be matched one-to-one with all (or some) of the integers. Thus we might let the sentence letters be A; B : : : Z, where these correlate with the integers 1 : : : 26. Or we might let there be infinitely many sentence letters, S0 ; S1 ; S2 : : : where the letters are correlated with the integers by their subscripts. 1 Sometimes

sentential languages are introduced with different symbols, for example, : for , or  for !. It should be easy to convert between presentations of the different sorts. And sometimes sentential languages include operators in addition to  and ! (for example, _, ^, $). Such symbols will be introduced in due time — but as abbreviations for complex official expressions.

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So there is room for different sentential languages. Having made this point, though, we immediately focus on a standard sentential language Ls whose sentence letters are Roman italics A : : : Z with or without integer subscripts. Thus, A

B

K

B3

K7

Z

are sentence letters of Ls . Similarly, A1

Z23

are sentence letters of Ls . We will not use the subscripts very often, but they guarantee that we never run out of sentence letters! Perhaps surprisingly, as described in the box on p. 36 (and E2.2), these letters too can be correlated with the integers. Official sentences of Ls are built out of this vocabulary. To proceed, we need some conventions for talking about expressions of a language like Ls . Here, Ls is an object language — the thing we want to talk about, and we require conventions for the metalanguage — for talking about the object language. In general, for any formal object language L, an expression is a sequence of one or more elements of its vocabulary. Thus .A ! B/ is an expression of Ls , but .A ? B/ is not. (What is the difference?) We shall use script characters A : : : Z as variables that range over expressions. ‘’, ‘!’, ‘.’, and ‘/’ represent themselves. Concatenated or joined symbols in the metalanguage represent the concatenation of the symbols they represent. To see how this works, think of metalinguistic expressions as “mapping” to objectlanguage ones. Thus, for example, where S represents an arbitrary sentence letter, S may represent any of, A, B, or Z. But .A ! B/ is not of that form, for it does not consist of a tilde followed by a sentence letter. With S restricted to sentence letters, there is a straightforward map from S onto A, B, or Z, but not from S onto .A ! B/. S

S

S

(A) ??

A

??

B

??

Z

S ‹ ?‚ …„ ƒ

 .A ! B/

In the first three cases,  maps to itself, and S to a sentence letter. In the last case there is no map. We might try mapping S to A or B; but this would leave the rest of the expression unmatched. While .A ! B/ is not of the form S, if we let P represent any arbitrary expression, then .A ! B/ is of the form P , for it consists of a tilde followed by an expression of some sort. An object-language expression has some metalinguistic form just when there is a complete map from the metalinguistic form to it.

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36

Countability To see the full range of languages which are allowed under VC, observe how multiple infinite series of sentence letters may satisfy the countability constraint. Thus, for example, suppose we have two series of sentence letters, A0 ; A1 : : : and B0 ; B1 : : : These can be correlated with the integers as follows, A0 B0 A1 B1 A2 B2 j j j j j j ::: 0 1 2 3 4 5 For any integer n, An is matched with 2n, and Bn with 2n 1. So each sentence letter is matched with some integer; so the sentence letters are countable. If there are three series, they may be correlated, A0 B0 C0 A1 B1 C1 j j j j j j ::: 0 1 2 3 4 5 so that every sentence letter is matched to some integer. And similarly for any finite number of series. And there might be 26 such series, as for our language Ls . In fact even this is not the most general case. If there are infinitely many series of sentence letters, we can still line them up and correlate them with the integers. Here is one way to proceed. Order the letters as follows, A0 B0 # C0

! .

A1

A2

A3

:::

B2

B3

:::

C1

C2

C3

:::

D1

D2

D3

:::

% B1

%

! .

.

. D0 :: :

And following the arrows, match them accordingly with the integers, A0 A1 B0 C0 B1 A2 j j j j j j ::: 0 1 2 3 4 5 so that, again, any sentence letter is matched with some integer. It may seem odd that we can line symbols up like this, but it is hard to dispute that we have done so. Thus we may say that VC is compatible with a wide variety of specifications, but also that all legitimate specifications have something in common: If a collection is countable, it is possible to sort its members into a series with a first member, a second member, and so forth.

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Say P represents any arbitrary expression. Then by similar reasoning, .A ! B/ ! .A ! B/ is of the form P ! P . P !P (B)



R ? @

‚ …„ ƒ ‚ …„ ƒ .A ! B/ ! .A ! B/ In this case, P maps to all of .A ! B/ and ! to itself. A constraint on our maps is that the use of the metavariables A : : : Z must be consistent within a given map. Thus .A ! B/ ! .B ! B/ is not of the form P ! P . P !P (C)



?

P !P ‹

or

‚ …„ ƒ ‚ …„ ƒ .A ! B/ ! .B ! B/



R ? @

‚ …„ ƒ ‚ …„ ƒ .A ! B/ ! .B ! B/

We are free to associate P with whatever we want. However, within a given map, once P is associated with some expression, we have to use it consistently within that map. Observe again that S and P ! P are not expressions of Ls . Rather, we use them to talk about expressions of Ls . And it is important to see how we can use the metalanguage to make claims about a range of expressions all at once. Given that A, B and Z are all of the form S, when we make some claim about expressions of the form S, we say something about each of them — but not about .A ! B/. Similarly, if we make some claim about expressions of the form P ! P , we say something with application to a range of expressions. In the next section, for the specification of formulas, we use the metalanguage in just this way. E2.1. Assuming that S may represent any sentence letter, and P any arbitrary expression of Ls , use maps to determine whether each of the following expressions is (i) of the form .S ! P / and then (ii) whether it is of the form .P ! P /. In each case, explain your answers. a. .A ! A/ b. .A ! .R ! Z// c. .A ! .R ! Z// d. ..R ! Z/ ! .R ! Z// *e. ..! / ! .! //

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38

E2.2. On the pattern of examples from the countability guide on p. 36, show that the sentence letters of Ls are countable — that is, that they can be correlated with the integers. On the scheme you produce, what integers correlate with A, B1 and C10 ? Hint: Supposing that A without subscript is like A0 , for any integer n, you should be able to produce a formula for the position of any An , and similarly for Bn , Cn and the like. Then it will be easy to find the position of any letter, even if the question is about, say, L125 .

2.2.2

Formulas

We are now in a position to say which expressions of a sentential language are its grammatical formulas and sentences. The specification itself is easy. We will spend a bit more time explaining how it works. For a given sentential language L, FR

(s) If S is a sentence letter, then S is a formula. () If P is a formula, then P is a formula. (!) If P and Q are formulas, then .P ! Q/ is a formula. (CL) Any formula may be formed by repeated application of these rules.

At this stage, we simply identify the formulas with the sentences. For any sentential language L, an expression is a sentence iff it is a formula. FR is a first example of a recursive definition. Such definitions always build from the parts to the whole. Frequently we can use “tree” diagrams to see how they work. Thus, for example, by repeated applications of the definition, .A ! .B ! A// is a formula and sentence of Ls . A

B

A A A A A A

(D)

A

B @ @ @ .B ! A/

A A A

These are formulas by FR(s)

Since B is a formula, this is a formula by FR()

Since B and A are formulas, this is a formula by FR(!)

AA

.A ! .B ! A//

Since A and .B ! A/ are formulas, this is a formula by FR(!)

.A ! .B ! A//

Since .A ! .B ! A// is a formula, this is a formula by FR()

CHAPTER 2. FORMAL LANGUAGES

39

By FR(s), the sentence letters, A, B and A are formulas; given this, clauses FR() and FR (!) let us conclude that other, more complex, expressions are formulas as well. Notice that, in the definition, P and Q may be any expressions that are formulas: By FR (), if B is a formula, then tilde followed by B is a formula; but similarly, if B and A are formulas, then an opening parenthesis followed by B, followed by ! followed by A and then a closing parenthesis is a formula; and so forth as on the tree above. You should follow through each step very carefully. In contrast, .AB/ for example, is not a formula. A is a formula and B is a formula; but there is no way to put them together, by the definition, without ! in between. A recursive definition always involves some “basic” starting elements, in this case, sentence letters. These occur across the top row of our tree. Other elements are constructed, by the definition, out of ones that come before. The last, closure, clause tells us that any formula is built this way. To demonstrate that an expression is a formula and a sentence, it is sufficient to construct it, according to the definition, on a tree. If an expression is not a formula, there will be no way to construct it according to the rules. Here are a couple of last examples which emphasize the point that you must maintain and respect parentheses in the way you construct a formula. Thus consider, A

B

These are formulas by FR(s)

@ @ @

(E)

.A ! B/

Since A and B are formulas, this is a formula by FR(!)

.A ! B/

Since .A ! B/ is a formula, this is a formula by FR()

And compare it with, A

(F)

B

A @ @ @ .A ! B/

These are formulas by FR(s)

Since A is a formula, this is a formula by FR()

Since A and B are formulas, this is a formula by FR(!)

Once you have .A ! B/ as in the first case, the only way to apply FR() puts the tilde on the outside. To get the tilde inside the parentheses, by the rules, it has to go on first, as in the second case. The significance of this point emerges immediately below. It will be helpful to have some additional definitions, each of which may be introduced in relation to the trees. First, for any formula P , each formula which

CHAPTER 2. FORMAL LANGUAGES

40

appears in the tree for P including P itself is a subformula of P . Thus .A ! B/ has subformulas, A

B

.A ! B/

.A ! B/

A

.A ! B/

In contrast, .A ! B/ has subformulas, A

B

So it matters for the subformulas how the tree is built. The immediate subformulas of a formula P are the subformulas to which P is directly connected by lines. Thus .A ! B/ has one immediate subformula, .A ! B/; .A ! B/ has two, A and B. The atomic subformulas of a formula P are the sentence letters that appear across the top row of its tree. Thus both .A ! B/ and .A ! B/ have A and B as their atomic subformulas. Finally, the main operator of a formula P is the last operator added in its tree. Thus  is the main operator of .A ! B/, and ! is the main operator of .A ! B/. So, again, it matters how the tree is built. We sometimes speak of a formula by means of its main operator: A formula of the form P is a negation; a formula of the form .P ! Q/ is a (material) conditional, where P is the antecedent of the conditional and Q is the consequent.

Parts of a Formula The parts of a formula are here defined in relation to its tree. SB

Each formula which appears in the tree for formula P including P itself is a subformula of P .

IS

The immediate subformulas of a formula P are the subformulas to which P is directly connected by lines.

AS

The atomic subformulas of a formula P are the sentence letters that appear across the top row of its tree.

MO

The main operator of a formula P is the last operator added in its tree.

E2.3. For each of the following expressions, demonstrate that it is a formula and a sentence of Ls with a tree. Then on the tree (i) bracket all the subformulas, (ii) box the immediate subformula(s), (iii) star the atomic subformulas, and (iv) circle the main operator. A first case for ..A ! B/ ! A/ is worked as an example.

CHAPTER 2. FORMAL LANGUAGES s u b f o r m u l a s

A?

B?

A @ @ @

A?

  

These are formulas by FR(s)

From A, formula by FR()

  

.A ! B/



41

  @  @  @  ..A ! B/ ! A/



From A and B, formula by FR(!)

From .A ! B/ and A, formula by FR(!)

*a. A b. A c. .A ! B/ d. .C ! .A ! B// e. ..A ! B/ ! .C ! A// E2.4. Explain why the following expressions are not formulas or sentences of Ls . Hint: you may find that an attempted tree will help you see what is wrong. a. .A  B/ *b. .P ! Q/ c. .B/ d. .A ! B ! C / e. ..A ! B/ ! .A ! C / ! D/ E2.5. For each of the following expressions, determine whether it is a formula and sentence of Ls . If it is, show it on a tree, and exhibit its parts as in E2.3. If it is not, explain why as in E2.4. *a. ..A ! B/ ! ..A ! B/ ! A// b. .A ! B ! ..A ! B/ ! A// *c. .A ! B/ ! ..A ! B/ ! A/

CHAPTER 2. FORMAL LANGUAGES

42

d. .A ! A/ e. ...A ! B/ ! .C ! D// ! ..E ! F / ! G//

2.2.3

Abbreviations

We have completed the official grammar for our sentential languages. So far, the languages are relatively simple. When we turn to reasoning about logic (in later parts), it will be good to have our languages as simple as we can. However, for applications of logic, it will be advantageous to have additional expressions which, though redundant with expressions of the language already introduced, simplify the work. I begin by introducing these additional expressions, and then turn to the question about how to understand the redundancy. Abbreviating. As may already be obvious, formulas of a sentential language like Ls can get complicated quickly. Abbreviated forms give us ways to manipulate official expressions without undue pain. First, for any formulas P and Q, AB

(_) .P _ Q/ abbreviates .P ! Q/ (^) .P ^ Q/ abbreviates .P ! Q/ ($) .P $ Q/ abbreviates ..P ! Q/ ! .Q ! P //

The last of these is easier than it looks; I say something about this below. _ is wedge, ^ is caret, and $ is double arrow. An expression of the form .P _ Q/ is a disjunction with P and Q as disjuncts; it has the standard reading, (P or Q). An expression of the form .P ^ Q/ is a conjunction with P and Q as conjuncts; it has the standard reading, (P and Q). An expression of the form .P $ Q/ is a (material) biconditional; it has the standard reading, (P iff Q).2 Again, we do not use ordinary English to define our symbols. All the same, this should suggest how the extra operators extend the range of what we are able to say in a natural way. With the abbreviations, we are in a position to introduce derived clauses for FR. Suppose P and Q are formulas; then by FR(), P is a formula; so by FR(!), .P ! Q/ is a formula; but this is just to say that .P _ Q/ is a formula. And similarly in the other cases. (If you are confused by such reasoning, work it out on a tree.) Thus we arrive at the following conditions. FR 0

(_) If P and Q are formulas, then .P _ Q/ is a formula.

2 Common

alternatives are & for ^, and  for $.

CHAPTER 2. FORMAL LANGUAGES

43

(^) If P and Q are formulas, then .P ^ Q/ is a formula. ($) If P and Q are formulas, then .P $ Q/ is a formula. Once FR is extended in this way, the additional conditions may be applied directly in trees. Thus, for example, if P is a formula and Q is a formula, we can safely move in a tree to the conclusion that .P _ Q/ is a formula by FR0 (_). Similarly, for a more complex case, ..A $ B/ ^ .A _ B// is a formula. A

B

A

@ @ @

.A $ B/

(G)

B

These are formulas by FR(s)

These are formulas by FR0 ($) and FR()

@ @ @

A

\

\ \ .A _ B/ \  \   \\ 

This is a formula by FR0 (_)

This is a formula by FR0 (^)

..A $ B/ ^ .A _ B//

In a derived sense, expressions with the new symbols have subformulas, atomic subformulas, immediate subformulas, and main operator all as before. Thus, with notation from exercises, with bracket for subformulas, star for atomic subformulas, box for immediate subformulas and circle for main operator, on the diagram immediately above,

A?

(H)

s u b f o r m u l a s

B?

@ @ @

.A $ B/



A?

B?

A @ @ @

\ \ \ .A _ B/ \  \   \  ..A $ B/ ^ .A _ B//



These are formulas by FR(s)

These are formulas by FR0 ($) and FR()

This is a formula by FR0 (_)

This is a formula by FR0 (^)

In the derived sense, ..A $ B/ ^ .A _ B// has immediate subformulas .A $ B/ and .A _ B/, and main operator ^. Return to the case of .P $ Q/ and observe that it can be thought of as based on a simple abbreviation of the sort we expect. That is, ..P ! Q/ ^ .Q ! P // is of the sort .A ^ B/; so by AB(^), it abbreviates .A ! B/; but with .P ! Q/ for A and .Q ! P / for B, this is just, ..P ! Q/ ! .Q ! P // as in AB($). So you

CHAPTER 2. FORMAL LANGUAGES

44

may think of .P $ Q/ as an abbreviation of ..P ! Q/ ^ .Q ! P //, which in turn abbreviates the more complex ..P ! Q/ ! .Q ! P //. This is what we expect: a double arrow is like an arrow going from P to Q and an arrow going from Q to P . A couple of additional abbreviations concern parentheses. First, it is sometimes convenient to use a pair of square brackets [ ] in place of parentheses ( ). This is purely for visual convenience; for example ((()())) may be more difficult to absorb than ([()()]). Second, if the very last step of a tree for some formula P is justified by FR (!), FR 0 (^), FR 0 (_), or FR 0 ($), we feel free to abbreviate P with the outermost set of parentheses or brackets dropped. Again, this is purely for visual convenience. Thus, for example, we might write, A ! .B ! C / in place of .A ! .B ! C //. As it turns out, where A, B, and C are formulas, there is a difference between ..A ! B/ ! C/ and .A ! .B ! C //, insofar as the main operator shifts from one case to the other. In .A ! B ! C/, however, it is not clear which arrow should be the main operator. That is why we do not count the latter as a grammatical formula or sentence. Similarly there is a difference between .A ! B/ and .A ! B/; again, the main operator shifts. However, there is no room for ambiguity when we drop just an outermost pair of parentheses and write .A ! B/ ! C for ..A ! B/ ! C/; and similarly when we write A ! .B ! C / for .A ! .B ! C//. The same reasoning applies for abbreviations with ^, _, or $. So dropping outermost parentheses counts as a legitimate abbreviation. An expression which uses the extra operators, square brackets, or drops outermost parentheses is a formula just insofar as it is a sort of shorthand for an official formula which does not. But we will not usually distinguish between the shorthand expressions and official formulas. Thus, again, the new conditions may be applied directly in trees and, for example, the following is a legitimate tree to demonstrate that A _ .ŒA ! B ^ B/ is a formula. A

A

S S S

B

ŒA ! B H HH H H H

S

(I)

B

Formulas by FR(s)

@ @ @ S S

.ŒA ! B ^ B/

S S S S  S

Formula by FR(!), with [ ]

Formula by FR0 (^)

 

A _ .ŒA ! B ^ B/

Formula by FR0 (_), with outer ( ) dropped

So we use our extra conditions for FR0 , introduce square brackets instead of parentheses, and drop parentheses in the very last step. Remember that the only case where

CHAPTER 2. FORMAL LANGUAGES

45

you can omit parentheses is if they would have been added in the very last step of the tree. So long as we do not distinguish between shorthand expressions and official formulas, we regard a tree of this sort as sufficient to demonstrate that an expression is a formula and a sentence. Unabbreviating. As we have suggested, there is a certain tension between the advantages of a simple language, and one that is more complex. When a language is simple, it is easier to reason about; when it has additional resources, it is easier to use. Expressions with ^, _ and $ are redundant with expressions that do not have them — though it is easier to work with a language that has ^, _ and $ than with one that does not (something like reciting the Pledge of Allegiance in English, and then in Morse code; you can do it in either, but it is easier in the former). If all we wanted was a simple language to reason about, we would forget about the extra operators. If all we wanted was a language easy to use, we would forget about keeping the language simple. To have the advantages of both, we have adopted the position that expressions with the extra operators abbreviate, or are a shorthand for, expressions of the original language. It will be convenient to work with abbreviations in many contexts. But, when it comes to reasoning about the language, we set the abbreviations to the side, and focus on the official language itself. For this to work, we have to be able to undo abbreviations when required. It is, of course, easy enough to substitute parentheses back for square brackets, or to replace outermost dropped parentheses. For formulas with the extra operators, it is always possible to work through trees, using AB to replace formulas with unabbreviated forms, one operator at a time. Consider an example. A

B

A

B

A

B

A

B

@ @ @

(J)

@ @ @ ..A ! B/ ! .B ! A// .A $ B/ A A \ @ \ @ \ @ \ @ @ @ \ \ .A _ B/ .A ! B/ \ \    \ \   \  \  ..A $ B/ ^ .A _ B//

...A ! B/ ! .B ! A// ! .A ! B//

The tree on the left is (G) from above. The tree on the right uses AB to “unpack” each of the expressions on the left. Atomics remain as before. Then, at each stage, given an unabbreviated version of the parts, we give an unabbreviated version of the whole. First, .A $ B/ abbreviates ..A ! B/ ! .B ! A//; this is a simple application of AB($). A is not an abbreviation and so remains as before. From AB(_), .P _ Q/

CHAPTER 2. FORMAL LANGUAGES

46

abbreviates .P ! Q/ so .A _ B/ abbreviates tilde the left disjunct, arrow the right (so that we get two tildes). For the final result, we combine the input formulas according to the unabbreviated form for ^. It is more a bookkeeping problem than anything: There is one formula P that is .A $ B/, another Q that is .A _ B/; these are combined into .P ^ Q/ and so, by AB(^), into .P ! Q/. You should be able to see that this is just what we have done. There is a tilde and a parenthesis; then the P ; then an arrow and a tilde; then the Q, and a closing parenthesis. Not only is the abbreviation more compact but, as we shall see, there is a corresponding advantage when it comes to grasping what an expression says. Here is a another example, this time from (I). In this case, we replace also square brackets and restore dropped outer parentheses. A

A

S

B

@ @ ŒA ! B HH HH H

S S S

(K)

B

@

S S

.ŒA ! B ^ B/

S S

 S  S

A _ .ŒA ! B ^ B/

A

A

S

B

B

@

@ @ .A ! B/ HH HH H

S S S S S

..A ! B/ ! B/

S S

S  S



.A ! ..A ! B/ ! B//

In the right hand tree, we reintroduce parentheses for the square brackets. Similarly, we apply AB(^) and AB(_) to unpack shorthand symbols. And outer parentheses are reintroduced at the very last step. Thus A _ .ŒA ! B ^ B/ is a shorthand for the unabbreviated expression, .A ! ..A ! B/ ! B//. Observe that right-hand trees are not ones of the sort you would use directly to show that an expression is a formula by FR! FR does not let you move directly from that .A ! B/ is a formula and B is a formula, to the result that ..A ! B/ ! B/ is a formula as just above. Of course, if .A ! B/ and B are formulas, then ..A ! B/ ! B/ is a formula, and nothing stops a tree to show it. This is the point of our derived clauses for FR0 . In fact, this is a good check on your unabbreviations: If the result is not a formula, you have made a mistake! But you should not think of trees as on the right as involving application of FR. Rather they are unabbreviating trees, with application of AB to shorthand expressions from trees as on the left. A fully unabbreviated expression always meets all the requirements from section 2.2.2. E2.6. For each of the following expressions, demonstrate that it is a formula and a sentence of Ls with a tree. Then on the tree (i) bracket all the subformulas,

CHAPTER 2. FORMAL LANGUAGES

47

(ii) box the immediate subformula(s), (iii) star the atomic subformulas, and (iv) circle the main operator. *a. .A ^ B/ ! C b. .ŒA ! K14  _ C3 / c. B ! .A $ B/ d. .B ! A/ ^ .C _ A/ e. .A _ B/ $ .C ^ A/ *E2.7. For each of the formulas in E2.6a - e, produce an unabbreviating tree to find the unabbreviated expression it represents. *E2.8. For each of the unabbreviated expressions from E2.7a - e, produce a complete tree to show by direct application of FR that it is an official formula. E2.9. In the text, we introduced derived clauses to FR by reasoning as follows, “Suppose P and Q are formulas; then by FR(), P is a formula; so by FR (!), .P ! Q/ is a formula; but this is just to say that .P _ Q/ is a formula. And similarly in the other cases” (p. 42). Supposing that P and Q are formulas, produce the similar reasoning to show that .P ^ Q/ and .P $ Q/ are formulas. Hint: Again, it may help to think about trees. E2.10. For each of the following concepts, explain in an essay of about two pages, so that (high-school age) Hannah could understand. In your essay, you should (i) identify the objects to which the concept applies, (ii) give and explain the definition, and give and explicate examples of your own construction (iii) where the concept applies, and (iv) where it does not. Your essay should exhibit an understanding of methods from the text. a. The vocabulary for a sentential language, and use of the metalanguage. b. A formula of a sentential language. c. The parts of a formula. d. The abbreviation and unabbreviation for an official formula of a sentential language.

CHAPTER 2. FORMAL LANGUAGES

2.3

48

Quantificational Languages

The methods by which we define the grammar of a quantificational language are very much the same as for a sentential language. Of course, in the quantificational case, additional expressive power is associated with additional complications. We will introduce a class of terms before we get to the formulas, and there will be a distinction between formulas and sentences — not all formulas are sentences. As before, however, we begin with the vocabulary; we then turn to the terms, formulas, and sentences. Again we conclude with some discussion of abbreviations. Here is a brief intuitive picture. At the start of section 2.2 we introduced ‘The butler is guilty’ and ‘The maid is guilty’ as atoms for sentential languages, and the rest of the section went on to fill out that picture. For the quantificational languages of this section, our atoms are certain sentence parts. Thus we introduce a class of individual terms which work to pick out objects. In the simplest case, we might introduce b and m to pick out the butler and the maid. Similarly, we introduce a class of predicate expressions as .x is guilty) and .x killed y/ indicating them by capitals as G 1 or K 2 (with the superscript to indicate the number of object places). Then G 1 b says that the butler is guilty, and K 2 bm that the butler killed the maid. We shall read 8xG 1 x to say for any thing x, it is guilty — that everything is guilty. (The upside-down ‘A’ for all is the universal quantifier.) As indicated by this reading, the variable x works very much like a pronoun in ordinary language. And, of course, our notions may be combined. Thus, 8xG 1 x ^ K 2 bm says that everything is guilty and the butler killed the maid. Thus we expose structure buried in sentence letters from before. Of course we have so far done nothing to define quantificational languages. But this should give you a picture of the direction in which we aim to go.

2.3.1

Vocabulary

We begin by specifying the vocabulary or symbols of our quantificational languages. For now, do not worry about what the symbols mean or how they are used. Our task is to identify the symbols and give some grammatical rules for the way they are combined. The vocabulary consists of infinitely many distinct symbols including, VC

(p) Punctuation symbols: . / (o) Operator symbols:  ! 8 (v) Variable symbols: i j : : : z with or without integer subscripts (s) A possibly-empty countable collection of sentence letters (c) A possibly-empty countable collection of constant symbols

CHAPTER 2. FORMAL LANGUAGES

49

(f) For any integer n function symbols

1, a possibly-empty countable collection of n-place

(r) For any integer n relation symbols

1, a possibly-empty countable collection of n-place

Unless otherwise noted, ‘D’ is always included among the 2-place relation symbols. Notice that all the punctuation symbols, operator symbols and sentence letters remain from before (except that the collection of sentence letters may be empty). There is one new operator symbol, with the new variable symbols, constant symbols, function symbols, and relation symbols. This definition VC is parallel to definition VC from section 2.2. For definitions with both sentential and quantificational versions, I adopt the convention of naming the initial sentential version in small caps, and then the quantificational version in large. To fully specify the vocabulary of any particular language, we need to specify its sentence letters, constant symbols, function symbols, and relation symbols. Our general definition VC leaves room for languages with different collections of these symbols. As before, the requirement that the collections be countable is compatible with multiple series; for example, there may be sentence letters A; A1 ; A2 : : : ; B; B1 ; B2 . . . (where we may think of the unsubscripted letter as with an implicit subscript zero). So, again VC is compatible with a wide variety of specifications, but legitimate specifications always require that sentence letters, constant symbols, function symbols, and relation symbols can be sorted into series with a first member, a second member, and so forth. Notice that the variable symbols may be sorted into such a series as well. i j k : : : z i1 j1 j j j j j j ::: 0 1 2 : : : 17 18 19 So every variable is matched with an integer, and the variables are countable. As a sample for the other symbols, we shall adopt a generic quantificational language Lq which includes the equality symbol ‘D’ along with, Sentence letters: uppercase Roman italics A : : : Z with or without integer subscripts Constant symbols: lowercase Roman italics a : : : h with or without integer subscripts

CHAPTER 2. FORMAL LANGUAGES

50

More on Countability Given what was said on p. 36, one might think that every collection is countable. However, this is not so. This amazing and simple result was proved by G. Cantor in 1873. Consider the collection which includes every countably infinite series of digits 0 through 9 (or, if you like, the collection of all real numbers between 0 and 1). Suppose that the members of this collection can be correlated one-to-one with the integers. Then there is some list, 0 1 2 3 4

a1 b1 c1 d1 e1

a0 b0 c0 d0 e0

a2 b2 c2 d2 e2

a3 b3 c3 d3 e3

a4 b4 c4 d4 e4

::: ::: ::: ::: :::

and so forth, which matches each series of digits with an integer. For any digit x, say x 0 is the digit after it in the standard ordering (where 0 follows 9). Now consider the digits along the diagonal, a0 , b1 , c2 , d3 , e4 : : : and ask: does the series a00 ; b10 ; c20 ; d30 ; e40 : : : appear anywhere in the list? It cannot be the first member, because a0 6 = a00 ; it cannot be the second, because b1 6 = b10 , and similarly for every member! So a10 ; b20 ; c30 ; d40 ; e50 . . . does not appear in the list. So we have failed to match all the infinite series of digits with integers — and similarly for any attempt! So the collection which contains every countably infinite series of digits is not countable. As an example, consider the following attempt to line up the integers with the series of digits: 0 1 2 3 4 5 6 7 8 9 10 11 12 13

0 1 2 3 4 5 6 7 8 9 1 0 1 1

0 1 2 3 4 5 6 7 8 9 0 1 2 3

0 1 2 3 4 5 6 7 8 9 1 1 1 1

0 1 2 3 4 5 6 7 8 9 0 1 2 3

0 1 2 3 4 5 6 7 8 9 1 1 1 1

0 1 2 3 4 5 6 7 8 9 0 1 2 3

0 1 2 3 4 5 6 7 8 9 1 1 1 1

0 1 2 3 4 5 6 7 8 9 0 1 2 3

0 1 2 3 4 5 6 7 8 9 1 1 1 1

0 1 2 3 4 5 6 7 8 9 0 1 2 3

0 1 2 3 4 5 6 7 8 9 1 1 1 1

0 1 2 3 4 5 6 7 8 9 0 1 2 3

0 1 2 3 4 5 6 7 8 9 1 1 1 1

0 1 2 3 4 5 6 7 8 9 0 1 2 3

::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: ::: :::

and so forth. For each integer, repeat its digits, except that for “duplicate” cases — 1 and 11, 2 and 22, 12 and 1212 — prefix enough 0s so that no later series duplicates an earlier one. Then, by the above method, from the diagonal, 1

2

3

4

5

6

7

8

9

0

2

2

2

4

:::

cannot appear anywhere on the list. And similarly, any list has some missing series.

CHAPTER 2. FORMAL LANGUAGES

51

Function symbols: for any integer n 1, superscripted lowercase Roman italics an : : : z n with or without integer subscripts Relation symbols: for any integer n 1, superscripted uppercase Roman italics An : : : Z n with or without integer subscripts. Observe that constant symbols and variable symbols partition the lowercase alphabet: a : : : h for constants, and i : : : z for variables. Sentence letters are distinguished from relation symbols by superscripts; similarly, constant and variable symbols are distinguished from function symbols by superscripts. Function symbols with a superscript 1 (a1 : : : z 1 ) are one-place function symbols; function symbols with a superscript 2 (a2 : : : z 2 ) are two-place function symbols; and so forth. Similarly, relation symbols with a superscript 1 (A1 : : : Z 1 ) are one-place relation symbols; relation symbols with a superscript 2 (A2 : : : Z 2 ) are two-place relation symbols; and so forth. Subscripts merely guarantee that we never run out of symbols of the different types. Notice that superscripts and subscripts suffice to distinguish all the different symbols from one another. Thus, for example A and A1 are different symbols — one a sentence letter, and the other a one-place relation symbol; A1 , A11 and A2 are distinct as well — the first two are one-place relation symbols, distinguished by the subscript; the latter is a completely distinct two-place relation symbol. In practice, again, we will not see subscripts very often. (And we shall even find ways to abbreviate away some superscripts.) The metalanguage works very much as before. We use script letters A : : : Z and a : : : z to represent expressions of an object language like Lq . Again, ‘’, ‘!’, ‘8’, ‘D’, ‘(’, and ‘)’ represent themselves. And concatenated or joined symbols of the metalanguage represent the concatenation of the symbols they represent. As before, the metalanguage lets us make general claims about ranges of expressions all at once. Thus, where x is a variable, 8x is a universal x-quantifier. Here, 8x is not an expression of an object language like Lq (Why?) Rather, we have said of object language expressions that 8x is a universal x-quantifier, 8y2 is a universal y2 -quantifier, and so forth. In the metalinguistic expression, ‘8’ stands for itself, and ‘x’ for the arbitrary variable. Again, as in section 2.2.1, it may help to use maps to see whether an expression is of a given form. Thus given that x maps to any variable, 8x and 8y are of the form 8x, but 8c and 8f 1 z are not. 8x

8x

8x

(L)

8x ‹

??

8x

??

8y

?

8c ‹

?‚…„ƒ

8 f 1z

CHAPTER 2. FORMAL LANGUAGES

52

In the leftmost two cases, 8 maps to itself, and x to a variable. In the next, ‘c’ is a constant so there is no variable to which x can map. In the rightmost case, there is a variable z in the object expression, but if x is mapped to it, the function symbol f 1 is left unmatched. So the rightmost two expressions are not of the form 8x. E2.11. Assuming that R1 may represent any one-place relation symbol, h2 any twoplace function symbol, x any variable, and c any constant of Lq , use maps to determine whether each of the following expressions is (i) of the form, 8x.R1 x ! R1 c/ and then (ii) of the form, 8x.R1 x ! R1 h2 xc/. a. 8k.A1 k ! A1 d / b. 8h.J 1 h ! J 1 b/ c. 8w.S 1 w ! S 1 g 2 wb/ d. 8w.S 1 w ! S 1 c 2 xc/ e. 8vL1 v ! L1 yh2

2.3.2

Terms

With the vocabulary of a language in place, we can turn to specification of its grammatical expressions. For this, in the quantificational case, we begin with terms. TR

(v) If t is a variable x, then t is a term. (c) If t is a constant c, then t is a term. (f) If hn is a n-place function symbol and t1 : : : tn are n terms, then hn t1 : : : tn is a term. (CL) Any term may be formed by repeated application of these rules.

TR is another example of a recursive definition. As before, we can use tree diagrams to see how it works. This time, basic elements are constants and variables. Complex elements are put together by clause (f). Thus, for example, f 1 g 2 h1 xc is a term of Lq .

CHAPTER 2. FORMAL LANGUAGES x

(M)

c

h1 x @ @ @ g 2 h1 xc

f 1 g 2 h1 xc

53

x is a term by TR(v), and c is a term by TR(c)

since x is a term, this is a term by TR(f)

since h1 x and c are terms, this is a term by TR(f)

since g 2 h1 xc is a term, this is a term by TR(f)

Notice how the superscripts of a function symbol indicate the number of places that take terms. Thus x is a term, and h1 followed by x to form h1 x is another term. But then, given that h1 x and c are terms, g 2 followed by h1 x and then c is another term. And so forth. Observe that neither g 2 h1 x nor g 2 c are terms — the function symbol g 2 must be followed by a pair of terms to form a new term. And neither is h1 xc a term — the function symbol h1 can only be followed by a single term to compose a term. You will find that there is always only one way to build a term on a tree. Here is another example. x

T T

(N)

c

z

x

,   ,, T  , 1 T h c  , T , TT , , f 4 xh1 czx

these are terms by TR(v), TR(c), TR(v), and TR(v)

since c is a term, this is a term by TR(f)

given the four input terms, this is a term by TR(f)

Again, there is always just one way to build a term by the definition. If you are confused about the makeup of a term, build it on a tree, and all will be revealed. To demonstrate that an expression is a term, it is sufficient to construct it, according to the definition, on such a tree. If an expression is not a term, there will be no way to construct it according to the rules. E2.12. For each of the following expressions, demonstrate that it is a term of Lq with a tree. a. f 1 c b. g 2 yf 1 c *c. h3 cf 1 yx d. g 2 h3 xyf 1 cx

CHAPTER 2. FORMAL LANGUAGES

54

e. h3 f 1 f 1 xcg 2 f 1 za E2.13. Explain why the following expressions are not terms of Lq . Hint: you may find that an attempted tree will help you see what is wrong. a. X b. g 2 c. zc *d. g 2 yf 1 xc e. h3 f 1 f 1 cg 2 f 1 za E2.14. For each of the following expressions, determine whether it is a term of Lq ; if it is, demonstrate with a tree; if not, explain why. *a. g 2 g 2 xyf 1 x *b. h3 cf 2 yx c. f 1 g 2 xh3 yf 2 yc d. f 1 g 2 xh3 yf 1 yc e. h3 g 2 f 1 xcg 2 f 1 zaf 1 b

2.3.3

Formulas

With the terms in place, we are ready for the central notion of a formula. Again, the definition is recursive. FR

(s) If S is a sentence letter, then S is a formula. (r) If Rn is an n-place relation symbol and t1 : : : tn are n terms, then Rn t1 : : : tn is a formula. () If P is a formula, then P is a formula. (!) If P and Q are formulas, then .P ! Q/ is a formula. (8) If P is a formula and x is a variable, then 8xP is a formula. (CL) Any formula can be formed by repeated application of these rules.

CHAPTER 2. FORMAL LANGUAGES

55

Again, we can use trees to see how it works. In this case, FR(r) depends on which expressions are terms. So it is natural to split the diagram into two, with applications of TR above a division, and FR below. Then, for example, 8x.A1 f 1 x ! 8yB 2 cy/ is a formula. x

y

c

D D f 1x

  

D D  . . . . . . . . . . . . . . . D. . . . . . . . . DD A1 f 1 x

Terms by TR(v), TR(c), and TR(v)

Term by TR(f)

B 2 cy

Formulas by FR(r)

8yB 2 cy

Formula by FR(8)

8yB 2 cy

Formula by FR()

.A1 f 1 x ! 8yB 2 cy/

Formula by FR(!)

8x.A1 f 1 x ! 8yB 2 cy/

Formula by FR(8)

C C

(O)

C C C C C C C CC

By now, the basic strategy should be clear. We construct terms by TR just as before. Given that f 1 x is a term, FR(r) gives us that A1 f 1 x is a formula, for it consists of a one-place relation symbol followed by a single term; and given that c and y are terms, FR(r) gives us that B 2 cy is a formula, for it consists of a two-place relation symbol followed by two terms. From the latter, by FR(8), 8yB 2 cy is a formula. Then FR() and FR(!) work just as before. The final step is another application of FR(8). Here is another example. By the following tree, 8x.L ! 8yB 3 f 1 ycx/ is a formula of Lq .

CHAPTER 2. FORMAL LANGUAGES y

c

56 x

f 1y . . . . . . . . . . . . . . . . . . . . .@ . . . . . . . . . . . . . . . @ @ L

(P)

B 3 f 1 ycx

\ \ \ 8yB 3 f 1 ycx \  \  \\ 

Terms by TR(v), TR(c), and TR(v)

Term by TR(f)

Formulas by FR(s), and FR(r)

Formula by FR(8)

.L ! 8yB 3 f 1 ycx/

Formula by FR(!)

.L ! 8yB 3 f 1 ycx/

Formula by FR()

8x.L ! 8yB 3 f 1 ycx/

Formula by FR(8)

The basic formulas appear in the top row of the formula part of the diagram. L is a sentence letter. So it does not require any terms to be a formula. B 3 is a three-place relation symbol, so by FR(r) it takes three terms to make a formula. After that, other formulas are constructed out of ones that come before. If an expression is not a formula, then there is no way to construct it by the rules. Thus, for example, .A1 x/ is not a formula of Lq . A1 x is a formula; but the only way parentheses are introduced is in association with !; the parentheses in .A1 x/ are not introduced that way; so there is no way to construct it by the rules, and it is not a formula. Similarly, A2 x and A2 f 2 xy are not formulas; in each case, the problem is that the two-place relation symbol is followed by just one term. You should be clear about these in your own mind, particularly for the second case. Before turning to the official notion of a sentence, we introduce some additional definitions, each directly related to the trees — and to notions you have seen before. First, where ‘!’, ‘’, and any quantifier count as operators, a formula’s main operator is the last operator added in its tree. Second, every formula in the formula portion of a diagram for P , including P itself, is a subformula of P . Notice that terms are not formulas, and so are not subformulas. An immediate subformula of P is a subformula to which P is directly connected by lines. A subformula is atomic iff it contains no operators and so appears in the top line of the formula part of the tree. Thus, with notation from exercises before, with star for atomic formulas, box for immediate

CHAPTER 2. FORMAL LANGUAGES

57

subformulas and circle for main operator, on the diagram immediately above we have, y

c

x

f 1y

. . . . . . . . . . . . . . . . . . . . .@ . . . . . . . . . . . . . . . @ @ L?

B 3 f 1 ycx ?

\

(Q)

s u b f o r m u l a s

\ \ 8yB 3 f 1 ycx \  \  \\  .L ! 8yB 3 f 1 ycx/

.L ! 8yB 3 f 1 ycx/

 8x .L ! 8yB 3 f 1 ycx/



The main operator is 8x, and the immediate subformula is .L ! 8yB 3 f 1 ycx/. The atomic subformulas are L and B 3 f 1 ycx. The atomic subformulas are the most basic formulas. Given this, everything is as one would expect from before. In general, if P and Q are some formulas and x is a variable, then the main operator of 8xP is the quantifier, and the immediate subformula is P ; the main operator of P is the tilde, and the immediate subformula is P ; the main operator of .P ! Q/ is the arrow, and the immediate subformulas are P and Q — for you would build these formulas by getting P , or P and Q, and then adding the quantifier, tilde, or arrow as the last operator. Now if a formula includes an operator, that operator’s scope is just the subformula in which the operator first appears. Using underlines to indicate quantifier scope,

CHAPTER 2. FORMAL LANGUAGES z

58

y

x

A  A  A

........................ B 2 xy

A1 z

C C C C

(R)

8xB 2 xy

The scope of the x-quantifier is 8xB 2 xy

C C C

8y8xB 2 xy

C C

The scope of the y-quantifier is 8y8xB 2 xy

C CC

.A1 z ! 8y8xB 2 xy/

8z.A1 z ! 8y8xB 2 xy/

The scope of the z-quantifier is the entire formula

A variable x is bound iff it appears in the scope of an x-quantifier, and a variable is free iff it is not bound. In the above diagram, each variable is bound. The xquantifier binds both instances of x; the y-quantifier binds both instances of y; and the z-quantifier binds both instances of z. In 8xR2 xy, however, both instances of x are bound, but the y is free. Finally, an expression is a sentence iff it is a formula and it has no free variables. To determine whether an expression is a sentence, use a tree to see if it is a formula. If it is a formula, use underlines to check whether any variable x has an instance that falls outside the scope of an x-quantifier. If it is a formula, and there is no such instance, then the expression is a sentence. From the above diagram, 8z.A1 z ! 8y8xB 2 xy/ is a formula and a sentence. But as follows, 8y.Q1 x ! 8xDxy/ is not.

CHAPTER 2. FORMAL LANGUAGES x

59

y

x

A  A  A

........................ Q1 x

(S)

Dxy

Q1 x

8xDxy

The scope of the x-quantifier is 8xDxy

@ @ @

.Q1 x

! 8xDxy/

8y.Q1 x ! 8xDxy/

The scope of the y-quantifier is the entire formula

Recall that ‘D’ is a two-place relation symbol. The expression has a tree, so it is a formula. The x-quantifier binds the last two instances of x, and the y-quantifier binds both instances of y. But the first instance of x is free. Since it has a free variable, although it is a formula, 8y.Q1 x ! 8xDxy/ is not a sentence. Notice that 8xR2 ax, for example, is a sentence, as the only variable is x (a being a constant) and all the instances of x are bound. E2.15. For each of the following expressions, (i) Demonstrate that it is a formula of Lq with a tree. (ii) On the tree bracket all the subformulas, box the immediate subformulas, star the atomic subformulas, circle the main operator, and indicate quantifier scope with underlines. Then (iii) say whether the formula is a sentence, and if it is not, explain why. a. H 1 x *b. .A1 x ! B 2 cf 1 x/ c. 8x.Dxc ! A1 g 2 ay/ d. 8x.B 2 xc ! 8yA1 g 2 ay/ e. .S ! .8wB 2 f 1 wh1 a ! 8z.H 1 w ! B 2 za/// E2.16. Explain why the following expressions are not formulas or sentences of Lq . Hint: You may find that an attempted tree will help you see what is wrong. a. H 1

CHAPTER 2. FORMAL LANGUAGES

60

b. g 2 ax *c. 8xB 2 xg 2 ax d. .8aA1 a ! .S ! B 2 zg 2 xa// e. 8x.Dax ! 8zK 2 zg 2 xa/ E2.17. For each of the following expressions, determine whether it is a formula and a sentence of Lq . If it is a formula, show it on a tree, and exhibit its parts as in E2.15. If it fails one or both, explain why. a. .L ! V / b. 8x.L ! K 1 h3 xb/ c. 8z8w.8xR2 wx ! K 2 zw/ ! M 2 zz/ *d. 8z.L1 z ! .8wR2 wf 3 axw ! 8wR2 f 3 azww// e. ..8w/B 2 f 1 wh1 a ! .8z/.H 1 w ! B 2 za//

2.3.4

Abbreviations

That is all there is to the official grammar. Having introduced the official grammar, though, it is nice to have in hand some abbreviated versions for official expressions. Abbreviated forms give us ways to manipulate official expressions without undue pain. First, for any variable x and formulas P and Q, AB (_) .P _ Q/ abbreviates .P ! Q/ (^) .P ^ Q/ abbreviates .P ! Q/ ($) .P $ Q/ abbreviates ..P ! Q/ ! .Q ! P // (9) 9xP abbreviates 8xP The first three are as from AB. The last is new. For any variable x, an expression of the form 9x is an existential quantifier — it is read, ‘there exists an x such that P ’. As before, these abbreviations make possible derived clauses to FR. Suppose P is a formula; then by FR(), P is a formula; so by FR(8), 8xP is a formula; so by FR() again, 8xP is a formula; but this is just to say that 9xP is a formula. With results from before, we are thus given,

CHAPTER 2. FORMAL LANGUAGES

61

FR0 (^) If P and Q are formulas, then .P ^ Q/ is a formula. (_) If P and Q are formulas, then .P _ Q/ is a formula. ($) If P and Q are formulas, then .P $ Q/ is a formula. (9) If P is a formula and x is a variable, then 9xP is a formula. The first three are from before. The last is new. And, as before, we can incorporate these conditions directly into trees for formulas. Thus 8x.A1 x ^ 9yA2 yx/ is a formula. x

y

x

These are terms by TR(v)

. . . . . . . . . . . A. . . . . . . .

(T)

A  A

A1 x

A2 yx

A1 x

9yA2 yx

@ @ @

These are formulas by FR(r)

These are formulas by FR() and FR0 (9)

.A1 x ^ 9yA2 yx/

This is a formula by FR0 (^)

8x.A1 x ^ 9yA2 yx/

This is a formula by FR(8)

In a derived sense, we carry over additional definitions from before. Thus, the main operator is the last operator added in its tree, subformulas are all the formulas in the formula part of a tree, atomic subformulas are the ones in the upper row of the formula part, and immediate subformulas are the one(s) to which a formula is directly connected by lines. Thus the main operator of 8x.A1 x ^ 9yA2 yx/ is the universal quantifier and the immediate subformula is .A1 x ^ 9yA2 yx/. In addition, a variable is in the scope of an existential quantifier iff it would be in the scope of the unabbreviated universal one. So it is possible to discover whether an expression is a sentence directly from diagrams of this sort. Thus, as indicated by underlines, 8x.A1 x ^ 9yA2 yx/ is a sentence. To see what it is an abbreviation for, we can reconstruct the formula on an unabbreviating tree, one operator at a time.

CHAPTER 2. FORMAL LANGUAGES y

x

x

. . . . . . . . . . . A. . . . . . . . A  A

(U)

A1 x

A2 yx

A1 x

9yA2 yx

@

62 y

x

x

. . . . . . . . . . . A. . . . . . . . A  A A1 x

A2 yx

A1 x

8yA2 yx

By AB(9)

@ @ @

.A1 x

^ 9yA2 yx/

8x.A1 x ^ 9yA2 yx/

.A1 x

@ @

! 8yA2 yx/

By AB(^)

8x.A1 x ! 8yA2 yx/

First the existential quantifier is replaced by the unabbreviated form. Then, where P and Q are joined by FR0 (^) to form .P ^ Q/, the corresponding unabbreviated expressions are combined into the unabbreviated form, .P ! Q/. At the last step, FR(8) applies as before. So 8x.A1 x ^ 9yA2 yx/ abbreviates 8x.A1 x ! 8yA2 yx/. Again, abbreviations are nice! Notice that the resultant expression is a formula and a sentence, as it should be. As before, it is sometimes convenient to use a pair of square brackets [ ] in place of parentheses ( ). And if the very last step of a tree for some formula is justified by FR(!), FR0 (_), FR0 (^), or FR0 ($), we may abbreviate that formula with the outermost set of parentheses or brackets dropped. In addition, for terms t1 and t2 we will frequently represent the formula Dt1 t2 as .t1 D t2 /. Notice the extra parentheses. This lets us see the equality symbol in its more usual “infix” form. When there is no danger of confusion, we will sometimes omit the parentheses and write, t1 D t2 . Also, where there is no potential for confusion, we sometimes omit superscripts. Thus in Lq we might omit superscripts on relation symbols — simply assuming that the terms following a relation symbol give its correct number of places. Thus Ax abbreviates A1 x; Axy abbreviates A2 xy; Axf 1 y abbreviates A2 xf 1 y; and so forth. Notice that Ax and Axy, for example, involve different relation symbols. In formulas of Lq , sentence letters are distinguished from relation symbols insofar as relation symbols are followed immediately by terms, where sentence letters are not. Notice, however, that we cannot drop superscripts on function symbols in Lq — thus, even given that f and g are function symbols rather than constants, apart from superscripts, there is no way to distinguish the terms in, say, Afgxyzw. As a final example, 9y.c D y/ _ 8xRxf 2 xd is a formula and a sentence.

CHAPTER 2. FORMAL LANGUAGES y

y

D D



x x

d

D A   D A  D  D A2  D  D f xd D  . . . . . . . . . . . . . . . D. . .. . . . . . DD DD .c D y/

Rxf 2 xd

.c D y/

8xRxf 2 xd

63 y

y

x x

d

 D A D A  D D A2  D  D  D f xd  D . . . . . . . . . . . . . . . D. . .. . . . . . DD DD  

D

Dcy

R2 xf 2 xd

Dcy

8xR2 xf 2 xd

(V)

9y.c D y/ @ @ @

9y.c D y/ _ 8xRxf 2 xd

8yDcy @ @ @

.8yDcy ! 8xR2 xf 2 xd /

The abbreviation drops a superscript, uses the infix notation for equality, uses the existential quantifier and wedge, and drops outermost parentheses. As before, the righthand diagram is not a direct demonstration that .8yDcy ! 8xR2 xf 2 xd / is a sentence. However, it unpacks the abbreviation, and we know that the result is an official sentence, insofar as the left-hand tree, with its application of derived rules, tells us that 9y.c D y/ _ 8xRxf 2 xd is an abbreviation of formula and a sentence, and the right-hand diagram tells us what that expression is. E2.18. For each of the following expressions, (i) Demonstrate that it is a formula of Lq with a tree. (ii) On the tree bracket all the subformulas, box the immediate subformulas, star the atomic subformulas, circle the main operator, and indicate quantifier scope with underlines. Then (iii) say whether the formula is a sentence, and if it is not, explain why. a. .A ! B/ $ .A ^ C / b. 9xF x ^ 8yGxy *c. 9xAf 1 g 2 ah3 zwf 1 x _ S d. 8x8y8z.Œ.x D y/ ^ .y D z/ ! .x D z// e. 9yŒc D y ^ 8xRxf 1 xy

CHAPTER 2. FORMAL LANGUAGES

64

Grammar Quick Reference VC

(p) Punctuation symbols: (, ) (o) Operator symbols: , !, 8 (v) Variable symbols: i : : : z with or without integer subscripts (s) A possibly-empty countable collection of sentence letters (c) A possibly-empty countable collection of constant symbols

TR

(f) For any integer n

1, a possibly-empty countable collection of n-place function symbols

(r) For any integer n

1, a possibly-empty countable collection of n-place relation symbols

(v) If t is a variable x, then t is a term. (c) If t is a constant c, then t is a term. (f) If hn is a n-place function symbol and t1 : : : tn are n terms, then hn t1 : : : tn is a term. (CL) Any term may be formed by repeated application of these rules.

FR

(s) If S is a sentence letter, then S is a formula. (r) If Rn is an n-place relation symbol and t1 : : : tn are n terms, Rn t1 : : : tn is a formula. () If P is a formula, then P is a formula. (!) If P and Q are formulas, then .P ! Q/ is a formula. (8) If P is a formula and x is a variable, then 8xP is a formula. (CL) Any formula can be formed by repeated application of these rules.

An operator’s scope includes just the formula in which it is introduced; a variable x is free iff it is not in the scope of an x-quantifier; an expression is a sentence iff it is a formula with no free variables. Subformulas are all the formulas in the tree; atomic subformulas appear in the top row; immediate subformulas are the ones to which a formula is directly connected by lines; the main operator is the last operator added. AB (_) .P _ Q/ abbreviates .P ! Q/ (^) .P ^ Q/ abbreviates .P ! Q/ ($) .P $ Q/ abbreviates ..P ! Q/ ! .Q ! P // (9) 9xP abbreviates 8xP FR0 (^) If P and Q are formulas, then .P ^ Q/ is a formula. (_) If P and Q are formulas, then .P _ Q/ is a formula. ($) If P and Q are formulas, then .P $ Q/ is a formula. (9) If P is a formula and x is a variable, then 9xP is a formula. The generic language Lq includes the equality symbol ‘D’ along with, Sentence letters: A : : : Z with or without integer subscripts Constant symbols: a : : : h with or without integer subscripts Function symbols: for any n

1, an : : : z n with or without integer subscripts

Relation symbols: for any n

1, An : : : Z n with or without integer subscripts.

CHAPTER 2. FORMAL LANGUAGES

65

*E2.19. For each of the formulas in E2.18, produce an unabbreviating tree to find the unabbreviated expression it represents.

*E2.20. For each of the unabbreviated expressions from E2.19, produce a complete tree to show by direct application of FR that it is an official formula. In each case, using underlines to indicate quantifier scope, is the expression a sentence? does this match with the result of E2.18?

2.3.5

Another Language

To emphasize the generality of our definitions VC, TR, and FR, let us introduce a < language like one with which we will be much concerned later in the text. LNT is like a minimal language we shall introduce later for number theory. Recall that VC leaves open what are the constant symbols, function symbols, sentence letters, and relation symbols of a quantificational language. So far, our generic language Lq fills these in < by certain conventions. LNT replaces these with, Constant symbol: ; one-place function symbol: S two-place function symbols: C;  two-place relation symbols: D; < < and that is all. Later we shall introduce a language like LNT except without the < symbol; for now, we leave it in. Notice that Lq uses capitals for sentence letters and lowercase for function symbols. But there is nothing sacred about this. Similarly, Lq indicates the number of places for function and relation symbols by superscripts, < where in LNT the number of places is simply built into the definition of the symbol. In < fact, LNT is an extremely simple language! Given the vocabulary, TR and FR apply in the usual way. Thus ;, S ; and S S; are terms — as is easy to see on a tree. And ; ^ b > ;/ ! G.a; b; S v/

CHAPTER 13. GÖDEL’S THEOREMS

665

Begin with b D ; _ b > ; and go for the existentially quantified goal. In the second case, there is some l such that b D S l and it is easy to show a  ; C b D b  1 and generalize. If a or b is not greater than ; then vacuously d.a; b/ is just ;. Otherwise, the notion is more significant. Again, PA proves a series of results. Observe again that if we are interested in whether a prime divides some b we are interested in whether Pr.Sa/ ^ ajb since it is the successor that is divided into b. *T13.23. The following result in PA: (a) PA ` Pr.;/ (b) PA ` Pr.1/ (c) PA ` Pr.2/ *(d) PA ` 8xŒx > 1 ! 9z.Pr.S z/ ^ zjx/ *(e) PA ` Rp.a; b/ $ 9xŒPr.S x/ ^ xja ^ xjb (f) PA ` 8x8yŒG.a; b; x/ ! G.a; b; x  y/ *(g) PA ` .a > ; ^ b > ;/ ! 8x8yŒ.G.a; b; x/ ^ G.a; b; y/ ^ x  y/ ! G.a; b; x : y/ *(h) PA ` ŒRp.a; b/ ^ a > ; ^ b > ; ! G.a; b; 1/ *(i) PA ` ŒPr.Sa/ ^ aj.b  c/ ! .ajb _ ajc/ T13.23h is important. But the argument is relatively complex; the main stages are as on p. 666. T13.23(a) - (c) are simple particular facts. From (d) every number greater than one is divided by some prime (which may or may not be itself). From (e), a and b are relatively prime iff there is no prime that divides them both; in one direction this is obvious — if a prime divides them both, then they are not relatively prime; in the other direction, if some number other than (the successor of) zero divides them both, then some prime of it divides them both. (f) and (g) let you manipulate G; they are required for (h) which is in turn required for (i). (h) is an instance of Bézout’s lemma according to which there are x and y such that ax C d D by when d is the greatest common divisor of a and b; if a and b are relatively prime, their greatest common

CHAPTER 13. GÖDEL’S THEOREMS

666

T13.23h 1. Œ.a > ; ^ b > ;/ ! G.a; b; Sd.a; b// ^ .8y < d.a; b//Œ.a > ; ^ ^b > ;/ ! G.a; b; Sy/

def d

2. .a > ; ^ b > ;/ ! G.a; b; Sd.a; b// 3. Rp.a; b/ ^ a > ; ^ b > ;

1 ^E A (g !I)

4. Rp.a; b/ 5. 8xŒ.xja ^ xjb/ ! x D ; 6. a > ; ^ b > ; 7. G.a; b; Sd.a; b// 8. G.a; b; a/ 9. G.a; b; b/ 10. 8xŒG.a; b; x/ ! d.a; b/jx 11. d.a; b/ja 12. d.a; b/jb 13. d.a; b/ja ^ d.a; b/jb 14. d.a; b/ D ; 15. G.a; b; 1/ 16. ŒRp.a; b/ ^ a > ; ^ b > ; ! G.a; b; 1/

So the argument reduces to [a] - [c]. For hints, see the associated E13.20.

divisor is one. (i) is sometimes known as Euclid’s lemma: if Sa is prime and Sa divides b  c then Sa divides b or c; if Sa is prime and divides b  c then it must appear in the factorization of b or the factorization of c — so that it divides one or the other. Now least common multiple. Given a function m.i /, lcmfm.i / j i < kg is the least y > ; such that for any i < k, S m.i / divides y. We avoid worries about the case when m.i / D ; by our usual account of factor. And since y > ; it is possible to define a predecessor to the least common multiple, helpful when switching between the numerator and denominator of fractions. *Def [lcm] lcmfm.i / j i < kg D vŒv > ; ^ .8i < k/m.i /jv (i) PA ` 9xŒx > ; ^ .8i < k/m.i /jx Hint: This is an argument by IN on k. For the basis, you may assert that 1 > ;; then the argument is trivial. For the main argument, under the assumptions 9xŒx > ; ^ .8i < j /m.i /jx for !I and a > ; ^ .8i < j /m.i /ja for 9E, set out to show a  S m.j / > ; ^ .8i < Sj /m.i /j.a  S m.j // and generalize.

3 ^E 4 def 3 ^E 2,6 !E [a] [b] [c] 8,10 8E 9,10 8E 11,12 ^I 5,13 8E 7,14 DE 3-15 !I

CHAPTER 13. GÖDEL’S THEOREMS

667

Because lcm is defined by minimization, only the existence condition is required. As a matter of notation, let lŒmk = lcmfm.i / j i < kg and, where m is clear from context, let lk = lcmfm.i / j i < kg. Def [plm] v D plmfm.i / j i < kg $ S v D lcmfm.i / j i < kg (i) PA ` 9v.S v D lk / (ii) PA ` 8x8yŒ.S x D lk ^ Sy D lk / ! x D y Again, let pŒmk = plmfm.i / j i < kg and, where m is clear from context, pk = plmfm.i / j i < kg. *T13.24. The following result in PA: (a) PA ` l; D 1 (b) PA ` j < k ! m.j /jlk *(c) PA ` .8i < k/m.i /jx ! pk jx *(d) PA ` 8nŒ.Pr.S n/ ^ njlk / ! .9i < k/njS m.i / (a) for any function m.i /, the least common multiple for i < 0 defaults to 1. (b) applies the definition for the result that when j < k, m.j / divides lcmfm.i / j i < kg. (c) is perhaps best conceived by prime factorization: the least common multiple of some collection has all the primes of its members and no more; but any number into which all the members of the collection divide must include all those primes; so the least common multiple divides it as well. (d) is the related result that if a prime divides the least common multiple of some collection, then it divides some member of the collection. Finally we arrive at the Chinese Remainder Theorem. Let m.i / be a function such that (successors of) its values are relatively prime; h.i / is a function whose values are to be matched by remainders. Then the theorem tells us that if for all i < k, m.i / is “big enough” (m.i / > ; and m.i /  h.i /), and if for all i < j < k, S m.i / and S m.j / are relatively prime, then 9p.8i < k/rm.p; m.i // D h.i /. So the remainder of p and m.i / matches the value of h.i /. *T13.25. PA ` Œ.8i < k/.m.i/ > ;^m.i/  h.i//^8i8j.i < j ^j < k ! Rp.Sm.i/; Sm.j /// ! 9p.8i < k/rm.p; m.i// D h.i/. Let, A.k/ = .8i < k/.m.i/ > ; ^ m.i/  h.i// ^ 8i8j.i < j ^ j < k ! Rp.Sm.i/; Sm.j ///

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B.k/ = 9p.8i < k/rm.p; m.i// D h.i/.

So we want PA ` A.k/ ! B.k/. By induction on n we show 8nŒn  k ! .A.n/ ! B.n//. The result follows immediately with k  k. For the overall structure of the argument, see p. 669. This gives us remainders to match h.i /. It remains to apply this result to the ˇfunction. For application to the ˇ-function we shall require a couple notions for maximum value. First maxp for the greatest of a pair of values, and then maxs for the maximum from a set. Def [maxp] PA ` maxp.x; y/ D vŒv  x ^ v  y (i) PA ` 9vŒv  x ^ v  y Hint: x  y _ y > x; in either case the result is easy. Def [maxs] PA ` maxsfm.i / j i < kg D vŒ.8i < k/m.i /  v (i) PA ` 9vŒ.8i < k/m.i /  v Hint: First obtain maxp and T13.26a. Then the argument is by IN on k. For the show you will have assumptions of the sort .8i < j /m.i /  l and a < Sj ; then a < j _ a D j ; in either case you will be able to show that m.a/  maxp.l; m.j //. So maxp.x; y/ is the maximum of x and y, and maxsfm.i / j i < kg is the maximum from m.i / with i < k. As a matter of notation, let maxsŒmk = maxsfm.i / j i < kg and where m is understood, maxsk = maxsfm.i / j i < kg. A couple of results are immediate with T13.17b. T13.26. The following result in PA. (a) PA ` maxp.x; y/  x ^ maxp.x; y/  y (b) PA ` .8i < k/m.i /  maxsk These simply state the obvious: that the maximum is greater than or equal to the rest. From (a) the maximum is the greater of the the two in the pair; from (b) the maximum is the greatest of the values of the function. Now we are in a position to generate some results for the ˇ function. With values of q and m.i / as below, we may demonstrate the antecedent to the CRT (T13.25), and so obtain its consequent — where this is a result for the ˇ-function.

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T13.25 1. ;  k ! .A.;/ ! B.;// 2.

a  k ! .A.a/ ! B.a//

[a] A (g !I)

3.

Sa  k

A (g !I)

4. 5. 6. 7.

a ; ^ m.i /  h.i // ^ 8i 8j..i < j ^ j < a/ ! Rp.S m.i /; S m.j /// ! 9p.8i < a/rm.p; m.i // D h.i / .8i < Sa/.m.i / > ; ^ m.i /  h.i // ^ 8i 8j..i < j ^ j < Sa/ ! Rp.S m.i /; S m.j /// .8i < Sa/.m.i / > ; ^ m.i /  h.i // 8i 8j..i < j ^ j < Sa/ ! Rp.S m.i /; S m.j /// 9p.8i < a/rm.p; m.i // D h.i / .8i < a/rm.r; m.i // D h.i / Rp.lŒma ; S m.a// S m.a/ > ; la > ; G.la ; S m.a/; 1/ : G.la ; S m.a/; r C .la 1/  h.a// : 9x9y.la  x C Œr C .la 1/  h.a/ D S m.a/  y/ : la  b C Œr C .la 1/  h.a/ D S m.a/  c s D la  .b C h.a// C r s D S m.a/  c C h.a/ .8i < Sa/rm.s; m.i // D h.i / 9p.8i < Sa/rm.p; m.i // D h.i / B.Sa/ B.Sa/ B.Sa/ A.Sa/ ! B.Sa/ Sa  k ! .A.Sa/ ! B.Sa// Œa  k ! .A.a/ ! B.a// ! ŒSa  k ! .A.Sa/ ! B.Sa// 8n.Œn  k ! .A.n/ ! B.n// ! ŒS n  k ! .A.S n/ ! B.S n/// .8n  k/.A.n/ ! B.n// kk A.k/ ! B.k/

6 abv 7 abv 9 ^E 9 ^E [b] A (g 129E) [c] T13.11e def la 14,15,16 T13.23h 17 T13.23f 18 def G A (g 199E) def [d] [e] 23 9I 24 abv 19,20-25 9E 12,13-26 9E 7-27 !I 3-28 !I 2-29 !I 30 8I 1,31 IN T13.11m 32,33 (8E)

The core of the derivation is to obtain (21) and (22) and from them (23). For a claim about all i < Sa, s appears in the forms from both (21) and (22). For any i < a and x, m.i / divides la x evenly; so m.i / divides the first term from (21) evenly; so the remainder of m.i / and s is the same as the remainder of m.i / with r — and with (13) this is just h.i /. But the multiplier b C h.a/ is chosen so that from (20) and (21), we get (22); so when i D a, m.i / divides the first term evenly, and since m.i /  h.a/ again the remainder of m.i / and s is h.i /. Putting these together, for any i < Sa, the remainder of m.i / and s is h.i /. The “trick” to this is in the construction of s so that remainders for i < a stay the same, but the remainder at a is h.a/. For this construction see Boolos, The Logic of Provability, 30-31.

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*T13.27. PA ` 9p9q.8i < k/ˇ.p; q; i / D h.i /. Let r = maxp.k; maxsŒhk /; s = S r; q = lcmfi j i < sg; m.i / = q  S i . Recall from Def [beta] that PA ` ˇ.p; q; i / D rm.p; q  S i /. And we may reason, 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

.8i < k/.m.i/ > ; ^ m.i/  h.i// 8i8j Œ.i < j ^ j < k/ ! Rp.Sm.i/; Sm.j // 9p.8i < k/rm.p; m.i// D h.i/ m.i/ D q  Si 9p.8i < k/rm.p; q  Si/ D h.i/ ˇ.p; q; i/ D rm.p; q  Si/ 9p.8i < k/ˇ.p; q; i/ D h.i/ .8i < k/ˇ.p; q; i/ D h.i/ 9q.8i < k/ˇ.p; q; i/ D h.i/ 9p9q.8i < k/ˇ.p; q; i/ D h.i/

11. 9p9q.8i < k/ˇ.p; q; i/ D h.i/

[i] [ii] 1,2 T13.25 def 3,4 DE def 5,6 DE A (g 79E) 8 9I 9 9I 7,8-10 9E

So the demonstration reduces to that of [i] and [ii], the two conjuncts to the antecedent of CRT (T13.25). [i]: Under the assumption j < k for (8I) it will be easy to show m.j / > ;; then you will be able to use T13.26 to show h.j / < s; but also with T13.24b that rjq and from this that s  q which gives s  q  Sj and the result you want. [ii]: For the main structure of the argument, see p. 671. Now a theorem that uses this result to show that a ˇ-function for values < k can always be extended to another like it but with an arbitrary k th value. We show that given ˇ.a; b; i / there are sure to be p and q such that ˇ.p; q; i / is like ˇ.a; b; i / for i < k and for arbitrary n, ˇ.p; q; k/ D n. This is because we may define a function h which is like ˇ.a; b; i / for i < k and otherwise n — and find p; q such that ˇ.p; q; i / matches it. As a preliminary, Def [h.i/] PA ` v D h.i / $ Œ.i < k ^ v D ˇ.a; b; i // _ .i  k ^ v D n/ (i) PA ` 9vŒ.i < k ^ v D ˇ.a; b; i // _ .i  k ^ v D n/ (ii) PA ` 8x8yŒ.Œ.i < k ^ x D ˇ.a; b; i // _ .i  k ^ x D n/ ^ Œ.i < k ^ y D ˇ.a; b; i // _ .i  k ^ y D n// ! x D y

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T13.27 1.

i 1; then S k is prime or not; if it is prime, the result is immediate; if it is not, you will be able to show Sj  k and apply the assumption. (e): From left to right, under the assumption for $I assume 9xŒPr.S x/ ^ xja ^ xjb and Pr.Sj / ^ j ja ^ j jb for I and 9E; then you should be able to show that 1 < Sj and 1 – Sj ; in the other direction, under the assumption for $I and then j ja ^ j jb for !I, j D ; _ j > ; by T13.11f; the latter is impossible, which gives the result you want. (g): Under the assumptions a > ; ^ b > ; and then G.a; b; i / ^ G.a; b; j / ^ i  j for !I and then ap C i D bq and ar C j D bs for 9E, starting with .bq C bar/ C .bsa : bs/ D .bq C bar/ C .bsa : bs/ by DI, with some effort, you will be able to show aŒ.p C bs/ C .br : r/ C .i : j / D bŒ.q C ar/ C .sa : s/ and generalize. (i): Under the assumption Pr.Sa/ ^ aj.b  c/ assume a − b with the idea of obtaining a − b ! ajc for Impl; set out to show Rp.b; Sa/ for an application of T13.23h to get 9x9yŒbx C 1 D Sa  y; with this, you will have bp C 1 D Sa  q by 9E; and you should be able to show ajcbp and aj.cbp C c/ for an application of T13.22g.

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*E13.21. Show the condition for Def [lcm] and provide a demonstration for T13.24d. Hard core: show all of the results for Def [lcm], Def [plm] and T13.24. Hints for T13.24. (c): Let q D qt.x; pk / and r D rm.x; pk /; assume .8i < k/m.i /jx for !I; you have .8y < lk /Œy > ; ^ .8i < k/m.i /jy from def lk with T13.17c; you should be able to apply this to show that r D ; and so that pk jx. (d): This is an induction on k. The basis is straightforward given l; D 1 from T13.24a; for the main argument, you have .8i < j /m.i /jlj from def lj ; under assumptions 8nŒ.Pr.S n/ ^ njlj / ! .9i < j /njS m.i / and Pr.Sa/ ^ ajlSj for !I, you should be able to use T13.24c to show pSj j.lj  S m.j //; and from this ajlj _ ajS m.j /; in either case, you have your result.

*E13.22. Provide derivations to show each of [a] - [e] to complete the derivation for T13.25. Hints. [c]: Suppose otherwise; with T13.23e there is a u such that Pr.S u/ ^ ujla ^ ujS m.a/; then with T13.24d there is a v < a such that ujS m.v/ so that with (11) Rp.S m.v/; S m.a//. But this is impossible with ujS m.a/, ujS m.v/ and T13.23e. [d]: By Def [lcm], la > ; so that h.a/la > h.a/. Then with T13.21a and T13.21p you can show s D .la b CŒr C.la : 1/h.a//Ch.a/ and apply (20). [e]: Suppose for (8I) u < Sa; then u < a _ u D a. In the first case, with T13.24b and T13.22d m.u/jla .b C h.a//; so that there is a v such that S m.u/v D la .b C h.a//; then using (21) and T13.22k, rm.d; m.u// D rm.s; m.u//; so that you can apply (13). In the second case, with (22) and T13.22k rm.d; m.u// D rm.h.u/; m.u//; but from (10), m.u/  h.u/ and you will be able to show that rm.h.u/; m.u// D h.u/. E13.23. Provide a derivation to show the condition of Def [maxs]. Hard core: Provide justifications for Def [maxs] and Def [maxp]; and show the results in T13.26.

*E13.24. Complete the demonstration for T13.27.

*E13.25. Show T13.28. Hard core: show the conditions for Def [h.i /].

*E13.26. Complete the demonstration of T13.29 by showing the zero case.

CHAPTER 13. GÖDEL’S THEOREMS

13.4

679

The Second Condition: .P ! Q/ ! .P ! Q/

We turn now to demonstration of the second derivability condition. Again there is some background — after which demonstration of the condition itself is straightforward. The overall idea is simple: Suppose both T ` .P ! Q/ and T ` P . Then there are j and k such that PRFT.j; pP ! Qq/ and PRFT.k; pP q/. Intuitively, then, ypQq numbers a proof of Q — for we prove P ! Q and P , so that Q l = j?k?2

b

4

b

b

follows immediately as the last line by MP. So PRFT.l; pQq/, and Q follows from the assumptions. The task is to prove all of this in PA. For this, having shown that PA defines recursive functions, we require some results about them.

13.4.1

Equivalencies

To start, observe that plus.x; y/, say, is defined by a complex expression through recursion, and so is not the same expression as our old friend x C y. Thus it is not obvious that our standard means for manipulation of C apply to plus. We could recover our ordinary results if we could show PA ` x C y D plus.x; y/. And similar comments apply to other ordinary functions, operators and relations. Thus initially we seek to show that defined functions, operators and relations are equivalent to ones with which we are familiar. Again many details are shifted to exercises and/or answers to exercises. We begin with equivalences between functions, operators and relations already defined in PA, and ones that result by T13.19 and T13.30. In particular, we begin with equivalences for S, C, , D, , ; ^ v D 1/ (ii) PA ` 8u8vfŒ..y D ; ^ u D ;/ _ .y > ; ^ u D 1// ^ ..y D ; ^ v D ;/ _ .y > ; ^ v D 1// ! u D vg Def [csg] PA ` v D csg.y/ $ .y D ; ^ v D 1/ _ .y > ; ^ v D ;/ (i) PA ` 9vŒ.y D ; ^ v D 1/ _ .y > ; ^ v D ;/ (ii) PA ` 8u8vfŒ..y D ; ^ u D 1/ _ .y > ; ^ u D ;// ^ ..y D ; ^ v D 1/ _ .y > ; ^ v D ;// ! u D vg And some basic results on these notions, T13.35. The following result in PA. (a) PA ` pred.;/ D ; (b) PA ` pred.1/ D ; (c) PA ` y > ; ! S pred.y/ D y

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(d) PA ` pred.Sy/ D y (e) PA ` y D ; $ sg.y/ D ; (f) PA ` y > ; $ sg.y/ D 1 (g) PA ` y D ; $ csg.y/ D 1 (h) PA ` y > ; $ csg.y/ D ; (a) - (d) recover from the definition some basic results for pred. (e) and (f) extract basic information for the behavior of sg; and then (g) and (h) for csg. And given these notions in PA, we can build on them for another set of equivalents. *T13.36. The following result in PA. (a) PA ` pred.y/ D pred.y/ *(b) PA ` subc.x; y/ D x : y (c) PA ` absval.x - y/ D .x : y/ C .y : x/ (d) PA ` sg.y/ D sg.y/ (e) PA ` csg.y/ D csg.y/ *(f) PA ` Eq.x; y/ $ x D y (g) PA ` Leq.x; y/ $ x  y (h) PA ` Less.x; y/ $ x < y *(i) PA ` Neg.P.x// E $ P.x/ E (j) PA ` Dsj.P.x/; E Q.y// E $ P.x/ E _ Q.y/ E Again, for hints see the associated exercise, E13.29. So this theorem delivers the equivalences we expect for pred, subc, absval, sg, csg, Eq, Leq, Less, Neg, and Dsj. Given this, we will typically move without comment from some PA ` Dsj.A; B/ given from T13.32 to PA ` A_ B. And similarly in other cases. We pause to remark on a simple consequence for characteristic functions. Recall from (Cf) that a characteristic function is (officially) of the sort sg.p.Ex// so that,

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T13.37. For any recursive characteristic function chR .Ex/, PA ` chR .x/ E D ;_ chR .x/ E D 1. From (Cf), chR .Ex/ is of the sort sg.p.Ex//; so with T13.32, PA ` chR .x/ E D sg.p.x//. E The result is nearly immediate with PA ` p.x/ E D ; _ p.x/ E >; and results for sg. It is worth observing that this theorem, which depends on results for functions through T13.36d, is independent of any applications of T13.30 or T13.32b for operator symbols. There is therefore no problem about the appeal to T13.37 for T13.30 on p. 674n12. Now reasoning for the bounded quantifiers, bounded minimization and a couple relations built on them. *T13.38. The following result in PA. E y/ $ .9y  z/P.x; E y/ *(a) PA ` .9 y  z/P.x; (b) PA ` .9 y < z/P.x; E y/ $ .9y < z/P.x; E y/ (c) PA ` . 8y  z/P.x; E y/ $ .8y  z/P.x; E y/ E y/ $ .8y < z/P.x; E y/ (d) PA ` . 8y < z/P.x; *(e) PA ` .y  z/P.x; E y/ $ .y  z/P.x; E y/ (f) PA ` Fctr.m; n/ $ mjn *(g) PA ` Prime.n/ $ Pr.n/ The argument for T13.38e is particularly involved. Recall from chapter 12 that m.Ex; z/ = .y  z/P.Ex; y/ is defined by means of R.Ex; n/ corresponding to .9y  n/P.Ex; y/ and q.Ex; n/ corresponding to .y  n/P.Ex; y/. The main reasoning is by IN to show q.x; E n/ D .y  n/P.x; E y/. See p. 686 for the main outlines of that part. T13.38 delivers the equivalences we expect for the bounded quantifiers, bounded minimization, factor and prime. At this stage, we have defined in PA functions, relations and operators corresponding to all the recursive functions, relations and operators. And in simple cases we have established equivalences to functions, relations and operators already defined. Thus supposing T is a theory including PA, we are in a position simply to write down the following.

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T13.38e 1. 2. 3. 4. 5.

q.x; E ;/ D .y  ;/P.x; E y/ chR .x; E j / D ; _ chR .x; E j/ D 1 chR .x; E j / D ; $ .9y  j /P.x; y/ q.x; E Sj / D hq.x; E j; q.x; E j // hq.x; E j; u/ D plus.u; chR .x; E j //

6. hq.x; E j; u/ D u C chR .x; E j/ 7. hq.x; E j; q.x; E j // D q.x; E j / C chR .x; E j/ 8. q.x; E Sj / D q.x; E j / C chR .x; E j/ 9. q.x; E j / D .y  j /P.x; E y/ 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.

a D q.x; E j/ b D q.x; E Sj / b D a C chR .x; E j/ a D .y  j /P.x; E y/ a D yŒy D j _ P.x; E y/ .8w < a/Œw ¤ j ^ P.x; E w/ a D j _ P.x; E a/ aDj P.x; E j / _ P.x; E j/ P.x; E j/

[a] T13.37 def, T13.38a T13.31b def from least, T13.32 5 T13.34d 6 8E 4,7 DE A (g !I) abv abv 8,10,11 DE 9,10 DE 13 def 14 T13.17c 14 T13.17b A (g 16_E) T3.1 A (g 18_E)

20.

Œb D Sj _ P.x; E b/ ^ .8w < b/.w ¤ Sj ^ P.x; E w//

[b]

21.

P.x; E j/

A (g18_E)

22.

Œb D Sj _ P.x; E b/ ^ .8w < b/.w ¤ Sj ^ P.x; E w//

[c]

23.

Œb D Sj _ P.x; E b/ ^ .8w < b/.w ¤ Sj ^ P.x; E w//

18,19-20,21-22 _E

24.

P.x; E a/

A (g 16_E)

Œb D Sj _ P.x; E b/ ^ .8w < b/.w ¤ Sj ^ P.x; E w//

[d]

25. 26. 27. 28. 29.

Œb D Sj _ P.x; E b/ ^ .8w < b/.w ¤ Sj ^ P.x; E w// b D yŒy D Sj _ P.x; E j / b D .y  Sj /P.x; E y/ q.x; E Sj / D .y  Sj /P.x; E y/

30. Œq.x; E j / D .y  j /P.x; E y/ ! Œq.x; E Sj / D .y  Sj /P.x; E y/ 31. 8n.Œq.x; E n/ D .y  n/P.x; E y/ ! Œq.x; E S n/ D .y  S n/P.x; E y// 32. q.x; E n/ D .y  n/P.x; E y/

16,17-23,24-25 _E 26 def  27 def 28 abv 9-29 !I 30 8I 1,31 IN

Hints: The zero case [a] is straightforward with T13.18a; for [b] you will be able to show that b D Sj ; for [c] and [d] you will be able to show b D a. And the final result is nearly automatic from this.

T13.39. The following are theorems of PA: (a) PA ` unv.v; n/ D cncat.cncat.p8q; v/; n/ and similarly for cnd and neg (b) PA ` Mp.m; n; o/ $ cnd.n; o/ D m

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(c) PA ` Gen.m; n/ $ .9v  n/ŒVar.v/ ^ n D unv.v; m/ (d) PA ` Icon.m; n; o/ $ Mp.m; n; o/ _ .m D n ^ Gen.n; o// (e) PA ` Axiomad1.n/ $ .9p  n/.9q  n/ŒWff .p/ ^ Wff .q/ ^ n D cnd.p; cnd.q; p/ and similarly for the other axioms (f) PA ` Axiompa.n/ $ Axiomad1.n/ _ : : : _ Axiomq1.n/ _ : : : _ Axiompa7.n/ (g) PA ` Prft.m; n/ $ exp.m; len.m/

:

1/ D n ^ m > 1 ^ .8k < len.m//ŒAxiomt.exp.m; k// _

.9i < k/.9j < k/Icon.exp.m; i/; exp.m; j /; exp.m; k//

These follow directly from our results with recursive definitions. So for example T13.33a (the definition MP, with T13.32) gives us, PA ` Mp.m; n; o/ $ Eq.cnd.n; o/; m/; then with T13.36f, we arrive at (b). And similarly in other cases. Where Mp, cnd and the like are defined relative to corresponding recursive functions, it is important that the operators in expressions above are the ordinary operators of LNT . Thus we shall be able to manipulate the expressions in the usual ways. We shall find these results useful for the following! E13.27. Produce derivations to show T13.31a and T13.34e. Hard core: show the remaining cases from T13.34.

E13.28. Show (i) of the condition for Def [pred] and then T13.35c. Hard core: Show each of the conditions for Def [pred], Def [sg] and Def [csg] and all of the results in T13.35.

*E13.29. Show a, g and j from T13.36. Hard core: Demonstrate each of the results in T13.36. Hints for T13.36. (b): This works in the usual way up to the point in the show stage where you get subc.x; Sj / D pred.x : j /; then it will take some work to show x : Sj D pred.x : j /; for this begin with x  j _ x > j by T13.11q; the first case is straightforward; for the second, you will be able to show S.x : Sj / D S pred.x : j / and apply T6.41. (f): For this relation, you have Eq.x; y/ $ sg.absval.x -y// D ; from the def EQ and T13.32; this gives Eq.x; y/ $ Œ.x : y/ C .y : x/ D ;; now for $I, the case from x D y is easy; from Eq.x; y/, you have x  y _ x < y from T13.11q; the cases are

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not hard and similar (since x < y gives y  x). (i): This is straightforward with P.x/ E $ chP .x/ E D ; and Neg.P.x// E $ csg.chP .x// E D ; from NEG with T13.32.

*E13.30. Show T13.38a. Hard core: show T13.37 along with each of the results in T13.38. Hints for T13.38. (a): Recall from chapter 12 that s.Ex; z/ = .9y  z/P.Ex; y/ is defined by means of a R.Ex; n/ corresponding to .9y  n/P.Ex; y/; the main argument is to show by IN that PA ` chR .x; E n/ D ; $ .9y  n/P.x; E y/. You have P.x; E y/ $ chP .x; E y/ D ; from T13.30. For the zero case, you have chR .x; E ;/ D gchR .x/, E and gchR .x/ E D chP .x; E ;/ from the definitions with T13.32; for the main reasoning, you have chR .x; E Sj / D hchR .x; E j; chR .x; E j //, and hchR .x; E j; u/ D timesŒu; chP .x; E suc.j // from the definitions with T13.32; once you have finished the induction, it is a simple matter of applying chS .x; E z/ D chR .x; E z/ from the definition, and where where S.x; E z/ just abbreviates .9 y  z/P.x; E y/, applying S.x; E z/ $ chS .x; E z/ D ; to get .9 y  z/P.x; E y/ $ .9y  z/P.x; E y/. (f) and (g): Given previous results, the left and right sides have nearly matching definitions except that the recursive side includes a bounded quantifier — so that you have to work to show the bound obtains for one direction of the biconditional.

13.4.2

Further Results

T13.39 gives us functions in PA corresponding to all the ones from chapter 12. Now we require the ability to manipulate them. Thus we begin with some results for exponentiation, factorial and the like, and continue through to complex notions including Wff and formsub. At this stage, we are acquiring results, not by demonstrating equivalence to expressions already defined (since there are no such expressions already defined), but by showing them directly for symbols defined for the recursive functions. *T13.40. The following are theorems of PA. (a) (i) PA ` m; D suc.zero.x// (ii) PA ` mS n D idnt33 .m; n; mn /  m (b) PA ` m1 D m 2

(c) PA ` 2 D 4

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(d) PA ` a > ; ! ;a D ; (e) PA ` ma  mb D maCb (f) PA ` m  n ! ma  na (g) PA ` pred.mb /jmaCb (h) PA ` .a > ; ^ m > 1/ ! pred.maCb / − mb (i) PA ` m > ; ! ma > ; (j) PA ` .m > ; ^ a  b/ ! ma  mb (k) PA ` .m > 1 ^ a > b/ ! ma > mb (l) PA ` a > ; ! ma  m *(m) PA ` m > 1 ! a < ma (n) PA ` m > 1 ! .ma D mb ! a D b/ (a) gives the recursive conditions from which the rest follow. Then (b) - (n) are basic results that should be accessible from ordinary arithmetic. *T13.41. The following are theorems of PA. (a) (i) PA ` fact.;/ D 1 (ii) PA ` fact.S n/ D fact.n/  S n (b) PA ` fact.1/ D 1 (c) PA ` fact.n/ > ; (d) PA ` .8y < n/yjfact.n/ *(e) .9y  S ffact.n/g/Œn < y ^ Pr.y/ These are some basic results for factorial. Again (a) gives the recursive conditions from which the rest follow. (b) is a simple particular fact; and the result from (c) is obvious. (d) is a consequence of the way the factorial includes successors of all the numbers less than it. We will be able to take advantage of (e) immediately below. *T13.42. The following are theorems of PA.

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(a) (i) PA ` pi.;/ D 2 (ii) PA ` pi.S n/ D idnt22 .n; .z  S ffact.pi.n//g/Œpi.n/ < z ^ Pr.z// (b) .9z  Sffact.pi.n//g/Œpi.n/ < z ^ Pr.z/ (c) PA ` pi.S n/ D zŒpi.n/ < z ^ Pr.z/ (d) PA ` pi.n/ < pi.S n/ ^ Pr.pi.S n// (e) PA ` .8w < pi.S n//Œpi.n/ < w ^ Pr.w/ (f) PA ` Pr.pi.n// (g) PA ` pi.n/ > 1 (h) PA ` pi.n/a > ; (i) PA ` a > ; ! pi.n/a > 1 (j) PA ` S pred.pi.n/a / D pi.n/a (k) PA ` .8m < n/pi.m/ < pi.n/ (l) PA ` .8m  n/S m < pi.n/ *(m) PA ` 8yŒPr.y/ ! 9j pi.j / D y *(n) PA ` m ¤ n ! pred.pi.m// − pi.n/a (o) PA ` m ¤ n ! pred.pi.m/S b / − pi.n/a *(p) PA ` Œm ¤ n ^ pred.pi.m/b /j.s  pi.n/a / ! pred.pi.m/b /js These are some basic results from prime sequences. (a) gives the basic recursive conditions. (b) is an existential result; then (c) extracts the successor condition from bounded to unbounded minimization; this allows application of the definition in (d) and (e). (f) - (j) are some simple consequences of the fact that pi.n/ is prime. Then the primes are ordered (k). And (l) each prime is greater than the successor of its index. (m) every prime appears as some pi.j /. And (n) - (p) echo results for factor except combined with primes and exponentiation. In this theorem (b) and then (c) - (e) are a first instance of a pattern we shall see repeatedly: Given a bounded condition a D .x  t /P .x/ of the sort that arises from a primitive recursive definition, we show there exists some P .x/ less than or

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equal to the bound; this allows application of T13.18.b to “extract” the bounded to an unbounded minimization, and then T13.17 to obtain P .a/; this forms the basis for further results. In order to manipulate exp, it will be convenient to introduce a function ex, that finds the least exponent x such that pi.i /x does not divide S n. Def [ex] ex.n; i / D xŒpred.pi.i /x / − S n (i) PA ` 9xŒpred.pi.i /x / − S n 1. 2. 3. 4. 5. 6. 7.

pi.i/ > 1 Sn < pi.i/S n Spred.pi.i/S n / D pi.i/S n Sn < Spred.pi.i/S n / n < pred.pi.i/S n / pred.pi.i/S n / − Sn 9xŒpred.pi.i/x / − Sn

T13.42g 1 T13.40m T13.42j 2,3 DE 4 T13.11k 5 T13.22i 6 9I

*T13.43. The following are theorems of PA. (a) PA ` exp.n; i / D .x  n/Œpred.pi.i /x /jn ^ pred.pi.i /S x / − n (b) PA ` exp.;; i / D ; *(c) PA ` exp.S n; i / D xŒpred.pi.i /x /jS n ^ pred.pi.i /S x / − S n (d) PA ` pred.pi.i /exp.S n;i / /jS n ^ pred.pi.i /S exp.S n;i / / − S n (e) PA ` .8w < exp.S n; i //Œpred.pi.i /w /jS n ^ pred.pi.i /S w / − S n (f) PA ` Œpred.pi.i /a jS n ^ pred.pi.i /Sa − S n ! exp.S n; i / D a (g) PA ` exp.m; j /  m (h) PA ` j  n ! exp.S n; j / D ; (i) PA ` exp.pi.i /p ; i / D p (j) PA ` i ¤ j ! exp.pi.i /p ; j / D ; (k) PA ` pred.pi.i //jS m $ exp.S m; i /  1 *(l) PA ` 9qŒpi.i /exp.S n;i /  q D S n ^ pred.pi.i // − q ^ 8y.y ¤ i ! exp.q; y/ D exp.S n; y//

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*(m) PA ` exp.S m  S n; i / D exp.S m; i / C exp.S n; i / (a) is from the definition. (b) is the standard result with bound ;. (c) extracts the successor case from the bounded to an unbounded minimization; this allows application of the definition in (d) and (e). From (f) the reasoning goes the other way around: not only does the condition apply to the exponent, but if the condition applies to some a, then a is the exponent. Then (g) the exponent of some prime in the factorization of m cannot be greater than m; and (h) a prime whose index is greater than or equal to n does not divide into S n. (i) and (j) make an obvious connection for the exponent of a prime, and (k) between exponent and factor. According (l) once you divide S n by pi.i / exp.S n; i / times you are left with a q such that pi.i / does not divide into it any more, and such that the exponents of all the other primes remain the same as in S n. From (m) the i t h exponent of a product sums the i t h exponents of its factors. *T13.44. The following are theorems of PA. (a) PA ` len.n/ D .y  n/.8z  n/Œz  y ! exp.n; z/ D 0 (b) PA ` len.;/ D ; (c) PA ` len.S n/ D y.8z  S n/Œz  y ! exp.S n; z/ D ; (d) PA ` .8z  S n/Œz  len.S n/ ! exp.S n; z/ D ; (e) PA ` .8w < len.S n//.8z  S n/Œz  w ! exp.S n; z/ D ; (f) PA ` len.1/ D ; (g) PA ` len.m/ > ; ! m > 1 *(h) PA ` exp.m; l/ > ; ! len.m/ > l (i) PA ` .8k > l/exp.S m; k/ D ; ! len.S m/  S l (j) PA ` m > 1 ! len.m/ > ; *(k) PA ` p > ; ! len.pi.i /p / D S i (l) PA ` .8z  len.n//exp.n; z/ D ; *(m) PA ` len.n/ D S l ! exp.n; l/  1

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Again (a) is from the definition and (b) gives the standard result for bound ;. (c) extracts the successor case from bounded to unbounded minimization; (d) and (e) then apply the definition. (f) is a simple particular result; and then (g) is an immediate consequence of (b) and (f). From (h) if an exponent of some prime in the factorization of m is greater than zero, that prime is involved in the factorization of m; (i) length is set up so that it finds the first prime such that it and all the ones after have exponent zero; so if all the primes after some point have exponent zero, then the length is no greater than the next; (j) gives the biconditional from (g); (k) gives the length for a prime to any power; and from (l) primes  the length of n must all have exponent ;; (m) the prime prior to the length has exponent  1. For the rest of this section including results for concatenation to follow, it will be helpful to introduce exc and val as auxiliary notions. First, exc.m; n; i / which (indirectly) takes the value of the i t h exponent in the concatenation of m and n. : PA ` exc.m; n; i / D .y  exp.m; i / C exp.n; i len.m/// : .Œi < len.m/ ^ y D exp.m; i / _ Œi  len.m/ ^ y D exp.n; i len.m///

Since the definition is by bounded minimization, no condition is required. The idea is simply to set y to one or the other of exp.m; i / or exp.n; i : len.m// so that y takes the value of the i t h exponent in the concatenation of m and n. Then val.n; i / returns the product of the first i primes in the factorization of n. PA ` val.n; ;/ D 1 PA ` val.n; Sy/ D val.n; y/  pi.y/exp.n;y/

Similarly val .m; n; i / is defined by recursion as follows. PA ` val .m; n; ;/ D 1 PA ` val .m; n; Sy/ D val .m; n; y/  pi.y/exc.m;n;y/

So val .m; n; i / returns the product of the first i primes in the factorization of the concatenation of m and n. Here are some results for these notions. Let l D len.m/ C len.n/. *T13.45. The following are theorems of PA. (a) PA ` exc.m; n; i / D y.Œi < len.m/ ^ y D exp.m; i / _ Œi  len.m/ ^ y D exp.n; i : len.m/// (b) PA ` i < len.m/ ! exc.m; n; i / D exp.m; i / (c) PA ` i  len.m/ ! exc.m; n; i / D exp.n; i : len.m// (d) PA ` val .m; n; i / > ;

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*(e) PA ` .8i  a/pred.pi.i // − val .m; n; a/ *(f) PA ` .8j < i /exp.val .m; n; i /; j / D exc.m; n; j / *(g) PA ` .8i < len.m//Œexp.val .m; n; l/; i / D exp.m; i / ^ .8i < len.n//Œexp.val .m; n; l/; i C len.m// D exp.n; i / *(h) PA ` Œpi.l/mCn l  val .m; n; l/ (i) PA ` val.m; i / > ; (j) PA ` len.val.a; j //  j (k) PA ` len.val.a; j //  len.a/ (l) PA ` .8i < k/exp.m; i / D exp.val.m; k/; i / (m) PA ` .8i < k/exp.a; i / D exp.b; i / ! val.a; k/ D val.b; k/ *(n) x  len.S n/ ! val.S n; x/ D S n corollary: PA ` val.S n; len.S n// D S n *(o) PA ` Œlen.n/  q^.8k < len.n//exp.n; k/  r ! Œpi.q/r q  val.n; len.n// (a) extracts exc from the bounded to unbounded minimization; (b) and (c) apply the definition. (d) is obvious. (e) results because val .m; n; a/ is a product of primes prior to pi.a/ so that greater primes do not divide it. Then (f) the exponents in in val are like the exponents in exc. This gives us (g) that the exponents in val are like the exponents in m and n. But (h) val is constructed so that an induction enables a natural comparison of exponents. Then (i) - (o) are related results for val. In cases to follow, the comparison of exponents from T13.45h and the closely related T13.45o will be crucial for finding bounds and so extracting results from bounded minimization. We are now ready for some results about concatenation. Say m  n is the defined correlate to m ? n and as above l D len.m/ C len.n/. *T13.46. The following are theorems of PA. (a) (i) PA ` m  n D .x  Bm;n /Œx  1 ^ .8i < len.m//fexp.x; i / D exp.m; i /g ^ .8i < len.n//fexp.x; i C len.m// D exp.n; i /g (ii) PA ` Bm;n D Œpi.l/mCn l

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(b) PA ` m  n D xŒx  1 ^ .8i < len.m//fexp.x; i / D exp.m; i /g ^ .8i < len.n//fexp.x; i C len.m// D exp.n; i /g (c) PA ` m  n  1 ^ .8i < len.m//fexp.m  n; i / D exp.m; i /g ^ .8i < len.n//fexp.m  n; i C len.m// D exp.n; i /g (d) PA ` .8w < m  n/Œw  1 ^ .8i < len.m//fexp.w; i / D exp.m; i /g ^ .8i < len.n//fexp.w; i C len.m// D exp.n; i /g *(e) PA ` len.m  n/  l *(f) PA ` len.m  n/ D l (g) PA ` exp.m  n; i C len.m// D exp.n; i / (h) PA ` .a  b/  c D a  .b  c/ (i) PA ` n  1 ! S m  n D S m (j) PA ` n  1 ! n  S m D S m (k) PA ` .len.c/ D len.d / ^ Sa  c D S b  d / ! Sa D S b corollary: PA ` Sa  c D S b  c ! Sa D S b (l) PA ` .len.c/ D len.d / ^ c  Sa D d  S b/ ! Sa D S b corolary: PA ` c  Sa D c  S b ! Sa D S b *(m) PA ` val.S m  S n; a/ D val.S m; a/  val.S n; a : len.S m// (n) PA ` .8y  len.n//Œval.m  n; y C len.m//  val.m; len.m// corollary: PA ` m  n  m (o) PA ` .8y  len.n//Œval.m  n; y C len.m//  val.n; y/ corollary: PA ` m  n  n (a) is from the definition. T13.45h enables us to extract m  n from bounded to unbounded minimization to get (b) and then (c) and (d). (e) and (f) establish that the length of m  n sums the lengths of m and n. (g) generalizes the last conjunct of (c). (h) is an association result — and with this, we typically ignore parentheses in concatenations much as we have done for association with addition. (i) and (j) concatenation with a number less than or equal to one results in no change. (k) and (l) enable a sort of cancellation law for concatenation. (m) distributes val over

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concatenation; then (n) and (o) apply results from T13.45m and T13.45n for relative values of m  n. The idea for application of T13.45h to get (b) is the same as behind the intuitive account of the bound from chapter 12: pi.l/mCn is greater than every term in the factorization of m  n; so Œpi.l/mCn i remains greater than val .m; n; i /; and val .m; n; l/ is therefore both under the bound and satisfies the condition for m  n — so that the existential condition is satisfied, and we may extract the bounded to an unbounded minimization. Once this is accomplished, we are most of the way home. To manipulate Termseq it will be convenient to let, A.s; x/ B.s; x/ C.s; x/ D.s; x/

= = = =

exp.s; x/ D p;q _ Var.exp.s; x// .9j < x/exp.s; x/ D pS q  exp.s; j / .9i < x/.9j < x/exp.s; x/ D pCq  exp.s; i/  exp.s; j / .9i < x/.9j < x/exp.s; x/ D pq  exp.s; i/  exp.s; j /

*T13.47. The following are theorems of PA. (a) PA ` Termseq.m; t / $ exp.m; len.m/ : 1/ D t ^ m > 1 ^ .8k < len.m//ŒA.m; k/ _ B.m; k/ _ C.m; k/ _ D.m; k/ (b) (i) PA ` Term.t / $ .9x  Bt /Termseq.x; t / (ii) PA ` Bt D Œpi.len.t //t len.t / 23C2x

(c) PA ` Var.t / $ .9x  t /.t D 2

/

(d) PA ` Var.t / ! len.t / D 1 (e) PA ` Termseq.m; t / ! .8k < len.m//exp.m; k/ > 1 (f) PA ` Term.t / ! t > 1 t

(g) PA ` t D p;q ! Termseq.2 ; t / t

(h) PA ` Var.t / ! Termseq.2 ; t / pS qt

*(i) PA ` Termseq.m; t / ! Termseq.m  2

; pSq  t / pCqt q

(j) PA ` ŒTermseq.m; t /^Termseq.n; q/ ! Termseq.mn 2 t  q/

; pCq

pqt q

(k) PA ` ŒTermseq.m; t / ^ Termseq.n; q/ ! Termseq.m  n  2 t  q/

; pq 

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*(l) PA ` Termseq.m; t / ! 8x.8k < len.m//flen.exp.m; k//  x ! 9nŒTermseq.n; exp.m; k//^.8i < len.n//exp.n; i /  exp.m; k/ ^len.n/  len.exp.m; k//g (m) PA ` Termseq.m; t / ! Term.t / *(n) PA ` Termseq.m; t / ! .8i < len.m//Term.exp.m; i // (o) PA ` Term.p;q/ (p) PA ` Var.v/ ! ŒTerm.v/ ^ Term.pS q  v/ (a), (b) and (c) are from the definitions term sequence and term and variable with prior results. (d), (e) and (f) are simple results. (g) - (k) generate term sequences. (l) yields (m), that anything with a term sequence is a term; the rest follow from that. From its definition, Term.t / does not immediately follow from Termseq.m; t / insofar as the sequence might build in extraneous terms not required for t — with the result that m is not less than Bn . The general idea for these theorems is that given a term sequence, there is a standard term sequence containing just the elements you would have included in a chapter 2 tree, adequate to yield Term.t /. Thus we move from the existence of a term sequence through (l) to a term sequence of the right sort, and so to (m). Something new happens in (l) insofar as the induction is not on the length of m but on the length of its exponents. We continue with some results for Formseq and Wff that are closely related to T13.47. Let, E.s; x/ F .s; x/ G.s; x/ H.p; s; x/

= = = =

Atomic.exp.s; x// .9j < x/Œexp.s; x/ D neg.exp.s; j // .9i < x/.9j < x/Œexp.s; x/ D cnd.exp.s; i/; exp.s; j // .9i < x/.9j < p/ŒVar.j / ^ exp.s; x/ D unv.j; exp.s; i//

*T13.48. The following are theorems of PA. (a) PA ` Formseq.m; p/ $ exp.m; len.m/ : 1/ D p ^ m > 1 ^ .8k < len.m//ŒE.m; k/ _ F .m; k/ _ G.m; k/ _ H.p; m; k/ (b) (i) PA ` Wff .p/ $ .9x  Bp /Formseq.x; p/ (ii) PA ` Bp D Œpi.len.p//p len.p/ (c) PA ` Atomic.p/ $ .9x  p/.9y  p/ŒTerm.x/ ^ Term.y/ ^ p D pDq  xy (d) PA ` Formseq.m; p/ ! .8k < len.m//exp.m; k/ > 1

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(e) PA ` Wff .p/ ! p > 1 p

(f) PA ` Atomic.p/ ! Formseq.2 ; p/ neg.p/

(g) PA ` Formseq.m; p/ ! Formseq.m  2

; neg.p// cnd.p;q/

(h) PA ` ŒFormseq.m; p/^Formseq.n; q/ ! Formseq.mn2 unv.v;p/

(i) PA ` ŒFormseq.m; p/ ^ var.v/ ! Formseq.m  2

; cnd.p; q//

; unv.v; p//

(j) PA ` Formseq.m; p/ ! 8x.8k < len.m//flen.exp.m; k//  x ! 9nŒFormseq.n; exp.m; k//^.8i < len.n//exp.n; i /  exp.m; k/^len.n/  len.exp.m; k//g (k) PA ` Formseq.m; p/ ! Wff .p/ (l) PA ` Formseq.m; p/ ! .8i < len.m//Wff .exp.m; i // (m) PA ` Atomic.p/ ! Wff .p/ (n) PA ` Wff .p/ ! Wff .neg.p// (o) PA ` ŒWff .p/ ^ Wff .q/ ! Wff .cnd.p; q// (p) PA ` ŒWff .p/ ^ Var.v/ ! Wff .unv.v; p// Hints: For each of (a) - (l), see the parallel theorems for T13.47. The others are nearly trivial. Again, from its definition, Wff .p/ does not immediately follow from Formseq.m; p/ insofar as the sequence might build in extraneous elements not required for p — with the result that m is not less than Bp . And again the general idea is that given a formula sequence, there is a standard formula sequence containing just the elements you would have included in a chapter 2 tree, adequate to yield Wff .n/. Thus we move from the existence of a formula sequence through (j) to a formula sequence of the required sort. Continuing roughly in the order of chapter 12 we move on to some substitution results for terms and atomics. Let, I.m; n; k/ J.v; m; n; k/ K.v; s; m; n; k/ L.m; n; k/ M.m; n; k/

= = = = =

N.m; n; k/

=

exp.m; k/ D p;q ^ exp.n; k/ D p;q Var.exp.m; k// ^ exp.m; k/ ¤ v ^ exp.n; k/ D exp.m; k/ Var.exp.m; k// ^ exp.m; k/ D v ^ exp.n; k/ D s .9i < k/Œexp.m; k/ D pSq  exp.m; i/ ^ exp.n; k/ D pSq  exp.n; i/ .9i < k/.9j < k/Œexp.m; k/ D pCq  exp.m; i/  exp.m; j / ^ exp.n; k/ D pCq  exp.n; i/  exp.n; j / .9i < k/.9j < k/Œexp.m; k/ D pq  exp.m; i/  exp.m; j / ^ exp.n; k/ D pq  exp.n; i/  exp.n; j /

CHAPTER 13. GÖDEL’S THEOREMS

699

*T13.49. The following are theorems of PA. (a) PA ` Tsubseq.m; n; t; v; s; u/ $ Termseq.m; t / ^ len.m/ D len.n/ ^ exp.n; len.n/ : 1/ D u ^ .8k < len.m//.I.m; n; k/ _ J.v; m; n; k/ _ K.v; s; m; n; k/ _ L.m; n; k/ _ M.m; n; k/ _ N.m; n; k// (b) (i) PA ` Termsub.t; v; s; u/ $ .9x  Xt /.9y  Yt;u /Tsubseq.x; y; t; v; s; u/ (ii) PA ` Xt D Œpi.len.t //t len.t / (iii) PA ` Yt;u D Œpi.len.t //u len.t / (c) PA ` Atomsub.p; v; s; q/ $ .9a  p/.9b  p/.9a0  q/.9b 0  q/ŒTerm.a/^ Term.b/^p D pDqab^Termsub.a; v; s; a0 /^Termsub.b; v; s; b 0 /^q D pDq  a0  b 0  (d) PA ` ŒTerm.s/^Tsubseq.m; n; t; v; s; u/ ! .8j < len.n//Term.exp.n; j // corollary: PA ` ŒTerm.s/ ^ Termsub.t; v; s; u/ ! Term.u/ (e) PA ` ŒTerm.s/ ^ Atomsub.p; v; s; q/ ! Atomic.q/ t

t

(f) PA ` t D p;q ! Tsubseq.2 ; 2 ; t; v; s; t / t

t

t

s

(g) PA ` .Var.t / ^ t ¤ v/ ! Tsubseq.2 ; 2 ; t; v; s; t / (h) PA ` .Var.t / ^ t D v/ ! Tsubseq.2 ; 2 ; t; v; s; s/ pS qt

*(i) PA ` Tsubseq.m; n; t; v; s; u/ ! Tsubseq.m  2 t; v; s; pSq  u/

pSqu

;n  2

; pSq 

(j) PA ` ŒTsubseq.m; n; t; v; s; u/^Tsubseq.m0 ; n0 ; t 0 ; v; s; u0 / ! Tsubseq.m pCqt t 0

m0  2

pCquu0

; n  n0  2

; pCq  t  t 0 ; v; s; pCq  u  u0 /

(k) PA ` ŒTsubseq.m; n; t; v; s; u/^Tsubseq.m0 ; n0 ; t 0 ; v; s; u0 / ! Tsubseq.m pqt t 0

m0  2

pquu0

; n  n0  2

; pq  t  t 0 ; v; s; pq  u  u0 /

*(l) PA ` Tsubseq.m; n; t; v; s; u/ ! Termsub.t; v; s; u/ *(m) PA ` ŒTerm.t / ^ Term.s/ ! 9uŒTermsub.t; v; s; u/ ^ len.u/  len.t /  len.s/ ^ .8k < len.u//exp.u; k/  t C s *(n) PA ` ŒAtomic.p/ ^ Term.s/ ! 9qŒAtomsub.p; v; s; q/ ^ len.q/  len.p/  len.s/ ^ .8k < len.q//exp.q; k/  p C s

CHAPTER 13. GÖDEL’S THEOREMS

700

Some substitution results for formulas are closely related to the previous theorem. Let, O.v; s; m; n; k/ P .m; n; k/ Q.m; n; k/

= = =

R.v; p; m; n; k/

=

S.v; p; m; n; k/

=

Atomic.exp.m; k// ^ Atomsub.exp.m; k/; v; s; exp.n; k// .9i < k/Œexp.m; k/ D neg.exp.m; i// ^ exp.n; k/ D neg.exp.n; i// .9i < k/.9j < k/Œexp.m; k/ D cnd.exp.m; i/; exp.m; j // ^ exp.n; k/ D cnd.exp.n; i/; exp.n; j // .9i < k/.9j < p/ŒVar.j / ^ j ¤ v ^ exp.m; k/ D unv.j; exp.m; i// ^ exp.n; k/ D unv.j; exp.n; i/ .9i < k/.9j < p/ŒVar.j / ^ j D v ^ exp.m; k/ D unv.j; exp.m; i// ^ exp.n; k/ D exp.m; k/

*T13.50. The following are theorems of PA. (a) PA ` Fsubseq.m; n; p; v; s; q/ $ ŒFormseq.m; p/ ^ len.m/ D len.n/ ^ exp.n; len.n/ : 1/ D q ^ .8k < len.m//.O.v; s; m; n; k/ _ P .m; n; k/ _ Q.m; n; k/ _ R.p; m; n; k/ _ S.p; m; n; k// (b) (i) PA ` Formsub.p; v; s; q/ $ .9x  Xp /.9y  Yp;q /Fsubseq.x; y; p; v; s; q/ (ii) PA ` Xp D Œpi.len.p//p len.p/ (iii) PA ` Yp;q D Œpi.len.p//q len.p/ (c) (i) PA ` formusb.p; v; s/ D .q  Zp;s /Formsub.p; v; s; q/ (ii) PA ` Zp;s D Œpi.len.p/  len.s//pCs len.p/len.s/ (d) PA ` ŒTerm.s/^Fsubseq.m; n; p; v; s; q/ ! .8j < len.n//Wff .exp.n; j // corollary: PA ` ŒTerm.s/ ^ Formsub.p; v; s; q/ ! Wff .q/ p

q

(e) PA ` ŒAtomic.p/ ^ Atomsub.p; v; s; q/ ! Fsubseq.2 ; 2 ; p; v; s; q/ neg.p/

(f) PA ` Fsubseq.m; n; p; v; s; q/ ! Fsubseq.m2

neg.q/

; n2

; neg.p/; v; s; neg.q//

(g) PA ` ŒFsubseq.m; n; p; v; s; q/^Fsubseq.m0 ; n0 ; p 0 ; v; s; q 0 / ! Fsubseq.m cnd.p;p 0 / cnd.q;q 0 / m0  2 ; n  n0  2 ; cnd.p; p 0 /; v; s; cnd.q; q 0 // unv.u;p/

(h) PA ` ŒFsubseq.m; n; p; v; s; q/^Var.u/^u ¤ v ! Fsubseq.m2 unv.u;q/ 2 ; unv.u; p/; v; s; unv.u; q//

unv.u;p/

(i) PA ` ŒFsubseq.m; n; p; v; s; q/^Var.u/^u D v ! Fsubseq.m2 unv.u;p/ 2 ; unv.u; p/; v; s; unv.u; p// (j) PA ` Fsubseq.m; n; p; v; s; q/ ! Formsub.p; v; s; q/

; n

; n

CHAPTER 13. GÖDEL’S THEOREMS

701

(k) PA ` ŒWff .p/ ^ Term.s/ ! 9qŒFormsub.p; v; s; q/ ^ len.q/  len.p/  len.s/ ^ .8k < len.q//exp.q; k/  p C s (l) PA ` ŒWff .p/ ^ Term.s/ ! Formsub.p; v; s; formsub.p; v; s// (m) PA ` ŒWff .p/ ^ Term.s/ ! Wff .formsub.p; v; s// Hints: For (a) - (k) see hints for the parallel results from T13.49. (l) follows easily with (k). Finally we extend our results by means of a pair of matched theorems whose results are related to unique readability for terms and then for formulas (see section 11.2). *T13.51. The following result in PA. First, as a preliminary to T13.51f and then T13.52g it will be helpful to show the following. We are thinking of c  a  c1  b  c2 as for example, p.q  a  p!q  b  p/q. Let, l1 = len.c/ l2 = len.c/ C len.a/ l3 = len.c/ C len.a/ C len.c1 / l4 = len.c/ C len.a/ C len.c1 / C len.b/ l = len.c/ C len.a/ C len.c1 / C len.b/ C len.c2 /

*(a)

a. b. c. d. e. f.

8uŒ.P .u/ ^ len.u/  x/ ! .8k < len.u//P .val.u; k// : : : : val.c; j /  val.a; j l1 /  val.c1 ; j l2 /  val.b; j l3 /  val.c2 ; j l4 / D c  d  c1  e  c2 P .a/ ^ P .b/ ^ P .d / ^ P .e/ 8v.P .v/ ! v > 1/ len.c/ D 1 ^ c1 > ; ^ c2 > ; ^ len.c1 /  1 ^ len.c2 /  1 j < l ^ Sx  l

P P P P P P

: : : g. ?

If we know by an assumption that an initial segment of a term (or formula) is not a term (or formula), then it is inconsistent to suppose that an an initial segment of a more complex concatenation is equal to a concatenation of terms (or formulas). As a corollary, when c1 D c2 D 1 their lengths go to zero and by T13.45n for any x, val.c1 ; x/ D val.c2 ; x/ D 1 so that these terms drop out of the concatenations and the theorem reduces to a version where (b) is val.c; j /  val.a; j : l1 /  val.b; j : l3 / D c  d  e, and the only substantive conjunct of (e) is the first. (b) PA ` ŒTerm.a/ ^ Term.b/ ! ŒpS q  a D pS q  b ! a D b (c) PA ` Term.pS q  a/ ! 9rŒpSq  a D pS q  r ^ Term.r/

CHAPTER 13. GÖDEL’S THEOREMS

702

(d) PA ` Term.pCq  a/ ! 9r9sŒpCq  a D pCq  r  s ^ Term.r/ ^ Term.s/ (e) PA ` Term.pq  a/ ! 9r9sŒpq  a D pq  r  s ^ Term.r/ ^ Term.s/ *(f) PA ` Term.t / ! .8k < len.t //Term.val.t; k// (g) PA ` ŒTerm.a/ ^ Term.b/ ^ Term.c/ ^ Term.d / ! ŒpCq  a  b D pCq  c  d ! .a D c ^ b D d / (h) PA ` ŒTerm.a/ ^ Term.b/ ^ Term.c/ ^ Term.d / ! Œpq  a  b D pq  c  d ! .a D c ^ b D d / (i) PA ` ŒTerm.a/ ^ Term.b/ ^ Term.c/ ^ Term.d / ! ŒpDq  a  b D pDq  c  d ! .a D c ^ b D d / Returning to our original results for unique readability, reasoning for (c) - (e) is like that for T11.6 - T11.4. Then (f) is like T11.5. And now there are the parallel results for formulas. *T13.52. The following are theorems of PA. (a) PA ` ŒWff .p/ ^ Wff .q/ ! Œneg.p/ D neg.q/ ! p D q (b) PA ` ŒWff .p// ^ Var.u/ ^ Wff .q/ ^ Var.v/ ! Œunv.u; p/ D unv.v; q/ ! .u D v ^ p D q/ (c) PA ` Wff .pDq  a/ ! 9r9sŒpDq  a D pDq  r  s ^ Term.r/ ^ Term.s/ (d) PA ` Wff .pq  p/ ! 9rŒpq  p D neg.r/ ^ Wff .r/ (e) PA ` Wff .p.q  p/ ! 9r9sŒp.q  p D cnd.r; s/ ^ Wff .r/ ^ Wff .s/ (f) PA ` Wff .p8q  p/ ! 9w9rŒp8q  p D unv.w; r/ ^ Var.w/ ^ Wff .r/ (g) PA ` Wff .p/ ! .8k < len.p//Wff .val.p; k// *(h) PA ` ŒWff .p// ^ Wff .q/ ^ Wff .a/ ^ Wff .b/ ! Œcnd.p; q/ D cnd.a; b/ ! .p D a ^ q D b/ (i) PA ` ŒWff .cnd.p; q// ^ Wff .p/ ! Wff .q/ *(j) PA ` Axiompa.p/ ! Wff .p/ (k) PA ` Prvpa.p/ ! Wff .p/

CHAPTER 13. GÖDEL’S THEOREMS

703

In the following we shall assume results like (j) - (k) for theories extending PA — though, of course, our prime example just is PA. Insofar as theories are recursively defined, some such results should be in the offing. *E13.31. Show (e) and (j) from T13.40. Hard core: show each of the results from T13.40. Hints for T13.40. (a) is from the the definition of power and prior results. For (c) take a look at E6.36e. (e) uses IN on the value of b and (f) uses IN on a. (g) is straightforward with cases for mb D ; and mb > ;. (i), (j), (k) and (m) are by IN. For (n), a < b _ a D b _ b < a; but the first and last are impossible. *E13.32. Show (d) and (e) from T13.41. Hard core: show each of the results from T13.41. Hints for T13.41. (a) is from the definition of fact and prior results. (c) and (d) are straightforward by IN. Reasoning for (e) is like (G2) in the arithmetic for Gödel numbering reference once you realize that all the primes less than n are included in fact.n/.

*E13.33. Show (k) and (l) from T13.42. Hard core: show each of the results from T13.42. Hints for T13.42. (a) is from definition pi and prior results. (b) is from T13.41e; (c) applies T13.18.b; and then (d) and (e) are by T13.17(b) and (c). (f), (k) and (l) are simple inductions. (m) is by using IN on i to show .8y  pi.i //ŒPr.y/ ! 9j pi.j / D y; the result then follows easily with (l). Under the assumption for !I, (n) is by IN on a. For (o) you will be able to show that if pred.pi.m/S b /jpi.n/a then pred.pi.m//jpi.n/a and use (n). For (p) under the assumption for !I you will be able to show i  b ! pred.pi.m/i /js by induction on i ; the result then follows easily with b  b. *E13.34. Show (c) and (f) from T13.43. Hard core: show each of the results from T13.43. Hints for T13.43. (a) is from definition exp and prior results. (c) is by PA ` .9x  S n/Œpred.pi.i /x /jS n ^ pred.pi.i /xC1 / − S n and then T13.18b; ex.n; i / D ; _ ex.n; i / > ;; in the latter case, the trick is to generalize on the number prior to ex.n; i /. (f) is by showing that a D xŒpred.pi.i /x /jS n ^

CHAPTER 13. GÖDEL’S THEOREMS

704

pred.pi.i /xC1 / − S n. (l): from pred.pi.i /exp.S n;i / /jS n there is a j such that pi.i /exp.S n;i /  j D S n; the hard part is to show k ¤ i ! exp.j; k/ D exp.S n; k/ — for this, it will be helpful to establish that j is a successor. (m): toward an application of T13.43f it will be easy to establish that pred.pi.i /exp.S m;i /Cexp.S n;i / /j.S m  S n/; for the other conjunct, it will be helpful to begin with a couple applications of T13.43l.

*E13.35. Show (f) and (l) from T13.44. Hard core: show each of the results from T13.44. Hints for T13.44. (a) is from definition length and prior results. (c) follows with T13.43h and existentially generalizing on S n itself. (f) is by application of (c). Under the assumption for !I, (h) divides into cases for m D ; and m > ;; for the latter, suppose len.m/  i ; then you will be able to make use of (d). (j) is straightforward with T13.23d and ultimately (h) above. For (k), begin with len.pi.i /p / < S i _ len.pi.i /p / D S i _ len.pi.i /p / > S i by T13.11p; the first is easily eliminated with T13.44h; then, supposing len.pi.i /p / > S i , you will be able to obtain a contradiction using T13.44e. (l): under the assumption a  len.n/ for (8I), either n D ; or n > ;; the first case is easy; for the second, there is some m such that n D S m; your main reasoning will be to show exp.S m; a/ D ;. (m): under the assumption for !I, the case when n D ; is impossible; so there is some m such that n D S m; with this, suppose exp.S m; l/ ž 1; then you you will be able to show, contrary to your assumption that len.S m/ D l. *E13.36. Show (a) and (b) from T13.45. Hard core: show each of the results from T13.45. Hints for T13.45. (e) is by IN on a. (f) is by IN on i ; in the show under .8j < i /exp.val .m; n; i /; j / D exc.m; n; j / and a < S i you will have separate cases for a < i and a D i. (g) is straightforward with applications of (f), (b) and (c). For (h) you may obtain i  l ! Œpi.l/mCn i  val .m; n; i / by induction on i ; in the show, the main task is to obtain exc.m; n; i /  m C n; the result then follows with previously established inequalities. (j) is easy with a result like (e). For (n) you will be able to show 8x8nŒlen.S n/  x ! val.S n; x/ D S n by induction on x: the ;-case is straightforward; then under the inductive assumption with len.Sa/  S x for !I you have len.Sa/  x_len.Sa/ D S x; the first case is straightforward; the second is an

CHAPTER 13. GÖDEL’S THEOREMS

705

extended argument — you will be able to apply T13.43l to obtain an S r whose prime factorization is like that of Sa but without pi.x/; show that len.S r/  x so that from the assumption, val.S r; x/ D S r; then val.Sa; S x/ D Sa is straightforward. For (o) under the assumption for !I, you will be able to get i  q ! Œpi.q/r i  val.n; i / by IN. *E13.37. Show (b) and (e) from T13.46. Hard core: show each of the results from T13.46. Hints for T13.46. (a) is from the definition concatenation with prior results. (b) uses T13.45h. (e) divides into cases for len.n/ D ; and len.n/ > ;; and within the first, again, cases for len.m/ D ; and len.m/ > ;. For (f) show len.m  n/  l and apply (e); for the main argument (which will be long!) assume len.m  n/ — l; then you will be able to apply T13.43l and show that the q so obtained contradicts T13.46d. (h) where l D len.a/Clen.b/Clen.c/, you will be able to show .8i < l/exp..a  b/  c/; i / D exp.a  .b  c/; i /. (k) and (l) are straightforward with T13.46c. For (m) you will be able to show .8i < a/exp.S m  S n; i / D exp.val.S m; a/  val.S n; a : len.S m//; i / and so val.S m  S n; a/ D val.val.S m; a/  val.S n; a : len.S m//; a/; and from this the result you want. (n) and (o) are by induction on y (with the bounded quantifier unabbreviated to the associated conditional).

*E13.38. Show (j) and the unfinished cases in the answers for the C disjunct in (l) and (n). Hard core: show each of the results from T13.47. Hints for T13.47. (e) is straightforward by an extended _E. (g) - (k) are disjunctive but straightforward. (l) is by induction on x: under the assumption Termseq.m; t / the basis is straightforward; then, under the inductive assumption along with a < len.m/ for (8I) and len.exp.m; a//  S x for !I, apply (a); the derivation is then a (long!) argument by cases where you will be able to apply (g)-(k). (m) follows easily with T13.45o. For (n) under the assumption for !I, you will be able to show 8kŒk < len.m/ ! 9x.Termseq.x; exp.m; k// by strong induction; the result follows easily. E13.39. Work (g) from T13.48 including at least the A and B cases. Hard core: show each of the results from T13.48.

CHAPTER 13. GÖDEL’S THEOREMS

706

*E13.40. Work the K and M cases from the answers for T13.49l. Hard core: show each of the results from T13.49. Hints for T13.49. For (l) let

P .m; n; v; s; k/ = 9a9bŒTsubseq.a; b; exp.m; k/; v; s; exp.n; k// ^

len.a/  len.exp.m; k// ^ .8i < len.a//.exp.a; i /  exp.m; k/ ^ exp.b; i /  exp.n; k//;

then under the assumption for !I, show 8x.8k < len.m//Œlen.exp.m; k/  x ! P  by IN; the result follows from this. Similarly, for (m) let P .m; i; v; s/ = 9x9y9uŒTsubseq.x; y; exp.m; i /; v; s; u/ ^ len.u/  len.exp.m; i //  len.s/ ^ .8k < len.u//exp.u; k/  exp.m; i / C s; under the assumption Term.t / ^ Term.s/ given Termseq.m; t / you will be able to show 8i Œi < len.m/ ! P  by strong induction on i (with extended disjunctions in both the basis and show); the result follows easily from this. E13.41. Work j from T13.50 including at least the O case from the answers. Hard core: show each of the results from T13.50. *E13.42. From the answers, work the case marked “similarly” on line 115 of T13.51a and the D case from T13.51f. Hard core: show each of the results from T13.51. Hints for T13.51. For (a) suppose j  l1 , this leads to contradiction so that j  l1 and you can “pick off” the first conjunct from premise (b) to get val.a; j : l1 /  val.c1 ; j : l2 /  val.b; j : l3 /  val.c2 ; j : l4 / D d  c1  e  c2 ; suppose j < l2 , again this leads to contradiction so that j  l2 ; either len.d / < len.a/ _ len.d / D len.a/ _ len.d / > len.a/; the first and last lead to contradiction and with the other you will be able to pick off another conjunct; continue to j  l, which contradicts the last premise. For (f) show 8t Œ.Term.t / ^ len.t /  x/ ! .8k < len.t //Term.val.t; k// by induction on x; the zero case is easy; then under the inductive assumption with Term.a/ ^ len.a/  S x for !I and j < len.a/ for (8I) you will be able to show j > ;; then with Termseq.m; a/ the argument is an extended disjunction from A.m; len.m/ : 1/ _ B.m; len.m/ : 1/ _ C.m; len.m/ : : 1/ _ D.m; len.m/ 1/; you can assume Term.val.a; j // and reach contradiction in each case. *E13.43. Show (g) including at least the A case from the answers, and (k) from T13.52. Hard core: show each of the results from T13.52. Hints for T13.52. Reasoning for (g) is like T13.51f. Reasoning for (i) is like the final uniqueness part of T11.6; the result is straightforward, starting with

CHAPTER 13. GÖDEL’S THEOREMS

707

(e) — though with p!q  q D p!q  s, for an application of T13.46l, you will need to worry about the case q D ;. Beginning with T13.39, (j) and (k) are not hard.

13.4.3

The Condition

After all our preparation, we are ready to turn to the second condition, that PA ` .P ! Q/ ! .P ! Q/. Again, given both T ` .P ! Q/ and T ` P the idea is that there are j and k such that PRFT.j; pP ! Qq/ and PRFT.k; pP q/ so that ypQq numbers a proof of Q. As it turns out, it will be convenient to have the l = j?k?2

b

4

b

result in a form with free variables, PA ` Prvt.cnd.p; q// ! .Prvt.p/ ! Prvt.q//; the second condition then follows as an immediate corollary. Observe that we have on the table expressions of the sort, C, Plus and plus — where the first is a primitive symbol of LNT , the second the original relation to capture the recursive function plus, and the last a function symbol defined from the recursive function. In view of demonstrated equivalences, we will tend to slide between them without notice. So, for example, given that hh2; 2i; 4i 2 plus, by capture PA ` Plus.2; 2; 4/; and by demonstrated equivalences, PA ` 2 C 2 D 4 and PA ` plus.2; 2/ D 4. In particular Prvt.n/ = 9xPrft.x; n/ is equivalent to Prvt.n/. *T13.53. PA ` Prvt.cnd.p; q// ! .Prvt.p/ ! Prvt.q//. Corollary: PA ` .P ! Q/ ! .P ! Q/. See the derivation on p. 708. The derivation is long, and skips steps; but it should be enough for you to see how the argument works — and to fill in the details if you choose. First, from the part up to the line labeled (a), under assumptions for !I, there are derivations numbered j , k and a longer sequence numbered l. And the the last member of this longer sequence is an immediate consequence of last members from the derivations numbered j and k. At (b) the results from (15) are all applied to the sequence numbered l; so the last sentence in the longer sequence is an immediate consequence of its earlier members. From lines up to (c), the different fragments of the longer sequence have the character of a proof. And at (d), the whole sequence numbered l has the character of a proof. Finally, from lines up to (e) we observe that this longer sequence yields Prvt.q/ and discharge the assumptions for the result that Prvt.cnd.p; q// ! ŒPrvt.p/ ! Prvt.q/ so that PA ` Prvt.cnd.p; q// ! .Prvt.p/ ! Prvt.q//. But then we have Prvt.cnd.pP q; pQq// ! ŒPrvt.pP q/ ! Prvt.pQq/ as an instance, and by capture, Prvt.pP ! Qq/ ! ŒPrvt.pP q/ ! Prvt.pQq/ so that

CHAPTER 13. GÖDEL’S THEOREMS

708

T13.53 1.

Prvt.cnd.p; q//

A (g !I)

2. 3.

Wff .cnd.p; q// Prvt.p/

1 T13.52k A (g !I)

4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15a 16. 17. 18. 19. 20b 21. 22. 23c 24. 25. 26.

Wff .p/ Wff .q/ Icon.cnd.p; q/; p; q/ 9vPrft.v; cnd.p; q// 9vPrft.v; p/ Prft.j; cnd.p; q// Prft.k; p/

3 T13.52k 2,4 T13.52i T13.39b,d 1 abv 3 abv A (g 79E) A (g 89E)

q

l Dj k2 : exp.j; len.j / 1/ D cnd.p; q/ : exp.k; len.k/ 1/ D p exp.l; len.j / C len.k// D q : : IconŒexp.j; len.j / 1/; exp.k; len.k/ 1/; exp.l; len.j / C len.k// .8i < len.j //Œexp.l; i / D exp.j; i / .8i < len.k//Œexp.l; len.j / C i / D exp.k; i / : : exp.l; len.j / 1/ D exp.j; len.j / 1/ : : exp.l; len.j / C len.k/ 1/ D exp.k; len.k/ 1/ : : IconŒexp.l; len.j / 1/; exp.l; len.j / C len.k/ 1/; exp.l; len.j / C len.k// .8i < len.j //ŒAxiomt.exp.l; i // _ .9m < i /.9n < i /Icon.exp.l; m/; exp.l; n/; exp.l; i // .8i < len.k//ŒAxiomt.exp.l; len.j / C i // _ .9m < i /.9n < i /Icon.exp.l; len.j / C m/; exp.l; len.j / C n/; exp.l; len.j / C i // .8i W len.j /  i < len.j / C len.k//ŒAxiomt.exp.l; i // _ .9m < i /.9n < i /Icon.exp.l; m/; exp.l; n/; exp.l; i // x < len.l/ x < len.j / _ len.j /  x < len.j / C len.k/ _ x D len.j / C len.k/ x < len.j /

def 9 T13.39g 10 T13.39g 11 T13.46c,f 6,12,13,14 DE 11 T13.46c 11 T13.46c 16 T13.44h (8E) 17 T13.44h (8E) 15,18,19 DE 9,16 T13.39g 10,17 T13.39g from 22 A (g (8I)) 11,24 T13.46f A (g 25_E)

27.

Axiomt.exp.l; x// _ .9m < x/.9n < x/Icon.exp.l; m/; exp.l; n/; exp.l; x//

21,26 (8E)

28.

len.j /  x < len.j / C len.k/

A (g 25_E)

29.

Axiomt.exp.l; x// _ .9m < x/.9n < x/Icon.exp.l; m/; exp.l; n/; exp.l; x//

23,28 (8E)

30.

x D len.j / C len.k/

A (g 25_E)

31. 32.

.9m < x/.9n < x/Icon.exp.l; m/; exp.l; n/; exp.l; x// Axiomt.exp.l; x// _ .9m < x/.9n < x/Icon.exp.l; m/; exp.l; n/; exp.l; x//

20,30 31 _I

33. 34d 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45.

Axiomt.exp.l; x// _ .9m < x/.9n < x/Icon.exp.l; m/; exp.l; n/; exp.l; x// .8x < len.l//ŒAxiomt.exp.l; x// _ .9m < x/.9n < x/Icon.exp.l; m/; exp.l; n/; exp.l; x// q>; q len.2 / D 1 len.l/  1 l >1 : exp.l; len.l/ 1/ D q : exp.l; len.l/ 1/ D q ^ l > 1 ^ .8x < len.l//ŒAxiomt.exp.l; x// _ .9m < x/.9n < x/Icon.exp.l; m/; exp.l; n/; exp.l; x// Prft.l; q/ Prvt.q/ Prvt.q/ Prvt.q/ Prvt.p/ ! Prvt.q/

46e Prvt.cnd.p; q// ! ŒPrvt.p/ ! Prvt.q/

25,26-31 _E 24-33 (8I) 5 T13.48e 35 T13.44k 11,36 T13.46f 37 T13.44g 14 T13.46f 39,38,34 ^I 40 T13.39g 41 9I 8,10-42 9E 7,9-43 9E 3-44 !I 1-45 !I

CHAPTER 13. GÖDEL’S THEOREMS PA ` .P ! Q/ ! .P ! established.

Q/.

709 Thus the second derivability condition is

*E13.44. As a start to a complete demonstration of T13.53, provide a demonstration through part (c) that does not skip any steps. You may find it helpful to divide your demonstration into separate parts for (a), (b) and then for lines (21), (22) and (23). Hard core: complete the entire derivation.

13.5

The Third Condition: P ! P

To show the third condition, that PA ` P ! P , it is sufficient to show PA ` Q ! Q. For when Q is P , the result is immediate. Further, P is Prvt.pP q/ and Prvt.pP q/ is †1 . So it is sufficient to show that for any †1 sentence Q, PA ` Q ! Q. We begin with some additional applications. Then we focus what needs to be shown by an alternate characterization of †1 formulas, along with some results about substitutions. Finally we will be in a position to show the third condition.

13.5.1

More Applications

Recall that where p = pP q, v = pvq, and s = psq, formsub.p; v; s/ returns the y y 2n y23 Gödel number of Psv . Let gvar.n/ = 2 be the Gödel number of variable xn . In addition, num.n/ returns the Gödel number of the standard numeral for n. So formsub.p; gvar.n/;num.y// is a function which returns the number of the formula that substitutes a numeral for the value (number) assigned to y into the place of xn . So, for example, if y is assigned the value of 2, then formsub.p; gvar.n/; num.y// returns pP xn q. And PA defines formsub.p; gvar.n/; num.y//. We require some results for 2 these notions. First, a pair of theorems with some results for substitutions into terms and then formulas. *T13.54. The following are theorems of PA. (a) PA ` Freet .t; v/ $ Termsub.t; v; v  4; t / (b) PA ` Var.v/ ! Term.v  4/ ^ v  4 ¤ v (c) PA ` exp.m; k/ D p;q ! ŒJ.v; m; n; k/ _ K.v; s; m; n; k/ _ L.m; n; k/ _ M.m; n; k/ _ N.m; n; k/

CHAPTER 13. GÖDEL’S THEOREMS

710

Second theorems of chapter 13 T13.19. For any friendly recursive function r.Ex/ and original formula R.x; E v/ by which it is expressed and captured, PA defines a function r.x/ E such that PA ` v D r.x/ E $ R.x; E v/. This theorem depends on conditions for the recursion clause and so on T13.20 and T13.29. T13.20. Where F .x; E y; v/ is the formula for recursion, PA ` 8m8nŒ.F .x; E y; m/ ^ F .x; E y; n// ! m D n. : T13.21 - T13.24. T13.21 Results for a b. T13.22 results for ajb. T13.23 results for Pr.a/ and Rp.a/. T13.24 results for lcm.a/. T13.25. PA ` Œ.8i < k/.m.i/ > ; ^ m.i/ > h.i// ^ 8i8j.i < j ^ j < k ! Rp.Sm.i/; Sm.j /// ! 9p.8i < k/rm.p; m.i// D h.i/ (CRT). T13.26 - T13.28. T13.26 results for maxp and maxs. T13.27 PA ` 9p9q.8i < k/ˇ.p; q; i/ D h.i/. T13.28 PA ` 9p9qŒ.8i < k/ˇ.p; q; i/ D ˇ.r; s; i/ ^ ˇ.p; q; k/ D n. T13.29. PA ` 9v9p9qŒˇ.p; q; ;/ D g.x/ E ^ .8i < y/h.x; E i; ˇ.p; q; i// D ˇ.p; q; Si/ ^ ˇ.p; q; y/ D v. T13.30. For any friendly recursive relation R.Ex/ with characteristic function chR .Ex/, PA ` R.x/ E $ chR .x/ E D ;. And for a recursive operator OP.P1 .Ex/ : : : Pn .Ex// with characteristic function f.chP1 .Ex/ : : : chPn .Ex//, PA ` Op.P1 .x/ E : : : Pn .x// E $ f.chP1 .x/ E : : : chPn .x// E D ;. Corollary: where R.Ex/ is originally captured by R.x; E ;/, PA ` R.x/ E $ R.x; E ;/. T13.31. Suppose f.x; E y/ is defined by g.x/ E and h.x; E y; u/ so that PA ` v D f.x; E y/ $ F .x; E y; v/; then, (i) PA ` f.x; E ;/ D g.x/ E and (ii) PA ` f.x; E S.y// D h.x; E y; f.x; E y//. E v/ by which it is T13.32. (a) For any friendly recursive function r.Ex/ and original formula R.x; expressed and captured, PA defines a coordinate function r.x/ E such that PA ` v D r.x/ E $ R.x; E v/. And (b) for any friendly recursive relation R.Ex/ with characteristic function chR .Ex/, PA defines a coordinate relation R.x/ E such that PA ` R.x/ E $ chR .x/ E D ;. j

T13.34 - T13.36. T13.34 equivalences for suc, zero, idntk , plus and times. T13.35 results for pred, sg and csg. T13.36 Equivalences for pred, subc, absval, sg, csg, Eq, Leq, Less, Neg, and Dsj. T13.37. PA proves a characteristic function takes the value ; or 1. T13.38. Equivalences for .9 y  z/, .9 y < z/, . 8y  z/, . 8y < z/, .y  z/, Fctr, and Prime. T13.39 - T13.43. T13.39 first applications to recursive functions. T13.40 Results for ma . T13.41 results for fact. T13.42 results for pi. T13.43 results for exp. T13.44 - T13.50. T13.44 results for len. T13.45 results for val. T13.46 results for m  n. T13.47 results for Termseq. T13.48 results for Formseq. T13.49 results for Tsubseq. T13.50 results for Fsubseq. T13.51 - T13.52. T13.51 on unique readability. T13.52 results for Wff and Prvpa. T13.53. PA ` .P ! Q/ ! .P ! Q/.

— D2

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711

(d) PA ` ŒVar.exp.m; k//^exp.m; k/ ¤ v ! ŒI.m; n; k/_K.v; s; m; n; k/_ L.m; n; k/ _ M.m; n; k/ _ N.m; n; k/ (e) PA ` ŒVar.exp.m; k// ^ exp.m; k/ D v ! ŒI.m; n; k/ _ J.v; m; n; k/ _ L.m; n; k/ _ M.m; n; k/ _ N.m; n; k/ (f) PA ` exp.m; k/ D pSqa ! ŒI.m; n; k/_J.v; m; n; k/_K.v; s; m; n; k/_ M.m; n; k/ _ N.m; n; k/ (g) PA ` exp.m; k/ D pCqa ! ŒI.m; n; k/_J.v; m; n; k/_K.v; s; m; n; k/_ L.m; n; k/ _ N.m; n; k/ (h) PA ` exp.m; k/ D pqa ! ŒI.m; n; k/_J.v; m; n; k/_K.v; s; m; n; k/_ L.m; n; k/ _ M.m; n; k/ *(i) PA ` ŒTermsub.t; v; s; q/ ^ Termsub.t; v; s; r/ ! q D r (j) PA ` ŒAtomsub.p; v; s; q/ ^ Atomsub.p; v; s; r/ ! q D r (k) PA ` ŒTerm.t / ^ Term.s/ ! ŒFreet .t; v/ ! Termsub.t; v; s; t / (l) PA ` Term.s/ ! ŒAtomsub.p; v; v  4; p/ ! Atomsub.p; v; s; p/ (m) PA ` ŒTerm.t / ^ Var.v/ ! Œ.Freet .t; v/ ^ Termsub.t; v; s; u// ! s  u *(n) PA ` Var.v/ ! Œ.Atomsub.p; v; v4; p/^Atomsub.p; v; s; q// ! s  q T13.55. The following are theorems of PA. (a) PA ` Freef .p; v/ $ Formsub.p; v; v  4; p/ (b) PA ` Atomic.exp.m; k/ ! ŒP .m; n; k/ _ Q.m; n; k/ _ R.v; p; m; n; k/ _ S.v; p; m; n; k/ (c) PA ` exp.m; k/ D pqa ! ŒO.v; s; m; n; k/_Q.m; n; k/_R.v; p; m; n; k/_ S.v; p; m; n; k/ (d) PA ` exp.m; k/ D p.qa ! ŒO.v; s; m; n; k/_P .m; n; k/_R.v; p; m; n; k/_ S.v; p; m; n; k/ (e) PA ` ŒVar.j / ^ exp.m; k/ D p8q  j  a ^ j ¤ v ! ŒO.v; s; m; n; k/ _ P .m; n; k/ _ Q.m; n; k/ _ S.v; p; m; n; k/

CHAPTER 13. GÖDEL’S THEOREMS

712

(f) PA ` ŒVar.j / ^ exp.m; k/ D p8q  j  a ^ j D v ! ŒO.v; s; m; n; k/ _ P .m; n; k/ _ Q.m; n; k/ _ R.v; p; m; n; k/ (g) PA ` ŒFormsub.p; v; s; q/ ^ Formsub.p; v; s; r/ ! q D r (h) PA ` ŒWff .p/ ^ Term.s/ ! ŒFormsub.p; v; s; q/ ! formusb.p; v; s/ D q (i) PA ` ŒWff .p/ ^ Term.s/ ! ŒFreef .p; v/ ! formsub.p; v; s/ D p corollary: If x is not free in P , then PA ` formsub.pP q; pxq; y/ D pP q (j) PA ` ŒWff .p/^Term.s/^Var.v/ ! ŒFreef .p; v/ ! s  formsub.p; v; s/ Hint: See the corresponding members of T13.54. We are now positioned for some results related to Gen and A4. Let numseq.n/ be as follows. PA ` numseq.;/ D pi.;/num.;/ PA ` numseq.Sy/ D numseq.y/  pi.Sy/num.Sy/

We shall be able to show that numseq.n/ numbers a term sequence for num.n/. In addition for a function coordinate to FFSEQ let, T .m; k/ U.m; k/ V .m; k/ W .u; v; m; k/ X.u; v; s; m; k/

= = = = =

Atomic.exp.m; k// .9j < k/Œexp.m; k/ D neg.exp.m; j // .9i < k/.9j < k/Œexp.m; k/ D cnd.exp.m; i/; exp.m; j // .9p  u/ŒWff .p/ ^ exp.m; k/ D unv.v; p/ .9i < k/.9j  u/ŒVar.j / ^ j ¤ v ^ .Freet .s; j / _ Freef .exp.m; i/; v// ^ exp.m; k/ D unv.j; exp.m; i//

T13.56. The following are theorems of PA. (a) PA ` Ffseq.m; s; v; u/ $ Œexp.m; len.m/ : 1/ D u ^ m > 1 ^ .8k < len.m//.T .m; k/ _ U.m; k/ _ V .m; k/ _ W .u; v; m; k/ _ X.u; v; s; m; k// (b) (i) PA ` Freefor.s; v; u/ $ .9x  Bu /Ffseq.x; s; v; u/ (ii) PA ` Bu D Œpi.len.u//u len.u/ (c) PA ` Axiomad4.n/ $ .9p  n/.9v  n/fWff .p/ ^ Var.v/ ^ Œ .Freef .p; v/ ^ n D cnd.unv.v; p/; p// _ .9s  n/.Freef .p; v/^Term.s/^Freefor.s; v; p/^n D cnd.unv.v; p/; formsub.p; v; s//g

CHAPTER 13. GÖDEL’S THEOREMS

713

(d) (i) PA ` num.;/ D p;q (ii) PA ` num.Sy/ D pSq  num.y/ 23C2n

(e) PA ` gvar.n/ D 2

(f) PA ` Var.gvar.n// (g) PA ` gvar.m/ D gvar.n/ ! m D n *(h) PA ` ŒPrvt.p/ ^ Var.v/ ! Prvt.unv.v; p// (i) PA ` Axiomt.n/ ! Prvt.n/ *(j) PA ` ŒWff .p/ ^ Var.v/ ! Freefor.v; v; p/ *(k) PA ` Axiomad4.n/ $ 9s.9p  n/.9v  n/ŒWff .p/ ^ Var.v/ ^ Term.s/ ^ Freefor.s; v; p/ ^ n D cnd.unv.v; p/; formsub.p; v; s// (l) PA ` num.x/ > ; (m) PA ` numseq.x/ > 1 (n) PA ` len.num.x// D S x *(o) PA ` len.numseq.x// D S x (p) PA ` 8yŒy  x ! exp.numseq.x/; y/ D num.y/ (q) PA ` Var.v/ ! v ¤ num.y/ (r) PA ` Termseq.numseq.x/; num.x// corollary: PA ` Term.num.x// (s) PA ` Termsub.num.n/; v; s; num.n// corollary: PA ` Freet .num.n/; v/ *(t) PA ` ŒWff .p/ ^ Var.v/ ! Freefor.num.x/; v; p/ (u) PA ` Wff .p/ ! Prvt.cnd.unv.gvar.n/; p/; formsub.p; gvar.n/; num.x////

CHAPTER 13. GÖDEL’S THEOREMS

714

Effectively, (h) is like Gen. (k) is like the intuitive version of A4 from p. 611. And (u) results with A4 when the substituted term is a numeral (so that associated restrictions are automatically met). Finally, a theorem with results first for substitution into a conditional, and then for substitution into other substitutions. The latter include matched results for Termsub, Atomsub and then Formsub. Suppose x = xi and y = xj . *T13.57. The following are theorems of PA. (a) PA ` ŒWff .p/ ^ Wff .q/ ^ Term.s/ ! formsub.cnd.p; q/; v; s/ D cnd.formsub.p; v; s/; formsub.q; v; s// *(b) PA ` ŒTerm.p/ ^ v ¤ w ! 9q9t9t 0 ŒTermsub.p; v; num.y/; t / ^ Termsub.p; w; num.z/; t 0 /^Termsub.t; w; num.z/; q/^Termsub.t 0 ; v; num.y/; q/ (c) PA ` ŒAtomic.p/^v ¤ w ! 9q9t 9t 0 ŒAtomsub.p; v; num.y/; t /^Atomsub.p; w; num.z/; t 0 / ^ Atomsub.t; w; num.z/; q/ ^ Atomsub.t 0 ; v; num.y/; q/ *(d) PA ` ŒWff .p/^v ¤ w ! formsub.formsub.p; v; num.y//; w; num.z// D formsub.formsub.p; w; num.z//; v; num.y// (e) PA ` ŒTerm.p/^Var.w/ ! 9q9t 9t 0 ŒTermsub.p; v; w; t /^Termsub.p; v; num.y/; t 0 / ^ Termsub.t; w; num.y/; q/ ^ Termsub.t 0 ; w; num.y/; q/ (f) PA ` ŒAtomic.p/^Var.w/ ! 9q9t 9t 0 ŒAtomsub.p; v; w; t /^Atomsub.p; v; num.y/; t 0 / ^ Atomsub.t; w; num.y/; q/ ^ Atomsub.t 0 ; w; num.y/; q/ (g) PA ` ŒWff .p/ ^ Var.w/ ! formsub.formsub.p; v; w/; w; num.y// D formsub.formsub.p; v; num.y//; w; num.y// (h) PA ` ŒTerm.p/ ^ Var.w/ ! 9q9t9t 0 ŒTermsub.p; v; pS q  w; t / ^ Termsub.p; v; num.Sy/; t 0 /^Termsub.t; w; num.y/; q/^Termsub.t 0 ; w; num.y/; q/ (i) PA ` ŒAtomic.p/ ^ Var.w/ ! 9q9t9t 0 ŒAtomsub.p; v; pSq  w; t / ^ Atomsub.p; v; num.Sy/; t 0 /^Atomsub.t; w; num.y/; q/^Atomsub.t 0 ; w; num.y/; q/ (j) PA ` ŒWff .p/^Var.w/ ! formsub.formsub.p; v; pSq  w/; w; num.y// D formsub.formsub.p; v; num.Sy//; w; num.y//.

CHAPTER 13. GÖDEL’S THEOREMS

715

Speaking loosely: From (a), .P ! Q/vs D Psv ! Qsv . From theorems leading up to (d), v w if v 6 = w then .Pnum.y/ /w D .Pnum.z/ /vnum.y/ . From ones leading to (g), .Pwv /w D num.z/ num.y/ v w v w v w .Pnum.y/ /num.y/ . And from ones leading to (j), .PS w /num.y/ D .Pnum.Sy/ /num.y/ . For these is important that num.y/ is a numeral and so has no variables to be replaced. Arguments combine methods we have seen before; reasoning is straightforward but long. *E13.45. Set up the argument for T13.54k including assertion of the main proposition to be shown by induction; then set up the show part working just the L case. Hard core: finish T13.54k and the rest of the results in T13.54. Hints for T13.54. (i) Under assumptions for !I and (9E) you have both Tsubseq.m; n; t; v; s; q/ and Tsubseq.m0 ; n0 ; t; v; s; r/; with this show 8kŒk < len.m/ ! .8x < len.m0 //.exp.m; k/ D exp.m0 ; x/ ! exp.n; k/ D exp.n0 ; x// by strong induction; the result follows easily from this. (k) Under assumptions for !I and then 9E, you have both Tsubseq.m; n; t; v; v  4; t / and Tsubseq.m0 ; n0 ; t; v; s; u/ with goal t D u; by strong induction show 8kŒk < len.m/ ! .8x < len.m0 //.exp.m; k/ D exp.m0 ; k/ ! .exp.m; k/ D exp.n; k/ ! exp.m0 ; x/ D exp.n0 ; x///; then the result follows easily. (m) Under assumptions for !I and 9E you have Termsub.m; n; t; v; v  4; r/ and Termsub.m0 ; n0 ; t; v; s; u/ where r ¤ t with goal s  u; by strong induction show 8k.k < len.m/ ! .8x < len.m0 //Œexp.m; k/ D exp.m0 ; x/ ! .exp.m; k/ ¤ exp.n; k/ ! s  exp.n0 x//; the result follows. E13.46. Set up the argument for T13.55i including assertion of the main proposition to be shown by induction; then set up the show part working just the P case. Hard core: finish T13.55c and the rest of the results in T13.55. E13.47. Beginning with PA ` Tsubseq.m; m; t; v; v; t /, show that PA ` formsub.p; v; v/ D p. Use this to show PA ` .Sent.p/ ^ Var.v// ! Freef .p; v/. *E13.48. Show (s) and (u) from T13.56. Hard core: show the rest of the results from T13.56. Hints for T13.56. (p) is by induction on the value of x. For (q) it may help to think about the length of v and num.y/. For (r) to show the bounded quantification for Termseq.numseq.x/; num.x// you assume j < len.numseq.x//; then j D ; _ j > ; and the cases are easy. (s) again, in the argument for the bounded quantifier, j D ; _ j > ;.

CHAPTER 13. GÖDEL’S THEOREMS

716

*E13.49. Show T13.57a; then set up the argument for T13.57g including assertion of the main proposition to be shown by induction; then set up the show part working just the P case. Hard core: finish T13.57g and the rest of the results in T13.57. Hints for T13.57. (b) Let P = 9q9a9b9c9d ŒTsubseq.a; b; exp.n; k/; w; num.z/; q/ ^ Tsubseq.c; d; show 8x.8k < len.m//.8k 0 < len.m0 //Œlen.exp.m; k//  x ! .exp.m; k/ D 0 0 exp.m ; k / ! P / by IN; the result follows. (c) Under the assumption for !I, apply T13.48c and then (b). For (e) let P = 9q9a9b9c9d ŒTsubseq.a; b; exp.n; k/; w; num.y/; q/ ^ Tsubseq.c; d; exp.n0 ; k 0 /; w; num.y/; q/; show 8x.8k < len.m//.8k 0 < len.m0 //Œlen.exp.m; k//  x ! .exp.m; k/ D exp.m0 ; k 0 / ! P / by IN. exp.n0 ; k 0 /; v; num.y/; q/;

13.5.2

Sigma Star

Our aim is to show PA ` Q ! Q for any †1 sentence Q. Given our minimal resources, the task is simplified if we can give a minimal specification of the †1 formulas themselves. Toward this end, we introduce a special class of formulas, the †? formulas; and show that every †1 formula is a †? formula. †? formulas are as follows. (†? ) For any variables x, y and :, (a) ; D :, y D :, Sy D :, x C y D : and x  y D : are strictly †? . (s) If P and Q are strictly †? , then so are .P _ Q/, and .P ^ Q/. (8) If P is strictly †? , then so is .8x  y/P where y does not occur in P . (9) If P is strictly †? , then so is 9xP . (c) Nothing else is strictly †? . A formula is †? iff it is equivalent to a strictly †? formula. Given that the existential quantifier comes to the front (as for T12.10), it is perhaps obvious that every †? formula is †1 . At any rate, we aim to show the other direction: that every †1 formula is provably equivalent a †? formula. Then results which apply to all the †? formulas immediately transfer to the †1 formulas. We begin showing that there are †? formulas equivalent to atomic equalities of the sort t D x. Then (depending on an extended notion of normal form and a result result according to which 0 formulas always have equivalent normal forms) we show that there are †? formulas equivalent to 0 formulas. From this it is a short step to the result that there are †? formulas equivalent to all the †1 formulas. First, then, the result for atomic equalities,

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T13.58. For any P of the form t D x, there is a †? formula P ? such that PA ` P $ P ?. By induction on the function symbols in t. Basis: If t has no function symbols, then it is the constant ; or a variable y, so P is of the form ; D x or y D x; but these are already †? formulas. So let P ? be the same as P . Then PA ` P $ P ? . Assp: For any i , 0 i k, if t has i function symbols, there is a P ? such that PA ` P $ P ? . Show: If t has k function symbols, there is a P ? such that PA ` P $ P ? . If t has k function symbols, then it is of the form S r, r C s or r  s for r and s with k function symbols. (S) t is Sr, so that P is Sr D x. Set P ? D 9zŒ.r D z/? ^ S z D x; then by assumption, PA ` r D z $ .r D z/? . So reason as follows, 1. r D z $ .r D z/?

assp

2.

Sr D x

A (g $I)

3. 4. 5.

r D r ^ Sr D x 9zŒr D z ^ Sz D x 9zŒ.r D z/? ^ Sz D x

from 2 3 9I 1,4 with T9.9

6.

9zŒ.r D z/? ^ Sz D x

A (g $I)

7.

.r D z/? ^ Sz D x

A (g 69E)

8. 9.

rDz Sr D x

1,7 $E from 7,8

10.

Sr D x

11. S r D x $ 9zŒ.r D z/? ^ Sz D x

6,7-9 9E 2-5,6-10 $I

So PA ` P $ P ? . (+) t D s C r, so that P is s C r D x. Set P ? = 9u9vŒ.s D u/? ^ .r D v/? ^ u C v D x. Then PA ` P $ P ? . () Similarly. Indct: For any P of the form t D x, there is a P ? such that PA ` P $ P ? . Now generalize some operations from T8.1. There we said a formula is in normal form iff its only operators are _, ^, and , and the only instances of  are immediately prefixed to atomics. Now a formula is in (extended) normal form iff its only operators are _, ^, , or a bounded quantifier, and the only instances of  are immediately

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prefixed to atomics (which may include inequalities). Again, generalizing from before, where P is a normal form, let P 0 be like P except that _ and ^, universal and existential quantifiers and, for an atomic A, A and A are interchanged. So, for example, .9x  p/.x D p_x 6> p/0 = .8x  p/.x ¤ p^x > p/. Still generalizing, for any 0 formula whose operators are , ! and the bounded quantifiers, for atomic A, let A = A; and ŒP  = ŒP  0 ; .P ! Q/ = .ŒP  0 _ Q /; Œ.9x  t/P  = .9x  t/P  and Œ.8x  t/P  = .8x  t/P  (and similarly for .9x < t/ and .8x < t/). Then as a simple extension to the result from E8.10, T13.59. For any 0 formula P , there is a normal formula P  such that ` P $ P  . The demonstration is straightforward extension of the reasoning from E8.9 and E8.10. We show our result as applied to these normal forms. Thus, *T13.60. For any 0 formula P there is a †? formula P ? such that PA ` P $ P ? . From T13.59, for any 0 formula P , there is a normal P  such that ` P $ P  . Now by induction on the number of operators in P  , we show there is a P ? such that PA ` P  $ P ? . Basis: If P  has no operators, then it is an atomic of the sort s D t, s  t or s < t. (D) P  is s D t. Set P ? = 9zŒ.s D z/? ^ .t D z/? . By T13.58, PA ` s D z $ .s D z/? and PA ` t D z $ .t D z/? ; so PA ` P  $ P ? . () P  is s  t, which is to say 9z.z C s D t/. By the case immediately above, PA ` .z C s D t/ $ .z C s D t/? . Set P ? = 9z.z C s D t/? . Then PA ` P  $ P ? . And similarly for eprfnc.n/ ^ RLWFF.z/

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So there is a recursive enumeration of the primitive recursive functions, there is an enumeration of the functions of one free variable, and so forth. Consider an enumeration of the primitive recursive functions of one free variable and an array as follows. 0

(M)

f0 f1 f2

1 f0 .1/

2 f0 .2/ f1 .2/

:::

f0 .0/ f1 .0/ f1 .1/ f2 .0/ f2 .1/ f2 .2/

:: : y. This function is computable; for any n: (i) And consider the function d.n/ = fn .n/ 1 run the enumeration to find fn ; (ii) run fn to find fn .n/; (iii) add one. Since each step is recursive, the whole is computable. But d.n/ is not primitive recursive: d.0/ 6 = f0 .0/; d.1/ 6 = f1 .1/; and in general, d.n/ 6 = fn .n/; so d is not identical to any of the primitive recursive functions. So there are computable functions that are not primitive recursive. It is natural to think that a related argument would show that not all computable functions are recursive: recursively enumerate the recursive functions; then diagonalize to find a computable function not on the list. But this does not work! It is an entirely “grammatical” matter to identify the primitive recursive functions — the function eprfnc.n/ results purely as a matter of form. But there is no parallel method for the recursive fuctions. This clear already by the halting and definition problems (for the latter see E14.7) — there is no recursive way to say in general whether a function is regular, and so to identify functions as recursive. But we may make the point by another diagonal argument (here applied to Turing machines). Suppose there is a recursive enumeration of Turing machines to compute recursive functions (of one free variable) and consider an array as follows. 0

(N)

1

2

:::

…0 …0 .0/ …0 .1/ …0 .2/ …1 …1 .0/ …1 .1/ …1 .2/ …2 …2 .0/ …2 .1/ …2 .2/ :: :

Let .n/ be …n .n/ 1. Since the enumeration is recursive, reasoning as from T14.2, …n .n/ computes the recursive f.n/ = decode.right.n; n; jŒstop.n; n; j/ y//; so f.n/ C 1 y computes .n/. And since f.n/ C 1 y is recursive, .n/ is D0 a Turing program of one free variable; so .n/ appears in the enumeration

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of Turing programs. But this is impossible: .0/ 6 = …0 .0/; .1/ 6 = …1 .1/; and in general .n/ 6 = …n .n/. Reject the assumption: there is no recursive enumeration of Turing machines to compute recursive functions. There is an enumeration of Turing machines; but as in the case of a machine that never halts, not every Turing machine computes a total function. Thus the enumeration of Turing machines does not automatically convert to an enumeration of Turing machines to compute recursive functions. And we are in fact blocked from recursively enumerating the recursive functions. So we are blocked from the proposed means of finding a computable function that is not a recursive function. So this attempt to find a counterexample to Church’s thesis fails. *E14.12. (i) Taking Plus2 to abbreviate Rec2 .I11 ; Comp3 .S 1 ; I33 // as above, write down the LR expression that corresponds to times. (ii) Taking Times2 to abbreviate the expression you have just found, write down the LR expression that corresponds to fact.

*E14.13. (i) Assign numbers to expressions of LR and produce the relation RLWFF to complete the demonstration that there is an enumeration of primitive recursive functions. (ii) Extend the demonstration that there is an enumeration of primitive recursive functions to an enumeration emurec of -recursive functions (as from E14.7). Hints: Take section 10.3.2 as a model for assigning numbers to symbols with superscripts and/or subscripts. For a RLSEQ like FORMSEQ, you will need to know the number of PLACES for members of the sequence (always contained in the first symbol). Also, for Comp, you will want a number for a sequence of m prior formulas each of which has 0 or n places. The nature of an algorithm. There are also reasons for Church’s thesis from the very nature of an algorithm.8 Perhaps the “received wisdom” with respect to Church’s thesis is as follows. The reason why Church’s [Thesis] is called a thesis is that it has not been rigorously proved and, in this sense, it is something like a “working hypothesis.” Its plausibility can be attested inductively — this time not in the sense of mathematical induction, but “on the basis of particular confirming cases.” The Thesis 8 Material

in this section is developed from Smith, An Introduction to Gödel’s Theorems, chapter 45; Smith, “Squeezing Arguments”; along with Kolmogorov and Uspenskii, “On the Definition of an Algorithm.” See also Black, “Proving Church’s Thesis.”

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is corroborated by the number of intuitively computable functions commonly used by mathematicians, which can be defined within recursion theory. But Church’s Thesis is believed by many to be destined to remain a thesis. The reason lies, again, in the fact that the notion of effectively computable function is a merely intuitive and somewhat fuzzy one. It is quite difficult to produce a completely rigorous proof of the equivalence between intuitively computable and recursive functions, precisely because one of the sides of the equivalence is not well-defined (Berto, There’s Something About Gödel, pp. 76-77). There are a couple of themes in this passage. First, that Church’s thesis is typically accepted on grounds of the sort we have already considered. Fair enough. But second that it is not, and perhaps cannot, be proved. The idea seems to be that the recursive functions are a precise mathematically defined class, while the algorithmically computable functions are not. Thus there is no hope of a demonstrable equivalence between the two. But we should be careful. Granted: If we start with an inchoate notion of computable function that includes, at once, calculations with pencil and paper, calculations on the latest and greatest supercomputer, and calculations on Zeno’s machine, there will be no saying whether the computable functions definitely are, or are not, identical to the Turing computable functions. But this is not the notion with which we are working. We have a relatively refined technical account of algorithmic computability. Of course, it is not yet a mathematical definition. But neither are our chapter 1 accounts of logical validity and soundness; yet we have been able to show in T9.1 that any argument that is quantificationally valid (in our mathematical sense) is logically valid. And similarly, the whole translation project of chapter 5 assumes the possibility of moving between ordinary and mathematical notions. It is at least possible that an informally defined predicate might pick out a precise object. The question is whether we can “translate” the notion of an algorithm to formal terms. So let us turn to the hard work of considering whether there is an argument for accepting Church’s thesis. A natural first suggestion is that the step-by-step and finite nature of any algorithm is always within the reach of, or reflected by, some Turing program or recursive function, so that the algorithmically computable functions are inevitably recursively computable.9 Already, this may amount to a consideration or reason in favor of accepting the Thesis. In chapter 45 of his An Introduction to Gödel’s Theorems, Peter Smith advances a proposal according to which such considerations amount to proof. 9 This idea is contained already in the foundational papers of Church, “An Unsolvable Problem,” and Turing, “On Computable Numbers.”

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Smith’s overall strategy involves “squeezing” algorithmic computability between a pair of mathematically precise notions. Even if a condition C (say, “being a tall person”) is vague, it might remain that there is some completely precise sufficient condition S (being over seven feet tall), such that anything that is S is C , and perfectly precise necessary condition N (being over five feet tall) such that anything that is C is N . So, S



C



N

If it should also happen that N implies S , then the loop is closed, so that, S



C



N

And the target condition C is equivalent to (squeezed between) the precise necessary and sufficient conditions. Of course, in our simple example, N does not imply C : being over five feet tall does not imply being over seven feet tall. For Church’s thesis, we already have that Turing computability is sufficient for algorithmic computability. So what is required is some necessary condition so that, T



A



N

Turing computability implies algorithmic computability and algorithmic computability implies the necessary condition. Church’s thesis follows if, in addition, N implies Turing computability. As it turns out, we shall be able to specify a condition N which (mathematically) implies T . It will be more controversial whether A implies N . The argument has three stages: The idea is that, (i) there are some necessary features of an algorithm, such that any algorithm has those features; (ii) any routine with those features is embodied in a modified Kolmogorov-Uspenskii (MKU) machine; (iii) every function that is MKU computable is recursive, and so Turing computable. Necessary features

- MKU computability

- Turing computability

The result is that MKU computability works as as the precise condition N in the squeezing argument: A implies N , and N implies T . So T iff A iff N , and Church’s thesis is established — or no less plausible than is the conclusion of this argument. Perhaps the following are necessary conditions on any algorithm, so that any algorithm satisfies the conditions. If, additionally, we hold that any routine which satisfies the constraints is an algorithm, then the conditions are necessary and sufficient — so we may see them as an extension or sharpening of our initial more sketchy account.

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At this stage, though, the important requirement is that any algorithm satisfies the conditions.10 AC

(1) There is some dataspace consisting of a finite array of “cells” which may stand in some relations R0 , R1 : : : Ra and contain some entities s0 , s1 : : : sb . (2) At every stage in a computation, there is some finite “active” portion of the dataspace upon which the algorithm operates. (3) The body of the algorithm includes finitely many instructions for modifying the active portion of the dataspace depending on its character, and for jumping to the next set of instructions. (4) For the calculation of a function f.Ex/ = y there is some finite initial representation of xE and some way to read off the value of y, after a finite number of steps.

So this sets up an algorithm abstractly described. It is hard to see how an algorithm would not involve some space, portions of which would stand in different relations. At any given stage, the algorithm operates on some portion of the space, where these operations may depend upon, and modify the arrangement of the active space. The algorithm itself consists of some instructions for operating on the dataspace, where these are generically of the sort, “if the active area is of type t, perform action a, and go to new instructions q.” The calculation of a function f.Ex/ somehow takes xE as an input, and gives a way to read off the value of y as an output. And an algorithm terminates in a finite number of steps. Observe that the squeezing argument is effective to the extent that we begin with the notion of an algorithm and show that for any algorithm there is a Turing machine equivalent to it. It is cast into doubt if we start with the notion of a Turing machine and force the notion of an algorithm to match. Thus it is important that we are simply spelling out the idea of an algorithm — of what is required of a rote, rule-based based procedure. Also the finite constraints on the dataspace, relations, symbols and area in (1) and (2) above seem to be consequences of (3) and (4): There is some upper bound 10 Smith

seems to grant that some such conditions are necessary, even though some method may satisfy the conditions yet fail to count as an algorithm. Perhaps this is because he is impressed by the initial examples of routines implemented by human agents with relatively limited computing power. This is not a problem for his squeezing argument, since the corresponding recursive function may yet be computable by some other method which satisfies more narrow constraints — for example, by a Turing machine.

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to the space modified by instructions from a finite collection, each member of which modifies at most a finite area. Then beginning with a finite initial representation of some xE, including finitely many cells of the dataspace standing in finitely many relations, filled with finitely many symbols and then modifying finite portions of the space finitely many times, all we are going to get are finitely many cells, standing in finitely many relations, filled with finitely many symbols. On the face of it, given their extreme simplicity, it is not obvious that Turing machines compute every algorithmically computable function. But a related device, the MKU machine (modified from Kolmogorov and Uspenskii, “On the Definition of an Algorithm”) purports to implement conditions along these lines. MKU

(1) There are some cells c0 , c1 : : : ca which may stand in relations R0 , R1 : : : Rb and contain symbols s0 , s1 : : : sc . In simple cases, we may think of such arrangements graphically as follows, R1

? a

(O)

b

c

R2

d

R2 is a binary relation and R1 tertiary. Each such relation constitutes an edge. (2) Among the one-place relations is an origin property O such that exactly one cell has it — as indicated by ? above. Then the active area includes all cells on paths n edges from the origin. From (O), cells other than the origin are all one edge from the origin cell. (3) Instructions are finitely many quadruples of the sort hqi ; Sa ; Sb ; qj i where qi and qj are instruction labels; Sa describes an active area; and Sb a state with which the active area is to be replaced. Associate each cell in Sa with the least number of edges between it and the origin; let n be the greatest such integer in Sa ; this n remains the same in every quadruple with label qi , though the value of n may vary as qi varies. Again, instructions are a function in the sense that no instruction has hqi ; Sa i the same but hSb ; qj i different.11 We may see Sa and Sb as follows. and Sa0 might both map onto a given dataspace in case one is included in the other. But the consistency requirement is satisfied when n is constant: for consistency, it is sufficient to require that so long as n.qi ; Sa / is a constant, there is no instruction with hqi ; Sa i the same but hSb ; qj i different. 11 S

a

CHAPTER 14. LOGIC AND COMPUTABILITY

778 (Sb )

(Sa )

R1

R1

? a

R2

c

b

e

R2

(P)

a



R2

c

b

e

R2

w

d

?

R2

d

R3

R3

x

f

In this case n = 2. The active area Sa is replaced by the configuration Sb . The concentric rectangles indicate the “boundary” cells which may themselves be related to cells not part of the active area; the replacing area must have a boundary with cells to match boundary cells of the active area. (4) There is some finite initial setup, and some means of reading off the final value of the function (for different relation and symbol sets, these may be different). We think of the origin cell as the “machine head,” where an algorithm always begins with an instruction label qi = 1 and terminates when qi = 0. So an MKU machine is a significant generalization of a Turing machine. We allow arbitrarily many symbols. And the dataspace is no longer a tape with cells in a fixed linear relation, but a space with cells in arbitrary relations which may themselves be modified by the program. Instructions respond to, and modify, not just individual cells, but arbitrarily large areas of the dataspace. Still, it remains that an instruction qi is of the sort, if Sa perform action A and go to instruction qj . So, the instruction (P) might be applied to get, (A)

(B)

R1

R1

? a

c

b

R2

e

R3

a

h

c

b

R2

e

R3

h

R2 R2

(Q)

d



w

R2

? d

R3

R3

x

f R3

g

R3

g

As indicated by the dotted line, the dataspace (A) has an active area of the sort required in instruction (P); so the active area is replaced according to the instruction for the

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resultant space (B). The example is arbitrary. But that is the point: The machine allows arbitrary rote modifications of a dataspace. Now every MKU computable function is recursive. T14.11. Every MKU computable function is a recursive function. We have been through this sort of thing before. And there are different ways to proceed. I indicate only some natural first steps. Begin assigning numbers to labels, symbols, cells and relations in some reasonable way. a. b.

gŒqi  = 3 8i gŒsi  = 5 8i

c. d.

gŒci  = 7 8i gŒrji  = 9 8.2i 3j / hr i i

Then the number for a page is  0ci  1sa : : :  nsb , and for an edge  0 j  1ca1 h i : : :  icai . So a page is a cell with some symbols, and an edge is an i -place relation applied to i cells. Some data is a sequence of pages with distinct cell numbers, and a structure is a sequence of distinct edges. Cells are (immediately) connected on a structure when it has an edge of which both are members, and connected on a structure when there is a sequence of cells from the structure, beginning with the one, ending with the other such that each is immediately connected to the next. A space is a structure with exactly one origin and every cell connected to all the others. A dataspace is of the sort  m  n1 where m numbers some data, n a space, and every 0 cell from m appears in n. After that, with considerable work, MKUMACH.n/ numbers the MKU machines. (Given the potential for arbitrarily many cells n edges from the origin,12 rather than supplementing the machine with repeating commands for every missing instruction, it is simplest to include a single label that loops on the origin, such that the machine defaults to it.) Then kumachs.i; m; n/ numbers an instruction as a function of the number for the machine, initial label, and dataspace. (Where cells are numbered, some Sa matches the active portion of a dataspace when there is a map on cells that makes Sa match the active area.) For machine i with input n, mkuspace.i; n; j/ and mkustate.i; n; j/ give the current number of the dataspace and state. And mkustop.i; n; j/ takes the value zero when the machine is stopped. Then, h

i

h

i

h

i

h

i

ı



ı ! !ı ! ı



12

ı ! ı ! ! ı

y// f.n/ = mkudecode.mkuspace.i; n; jŒmkustop.i; n; j/ = 0 So, for example, a command might replace ı ! ı with ı ! ! ı. After ı multiple iterations (and a shift of origin), the result might be like this.

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It is a chore to work this out (and you have an opportunity to do so in exercises). But it should be clear that it can be done. Then any MKU computable function is recursive, and therefore every MKU computable function is Turing computable. Given this, the squeezing argument is complete: Turing computability implies algorithmic computability and algorithmic computability implies MKU and so Turing computability. So the algorithmically computable functions are the same as the Turing computable functions. So Church’s thesis! This argument is just as strong as the premise that algorithmic computability implies MKU and so Turing computability. For this, we have translated an informal notion into a formal one. Insofar as translation is not itself a formal procedure, the result is not formal proof of Church’s thesis. But not all proof is formal, and it may be that the argument is sufficient to demonstrate its conclusion. Perhaps it is difficult to imagine an algorithmic method that does not conform to AC and then MKU. But failure of imagination is not the same as proof. So there is space for different objections: First, one might worry that the account AC of an algorithm is insufficient in some respect. But AC is offered as a further exposition or sharpening of what it is to be an algorithm. Given this, our version of Church’s thesis applies to it. An argument about whether Church’s thesis applies to a class C of functions is not undercut by observing that there are classes other than C . Still, one might worry that the MKU machine does not compute every algorithm from AC. Against this, there are a couple of replies. First, careful about what the MKU machine can do. Say we are interested in parallel computing, whether by persons following instructions or by computing devices. An MKU machine has but a single origin; this might seem to be a problem. Still, an active area might have many “shapes” — and things might be set up as follows, @ @ m 

(R)

6 @ @ m   ?m- m @ @ ? m @ @

with “satellite” centers, to achieve the effect of parallel computing. Similarly, with a bit of thought, one can see how the MKU machine might achieve the effect of absolute addressing or the like. So it is important to recognize the generality already built into the MKU machine.

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Perhaps, though, the objection goes through and some algorithmic method really is beyond the reach of the MKU machine. So for example some algorithm might require physical actions other than symbol manipulation. Consider a method for truth table construction with the instruction, “whack yourself in the head three times and write a T in the first row of the first column.” An MKU machine does not have a head, and so cannot perform this action. More seriously, we might consider actions as applied to, say, a physical abacus — as “move the bead on the second wire to the leftmost available position.” The MKU machine does not move physical beads on a wire, so it does not perform addition on an abacus. Still, it should be possible to number the states of an abacus, and to represent the successive states so as to calculate any function that can be worked on the physical device. In this case, the claim is not that the MKU machine effectuates every algorithm, but rather that it models every algorithm. Supposing this is sustained, the argument for Church’s thesis stands. So we are not left with a formal proof of Church’s thesis. Rather we have a (powerful) case from the independent definitions, the failure of counterexamples and the nature of an algorithm for the result that Church’s thesis is true. Plausibly, there is no formal proof that you have a head. Still, there is a strong case to establish that you do! Similarly our case may seem sufficient to establish Church’s thesis. To the extent that Church’s thesis is either plausible or established, our limiting results become full-fledged incomputability results with applications to logic and computing more generally. In addition, from Church’s thesis, the computability of a function implies that it is recursive. Having attained Church’s thesis only at the very end, we have not applied the thesis in this way. But one might move from the observation that some function is computable, through the thesis, to the result that the function is recursive. And this is frequently done! *E14.14. Work out codes for the MKU machine through dataspace. Very hard core: Assuming functions code.n/ and decode.d/, complete the demonstration that any MKU computable function f.n/ is recursive.

E14.15. For each of the following concepts, explain in an essay of about two pages, so that (college freshman) Hannah could understand. In your essay, you should (i) identify the objects to which the concept applies, (ii) give and explain the definition, and give and explicate examples (iii) where the concept applies, and (iv) where it does not. Your essay should exhibit an understanding of methods from the text.

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Theorems of chapter 14 T14.1 There is a recursive enumeration of the Turing machines. T14.2 Every Turing computable function is a recursive function. T14.3 Every recursive function is Turing computable. T14.4 There is no Turing machine H.i/ such that H.i/ = 0 if …i .i/ halts and H.i/ = 1 if it does not. T14.5 There is no Turing machine H.i; n/ such that H.i; n/ = 0 if …i .n/ halts and H.i; n/ = 1 if it does not. T14.6 If Q is !-consistent, then no Turing computable function f.n/ is such that f.n/ = 0 just in case n numbers a theorem of predicate logic. T14.7 The set of truths of LNT is not recursively enumerable. T14.8 If T is a recursively axiomatized sound theory whose language includes LNT , then T is negation incomplete. T14.9 If T is a recursively axiomatized theory extending Q, then there is a sentence P such that if T is consistent T ° P , and if T is !-consistent, T ° P . T14.10 Every (total) function that can be captured by a recursively axiomatized consistent theory extending Q is recursive. T14.11 Every MKU computable function is a recursive function. And we mention, CT Church’s Thesis: The total numerical functions that are effectively computable by some algorithmic method are just the recursive functions.

a. The Turing computable functions, and their relation to the recursive functions. b. The essential elements from the chapter contributing to a demonstration of the decision problem, along with the significance of Church’s thesis for this result. c. The essential elements from this chapter contributing to a demonstration of (the semantic version of) the incompleteness of arithmetic. d. Church’s thesis, along with reasons for thinking it is true, including the possibility of demonstrating its truth.

Concluding Remarks

783

Looking Forward and Back We began this text in Part I setting up the elements of classical symbolic logic. Thus we began with four notions of validity: logical validity, validity in our derivation systems AD and ND, along with semantic (sentential and) quantificational validity. After a parenthesis in Part II to think about techniques for reasoning about logic, we began to put those techniques to work. The main burden of Part III was to show soundness and adequacy of our classical logic, that € ` P iff €  P . This is the good news. In Part IV we established some limiting results. These include Gödel’s first and second theorems, that no consistent, recursively axiomatizable extension of Q is negation complete, and that no consistent recursively axiomatized theory extending PA proves its own consistency. Results about derivations are associated with computations, and the significance of this association extended by means of Church’s thesis. This much constitutes a solid introduction to classical logic, and should position you make progress in logic and philosophy, along with related areas of mathematics and computer science. Excellent texts which mostly overlap the content of one, but extend it in different ways are Mendelson, Introduction to Mathematical Logic; Enderton, Introduction to Mathematical Logic; and Boolos, Burgess and Jeffrey, Computability and Logic; these put increased demands on the reader (and such demands are one motivation for our text), but should be accessible to you now; Shoenfield, Mathematical Logic is excellent yet still more difficult. Smith, An Introduction to Gödel’s Theorems extends the material of Part IV; Cooper, Computability Theory develops it especially from the perspective of chapter 14. Much of what we have done presumes some set theory as Enderton, Elements of Set Theory, or model theory as Manzano, Model Theory and, more advanced, Hodges, A Shorter Model Theory. In places, we have touched on logics alternative to classical logic, including multivalued logic, modal logic, and logics with alternative accounts of the conditional. A good place to start is Priest, Non-Classical Logics, which is profitably read with Roy, “Natural Derivations for Priest” which introduces derivations in a style much 784

CONCLUSION

785

like our own. Our logic is first-order insofar as quantifiers bind just variables for objects. Second-order logic lets quantifiers bind variables for predicates as well (so 8x8yŒx D y ! 8F .F x $ F y/ expresses the indiscernibility of identicals). Second-order logic has important applications in mathematics, and raises important issues in metalogic. For this, see Shapiro, Foundations Without Foundationalism, and Manzano, Extensions of First Order Logic. Philosophy of logic and mathematics is a subject matter of its own. Shapiro, “Philosophy of Mathematics and Its Logic” (along with the rest of the articles in the Oxford Handbook, and Shapiro, Thinking About Mathematics are a good place to start. Benacerraf and Putnam, Philosophy of Mathematics and Marcus and McEvoy, Philosophy of Mathematics are collections of classic articles. Smith’s online, “Teach Yourself Logic” is an excellent comprehensive guide to further resources. Have fun!

Answers to Selected Exercises

786

787

ANSWERS FOR CHAPTER 1

Chapter One E1.1. Say whether each of the following stories is internally consistent or inconsistent. In either case, explain why. a. Smoking cigarettes greatly increases the risk of lung cancer, although most people who smoke cigarettes do not get lung cancer. Consistent. Even though the risk of cancer goes up with smoking, it may be that most people who smoke do not have cancer. Perhaps 49% of people who smoke get cancer, and 1% of people who do not smoke get cancer. Then smoking greatly increases the risk, even though most people who smoke do not get it. c. Abortion is always morally wrong, though abortion is morally right in order to save a woman’s life. Inconsistent. Suppose (whether you agree or not) that abortion is always morally wrong. Then abortion is wrong even in the case when it would save a woman’s life. So the story requires that abortion is and is not wrong. e. No rabbits are nearsighted, though some rabbits wear glasses. Consistent. One reason for wearing glasses is to correct nearsightedness. But glasses may be worn for other reasons. It might be that rabbits who wear glasses are farsighted, or have astigmatism, or think that glasses are stylish. Or maybe they wear sunglasses just to look cool. g. Barack Obama was never president of the United States, although Michelle is president right now. Consistent. Do not get confused by the facts! In a story it may be that Barack was never president and his wife was. Thus this story does not contradict itself and is consistent. i. The death star is a weapon more powerful than that in any galaxy, though there is, in a galaxy far, far away, a weapon more powerful than it. Inconsistent. If the death star is more powerful than any weapon in any galaxy, then according to this story it is and is not more powerful than the weapon in the galaxy far far away. E1.2. For each of the following sentences, (i) say whether it is true or false in the real world and then (ii) say, if you can, whether it is true or false according to the accompanying story. In each case, explain your answers. Exercise 1.2

788

ANSWERS FOR CHAPTER 1

c. Sentence: After overrunning Phoenix in early 2006, a herd of buffalo overran Newark, New Jersey. Story: A thundering herd of buffalo overran Phoenix, Arizona in early 2006. The city no longer exists. (i) It is false in the real world that any herd of buffalo overran Newark anytime after 2006. (ii) And, though the story says something about Phoenix, the story does not contain enough information to say whether the sentence regarding Newark is true or false. e. Sentence: Jack Nicholson has swum the Atlantic. Story: No human being has swum the Atlantic. Jack Nicholson and Bill Clinton and you are all human beings, and at least one of you swam all the way across! (i) It is false in the real world that Jack Nicholson has swum the Atlantic. (ii) This story is inconsistent! It requires that some human both has and has not swum the Atlantic. Thus we refuse to say that it makes the sentence true or false. g. Sentence: Your instructor is not a human being. Story: No beings from other planets have ever made it to this country. However, your instructor made it to this country from another planet. (i) Presumably, the claim that your instructor is not a human being is false in the real world (assuming that you are not working by independent, or computeraided study). (ii) But this story is inconsistent! It says both that no beings from other planets have made it to this country and that some being has. Thus we refuse to say that it makes any sentence true or false. i. Sentence: The Yugo is the most expensive car in the world. Story: Jaguar and Rolls Royce are expensive cars. But the Yugo is more expensive than either of them. (i) The Yugo is a famously cheap automobile. So the sentence is false in the real world. (ii) According to the story, the Yugo is more expensive than some expensive cars. But this is not enough information to say whether it is the most expensive car in the world. So there is not enough information to say whether the sentence is true or false. E1.3. Use our invalidity test to show that each of the following arguments is not logically valid, and so not logically sound. Exercise 1.3

789

ANSWERS FOR CHAPTER 1 *For each of these problems, different stories might do the job. a. If Joe works hard, then he will get an ‘A’ Joe will get an ‘A’ Joe works hard a. In any story with premises true and conclusion false, 1. If Joe works hard, then he will get an ‘A’ 2. Joe will get an ‘A’ 3. Joe does not work hard

b. Story: Joe is very smart, and if he works hard, then he will get an ‘A’. Joe will get an ‘A’; however, Joe cheats and gets the ‘A’ without working hard. c. This is a consistent story that makes the premises true and the conclusion false; thus, by definition, the argument is not logically valid. d. Since the argument is not logically valid, by definition, it is not logically sound. E1.4. Use our validity procedure to show that each of the following is logically valid, and decide (if you can) whether it is logically sound. *For each of these problems, particular reasonings might take different forms. a. If Bill is president, then Hillary is first lady Hillary is not first lady Bill is not president a. In any story with premises true and conclusion false, 1. If Bill is president, then Hillary is first lady 2. Hillary is not first lady 3. Bill is president b. In any such story, Given (1) and (3), 4. Hillary is first lady Given (2) and (4), 5. Hillary is and is not first lady

Exercise 1.4.a

790

ANSWERS FOR CHAPTER 1

c. So no story with the premises true and conclusion false is a consistent story; so by definition, the argument is logically valid. d. In the real world Hillary is not first lady and Bill and Hillary are married so it is true that if Bill is president, then Hillary is first lady; so all the premises are true and by definition the argument is logically sound. E1.5. Use our procedures to say whether the following are logically valid or invalid, and sound or unsound. Hint: You may have to do some experimenting to decide whether the arguments are logically valid or invalid — and so decide which procedure applies. c. Some dogs have red hair Some dogs have long hair Some dogs have long, red hair a. In any story with the premise true and conclusion false, 1. Some dogs have red hair 2. Some dogs have long hair 3. No dogs have long, red hair b. Story: There are dogs with red hair, and there are dogs with long hair. However, due to a genetic defect, no dogs have long, red hair. c. This is a consistent story that makes the premise true and the conclusion false; thus, by definition, the argument is not logically valid. d. Since the argument is not logically valid, by definition, it is not logically sound. E1.6. Use our procedures to say whether the following are logically valid or invalid, and sound or unsound. d. Cheerios are square Chex are round There is no round square a. In any story with the premises true and conclusion false, 1. Cheerios are square 2. Chex are round 3. There is a round square Exercise 1.6.d

791

ANSWERS FOR CHAPTER 1 b. In any such story, given (3), 4. Something is round and not round

c. So no story with the premises true and conclusion false is a consistent story; so by definition, the argument is logically valid. d. In the real world Cheerios are not square and Chex are not round, so the premises are not true; so though the argument is valid, by definition it is not logically sound. E1.8. Which of the following are true, and which are false? In each case, explain your answers, with reference to the relevant definitions. c. If the conclusion of an argument is true in the real world, then the argument must be logically valid. False. An argument is logically valid iff there is no consistent story that makes the premises true and the conclusion false. Though the conclusion is true in the real world (and so in the real story), there may be some other story that makes the premises true and the conclusion false. If this is so, then the argument is not logically valid. e. If a premise of an argument is false in the real world, then the argument cannot be logically valid. False. An argument is logically valid iff there is no consistent story that makes the premises true and the conclusion false. For logical validity, there is no requirement that every story have true premises — only that ones that do, also have true conclusions. So an argument might be logically valid, and have premises that are false in many stories, including the real story. g. If an argument is logically sound, then its conclusion is true in the real world. True. An argument is logically valid iff there is no consistent story that makes the premises true and the conclusion false. An argument is logically sound iff it is logically valid and its premises are true in the real world. Since the premises are true in the real world, they hold in the real story; since the argument is valid, this story cannot be one where the conclusion is false. So the conclusion of a sound argument is true in the real world. i. If the conclusion of an argument cannot be false (is false in no consistent story), then the argument is logically valid. Exercise 1.8.i

792

ANSWERS FOR CHAPTER 2

True. If there is no consistent story where the conclusion is false, then there is no consistent story where the premises are true and the conclusion is false; but an argument is logically valid iff there is no consistent story where the premises are true and the conclusion is false. So the argument is logically valid.

Chapter Two E2.1. Assuming that S may represent any sentence letter, and P any arbitrary expression of Ls , use maps to determine whether each of the following expressions is (i) of the form .S ! P / and then (ii) whether it is of the form .P ! P /. In each case, explain your answers. e. ..! / ! .! // .S ! P /

.P ! P /

H R…„Hƒ j H ‚ …„ ƒ ? ?‚@

 ? ? @HH /ƒ   j H ‚R…„ ƒ ‚…„



. .! / !  .! / /

. .! / !  .! / /

(i) Since .! / is not a sentence letter, there is nothing to which S maps, and ..! / ! .! // is not of the form .S ! P /. (ii) Since P maps to any expression, ..! / ! .! // is of the form .P ! P / by the above map. E2.3. For each of the following expressions, demonstrate that it is a formula and a sentence of Ls with a tree. Then on the tree (i) bracket all the subformulas, (ii) box the immediate subformula(s), (iii) star the atomic subformulas, and (iv) circle the main operator. a. A subformula: [ A?

This is a formula by FR(s)

In this case, the “tree” is very simple. There are no operators, and so no main operator. There are no immediate subformulas. E2.4. Explain why the following expressions are not formulas or sentences of Ls . Hint: you may find that an attempted tree will help you see what is wrong. b. .P ! Q/ This is not a formula because P and Q are not sentence letters of Ls . They are part of the metalanguage by which we describe Ls , but are not among the Exercise 2.4.b

793

ANSWERS FOR CHAPTER 2

Roman italics (with or without subscripts) that are the sentence letters. Since it is not a formula, it is not a sentence. E2.5. For each of the following expressions, determine whether it is a formula and sentence of Ls . If it is, show it on a tree, and exhibit its parts as in E2.3. If it is not, explain why as in E2.4. a. ..A ! B/ ! ..A ! B/ ! A// This is a formula and a sentence.

A?

B?

A?

@ @ @

A?

@

By FR(s)

 @ @

 

.A ! B/

.A ! B/

s u b f o r m u l a s

B?

By FR(!)

 S

 

S .A ! B/

S S S

..A ! B/ ! A/

S S S  S S



By FR()

  HH HH  H H

S

 

By FR(!)



..A ! B/ ! ..A ! B/ ! A//

By FR(!)

  ..A ! B/ ! ..A ! B/ ! A//



By FR()

c. .A ! B/ ! ..A ! B/ ! A/ A

B

A

@ @ @

B

A

@ @ @

 

.A ! B/

.A ! B/

By FR(s)

 By FR(!)

   .A ! B/

.A ! B/

H

l l l l

HH H

 H H

..A ! B/ ! A/

l

 l  l  l

.A ! B/ ! ..A ! B/ ! A/

Exercise 2.5.c

By FR()

  By FR(!)



Mistake!

794

ANSWERS FOR CHAPTER 2

Not a formula or sentence. The attempt to apply FR(!) at the last step fails, insofar as the outer parentheses are missing. E2.6. For each of the following expressions, demonstrate that it is a formula and a sentence of Ls with a tree. Then on the tree (i) bracket all the subformulas, (ii) box the immediate subformula(s), (iii) star the atomic subformulas, and (iv) circle the main operator. a. .A ^ B/ ! C s u b f o r m u l a s



A?

C?

B?

@ @ @



   .A ^ B/  @  @   @ .A ^ B/ ! C



Formulas by FR(s)

Formula by FR0 (^)

Formula by FR(!), outer parentheses dropped

E2.7. For each of the formulas in E2.6a - e, produce an unabbreviating tree to find the unabbreviated expression it represents. a. .A ^ B/ ! C A

B

C

@

 @  @  .A ! B/  @  @  @ ..A ! B/ ! C /

By AB(^)

Adding outer ( )

E2.8. For each of the unabbreviated expressions from E2.7a - e, produce a complete tree to show by direct application FR that it is an official formula.

Exercise 2.8

795

ANSWERS FOR CHAPTER 2 a. ..A ! B/ ! C / B

A

L L L L L LL

C



B

.A ! B/ .A ! B/ \ \ \

Formulas by FR(s)

Formula by FR()

Formula by FR(!)

Formula by FR()

..A ! B/ ! C /

Formula by FR(!)

E2.12. For each of the following expressions, demonstrate that it is a term of Lq with a tree. c. h3 cf 1 yx This is a term as follows. y

c

T T T

x

these are terms by TR(c), TR(v), and TR(v)

   1 T f y  T  TT h3 cf 1 yx

since y is a term, this is a term by TR(f)

given the three input terms, this is a term by TR(f)

E2.13. Explain why the following expressions are not terms of Lq . d. g 2 yf 1 xc. y is a term, f 1 x is a term and c is a term; but g 2 followed by these three terms is not a term. g 2 yf 1 x is a term, but not g 2 yf 1 xc. E2.14. For each of the following expressions, determine whether it is a term of Lq ; if it is, demonstrate with a tree; if not, explain why. a. g 2 g 2 xyf 1 x This is a term as follows.

Exercise 2.14.a

796

ANSWERS FOR CHAPTER 2 y

x

x

these are terms by TR(v), TR(v), and TR(v)

@ @ @

f 1x

g 2 xy

@ @  @

these are terms by TR(f) and TR(f)

 

g 2 g 2 xyf 1 x

this is a term by TR(f)

b. h3 cf 2 yx This is not a term. c is a term, and f 2 yx is a term; but h3 followed by these two terms is not a term. E2.15. For each of the following expressions, (i) Demonstrate that it is a formula of Lq with a tree. (ii) On the tree bracket all the subformulas, box the immediate subformulas, star the atomic subformulas, circle the main operator, and indicate quantifier scope with underlines. Then (iii) say whether the formula is a sentence, and if it is not, explain why. b. .A1 x ! B 2 cf 1 x/ x

c

x

These are terms by TR(v), TR(c), and TR(v)

L L L f 1x L L . . . . . . . . . . . . . . .L . . . . . . . . . L A1 x ?

subformulas

B 2 cf 1 x ?

This is a term by TR(f)

These are formulas by FR(r) and FR(r)

@

.A1 x

@ @ 2 1 B cf x/ ! 

This is a formula by FR(!)

The x is free; so it is not a sentence. E2.16. Explain why the following expressions are not formulas or sentences of Lq . c. 8xB 2 xg 2 ax This is not a formula because x is not a variable and a is not a constant. These are symbols of the metalanguage, rather than symbols of Lq . E2.17. For each of the following expressions, determine whether it is a formula and a sentence of Lq . If it is a formula, show it on a tree, and exhibit its parts as in E2.15. If it fails one or both, explain why. Exercise 2.17

797

ANSWERS FOR CHAPTER 2 d. 8z.L1 z ! .8wR2 wf 3 axw ! 8wR2 f 3 azww// This has a tree, so it is a formula. But x is free, so it is not a sentence. w a

z

A

D

x w

a

z w w

A   A   A  3 3 f azw  D f axw D  A  . . . . . . . . . . . . D. . . . . . . . . . . . . . . . . . . . . . . .  A D A D D

 A  A

L1 z ? R2 wf 3 axw ?

R2 f 3 azww ?

Terms by TR(v) and TR(c)

Terms by TR(f)

Formulas by FR(r)

C C s u b f o r m u l a s

C 2 3 8wR2 f 3 azww C 8wR wf axw Q  C Q  C Q  Q  C Q 2 3 C .8wR wf axw ! 8wR2 f 3 azww/ C   C  CC

.L1 z ! .8wR2 wf 3 axw ! 8wR2 f 3 azww//

Formulas by FR(8)

Formula by FR(!)

Formula by FR(!)

 8z .L1 z ! .8wR2 wf 3 axw ! 8wR2 f 3 azww// Formula by FR(8)



E2.18. For each of the following expressions, (i) Demonstrate that it is a formula of Lq with a tree. (ii) On the tree bracket all the subformulas, box the immediate subformulas, star the atomic subformulas, circle the main operator, and indicate quantifier scope with underlines. Then (iii) say whether the formula is a sentence, and if it is not, explain why.

Exercise 2.18

798

ANSWERS FOR CHAPTER 2 c. 9xAf 1 g 2 ah3 zwf 1 x _ S a

z w x

C C

D D D

C C C C

These are terms by TR(c) and TR(v)

1 D f x D  DD

This is a term by TR(f)

C h3 zwf 1 x C C  CC

This is a term by TR(f)

g 2 ah3 zwf 1 x

This is a term by TR(f)

f 1 g 2 ah3 zwf 1 x

This is a term by TR(f)

.............................



s Af 1 g 2 ah3 zwf 1 x ? S ? u b f o 1 2 3 1 r 9xAf g ah zwf x HH m u HH l H H   a _ S 9xAf 1 g 2 ah3 zwf 1 x  s

These are formulas by FR(r) and FR(s)

This is a formula by FR0 (9)

This is a formula by FR0 (_)

This has a tree, so it is a formula, but z and w are free, so it is not a sentence. E2.19. For each of the formulas in E2.18, produce an unabbreviating tree to find the unabbreviated expression it represents.

Exercise 2.19

799

ANSWERS FOR CHAPTER 2 c. 9xAf 1 g 2 ah3 zwf 1 x _ S a

z w x

C C

D D D

C C C C

1 D f x D  DD

C h3 zwf 1 x C C  CC

g 2 ah3 zwf 1 x

f 1 g 2 ah3 zwf 1 x ............................. A1 f 1 g 2 ah3 zwf 1 x

Superscript replaced

S

8xA1 f 1 g 2 ah3 zwf 1 x

HH

HH

H H

By AB(9)

.8xA1 f 1 g 2 ah3 zwf 1 x ! S/

By AB(_), with outer ( )

So 9xAf 1 g 2 ah3 zwf 1 x _ S abbreviates .8xA1 f 1 g 2 ah3 zwf 1 x ! S /. E2.20. For each of the unabbreviated expressions from E2.19, produce a compete tree to show by direct application of FR that it is an official formula. In each case, using underlines to indicate quantifier scope, is the expression a sentence? does this match with the result of E2.18?

Exercise 2.20

800

ANSWERS FOR CHAPTER 2 c. .8xA1 f 1 g 2 ah3 zwf 1 x ! S / a

z w x

C C

D D D

C C C C

Terms by TR(c) and TR(v)

1 D f x D  DD

Term by TR(f)

C h3 zwf 1 x C C  CC

Term by TR(f)

g 2 ah3 zwf 1 x

Term by TR(f)

f 1 g 2 ah3 zwf 1 x

Term by TR(f)

............................. A1 f 1 g 2 ah3 zwf 1 x

S

Formulas by FR(r) and FR(s)

A1 f 1 g 2 ah3 zwf 1 x

Formula by FR()

8xA1 f 1 g 2 ah3 zwf 1 x

Formula by FR(8)

8xA1 f 1 g 2 ah3 zwf 1 x

Formula by FR()

8xA1 f 1 g 2 ah3 zwf 1 x

Formula by FR()

PP

PP P

PP P

.8xA1 f 1 g 2 ah3 zwf 1 x ! S/

Formula by FR(!)

Since it has a tree it is a formula. But z and w are free so it is not a sentence. This is exactly the same situation as for E2.18(c). E2.21. For each of the following expressions, (i) Demonstrate that it is a formula of LNT with a tree. (ii) On the tree bracket all the subformulas, box the immediate subformulas, star the atomic subformulas, circle the main operator, and indicate quantifier scope with underlines. Then (iii) say whether the formula is a sentence, and if it is not, explain why. Exercise 2.21

801

ANSWERS FOR CHAPTER 3 b. 9x8y.x  y D x/ Both a formula and a sentence. y

x

 A A  A



x

xy

. . . . . .@ . . . . . . . . . . . . . @ @

s u b f o r m u l a s

Terms by TR(v)

Term by TR(f)

.x  y D x/?

Formula by FR(r)

8y.x  y D x/

Formula by FR(8)

 9x 8y.x  y D x/

Formula by FR0 (9)

E2.22. For each of the formulas in E2.21, produce an unabbreviating tree to find the unabbreviated expression it represents. b. 9x8y.x  y D x/ y

x

A  A  A



x

xy

. . . . . .@ . . . . . . . . . . . . . @ @ Dxyx

The function symbol followed by two terms

The relation symbol followed by two terms

8yDxyx

8x8yDxyx

The existential unabbreviated

So 9x8y.x  y D x/ abbreviates 8x8yDxyx.

Chapter Three E3.1. Where A1 is as above, construct derivations to show each of the following.

Exercise 3.1

802

ANSWERS FOR CHAPTER 3 a. A ^ .B ^ C/ `A1 B 1. 2. 3. 4. 5.

A ^ .B ^ C / ŒA ^ .B ^ C/ ! .B ^ C/ B^C .B ^ C / ! B B

prem ^2 2,1 MP ^1 4,3 MP

E3.2. Provide derivations for T3.6, T3.7, T3.9, T3.10, T3.11, T3.12, T3.13, T3.14, T3.15, T3.16, T3.18, T3.19, T3.20, T3.21, T3.22, T3.23, T3.24, T3.25, and T3.26. As you are working these problems, you may find it helpful to refer to the AD summary on p. 89. T3.12. `AD .A ! B/ ! .A ! B/ 1. 2. 3. 4. 5. 6. 7.

A ! A .A ! A/ ! Œ.A ! B/ ! .A ! B/ .A ! B/ ! .A ! B/ B ! B .A ! B/ ! Œ.A ! B/ ! .A ! B/ .A ! B/ ! .A ! B/ .A ! B/ ! .A ! B/

T3.10 T3.5 2,1 MP T3.11 T3.4 5,4 MP 3,6 T3.2

T3.16. `AD A ! ŒB ! .A ! B/ 1. 2. 3. 4.

.A ! B/ ! .A ! B/ A ! Œ.A ! B/ ! B Œ.A ! B/ ! B ! ŒB ! .A ! B/ A ! ŒB ! .A ! B/

T3.1 1 T3.3 T3.13 2,3 T3.2

T3.21. A ! .B ! C/ `AD .A ^ B/ ! C 1. 2. 3. 4. 5. 6. 7.

A ! .B ! C/ .B ! C/ ! .C ! B/ A ! .C ! B/ C ! .A ! B/ ŒC ! .A ! B/ ! Œ.A ! B/ ! C .A ! B/ ! C .A ^ B/ ! C

prem T3.13 1,2 T3.2 3, T3.3 T3.14 5,4 MP 6 abv

E3.3. For each of the following, expand the derivations to include all the steps from theorems. The result should be a derivation in which each step is either a premise, an axiom, or follows from previous lines by a rule.

Exercise 3.3

803

ANSWERS FOR CHAPTER 3 b. Expand the derivation for T3.4 1. .B ! C / ! ŒA ! .B ! C/ 2. ŒA ! .B ! C/ ! Œ.A ! B/ ! .A ! C / 3. .ŒA ! .B ! C / ! Œ.A ! B/ ! .A ! C // ! Œ.B ! C / ! .ŒA ! .B ! C / ! Œ.A ! B/ ! .A ! C // 4. .B ! C / ! .ŒA ! .B ! C / ! Œ.A ! B/ ! .A ! C // 5. Œ.B ! C / ! .ŒA ! .B ! C / ! Œ.A ! B/ ! .A ! C // ! Œ..B ! C/ ! ŒA ! .B ! C // ! ..B ! C / ! Œ.A ! B/ ! .A ! C// 6. ..B ! C / ! ŒA ! .B ! C // ! ..B ! C / ! Œ.A ! B/ ! .A ! C// 7. .B ! C / ! Œ.A ! B/ ! .A ! C /

A1 A2 A1 3,2 MP A2 5,4 MP 6,1 MP

E3.4. Consider an axiomatic system A2 as described in the main problem. Provide derivations for each of the following, where derivations may appeal to any prior result (no matter what you have done). a. A ! B; B ! C `A2 .C ^ A/ 1. 2. 3. 4. 5. 6.

A!B .A ! B/ ! Œ.B ^ C / ! .C ^ A/ .B ^ C/ ! .C ^ A/ B!C .B ^ C/ .C ^ A/

prem A3 2,1 MP prem 4 abv 5,3 MP

d. `A2 .A ^ B/ ! .B ! A/ 1. 2. 3. 4.

A ! A .A ! A/ ! Œ.A ^ B/ ! .B ^ A/ .A ^ B/ ! .B ^ A/ .A ^ B/ ! .B ! A/

(c) A3 2,1 MP 3 abv

g. A ! B `A2 B ! A 1. 2. 3. 4. 5. 6.

A ! B .A ! B/ ! Œ.B ^ B/ ! .B ^ A/ .B ^ B/ ! .B ^ A/ .B ^ B/ .B ^ A/ B!A

Exercise 3.4.g

prem A3 2,1 MP (b) 3,4 MP 5 abv

804

ANSWERS FOR CHAPTER 3 i. A ! B; B ! C; C ! D `A2 A ! D 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

A!B B!C .C ^ A/ C !D .C ! D/ ! .D ! C/ D ! C .D ! C/ ! Œ.C ^ A/ ! .A ^ D/ .C ^ A/ ! .A ^ D/ .A ^ D/ A!D

prem prem 1,2 (a) prem (f) 5,4 MP A3 7,6 MP 8,3 MP 9 abv

u. `A2 ŒA ! .B ! C/ ! Œ.A ^ B/ ! C Œ.A ^ B/ ^ C ! ŒA ^ .B ^ C/ .B ^ C/ ! .B ^ C / ŒA ^ .B ^ C / ! ŒA ^ .B ^ C / Œ.A ^ B/ ^ C ! ŒA ^ .B ^ C / .Œ.A ^ B/ ^ C  ! ŒA ^ .B ^ C// ! .ŒA ^ .B ^ C / ! Œ.A ^ B/ ^ C/ 6. ŒA ^ .B ^ C/ ! Œ.A ^ B/ ^ C  7. ŒA ! .B ! C/ ! Œ.A ^ B/ ! C 1. 2. 3. 4. 5.

(s) (e) 2 (q) 1,3 (l) (f) 5,4 MP 6 abv

w. A ! B; A ! .B ! C/ `A2 A ! C 1. 2. 3. 4. 5. 6. 7.

A ! .B ! C / ŒA ! .B ! C/ ! Œ.A ^ B/ ! C .A ^ B/ ! C A!A A!B A ! .A ^ B/ A!C

prem (u) 2,1 MP (j) prem 4,5 (r) 6,3 (l)

E3.5. Provide derivations for T3.29, T3.31 and T3.30, explaining in words for every step that has a restriction, how you know that that restriction is met. T3.29. `AD Axt ! 9xA 1. 2. 3. 4. 5. 6. 7.

— for any term t free for x in A

8xA ! ŒAx t 8xA ! ŒAx t x .8xA ! ŒAx t / ! .ŒAt  ! 8xA/ x ŒAt  ! 8xA x Ax t ! ŒAt  x At ! 8xA Ax t ! 9xA

Exercise 3.5 T3.29

A4 same expression T3.13 3,2 MP T3.11 4,5 T3.2 6 abv

805

ANSWERS FOR CHAPTER 3

For (1): Since t is free for x in A, we can be sure that t is free for x in A and so that (1) is an instance of A4. Also for line (2) — not strictly necessary as it involves no change — observe that ŒAxt is the same expression as ŒAxt ; this shift is tracked by square brackets; it matters when it comes time to apply T3.11. E3.6. Provide derivations to show each of the following. a. 8x.H x ! Rx/; 8yHy `AD 8zRz 1. 2. 3. 4. 5. 6. 7. 8.

8x.H x ! Rx/ 8yHy 8x.H x ! Rx/ ! .H z ! Rz/ H z ! Rz 8yHy ! H z Hz Rz 8zRz

prem prem A4 3,1 MP A4 5,2 MP 4,6 MP 7 Gen

c. `AD 9x8yRxy ! 8y9xRxy 1. 2. 3. 4. 5.

8yRxy ! Rxy Rxy ! 9xRxy 8yRxy ! 9xRxy 9x8yRxy ! 9xRxy 9x8yRxy ! 8y9xRxy

A4 T3.29 1,2 T3.2 3 T3.31 4 T3.28

E3.9. Provide demonstrations for the following instances of T3.36 and T3.37. Then, in each case, say in words how you would go about showing the results for an arbitrary number of places. b. .s D t/ ! .A2 rs ! A2 rt/ 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

.y D u/ ! .A2 xy ! A2 xu/ 8xŒ.y D u/ ! .A2 xy ! A2 xu/ 8xŒ.y D u/ ! .A2 xy ! A2 xu/ ! Œ.y D u/ ! .A2 ry ! A2 ru/ .y D u/ ! .A2 ry ! A2 ru/ 8yŒ.y D u/ ! .A2 ry ! A2 ru/ 8yŒ.y D u/ ! .A2 ry ! A2 ru/ ! Œ.s D u/ ! .A2 rs ! A2 ru/ .s D u/ ! .A2 rs ! A2 ru/ 8uŒ.s D u/ ! .A2 rs ! A2 ru/ 8uŒ.s D u/ ! .A2 rs ! A2 ru/ ! Œ.s D t/ ! .A2 rs ! A2 rt/ .s D t/ ! .A2 rs ! A2 rt/

Exercise 3.9.b

A8 1 Gen A4 3,2 MP 4 Gen A4 6,5 MP 7 Gen A4 9,8 MP

806

ANSWERS FOR CHAPTER 3

For an arbitrary ti D s and Rn t1 : : : tn begin with an instance of A8 that has xi D y and Rn x1 : : : xn ; then apply the Gen / A4 / MP pattern n times to convert x1 : : : xn to t1 : : : tn , and then once more to convert y to s. E3.10. Provide derivations to show each of T3.40, T3.41, T3.42, T3.43, T3.44, T3.49, T3.50, T3.51, T3.52, T3.53, and T3.54. T3.40. `PA .St D Ss/ ! .t D s/ 1. 2. 3. 4. 5. 6. 7.

.Sx D Sy/ ! .x D y/ 8xŒ.Sx D Sy/ ! .x D y/ 8xŒ.Sx D Sy/ ! .x D y/ ! Œ.St D Sy/ ! .t D y/ .St D Sy/ ! .t D y/ 8yŒ.St D Sy/ ! .t D y/ 8yŒ.St D Sy/ ! .t D y/ ! Œ.St D S s/ ! .t D s/ .St D Ss/ ! .t D s/

P2 1 Gen A4 3,2 MP 4 Gen A4 6,5 MP

T3.50. `PA Œ..r C s/ C t/ D .r C .s C t// 1. Œ..r C s/ C 0/ D .r C .s C 0// 2. Œ..r C s/ C x/ D .r C .s C x// ! ŒS..r C s/ C x/ D S.r C .s C x//

T3.49 T3.36

3. ŒS..r C s/ C x/ D ..r C s/ C Sx/ 4. ŒS..r C s/ C x/ D ..r C s/ C Sx/ ! .ŒS..r C s/ C x/ D S.r C .s C x// ! Œ..r C s/ C Sx/ D S.r C .s C x/// 5. ŒS..r C s/ C x/ D S.r C .s C x// ! Œ..r C s/ C Sx/ D S.r C .s C x// 6. Œ..r C s/ C x/ D .r C .s C x// ! Œ..r C s/ C Sx/ D S.r C .s C x//

T3.42*

ŒS.r C .s C x// D .r C S.s C x// ŒS.s C x/ D .s C Sx/ ŒS.s C x/ D .s C Sx/ ! Œ.r C S.s C x// D .r C .s C Sx// Œ.r C S.s C x// D .r C .s C Sx// ŒS.r C .s C x// D .r C .s C Sx// ŒS.r C .s C x// D .r C .s C Sx// ! .Œ..r C s/ C Sx/ D S.r C .s C x// ! Œ..r C s/ C Sx/ D .r C .s C Sx/// 13. Œ..r C s/ C Sx/ D S.r C .s C x// ! Œ..r C s/ C Sx/ D .r C .s C Sx// 14. Œ..r C s/ C x/ D .r C .s C x// ! Œ..r C s/ C Sx/ D .r C .s C Sx// 7. 8. 9. 10. 11. 12.

15. 16. 17. 18.

8x.Œ..r C s/ C x/ D .r C .s C x// ! Œ..r C s/ C Sx/ D .r C .s C Sx/// 8xŒ..r C s/ C x/ D .r C .s C x// 8xŒ..r C s/ C x/ D .r C .s C x// ! Œ..r C s/ C t/ D .r C .s C t// Œ..r C s/ C t/ D .r C .s C t//

Exercise 3.10 T3.50

T3.37 4,3 MP 2,5 T3.2 T3.42* T3.42* T3.36 9,8 MP 7,10 T3.35 T3.37 12,11 MP 6,13 T3.2 14 Gen 1,15 Ind* A4 17,16 MP

807

ANSWERS FOR CHAPTER 4 T3.53. `PA Œ.St  s/ D ..t  s/ C s/ first (a): `PA Œ..t  x/ C .x C S t// D ..t  S x/ C S x/ 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

Œ.x C St/ D S.x C t/ ŒS.x C t/ D .Sx C t/ Œ.x C St/ D .Sx C t/ Œ.Sx C t/ D .t C Sx/ Œ.x C St/ D .t C Sx/

T3.42 T3.47* 1,2 T3.35 T3.48 3,4 T3.35

Œ.x C St/ D .t C Sx/ ! Œ..t  x/ C .x C St// D ..t  x/ C .t C Sx// Œ..t  x/ C .x C St// D ..t  x/ C .t C Sx// Œ..t  x/ C .t C Sx// D ...t  x/ C t/ C Sx/ Œ..t  x/ C .x C St// D ...t  x/ C t/ C Sx/ Œ..t  x/ C t/ D .t  Sx/ Œ..t  x/ C t/ D .t  Sx/ ! Œ...t  x/ C t/ C Sx/ D ..t  Sx/ C Sx/ Œ...t  x/ C t/ C Sx/ D ..t  Sx/ C Sx/ Œ..t  x/ C .x C St// D ..t  Sx/ C Sx/

T3.36 6,5 MP T3.50* 7,8 T3.35 T3.44* T3.36 11,10 MP 9,12 T3.35

main result: 1. ŒSt  ;/ D ..t  ;/ C ;/ 2. Œ.St  x/ D ..t  x/ C x/ ! Œ..St  x/ C S t/ D ...t  x/ C x/ C St/

T3.52 T3.36

3. Œ..St  x/ C St/ D .St  Sx/ 4. Œ..St  x/ C St/ D .St  Sx/ ! .Œ..St  x/ C St/ D ...t  x/ C x/ C S t/ ! Œ.St  Sx/ D ...t  x/ C x/ C S t// 5. Œ..St  x/ C St/ D ...t  x/ C x/ C St/ ! Œ.S t  Sx/ D ...t  x/ C x/ C S t/ 6. Œ.S t  x/ D ..t  x/ C x/ ! Œ.St  Sx/ D ...t  x/ C x/ C S t/

T3.44*

Œ...t  x/ C x/ C St/ D ..t  x/ C .x C S t// Œ..t  x/ C .x C S t// D ..t  Sx/ C Sx/ Œ...t  x/ C x/ C St/ D ..t  Sx/ C Sx/ Œ...t  x/ C x/ C St/ D ..t  Sx/ C Sx/ ! .Œ.St  Sx/ D ...t  x/ C x/ C S t/ ! Œ.St  Sx/ D ..t  Sx/ C Sx// 11. Œ.S t  Sx/ D ...t  x/ C x/ C S t/ ! Œ.St  Sx/ D ..t  Sx/ C Sx/ 12. Œ.St  x/ D ..t  x/ C x/ ! Œ.S t  Sx/ D ..t  Sx/ C Sx/ 7. 8. 9. 10.

13. 14. 15. 16.

8x.Œ.St  x/ D ..t  x/ C x/ ! Œ.St  Sx/ D ..t  Sx/ C Sx// 8xŒ.St  x/ D ..t  x/ C x/ 8xŒ.St  x/ D ..t  x/ C x/ ! Œ.St  s/ D ..t  s/ C s/ Œ.St  s/ D ..t  s/ C s/

Exercise 3.10 T3.53

T3.37 4,3 MP 2,5 T3.2 T3.50 (a) T3.35 T3.36 10,9 MP 6,11 T3.2 12 Gen 1,13 Ind* A4 15,14 MP

808

ANSWERS FOR CHAPTER 4

Chapter Four E4.1. Where the interpretation is as in J from p. 100, use trees to decide whether the following sentences of Ls are T or F. a. A A.T/

A.F/

false From J

By T(), row 1

e. .A ! A/ A.T/

false A.T/

@ @ @

From J

.A ! A/.T/

By T(!), row 1

.A ! A/.F/

By T(), row 1

f. .A ! A/ A.T/

true A.T/

A.F/ @ @ @ .A ! A/.T/

From J

By T(), row 1

By T(!), row 3

Exercise 4.1.f

809

ANSWERS FOR CHAPTER 4 i. .A ! B/ ! .B ! A/ B .T/

A.T/

L L L L L LL

true A.T/

B .T/

L L L L L LL

B .F/

.A ! B/.F/

\ \ \ \ \ \\ 

A.F/

From J

By T(), row 1

.B ! A/.F/

By T(!), row 2

.B ! A/.T/

By T(), row 2

 

.A ! B/ ! .B ! A/.T/

By T(!), row 3

E4.2. For each of the following sentences of Ls construct a truth table to determine its truth value for each of the possible interpretations of its basic sentences. a. A A A T T F F F T

d. .B ! A/ ! B A B . B ! A/ ! B T T F F

T F T F

F T F T

T T T F

T F T T

g. C ! .A ! B/ A B C C ! .A ! B/ T T T T

T T F F

T F T F

T T F T

T T F F

F F F F

T T F F

T F T F

T T T T

T T T T

Exercise 4.2.g

810

ANSWERS FOR CHAPTER 4 i. .A ! B/ ! .C ! D/ A B C D . A ! B/ ! . C ! D/ T T T T

T T T T

T T F F

T F T F

F F F F

T T T T

T T T F

F F T T

T T T F

T T T T

F F F F

T T F F

T F T F

F F F F

T T T T

T T T F

F F T T

T T T F

F F F F

T T T T

T T F F

T F T F

T T T T

T T T T

T T T F

F F T T

T T T F

F F F F

F F F F

T T F F

T F T F

T T T T

F F F F

T T T T

F F T T

T T T F

E4.3. For each of the following, use truth tables to decide whether the entialment claims hold. a. A ! A s A valid A A ! A / A T T F F T F F T T F

F T T F

c. A ! B, A s B AB A!B T T F F

T F T T

T F T F

invalid

A /  B F F T T

F T F( T

g. s ŒA ! .C ! B/ ! Œ.A ! C / ! .A ! B/ A B C ŒA ! .C ! B/ ! Œ.A ! C / ! .A ! B/ T T T T

T T F F

T F T F

T T F T

T T F T

T T T T

T F T F

T T F T

T T F F

F F F F

T T F F

T F T F

T T T T

T T F T

T T T T

T T T T

T T T T

T T T T

Exercise 4.3.g

valid

811

ANSWERS FOR CHAPTER 4

E4.4. For each of the following, use truth tables to decide whether the entialment claims hold. c. B _ C s B ! C

invalid

B C B _ C / B ! C T T F F

T T F T

T F T F

T F T T

F T F T

(

d. A _ B, C ! A, .B ^ C / s C AB C A_B

C ! A

valid

 .B ^  C / / C

T T T T

T T F F

T F T F

T T T T

F T F T

T F T F

F F F F

T F T T

F T F F

F T F T

T F T F

F F F F

T T F F

T F T F

T T F F

F T F T

T T T T

T T T T

T F T T

F T F F

F T F T

T F T F

h. s .A $ B/ $ .A ^ B/ invalid A B  .A $ B/ $ .A ^  B/ T T F F

T F T F

F T T F

T F F T

T T F T

F T F F

F T F T

(

E4.5. For each of the following, use truth tables to decide whether the entailment claims hold. a. 9xAx ! 9xBx, 9xAx s 9xBx 9xAx 9xBx 9xAx ! 9xBx T T F F

T F T F

T F T T

invalid

 9xAx / 9xBx F F T T

T F T F

(

E4.8. For LNT and interpretation N as on p. 118, with d as described in the main problem, use trees to determine each of the following.

Exercise 4.8

812

ANSWERS FOR CHAPTER 4 a. Nd ŒCxS; = 3 ;Œ0

x Œ2

L L L L L LL

By TA(v) and TA(c)

S;Œ1

Since h0; 1i 2 NŒS, by TA(f)

CxS;Œ3

Since hh2; 1i; 3i 2 NŒC, by TA(f)

d. Nd.xj4/ Œx C .S S ;  x/ = 12 x Œ4

;Œ0

x Œ4

C C

By TA(v), TA(c) and TA(v)

  S ;Œ1

C C C

 

Since h0; 1i 2 NŒS, by TA(f)

 C

 

SS;Œ2

C C

Since h1; 2i 2 NŒS, by TA(f)

 @

C C C

 @  @

.SS;  x/Œ8

C

Since hh2; 4i; 8i 2 NŒ, by TA(f)

  C CC  Œ12

x C .SS ;  x/

Since hh4; 8i; 12i 2 NŒC, by TA(f)

E4.9. For LNT and interpretation I as above on p. 119, with d as described in the main problem, use trees to determine each of the following. a. Id ŒCxS ; = Hill x ŒHill

L L L L L LL

;ŒBill

S;ŒBill

CxS;ŒHill

By TA(v) and TA(c)

Since hBill, Billi 2 IŒS, by TA(f)

Since hhHill, Billi, Hilli 2 IŒC, by TA(f)

Exercise 4.9.a

813

ANSWERS FOR CHAPTER 4 d. Id.xjBill/ Œx C .S S ;  x/ = Bill ;ŒBill

x ŒBill

x ŒBill

C C

By TA(v), TA(c) and TA(v)

  S;ŒBill

C

 

C C

Since hBill, Billi 2 IŒS, by TA(f)

 C

 

SS;ŒBill

C C

Since hBill,Billi 2 IŒS, by TA(f)

 @

C C C

 @  @

.SS;  x/ŒHill

C

Since hhBill, Billi, Hilli 2 IŒ, by TA(f)

  C CC  ŒBill

x C .SS ;  x/

Since hhBill, Hilli, Billi 2 IŒC, by TA(f)

E4.11. For Lq and interpretation K with variable assignment d as described in the main problem, use trees to determine each of the following. b. Kd Œg 2 yf 1 c = Amy y ŒAmy

c ŒChris

L L L f 1 c ŒAmy L L LL g 2 yf 1 c ŒAmy

By TA(v) and TA(c)

Since hChris, Amyi 2 KŒf 1 , by TA(f)

Since hhAmy, Amyi, Amyi 2 KŒg 2 , by TA(f)

E4.12. Where the interpretation K and variable assignment d are as described in the main problem, use trees to determine whether the following formulas are satisfied on K with d. a. H x

Satisfied

Kd ŒH x.S/ ..

. .

x [Amy]

Exercise 4.12.a

Satisfied f. 8x.H x ! S /

814 ANSWERS FOR CHAPTER 4

Kd Œ8x.H x ! S/.S/



Kd Œ8x.H x ! S/.N/

8x

Kd.xjAmy/ ŒH x ! S.N/

!

!

! S.S/

Kd.xjBob/ ŒH x ! S.N/

Kd.xjChris/ ŒH x

!

. .

Kd.xjAmy/ ŒH x.S/..



Kd.xjAmy/ ŒS .N/



. .

Kd.xjBob/ ŒH x.S/ .. Kd.xjBob/ ŒS .N/

. .

Kd.xjChris/ ŒH x.N/..



Kd.xjChris/ ŒS .N/

x [Amy] Kd.xjAmy/ ŒS .S/

x [Bob] Kd.xjBob/ ŒS .S/

x [Chris] Kd.xjChris/ ŒS .S/

Exercise 4.12.f

Satisfied g. 8y8xLxy

815 ANSWERS FOR CHAPTER 4

Kd Œ8y8xLxy.S/

8y

Kd.yjAmy/ Œ8xLxy.S/

Kd.yjBob/ Œ8xLxy.S/





Kd.yjAmy/ Œ8xLxy.N/

Kd.yjBob/ Œ8xLxy.N/

8x

8x

[Amy] x .  [Amy] Hy .H

Kd.yjAmy;xjAmy/ ŒLxy.S/..

[Bob] x .  [Amy] Hy .H

Kd.yjAmy;xjBob/ ŒLxy.N/..

[Chris] x .  [Amy] Hy .H

Exercise 4.14

Kd.yjAmy;xjChris/ ŒLxy.N../

[Amy] x . H y [Bob] .H

Kd.yjBob;xjAmy/ ŒLxy.S/..

[Bob] x . H y [Bob] .H

Kd.yjBob;xjBob/ ŒLxy.S/ ..

[Chris] x . H y [Bob] .H

Kd.yjBob;xjChris/ ŒLxy.N/..

[Amy] x .  [Chris] Hy .H

Kd.yjChris;xjAmy/ ŒLxy.S../

[Chris]

[Bob] Kd.yjChris/ Œ8xLxy.S/ Kd.yjChris/ Œ8xLxy.N/ Kd.yjChris;xjBob/ ŒLxy.S/..  x  .  [Chris] 8x Hy .H

.  [Chris] Hy .H

Kd.yjChris;xjChris/ ŒLxy.N.. /  x

E4.14. For language Lq consider an interpretation I such that U = f1; 2g, IŒa = 1, IŒA = T, IŒP 1  = f1g, IŒf 1  = fh1; 2i; h2; 1ig. Use interpretation I and trees to show that (a) is not quantificationally valid. Each of the others can be shown to be invalid on an interpretation I that modifies just one of the main parts of I. Produce the modified interpretations, and use them to show that the other

816

ANSWERS FOR CHAPTER 4 arguments also are invalid. c. 8xPf 1 x 6ˆ 8xP x Set I Œf 1  = fh1; 1i; h2; 1ig. I Œ8xPf 1 x.S/ d

I ŒPf 1 x.S/ .. d.xj1/

f 1 x Œ1

x Œ1

ŒPf 1 x.S/ .. I d.xj2/

f 1 x Œ1

x Œ2

I ŒP x.S/ d.xj1/

. . . .

x Œ1

ŒP x.N/ I d.xj2/

. . . .

x Œ2

. .

8x

. .

.N/ I d Œ8xP x

8x

Since the premise is satisfied and a sentence, it is true; since the conclusion is not satisfied, it is not true. Since I makes the premise T and the conclusion not, the argument is not quantificationally valid. E4.15. Find interpretations and use trees to demonstrate each of the following. Be sure to explain why your interpretations and trees have the desired result. For these exercises, other interpretations might do the job! a. 8x.Qx ! P x/ 6ˆ 8x.P x ! Qx/ For an interpretation I set U = f1g, IŒP  = f1g, IŒQ = fg.

Id Œ8x.Qx !

P x/.S/

8x

Id.xj1/ ŒQx !

P x.S/

Id.xj1/ ŒQx.N/ ..

x Œ1

Id.xj1/ ŒP x.S/ ..

x Œ1

Id.xj1/ ŒP x.S/ ..

x Œ1

Id.xj1/ ŒQx.N/ ..

x Œ1

. .

!

. .

Id Œ8x.P x ! Qx/.N/

8x

Id.xj1/ ŒP x ! Qx.N/

. .

!

. .

Since the premise is satisfied and a sentence, it is true; since the conclusion is not satisfied, it is not true. Since I makes the premise T and the conclusion not, the argument is not quantificationally valid.

Exercise 4.15.a

817

ANSWERS FOR CHAPTER 4 c. 8xP x 6ˆ P a For an interpretation I, set U = f1; 2g, IŒa = 1, and IŒP  = f1g. Id Œ8xP x.S/



Id Œ8xP x.N/

Id.xj1/ ŒP x.S/ ..

x Œ1

Id.xj2/ ŒP x.N/ ..

x Œ2

. .

8x

. .

Id ŒP a.N/



Id ŒP a.S/ ..

. .

aŒ1

Since the premise is satisfied and a sentence, it is true; since the conclusion is not satisfied, it is not true. Since I makes the premise T and the conclusion not, the argument is not quantificationally valid. h. 8y8xRxy 6ˆ 8x8yRxy For an interpretation I, set U = f1; 2g, IŒR = fh1; 1i; h1; 2ig. Œ1 x . H y Œ1 .H

Id.xj1;yj1/ ŒRxy.S/ .. Id.xj1/ Œ8yRxy.S/

Id Œ8x8yRxy.S/



Id Œ8x8yRxy.N/

8y

Œ1 x . H y Œ2 .H

Id.xj1;yj2/ ŒRxy.S/ ..

8x Œ2 x . H . H y Œ1

Id.xj2;yj1/ ŒRxy.N/ .. Id.xj2/ Œ8yRxy.N/

8y

Œ2 x . H . H y Œ2

Id.xj2;yj2/ ŒRxy.N/ ..

Œ1 x . H . H y Œ1

Id.xj1;yj1/ ŒRxy.S/ .. Id.xj1/ Œ8yRxy.S/



Id Œ8x8yRxy.N/

Id.xj1/ Œ8yRxy.S/

8y

Œ1 x . H . H y Œ2

Id.xj1;yj2/ ŒRxy.S/ ..

8x Œ2 x . H . H y Œ1

Id.xj2;yj1/ ŒRxy.N/ .. Id.xj2/

Œ8yRxy.N/



Id.xj2/

Œ8yRxy.N/

8y

Œ2 x . H . H y Œ2

Id.xj2;yj2/ ŒRxy.N/ ..

Since the premise is satisfied and a sentence, it is true; since the conclusion is not satisfied, it is not true. Since I makes the premise T and the conclusion not, the argument is not quantificationally valid. Exercise 4.15.h

818

ANSWERS FOR CHAPTER 4

E4.17. Produce interpretations to demonstrate each of the following. Use trees, with derived clauses as necessary, to demonstrate your results. Be sure to explain why your interpretations and trees have the results they do. a. 9xP x 6ˆ 8yP y For an interpretation I, set U = f1; 2g, and IŒP  = f1g. Id Œ9xP x.S/

Id.xj1/ ŒP x.S/ ..

x Œ1

Id.xj2/ ŒP x.N/ ..

x Œ2

Id.yj1/ ŒP y.S/ ..

y Œ1

Id.yj2/ ŒP y.N/ ..

y Œ2

. .

9x

. .

Id Œ8yP y.N/

. .

8y

. .

Since the premise is satisfied and a sentence, it is true; since the conclusion is not satisfied, it is not true. Since I makes the premise T and the conclusion not, the argument is not quantificationally valid. g. 8x.9yRxy $ A/ 6ˆ 9xRxx _ A For an interpretation I, set U = f1; 2g, IŒA = F, IŒR = fh1; 2i; h2; 1ig. Œ1 x . H y Œ1 .H

Id.xj1;yj1/ ŒRxy.N/ .. Id.xj1/ Œ9yRxy.S/ Id.xj1/ Œ9yRxy $

A.S/

Œ1 x . H y Œ2 .H

Id.xj1;yj2/ ŒRxy.S/ ..

$ Id.xj1/ ŒA.S/

Id Œ8x.9yRxy $ A/.S/

9y



Id.xj1/ ŒA.N/ Œ2 x . H y Œ1 .H

Id.xj2;yj1/ ŒRxy.S/ ..

8y Id.xj2/ Œ9yRxy.S/ Id.xj2/ Œ9yRxy $ A.S/



Œ2 x . H y Œ2 .H

Id.xj2;yj2/ ŒRxy.N/ ..

$ Id.xj2/ ŒA.S/

Exercise 4.17.g

9y

Id.xj2/ ŒA.N/

819

ANSWERS FOR CHAPTER 5

Œ1 x . H . H x Œ1

Id.xj1/ ŒRxx.N/ .. Id Œ9xRxx.N/ Id Œ9xRxx

_ A.N/

9x

Œ2 x . H . H x Œ2

Id.xj2/ ŒRxx.N/ ..

_ Id ŒA.N/

Since the premise is satisfied and a sentence, it is true; since the conclusion is not satisfied, it is not true. Since I makes the premise T and the conclusion not, the argument is not quantificationally valid. E4.18. Produce an interpretation to demonstrate each of the following (now in LNT ). Use trees to demonstrate your results. Be sure to explain why your interpretations and trees have the results they do. d. 6ˆ 8x8yŒ.x D y/ ! .x < y _ y < x/ For an interpretation I, set U = f1; 2g, and IŒ ; ^ r < s/ ! r  t < s  t Hint: This this combines strategies from previous problems. T13.11.ae. PA ` .r < s ^ t < u/ ! r  t < s  u Hint: This does not require IN. It is straightforward with T6.66. T13.11.af. PA ` 8xŒ.8: < x/P:x ! P  ! 8xP

strong induction (a)

Hint: Under the assumption for !I, you will have a goal like P .j /; you can get .8z < j /P .z/ ! P .j / from the assumption; go for .8z < j /P .z/ by IN (where the induction is on j ). Then the goal follows immediately by !E. T13.11.ag. PA ` P;x ^ 8xŒ.8:  x/P:x ! PSxx  ! 8xP

strong induction (b)

Again under the assumption for !I, you will be able to obtain 8xP , this time by (af). T13.11.ah. PA ` 9xP ! 9xŒP ^ .8: < x/P:x 

least number principle

Hint: This follows immediately from T13.11af applied to P . E13.13. Produce the quick derivation to show TT13.17d.

Exercise 13.13 T13.17

907

ANSWERS FOR CHAPTER 13 1. .8z < m.x//Q. E x; E z/ 2.

Q.x; E v/

T13.17c A (g !I)

3.

v < m.x/ E

A (c I)

4. 5.

Q.x; E v/ ?

1,3 (8E) 2,4 ?I

6. 7.

v – m.x/ E m.x/ E v

8. Q.x; E v/ ! m.x/ E v

3-5 I 6 T13.11r 2-7 !I

E13.15. Complete the justifications for Def [rm] and Def [qt]. Def [rm]. (i) PA ` 9x.9w  ;/Œ; D S n  w C x ^ x < S n. Supposing the zero case is done,

Exercise 13.15 Def [rm]

ANSWERS FOR CHAPTER 13

908

1. 9x.9w  ;/Œ; D S n  w C x ^ x < S n

zero case

2.

9x.9w  j /Œj D S n  w C x ^ x < S n

A (g !I)

3.

.9w  j /Œj D S n  w C k ^ k < S n

A (g 29E)

4. 5. 6. 7. 8. 9. 10. 11. 12.

j D Sn  l C k ^ k < Sn l j

A (g 3 (9E))

j D Sn  l C k k < Sn Sj D S ŒS n  l C k S n  l C S k D S ŒS n  l C k Sj D S n  l C S k k 1 ! 9z.Pr.S z/ ^ zjS k/ 46. 8xf.8y  x/Œy > 1 ! 9z.Pr.S z/ ^ zjy/ ! ŒS x > 1 ! 9z.Pr.S z/ ^ zjS x/g 47. 8xŒx > 1 ! 9z.Pr.S z/ ^ zjx/

T13.23.e. PA ` Rp.a; b/ $ 9xŒPr.S x/ ^ xja ^ xjb.

Exercise 13.20 T13.23.e

3-43 !I 2-44 !I 45 8I 1,46 T13.11ag

920

ANSWERS FOR CHAPTER 13

1.

Rp.a; b/

A (g $I)

2. 3.

8d Œ.d ja ^ d jb/ ! d D ; 9xŒPr.S x/ ^ xja ^ xjb

1 def A (c I)

4.

Pr.Sj / ^ j ja ^ j jb

A (c 39E)

j ja ^ j jb j D; 11 Sj  1 1 – Sj Pr.Sj / 1 < Sj ^ 8d Œd jSj ! .d D ; _ Sd D Sj / 1 < Sj ?

4 ^E 2,5 8E T13.11m 6,7 DE 8 T13.11r 4 ^E 10 def 11 ^E 9,12 ?I

5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

?

3,4-13 9E

15.

9xŒPr.S x/ ^ xja ^ xjb

3-14 I

16.

9xŒPr.S x/ ^ xja ^ xjb

A (g $I)

17. 18.

8xŒPr.S x/ ! .xja ^ xjb/ j ja ^ j jb

16 QN,DeM A (g !I)

19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34.

j D;_j >; j >; Sj > 1 9z.Pr.S z/ ^ zjSj / Pr.S k/ ^ kjSj kjSj j ja kjSj ^ j ja kja j jb kjSj ^ j jb kjb kja ^ kjb Pr.S k/ .kja ^ kjb/ ?

35.

?

36. 37.

j ; j D;

38. 39. 40.

T13.11f A (c I) 20 T13.11k 21 T13.23d A (c 229E) 23 ^E 18 ^E 24,25 ^I 26 T13.22f 18 ^E 26,28 ^E 29 T13.22f 27,30 ^I 23 ^E 17,32 8E 31,33 ?I 22,23-34 9E 20-35 I 19,36 DS

.j ja ^ j jb/ ! j D ; 8d Œ.d ja ^ d jb/ ! d D ; Rp.a; b/

41. Rp.a; b/ $ 9xŒPr.S x/ ^ xja ^ xjb

18-37 !I 38 8I 39 def 1-15,16-40 $I

T13.23.f. PA ` 8x8yŒG.a; b; x/ ! G.a; b; x  y/ With the assumptions G.a; b; j / and then au C j D bv for !I and 9E, you Exercise 13.20 T13.23.f

ANSWERS FOR CHAPTER 13

921

can show auk C j k D bkv and generalize. T13.23.g. PA ` .a > ; ^ b > ;/ ! 8x8yŒ.G.a; b; x/ ^ G.a; b; y/ ^ x  y/ ! G.a; b; x : y/ 1.

a >;^b >;

A (g !I)

2. 3. 4.

a>; b>; G.a; b; i / ^ G.a; b; j / ^ i  j

1 ^E 1 ^E A (g !I)

5. 6. 7. 8. 9.

G.a; b; i / 9x9y.ax C i D by/ G.a; b; j / 9x9y.ax C j D by/ ap C i D bq

4 ^E 5 def 4 ^E 7 def A (g 69E)

10.

ar C j D bs

A (g 89E)

11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38.

i j bar  ar abs  bs ap C i  i bq  i bq  j bar C bq  ar C j bar C bq  bs : : .bq C bar/ C .bsa bs/ D .bq C bar/ C .bsa bs/ : : Œbsa C .bq C bar/ bs D .bq C bar/ C .bsa bs/ : : Œ.bq C bar/ bs C bsa D .bq C bar/ C .bsa bs/ : : Œ.bq C bar/ .ar C j / C bsa D .bq C bar/ C .bsa bs/ : : : Œ..bq C bar/ j / ar C bsa D .bq C bar/ C .bsa bs/ : : : Œ..bq j / C bar/ ar C bsa D .bq C bar/ C .bsa bs/ : : : Œ.bar ar/ C .bq j / C bsa D .bq C bar/ C .bsa bs/ : : : Œ.bar ar/ C ..ap C i / j / C bsa D .bq C bar/ C .bsa bs/ : : : Œ.bar ar/ C ..i j / C ap C bsa D .bq C bar/ C .bsa bs/ : : : .ap C abs/ C .bar ar/ C .i j / D .bq C bar/ C .bsa bs/ : : : a.p C bs/ C .bar ar/ C .i j / D b.q C ar/ C .bsa bs/ : : : a.p C bs/ C a.br r/ C .i j / D b.q C ar/ C b.sa s/ : : : aŒ.p C bs/ C .br r/ C .i j / D bŒ.q C ar/ C .sa s/ : 9x9yŒax C .i j / D by : G.a; b; i j / : G.a; b; i j / : G.a; b; i j / : ŒG.a; b; i / ^ G.a; b; j / ^ i  j  ! G.a; b; i j / : 8x8y.ŒG.a; b; x/ ^ G.a; b; y/ ^ x  y ! G.a; b; x y// : .a > ; ^ b > ;/ ! 8x8y.ŒG.a; b; x/ ^ G.a; b; y/ ^ x  y ! G.a; b; x y//

T13.23.h. PA ` ŒRp.a; b/ ^ a > 1 ^ b > 1 ! 9x9y.ax C 1 D by/ (a) Show a  .b : 1/ C a D b  a and generalize. Exercise 13.20 T13.23.h

4 ^E 3 T13.11z 2 T13.11z T13.11u 9,14 DE 11,15 T13.11a 12,16 T13.11x 10,17 DE DI 13,19 T13.21l 18,20 T13.21l 10,21 DE 22 T13.21n 16,23 T13.21l 12,24 T13.21l 9,25 DE 11,26 T13.21l 27 assoc com 28 T6.64 29 T13.21p 30 T6.64 31 9I 32 def 8,10-33 9E 6,9-34 9E 4-35 !I 36 8I 1-37 !I

922

ANSWERS FOR CHAPTER 13 (b) Show a  ; C b D b  1 and generalize. (c) Let q = qt.i; d.a; b// and r = rm.i; d.a; b//. c1. i D .Sd.a; b/  q/ C r c2. r < Sd.a; b/

def qt from def rm

c3. .8y < d.a; b//Œ.a > ; ^ b > ;/ ! G.a; b; Sy/ c4. G.a; b; i /

1 ^E A (g !I)

c5. c6. c7. c8. c9. c10. c11. c12. c13. c14.

G.a; b; Sd.a; b/  q/ Sd.a; b/  q  .Sd.a; b/  q/ C r Sd.a; b/  q  i : 8x8yŒ.G.a; b; x/ ^ G.a; b; y/ ^ x  y/ ! G.a; b; x y/ : G.a; b; i .Sd.a; b/  q// : i D Sd.a; b/  q C Œi .Sd.a; b/  q/ : Sd.a; b/  q C Œi .Sd.a; b/  q/ D .Sd.a; b/  q/ C r : i .Sd.a; b/  q/ D r G.a; b; r/ 9y.r D Sy/

7 T13.23f T13.11u c1,c6 DE 6 T13.23g c4,c5,c7,c8 8E c7 T13.21a c1,c10 DE c11 T6.69 c9,c11 DE A (c I)

c15.

r D Sk

A (c c149E)

c16. c17. c18. c19. c20. c21. c22.

S k < Sd.a; b/ k < d.a; b/ Œ.a > ; ^ b > ;/ ! G.a; b; S k/ .a > ; ^ b > ;/ ^ G.a; b; S k/ G.a; b; S k/ G.a; b; r/ ?

c2,c16 DE c16 T13.11k c3,c17 (8E) c18 Impl, Dem c19 ^E c20,c15 DE c13,c21 ?I

c23. c24. c25. c26.

?

c14,c15-c22 9E

9y.r D Sy/ r D; d.a; b/ji

c14-c23 I c24 T6.46 c25 T13.22j

c27. G.a; b; i / ! d.a; b/ji c28. 8xŒG.a; b; x/ ! d.a; b/jx

T13.23.i. PA ` Pr.Sa/ ^ aj.b  c/ ! .ajb _ ajc/

Exercise 13.20 T13.23.i

c4-c24 !I c27 8I

923

ANSWERS FOR CHAPTER 13

1.

Pr.Sa/ ^ aj.b  c/

A (g !I)

2. 3. 4. 5. 6.

Pr.Sa/ 1 < Sa ^ 8xŒxjSa ! .x D ; _ S x D Sa/ 8xŒxjSa ! .x D ; _ S x D Sa/ aj.b  c/ a−b

1 ^E 2 def 3 ^E 1 ^E A (g !I)

7. 8. 9. 10.

j jb ^ j jSa

A (g !I)

j jSa j D ; _ Sj D Sa Sj D Sa

7 ^E 4,8 8E A (c I)

11. 12. 13. 14.

j Da j jb ajb ?

10 T6.41 7 ^E 12,11 DE 6,13 ?I

15. 16.

Sj ¤ Sa j D;

10-14 I 9,15 DS

17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39.

.j jb ^ j jSa/ ! j D ; 8xŒ.xjb ^ xjSa/ ! x D ; Rp.b; Sa/ Sa > ; b; bD; aj; ajb ?

7-16 !I 17 8I 18 def T13.11e A (c E) 21 T13.11f T13.22c 22,23 DE 6,24 ?I

b>; 9x9yŒbx C 1 D Sa  y bp C 1 D Sa  q c.Sa  q/ D c.Sa  q/ c.bp C 1/ D c.Sa  q/ cbp C c D c.Sa  q/ ajcbp ajSa ajc.Sa  q/ aj.cbp C c/ ajc ajc

21-25 E 19,20,26 T13.23h A (g 279E) DI 28,29 DE 30 T6.64 5 T13.22d T13.22b 33 T13.22d 31,34 DE 32,35 T13.22g 27,28-36 9E

a − b ! ajc ajb _ ajc

6-37 !I 38 Impl

40. ŒPr.Sa/ ^ aj.b  c/ ! .ajb _ ajc/

1-39 !I

E13.21. Show the conditions for Def [lcm] and Def [plm]. Then show each of the results in T13.24. Def [lcm]. Exercise 13.21 Def [lcm]

924

ANSWERS FOR CHAPTER 13 (i) PA ` 9xŒx > ; ^ .8i < k/m.i /jx Supposing the zero case is done. 1. 9xŒx > ; ^ .8i < ;/m.i /jx

zero case

2.

9xŒx > ; ^ .8i < j /m.i /jx

A (g !I)

3.

a > ; ^ .8i < j /m.i /ja

A (g 29E)

a>; .8i < j /m.i /ja S m.j / > ; a  S m.j /  S m.j / a  S m.j / > ; l < Sj

3 ^E 3 ^E T13.11e 4 T13.11z 6,8 T13.11c A (g (8I))

4. 5. 6. 8. 9. 10. 11. 12.

l ; ^ .8i < Sj /m.i /jx 9xŒx > ; ^ .8i < Sj /m.i /jx

24. 9xŒx > ; ^ .8i < j /m.i /jx ! 9xŒx > ; ^ .8i < Sj /m.i /jx 25. 8y.9xŒx > ; ^ .8i < y/m.i /jx ! 9xŒx > ; ^ .8i < Sy/m.i /jx/ 26. 9xŒx > ; ^ .8i < k/m.i /jx

10-19 (8I) 9,20 ^I 21 9I 2,3-22 9E 2-23!I 248I 1,25 IN

Def [plm]. These are straightforward. T13.24. T13.24.a. Show 1 > ; ^ .8i < ;/m.i /j1 ^ .8z < 1/Œz > ; ^ .8i < ;/m.i /jz and apply the definition. T13.24.b. This is straightforward. T13.24.c. PA ` .8i < k/m.i /jx ! pk jx Let q = qt.x; pk / and r = rm.x; pk /.

Exercise 13.21 T13.24.c

925

ANSWERS FOR CHAPTER 13

1. 2. 3. 4.

.8y < lk /Œy > ; ^ .8i < k/m.i /jy Spk D lk x D .Spk  q/ C r r < Spk

def lk T13.17c def pk def q from def r

5.

.8i < k/m.i /jx

A (g !I)

6. 7.

r < lk a ; ^ .8i < k/m.i /jr r  ; _ .8i < k/m.i /jr r ; r D; pk jx

20. .8i < k/m.i /jx ! pk jx

5,7 (8E) 8,3 DE 7 T13.24b 2,10 DE 11 T13.22d 9,12 T13.22g 7-13 (8I) 1,6 (8E) 15 DeM 14,16 DS 17 T13.11f 18 T13.22j 5-19 !I

T13.24.d. PA ` 8nŒ.Pr.S n/ ^ njlk / ! .9i < k/njS m.i / Supposing the zero case is done.

Exercise 13.21 T13.24.d

ANSWERS FOR CHAPTER 13

926

1. 8nŒ.Pr.S n/ ^ njl; / ! .9i < ;/njS m.i / 2. lj > ; ^ .8i < j /m.i /jlj

zero case def lj T13.17b

3. .8i < j /m.i /jlj 4. 8nŒ.Pr.S n/ ^ njlj / ! .9i < j /njS m.i /

2 ^E A (g !I)

5.

Pr.Sa/ ^ ajlSj

A (g !I)

6. 7.

Pr.Sa/ b < Sj

5 ^E A (g (8I))

8. 9.

b h.j / m.j / > h.j / m.j /  h.j /

21-26 I 20,27 DS 28 T13.11z 19,29 DE 30,3 T13.11a 15,31 T13.11c 32 def m 33 T13.11m

35.

m.j /  h.j /

17,18-34 9E

36.

m.j / > ; ^ m.j /  h.j /

6,35 ^I

37. .8i < k/.m.i / > ; ^ m.i /  h.i //

1-36 (8I)

Exercise 13.24 T13.27

930

ANSWERS FOR CHAPTER 13 (ii) (a)

(b)

a1. a2. a3. a4. a5. a6. a7. a8. a9. a10. a11. a12. a13. a14. a15. a16. a17. a18. a19.

i j S i  Sj q  S i  q  Sj S.q  S i /  S.q  Sj / : aj.S.q  Sj / S.q  S i // : : S.q  Sj / S.q  S i / D S.q  Sj / S.q  S i / S.q  Sj / D .q  Sj / C 1 S.q  S i / D .q  S i / C 1 : : S.q  Sj / S.q  S i / D Œ.q  Sj / C 1 Œ.q  S i / C 1 : : Œ.q  Sj / C 1 Œ.q  S i / C 1 D .q  Sj / .q  S i / : : .q  Sj / .q  S i / D q.Sj Si/ : : q.Sj S i / D q.Sj Si/ Sj D j C 1 Si D i C 1 : : q.Sj S i / D q..j C 1/ .i C 1// : : : .j C 1/ .i 1/ D j i : : q.Sj S i / D q.j i/ : : S.q  Sj / S.q  S i / D q.j i/ : ajq.j i/

: b1. j i >; : i D S v b2. 9vŒj : b3. j i D Sl b4. b5. b6. b7. b8. b9. b10. b11. b12. b13. b14. b15.

2 T13.21f b1 T13.11g A (g b29E)

ajS l : j i j : j i r k ; chR .x; E j/ D ;

7,9 DE T13.11f A (g 11_E)

.9y  j /P.x; E y/ P.x; E a/ aj

8,12 $E A (g 13(9E))

a  Sj .9y  Sj /P.x; E y/

15 T13.11o 14,16 (9I)

18.

.9y  Sj /P.x; E y/

13,14-17 (9E)

19.

chR .x; E j/ > ;

A (g 11_E)

20. 22. 23. 24. 25. 26. 27.

chR .x; E j/ ¤ ; chR .x; E j/  ; D ; chR .x; E j /  chP .x; E Sj / D chR .x; E j/  ; chP .x; E Sj / D ; P.x; E Sj / Sj  Sj .9y  Sj /P.x; E y/

19 T13.11f T6.44 10,22 DE 23,20 T6.70 2,24 8E,$E T13.11m 15,26 (9I)

28.

.9y  Sj /P.x; E y/

29.

.9y  Sj /P.x; E y/

11,12-18,19-27 _E A (g $I)

30. 31.

P.x; E a/ a  Sj

A (g 29(9E))

32. 33.

a  j _ a D Sj aj

31 T13.11o A (g 32_E)

34. 35. 36. 37.

.9y  j /P.x; E y/ chR .x; E j/ D ; chR .x; E j /  chP .x; E Sj / D ; chR .x; E Sj / D ;

38.

a D Sj

A (g 32_E)

39. 40. 41. 42.

P.x; E Sj / chP .x; E Sj / D ; chR .x; E j /  chP .x; E Sj / D ; chR .x; E Sj / D ;

30,38 DE 2,39 8E, $E 40 T6.59 7,41 DE

43. 44. 45.

chR .x; E Sj / D ;

30,33 (9I) 8,34 $E 35 T6.59 7,36 DE

32,33-37,38-42 _E

chR .x; E Sj / D ;

29,30-43 (9E)

chR .x; E Sj / D ; $ .9y  Sj /P.x; E y/

46. ŒchR .x; E j / D ; $ .9y  j /P.x; E y/ ! ŒchR .x; E Sj / D ; $ .9y  Sj /P.x; E y/ 47. 8w.ŒchR .x; E w/ D ; $ .9y  w/P.x; E y/ ! ŒchR .x; E S w/ D ; $ .9y  S w/P.x; E y// 48. chR .x; E n/ D ; $ .9y  n/P.x; E y/

Exercise 13.30 T13.38.a

9-28,29-44 $I 8-45 !I 46 8I 1,47 IN

936

ANSWERS FOR CHAPTER 13

1. 2. 3. 4. 5. 6. 7.

chR .x; E n/ D ; $ .9y  n/P.x; E y/ chS .x; E z/ D chR .x; E z/ S.x; E z/ $ chS .x; E z/ D ; chR .x; E z/ D ; $ .9y  z/P.x; E y/ chS .x; E z/ D ; $ .9y  z/P.x; E y/ S.x; E z/ $ .9y  z/P.x; E y/ .9 y  z/P.x; E y/ $ .9y  z/P.x; E y/

from above def from ELEQ, T13.32 T13.30 1 8E 2,4 DE from 3,5 6 abv

T13.38.e. PA ` .y  z/P.x; E y/ $ .y  z/P.x; E y/ (a)

(b)

a1. a2. a3. a4. a5. a6.

q.x; E ;/ D gq.x/ E gq.x/ E D zero.chR .x; E ;// gq.x/ E D; q.x; E ;/ D ; .y  ;/P.x; E y/ D ; q.x; E ;/ D .y  ;/P.x; E y/

T13.31a def from least, T13.32 a2 T13.34b a1,a3 DE T13.18a a4,a5 DE

b1.

kj

A (g (8I))

b2. b3.

k ; pred.pi.i /exp.Sm;i/ /jS m pred.pi.i /exp.Sn;i/ /jS n pred.pi.i /exp.Sm;i/  pi.i /exp.Sn;i/ /j.S m  S n/ pred.pi.i /exp.Sm;i/Cexp.Sn;i/ /j.S m  S n/ 9qŒpi.i /exp.Sm;i/  q D S m ^ pred.pi.i // − q 9rŒpi.i /exp.Sn;i/  r D S n ^ pred.pi.i // − r pi.i /exp.m;i/  q D S m ^ pred.pi.i // − q pi.i /exp.Sn;i/  r D S n ^ pred.pi.i // − r S m  S n D pi.i /exp.Sm;i/  q  pi.i /exp.Sn;i/  r S m  S n D pi.i /exp.Sm;i/Cexp.Sn;i/  q  r pred.pi.i /exp.Sm;i/Cexp.Sn;i/C1 /j.S m  S n/ 9sŒSpred.pi.i /exp.Sm;i/Cexp.Sn;i/C1 /  s D S m  S n Spred.pi.i /exp.Sm;i /Cexp.Sn;i/C1 /  s D S m  S n pi.i /exp.Sm;i/Cexp.Sn;i/C1  s D S m  S n pi.i /exp.Sm;i/Cexp.Sn;i/  pi.i /  s D S m  S n pi.i /exp.Sm;i/Cexp.Sn;i/  pi.i /  s D pi.i /exp.Sm;i/Cexp.Sn;i/  q  r pi.i /  s D q  r Spred.pi.i //  s D q  r 9sŒSpred.pi.i //  s D q  r pred.pi.i //j.q  r/ pred.pi.i //jq _ pred.pi.i //jr ? ?

T13.42h T13.43d T13.43d 1,2,3 T13.22e 4 T13.40e T13.43l T13.43l A (g 6,7 9E) 8,9 ^E, etc. 10 T13.40e A (c I) 12 def A (g 13 9E) 1,14 T13.35c 15 T13.40a 11,16 DE 1,17 T6.70 18 T13.35c 19 9I 20 def T13.42f,13.23i 8,9,22 ?I 13,14-23 9E

pred.pi.i /exp.Sm;i/Cexp.Sn;i/C1 / − .S m  S n/ pred.pi.i /exp.Sm;i/Cexp.Sn;i/ /j.S m  S n/ ^ pi.i /exp.Sm;i/Cexp.Sn;i/C1 / − .S m  S n/ exp.S m  S n; i / D exp.S m; i / C exp.S n; i /

28. exp.S m  S n; i / D exp.S m; i / C exp.S n; i /

E13.35. Show each of the results from T13.44. T13.44.h. PA ` exp.m; i / > ; ! len.m/ > i

Exercise 13.35 T13.44.h

12-24 I 5,25 ^I 26 T13.43f 6,7,8-27 9E

949

ANSWERS FOR CHAPTER 13 1.

exp.m; i / > ;

A (g !I)

2. 3. 4.

exp.m; i / ¤ ; mD;_m>; mD;

1 T13.11f T13.11f A (g 3_E)

5.

len.m/  i

A (c E)

6. 7. 8.

exp.;; i / D ; exp.m; i / D ; ?

T13.43b 6,4 DE 2,7 ?I

9. 10.

len.m/ > i

5-8 E

m>;

A (g 3_E)

11.

len.m/  i

A (c E)

12. 13. 14.

len.m/  i 9v.m D S v/ m D Sa

11 T13.11r 10 T13.11g A (g 139E)

15. 16. 17.

exp.Sa; i / ¤ ; len.Sa/  i i > Sa i a exp.Sa; i / D ; ?

18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29.

2,14 DE 12,14 DE A (g I) 17 T13.11m,n 18 T13.43h 15,19 ?I

i  Sa i  Sa .8z  Sa/Œz  len.Sa/ ! exp.Sa; z/ D ; i  len.Sa/ ! exp.Sa; i / D ; exp.Sa; i / D ; ? ?

17-20 I T13.11r T13.44d 23,22 (8E) 24,16 !E 15,25 ?I 13,14-26 9E

len.m/ > i

11-27 E

len.m/ > i

3,4-9,10-28 _E

30. exp.m; i / > ; ! len.m/ > i

T13.44.k. PA ` p > ; ! len.pi.i /p / D S i

Exercise 13.35 T13.44.k

1-29 !I

950

ANSWERS FOR CHAPTER 13 1.

p>;

A (g !I)

2. 3. 4. 5. 6. 7. 8.

len.pi.i /p / < S i _ len.pi.i /p / D S i _ len.pi.i /p / > S i exp.pi.i /p ; i / D p exp.pi.i /p ; i / > ; len.pi.i /p / > i len.pi.i /p /  S i len.pi.i /p / – S i len.pi.i /p / > S i

T13.11p T13.43i 1,3 DE 4 T13.44h 5 T13.11l 6 T13.11r A (c I)

9. 10. 11. 12. 13. 14. 15. 16.

pi.i /p > ; 9yŒpi.i /p D Sy pi.i /p D Sj

T13.42h 9 T6.51 A (g 109E)

.8w < len.Sj //.8z < Sj /Œz  w ! exp.Sj; z/ D ; len.Sj / > S i .8z < Sj /Œz  S i ! exp.Sj; z/ D ; .8z < pi.i /p /Œz  S i ! exp.pi.i /p ; z/ D ; k < pi.i /p

T13.44e 8,11 DE 12,13 (8E) 11,14 DE A (g (8I))

17.

k  Si

A (g !I)

18. 19. 20. 21. 22. 23. 24. 25.

k>i pred.pi.k// − pi.i /p pred.pi.k// − Sj exp.Sj; k/ 6 1 exp.Sj; k/ < S ; exp.Sj; k/ < ; _ exp.Sj; k/ D ; exp.Sj; k/ D ; exp.pi.i /p ; k/ D ;

17 T13.11h,c 18 T13.42n 11,19 DE 20 T13.43k 21 T13.11r 22 T13.11n 23 with T6.50 24,11 DE

k  S i ! exp.pi.i /p ; k/ D ;

26. 27. 28. 29. 30. 31.

.8z < ?

pi.i /p /Œz

 Si !

exp.pi.i /p ; z/

17-25 !I D ;

?

16-26 (8I) 15,27 ?I 10,11-28 9E

len.pi.i /p / 6> S i len.pi.i /p / D S i

8-29 I 2,7,30 DS

32. p > ; ! len.pi.i /p / D S i

T13.44.m. PA ` len.n/ D S l ! exp.n; l/  1

Exercise 13.35 T13.44.m

1-31 !I

951

ANSWERS FOR CHAPTER 13 1.

len.n/ D S l

A (g !I)

2. 3.

nD;_n>; nD;

T13.11f A (c I)

4. 5. 6. 7. 8.

len.;/ D ; len.n/ D ; ; D Sl ; ¤ Sl ?

T13.44b 3,4 DE 1,5 DE T6.40 6,7 ?I

9. 10. 11. 12.

n¤; n>; 9v.n D S v/ n D Sm

13. 14. 15. 16. 17. 18. 19.

3-8 I 2,9 DS 10 T13.11g A (g 119E)

len.S m/ D S l .8z  S m/Œz  S l ! exp.S m; z/ D ; .8w < S l/.8z  S m/Œz  w ! exp.S m; z/ D ; exp.S m; l/ ž 1 exp.S m; l/ < 1 exp.S m; l/ D ; a  Sm

1,12 DE T13.44d 13 T13.44e A (c E) 16 T13.11r 17 T8.16 A (g (8I))

20.

al

A (g !I)

21. 22.

l Da_l S; pred.pi.j // > ; pred.pi.j // − 1 pred.pi.j // − val .m; n; ;/

10. .8i  ;/pred.pi.i // − val .m; n; ;/ .8i  a/pred.pi.i // − val .m; n; a/ 11.

1-9 (8I) A (g !I)

12.

j  Sa

A (g (8I))

13. 14. 15. 16. 17.

j >a j a pred.pi.j // − val .m; n; a/ val .m; n; Sa/ D val .m; n; a/  pi.a/exc.m;n;a/ pred.pi.j //jval .m; n; Sa/

12 T13.11l 13 T13.11m 11,14 (8E) def val A (c I)

18. 19. 20. 21. 22. 23.

pred.pi.j //jval .m; n; a/  pi.a/exc.m;n;a/ j ¤a pred.pi.j //jval .m; n; a/ ? pred.pi.j // − val .m; n; Sa/

17-21 I

.8i  Sa/pred.pi.i // − val .m; n; Sa/

12-22 (8I)

24. Œ.8i  a/pred.pi.i // − val .m; n; a/ ! Œ.8i  Sa/pred.pi.i // − val .m; n; Sa/ 25. .8i  a/pred.pi.i // − val .m; n; a/

T13.45.f. PA ` .8j < i /exp.val .m; n; i /; j / D exc.m; n; j / 1.

ax

37 T13.13 A (g 38_E)

40. 41. 42. 43. 44. 45.

l ¤x exp.S r; l/ D exp.Sa; l/ l  Sx l  len.Sa/ exp.Sa; l/ D ; exp.S r; l/ D ;

46.

l Dx

A (g 38_E)

47. 48. 49.

pred.pi.x// − S r exp.S r; x/ D ; exp.S r; l/ D ;

28,32 DE 47 T13.43k 48,46 DE

50. 51. 52. 53. 54. 55.

exp.S r; l/ D ; l  x ! exp.S r; l/ D ; .8z  S r/Œz  x ! exp.S r; z/ D ; ? len.S r/  x len.S r/  x

39 T13.11s 33,40 8E 39 T13.11l 42,24 DE 43 T13.44l 44,41 DE

38,39-45,46-49 _E 37-50 !I 36-51 (8I) 35,52 ?I 34-53 I 54 T13.11r

Exercise 13.36 T13.45.n

957

ANSWERS FOR CHAPTER 13

56. 57.

val.S r; x/ D S r l 1 24. pi.i /exp.n;i/  pi.i /r pi.i /r  pi.q/r 25. pi.i /exr.n;i/  pi.q/r 26. 27. val.n; i /  pi.i /exp.n;i/  val.n; i /  pi.q/r 28. val.n; i /  pi.q/r  Œpi.q/r i  pi.q/r val.n; i /  pi.i /exp.n;i/  Œpi.q/r i  pi.q/r 29. 30. Œpi.q/r Si  val.n; S i / 31. S i  q ! Œpi.q/r Si  val.n; S i / 32. fi  q ! Œpi.q/r i  val.n; i /g ! fS i  q ! Œpi.q/r Si  val.n; S i /g 33. i  q ! Œpi.q/r i  val.n; i / Œpi.q/r len.n/  val.n; len.n// 34. 35. pi.q/r > ; 36. Œpi.q/r len.n/  Œpi.q/r q 37. Œpi.q/r q  val.n; len.n// 38. Œlen.n/  q ^ .8k < le n.n//exp.n; k/  r ! Œpi.q/r q  val.n; len.n//

E13.37. Show each of the results from T13.46. T13.46.e. PA ` len.m  n/  l

Exercise 13.37 T13.46.e

15,16-17,18-21 _E T13.42g 22,23 T13.40j 14 T13.40f 24,25 T13.11a 26 T13.11aa 13 T13.11aa 27,28 T13.11a 29,11,12 DE 9-30 !I 8-31 !I 7,32 IN 2,33 !E T13.42h 35,2 T13.40j 34,36 T13.11a 1-37 !I

959

ANSWERS FOR CHAPTER 13 1. l D len.m/ C len.n/ 2. len.n/ D ; _ len.n/ > ; len.n/ D ; 3. 4. 5.

abv T13.11f A (g 2_E)

len.m/ D ; _ len.m/ > ; len.m/ D ;

T13.11f A (g 4 _E)

;C;D; len.m  n/  ; len.m  n/  ; C ; len.m  n/  l

T6.42 T13.11d 6,7 DE 8,5,3 DE

10.

len.m/ > ;

A (g 4_E)

11. 12.

9vŒlen.m/ D S v len.m/ D Sa

10 T13.11g A (g 119E)

6. 7. 8. 9.

13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24.

exp.m; a/ > ; a < len.m/ .8i < len.m//exp.m  n; i / D exp.m; i / exp.m  n; a/ D exp.m; a/ exp.m  n; a/ > ; len.m  n/ > a len.m  n/  Sa len.m  n/  len.m/ len.m/ C ; D len.m/ l D len.m/ len.m  n/  l len.m  n/  l

12 with T13.44m 12 T13.11i T13.46c 15,14 (8E) 13,16 DE 17 T13.44h 18 T13.11l 19,12 DE T6.42 21,3 DE 20,22 DE 11,12-23 9E

25.

len.m  n/  l

4,5-9,10-24 _E

26.

len.n/ > ;

A (g 2_E)

27. 28.

9vŒlen.n/ D S v len.n/ D Sa

26 T13.11g A (g 279E)

29. 30. 31. 32. 33. 34. 35. 36. 37. 38.

exp.n; a/ > ; a < len.n/ .8i < len.n//exp.m  n; i C len.m// D exp.n; i / exp.m  n; a C len.m// D exp.n; a/ exp.m  n; a C len.m// > ; len.m  n/ > a C len.m/ len.m  n/  S.a C len.m// S.a C len.m// D Sa C len.m/ S.a C len.m// D l len.m  n/  l

39. len.m  n/  l 40. len.m  n/  l

28 with T13.44m 28 T13.11i T13.46c 31,30 (8E) 29,32 DE 33 T13.44h 34 T13.11l T6.54 36,28 DE 35,37 DE 27,28-38 9E 2,3-25,26-39 _E

T13.46.f. PA ` len.m  n/ D l

Exercise 13.37 T13.46.f

ANSWERS FOR CHAPTER 13 1. l D len.m/ C len.n/ 2. len.m  n/  l len.m  n/ — l 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.

960 abv T13.46e A (c E)

len.m  n/ > l l ; len.m  n/ > ; mn>1 1>; mn>; 9vŒm  n D S v m  n D Sp

3 T13.11r T13.11d 4,5 T13.11c 6 T13.44g T8.14 7,8 T13.11b 9 T13.11g A (c 109E)

len.Sp/ > l 9vŒS v C l D len.Sp/ Sa C l D len.Sp/ S.a C l/ D Sa C l S.a C l/ D len.Sp/ exp.Sp; a C l/  1 9qŒpi.a C l/exp.Sp;aCl/  q D Sp ^ 8y.y ¤ a C l ! exp.q; y/ D exp.Sp; y// pi.a C l/exp.Sp;aCl/  j D Sp ^ 8y.y ¤ a C l ! exp.j; y/ D exp.Sp; y// pi.a C l/exp.Sp;aCl/  j D Sp pi.a C l/exp.Sp;aCl/  pi.a C l/ pi.a C l/ > 1 pi.a C l/exp.Sp;aCl/ > 1 Sp > ; pi.a C l/exp.Sp;aCl/  j > ; j >; pi.a C l/exp.Sp;aCl/  j > j j < Sp

Exercise 13.37 T13.46.f

4,11 DE 12 def A (c 139E) T6.54 14,15 DE 16 T13.44m T13.43l A (c 189E) 19 ^E 17 with T13.40l T13.42g 21,22 T13.11c T13.11e 24,20 DE 25 T13.11ab 23,26 T13.11ac 20,27 DE

ANSWERS FOR CHAPTER 13

29. 30.

8y.y ¤ a C l ! exp.j; y/ D exp.Sp; y// b < len.m/

31. 32. 33. 34. 35. 36. 37. 38. 39.

.8i < len.m//exp.m  n; i / D exp.m; i / exp.m  n; b/ D exp.m; b/ exp.Sp; b/ D exp.m; b/ len.m/  a C l b i len.S m/ : : : exp.val.S n; a len.S m//; i len.S m// D exp.S n; i len.S m// : exp.S m  S n; i / D exp.val.S m; a/  val.S n; a len.S m//; i / : exp.S m  S n; i / D exp.val.S m; a/  val.S n; a len.S m//; i / : exp.S m  S n; i / D exp.val.S m; a/  val.S n; a len.S m//; i / : .8i < a/exp.S m  S n; i / D exp.val.S m; a/  val.S n; a len.S m//; i / : val.S m  S n; a/ D val.val.S m; a/  val.S n; a len.S m//; a/ : : len.val.S m; a/  val.S n; a len.S m/// D len.val.S m; a// C len.val.S n; a len.S m/// a < len.S m/ _ a  len.S m/ a < len.S m/

T13.46g 22 T13.21a 23,24 DE T13.46g 26,24,14 DE 22,1 T13.21e 28 T13.45l 25,29,27 DE 15,16-21,22-30 _E 2,3-11,12-31 _E 1-32 (8I) 33 T13.45m T13.46f T13.11q A (g 36_E)

38. 39. 40. 41. 42. 43.

len.val.S m; a//  a : a len.S m/ D ; : val.S n; a len.S m// D 1 : len.val.S n; a len.S m/// D ; : len.val.S m; a/  val.S n; a len.S m/// D len.val.S m; a// : len.val.S m; a/  val.S n; a len.S m///  a

T13.45j 37 T13.21b 39 def 40 T13.44f 35,41 DE 38,42 DE

44.

a  len.S m/

A (g 36_E)

45. 46. 47. 48. 49. 50. 51. 52. 53. 54.

len.val.S m; a//  len.S m/ : : len.val.S n; a len.S m///  a len.S m/ : : len.val.S m; a// C len.val.S n; a len.S m///  len.S m/ C .a len.S m// : a D len.S m/ C .a len.S m// : len.val.S m; a// C len.val.S n; a len.S m///  a : len.val.S m; a/  val.S n; a len.S m///  a : len.val.S m; a/  val.S n; a len.S m///  a : val.S m; a/  val.S n; a len.S m//  1 : : val.val.S m; a/  val.S n; a len.S m//; a/ D val.S m; a/  val.S n; a len.S m// : val.S m  S n; a/ D val.S m; a/  val.S n; a len.S m//

Exercise 13.37 T13.46.m

T13.45k T13.45j 45,46 T13.11v 44 T13.21a 47,48 DE 35,49 DE 36,37-43,44-50 _E T13.46c 51 T13.45n 34,53 DE

963

ANSWERS FOR CHAPTER 13

E13.38. Show (j) and the unfinished cases for the C disjunct in (l) and (n). Hard core: show each of the results from T13.47. pSqt

T13.47.i. PA ` Termseq.m; t / ! Termseq.m  2

Exercise 13.38 T13.47.i

; pS q  t /

964

ANSWERS FOR CHAPTER 13

1.

Termseq.m; t /

2. 3. 4.

: exp.m; len.m/ 1/ D t m>1 .8k < len.m//ŒA.m; k/ _ B.m; k/ _ C.m; k/ _ D.m; k/

5.

len.2

6.

exp.2

7. 8. 9. 10.

len.m  2 / D len.m/ C 1 len.m/ C 1 D S len.m/ : len.m/ D S len.m/ 1 pSqt : len.m/ D len.m  2 / 1

11.

exp.m  2

pSqt pSqt

T13.44k

; ;/ D pS q  t

T13.43i

pSqt

5 T13.46f T6.48 T13.21k 9,7,8 DE

pS qt

12. 13.

len.m  2

pSqt

pSqt

m2

/>;

7 T13.11u 13 T13.44g

pSqt

a < len.m  2

16. 17.

a < len.m/ _ a D len.m/ a < len.m/

21. 22.

6 T13.46c 11,10 DE

>1

15.

18. 19. 20.

1 T13.47a 1 T13.47a 1 T13.47a

/D1

; len.m// D pS q  t pS qt pSqt : ; len.m  2 / 1/ D pS q  t exp.m  2

14.

A (g !I)

/

A (g (8I)) 15,7 T13.11n A (g 16_E)

pSqt

exp.m  2 ; a/ D exp.m; a// A.m; a/ _ B.m; a/ _ C.m; a/ _ D.m; a/ A.m; a/

17 T13.46c 4,17 (8E) A (g 19_E)

exp.m; a/ D p;q _ Var.exp.m; a// exp.m; a/ D p;q pSqt

23.

exp.m  2

24.

exp.m  2

25.

A.m  2

26.

Var.exp.m; a//

pSqt

pSqt

; a/ D p;q ; a//

23 _I

; a/

27.

Var.exp.m  2

28.

exp.m  2

29.

A.m  2

pSqt

pSqt

18,22 DE pS qt

; a/ D p;q _ Var.exp.m  2

24 abv A (g 21_E)

pSqt

pSqt

20 abv A (g 21_E)

; a//

18,26 DE pS qt

; a/ D p;q _ Var.exp.m  2

; a//

27 _I

; a/

28 abv

30.

A.m  2

31.

A.m  2

32.

B.m; a/

A (g 19_E)

33. 34. 35.

.9j < a/Œexp.m; a/ D pSq  exp.m; j / exp.m; a/ D pS q  exp.m; j / j ; : len.m/ 1 < len.m/

3 T13.44j 49 T13.21i

51.

exp.m  2

pSqt pSqt

; len.m/ ; len.m/

52.

exp.m  2

53.

exp.m  2

54.

exp.m  2

55.

.9j < a/Œexp.m  2

56.

B.m  2

57. 58.

pSqt pSqt

: :

pSqt

A.m  2

pSqt

A.m  2

60.

Termseq.m  2

50 T13.46c 51,2 DE 11,48 DE

pSqt

; a/ D pSq  exp.m  2

; len.m/

:

pSqt

; a/ D pSq  exp.m  2

1/

52,53 DE

; j /

50,54 (9I)

; a/

55 abv pSqt

; a/ _ B.m  2

; a/ _ B.m  2

.8k < len.m  2

1/

1/ D t

pSqt

59.

:

; a/ D pSq  t pSqt

pSqt

1/ D exp.m; len.m/

pS qt

pSqt

; a/ _ C.m  2

pSqt

pSqt

; a/ _ C.m  2

pSqt

//ŒA.m  2

; k/ _ B.m  2

pSqt

; a/ _ D.m  2

pSqt

; a/ _ D.m  2

pS qt

; a/

; a/

pSqt

; k/ _ C.m  2

56 _I 16,17-47,48-57 _E pSqt

; k/ _ D.m  2

; k/

; pS q  t /

15-58 (8I) 12,14,59 T13.47a

pSqt

61. Termseq.m; t / ! Termseq.m  2

; pS q  t /

1-60 !I

T13.47.l. PA ` Termseq.m; t / ! 8x.8k < len.m//flen.exp.m; k//  x ! 9nŒTermseq.n; exp.m; k// ^ .8i < len.n//exp.n; i /  exp.m; k/ ^ len.n/  len.exp.m; k/g Let P be the formula, .8k < len.m//flen.exp.m; k//  x ! 9nŒTermseq.n; exp.m; k// ^ .8i < len.n//exp.n; i /  exp.m; k/ ^ len.n/  len.exp.m; k/g 1.

Termseq.m; t /

A (g !I)

2. 3. 4.

P;x .8k < len.m//exp.m; k/ > 1 P

basis 1 T13.47e A g !I

5. 6.

.8k < len.m//flen.exp.m; k//  x ! 9nŒTermseq.n; exp.m; k// ^ .8i < len.n//exp.n; i /  exp.m; k/ ^ len.n/  len.exp.m; k/g a < len.m/

7.

exp.m; a/ > 1

8.

len.2

9. 10. 11. 12. 13.

exp.m;a/

4 abv A (g (8I)) 3,6 (8E)

/D1

7 with T13.44k

exp.m;a/

exp.2 ; ;/ D exp.m; a/ len.exp.m; a//  S x

T13.43i A (g !I)

.8k < len.m/ŒA.m; k/ _ B.m; k/ _ C.m; k/ _ D.m; k/ A.m; a/ _ B.m; a/ _ C.m; a/ _ D.m; a/ A.m; a/

14.

exp.m; a/ D p;q _ Var.exp.m; a//

15. 16.

Termseq.2 b ;

9.

len.2

pSqt

pSqt

pSqu

/ D 1 ^ len.2

pSqt

10.

len.m  2

11.

len.m  2

pSqu

12.

/ D len.n  2 pSqu : len.n  2 / 1 D len.n/

13.

exp.n  2

14.

exp.2

15.

exp.n  2

pS qu

8 T13.44k pS qu

/ D len.n/ C 1

9 T13.46f

/

10,3 DE 10 T13.21k

pSqu

; len.n// D exp.2

; ;/

T13.46g

; ;/ D pS q  u

pS qu

T13.43i

pSqu

; len.n  2 pSqt

16.

l < len.m  2

17. 18. 19.

l < Slen.m/ l < len.m/ _ l D len.m/ l < len.m/

20. 21. 22.

/D1

/ D len.m/ C 1 ^ len.n  2

pSqt

pSqu

2 T13.47i cap 7 T13.46n

/

:

1/ D pSq  u

12,13,14 DE

/

A (g (8I)) 10,16 DE 17 T13.11n A (g 18_E)

pSqt

pSqu

; l/ D exp.m; l/ ^ exp.n  2 ; l/ D exp.n; l/ exp.m  2 I.m; n; l/ _ J.v; m; n; l/ _ K.v; s; m; n; l/ _ L.m; n; l/ _ M.m; n; l/ _ N.m; n; l/ I.m; n; l/

19,3 T13.46c 5,19 (8E) A (g 21_E)

23.

exp.m; l/ D p;q ^ exp.n; l/ D p;q

24.

exp.m  2

25.

I.m  2

26.

I _ J _ K _ L _ M _ N.v; s; m  2

27.

J.v; m; n; l/ _ K.v; s; m; n; l/

28.

I _ J _ K _ L _ M _ N.v; s; m  2

29.

L.m; n; l/

A (g 21_E)

30. 31. 32.

.9i < l/Œexp.m; l/ D pSq  exp.m; i / ^ exp.n; l/ D pSq  exp.n; i / i 1 len.m/ > ; : len.m/ 1 < len.m/ : len.m/ 1 < len.n/ : len.m/ 1 < l

43,3 DE 2 T13.47a 45 T13.44j 46 T13.21i 47,3 DE 47,43 DE

50. 51.

exp.m  2 ; len.m/ : exp.m; len.m/ 1/ D t

52.

exp.m  2

pSqt

pSqt pSqt

53.

exp.m  2

54.

exp.2

55.

exp.m  2

56.

exp.m  2

pSqt

; len.m/

: :

1/ D exp.m; len.m/

:

1/

1/ D t

50,51 DE pSqt

; len.m// D exp.2

; ;/

T13.46g

; ;/ D pS q  t

pSqt

T13.43i

; len.m// D pS q  t

53,54 DE

pSqt

59. 60.

exp.n  2

61.

exp.2

62.

exp.n  2

63.

exp.n  2

64.

.9i < l/Œexp.m  2

65.

L.m  2

66.

I _ J _ K _ L _ M _ N.v; s; m  2

67.

pSqu

pSqu

:

pSqt

; l/ D pS q  exp.m  2 ; len.m/ : : pSqu ; len.m/ 1/ D exp.n; len.m/ 1/ exp.n  2 : exp.n; len.m/ 1/ D u : pSqu exp.n  2 ; len.m/ 1/ D u

57. 58.

47 T13.46c 2 T13.47a

pSqu

; len.n// D exp.2

1/

55,43,52 DE 48 T13.46c 3,4 DE 57,58 DE

; ;/

T13.46g

; ;/ D pSq  u

pSqt

pSqu

3,60,61 DE pSqu

; l/ D pSq  exp.n  2 pSqt

pSqt

T13.43i

; len.m// D pS q  u

69.

Tsubseq.m  2

pS qt

pSqt

62,43,59 DE pSqu

; i / ^ exp.n  2

;n  2

pS qu

pSqu

;n  2

; l/

56,63,49 (9I)

18,19-66 _E pS qt

pSqu

; i /

65 _I

; l/

//I _ J _ K _ L _ M _ N.v; s; m  2

;n  2

pSqu

; l/ D pSq  exp.n  2

64 abv pS qt

pSqt

.8k < len.m  2

1/

; l/

I _ J _ K _ L _ M _ N.v; s; m  2

68.

:

pSqt

; l/ D pSq  exp.m  2

pSqu

;n  2

; len.m/

;n  2

pS qu

; k/

; pS q  t; v; s; pS q  u/ pSqt

70. Tsubseq.m; n; t; v; s; u/ ! Tsubseq.m  2

pSqu

;n  2

16-67 (8I) 6,11,15,68 T13.49a

; pSq  t; v; s; pS q  u/

T13.49.l. PA ` Tsubseq.m; n; t; v; s; u/ ! Termsub.t; v; s; u/ Let P .m; n; v; s; k/ = 9a9bŒTsubseq.a; b; exp.m; k/; v; s; exp.n; k//^len.a/  len.exp.m; k//^.8i < len.a//.exp.a; i /  exp.m; k/^exp.b; i /  exp.n; k//

Exercise 13.40 T13.49.l

1-69 !I

972

ANSWERS FOR CHAPTER 13

1.

Tsubseq.m; n; t; v; s; u/

A (g !I)

2. 3. 4. 5. 6. 7. 8. 9.

Termseq.m; t / len.m/ D len.n/ : exp.n; len.n/ 1/ D u .8k < len.m//.I.m; n; k/ _ J.v; m; n; k/ _ K.v; s; m; n; k/ _ L.m; n; k/ _ M.m; n; k/ _ N.m; n; k// : exp.m; len.m/ 1/ D t m>1 .8k < len.m//ŒA.m; k/ _ B.m; k/ _ C.m; k/ _ D.m; k/ k < len.m/

1 T13.49a 1 T13.49a 1 T13.49a 1 T13.49a 2 T13.47a 2 T13.47a 2 T13.47a A (g (8I))

10.

len.exp.m; k//  ;

A (g !I)

11.

P

A (g E)

12. 13. 14.

exp.m; k/ > 1 exp.m; k/  1 ?

2,9 T13.47e 10 T13.44j 12,13 ?I

15. 16. 17. 18. 19.

11-14 E

P len.exp.m; k//  ; ! P

10-15 !I

.8k < len.m//Œlen.exp.m; k//  ; ! P  .8k < len.m//Œlen.exp.m; k//  x ! P 

9-16 (8I) A (g !I)

j < len.m/

A (g (8I))

20.

len.exp.m; j //  S x

A (g !I)

21. 22. 23. 24.

exp.m; j / > 1 len.exp.m; j //  1 I.m; n; j / _ J.v; m; n; j / _ K.v; s; m; n; j / _ L.m; n; j / _ M.m; n; j / _ N.m; n; j / I.m; n; j /

2,19 T13.47e 21 T13.44j 5,19 (8E) A (g 23_E)

25.

exp.m; j / D p;q ^ exp.n; j / D p;q

26.

Tsubseq.2

27.

len.2

28.

len.2

p;q

p;q p;q

24 abv

; exp.m; j /; v; s; exp.n; j //

25 T13.49f

/D1

cap

/  len.exp.m; j //

29.

l < len.2

30.

l D;

31.

exp.2

32.

exp.2

33.

exp.2

34.

exp.2

35. 36.

p;q

;2

p;q p;q p;q p;q

p;q

22,27 DE

/

A (g (8I)) 27,29 T8.16

; ;/ D p;q

T13.43i

; l/  exp.m; j /

25,31 DE

; l/  exp.n; j /

25,31 DE p;q

; l/  exp.m; j / ^ exp.2 p;q

.8i < len.2 Pjk

//.exp.2

p;q

; l/  exp.n; j /

; i /  exp.m; j / ^ exp.2

p;q

32,33 ^I ; i /  exp.n; j /

Exercise 13.40 T13.49.l

29-34 (8I) 26,28,35 9I

973

ANSWERS FOR CHAPTER 13

37.

J.v; m; n; j /

38.

Var.exp.m; j // ^ exp.m; j / ¤ v ^ exp.n; j / D exp.m; j /

39.

Tsubseq.2

40.

len.2

41.

len.2

A (g 23_E)

exp.m;j /

exp.m;j / exp.m;j /

;2

l < len.2

43.

l D;

44.

exp.2

45.

exp.2

46.

exp.2

37 abv

; exp.m; j /; v; s; exp.n; j //

38 T13.49g

/D1

21 T13.44k

/  len.exp.m; j //

exp.m;j /

42.

exp.n;j /

22,40 DE

/

A (g (8I)) 40,42 T8.16

exp.m;j / exp.m;j / exp.m;j /

exp.n;j /

; ;/ D exp.m; j / ^ exp.2

exp.n;j /

; ;/  exp.m; j / ^ exp.2

exp.n;j /

; l/  exp.m; j / ^ exp.2

exp.m;j /

exp.m;j /

T13.43i 44 T13.11m

; l/  exp.n; j /

43,45 DE

exp.n;j /

47. 48.

.8i < len.2 Pjk

49.

K.v; s; m; n; j /

A (g 23_E)

50.

Pjk

similarly

51.

L.m; n; j /

A (g 23_E)

52. 53. 54. 55.

.9i < j /Œexp.m; j / D pS q  exp.m; i / ^ exp.n; j / D pSq  exp.n; i / l 1 exp.n; j / > 1

71.

len.2

exp.m;j /

exp.n;j /

exp.n;j /

/ D 1 ^ len.2

exp.m;j /

21,70 T13.44k exp.n;j /

73. 74. 75.

len.c  2 / D len.d  2 / len.exp.m; j // D 1 C len.exp.m; l// len.c/ C 1  len.exp.m; l// C 1

76.

len.c  2

exp.m;j /

78. 79. 80.

q < Slen.c/ q < len.c/ _ q D len.c/ q < len.c/

65,72 DE 54,58 DE 63 T13.11v 72,74,75 DE

/

A (g (8I)) 72,77 DE 78 T13.11n A (g 79_E)

81. 82. 83.

q < len.d / exp.c; q/  exp.m; l/ ^ exp.d; q/  exp.n; l// exp.c; q/  exp.m; j / ^ exp.d; q/  exp.n; j /

84.

exp.c  2

exp.m;j / exp.m;j /

86.

q D len.c/

87.

exp.c  2

88.

exp.2

exp.c  2

exp.n;j /

; q/ D exp.c; q/ ^ exp.d  2

; q/  exp.m; j / ^ exp.d  2

80,65 DE 64,80 (8E) 54,55,82 T13.46o ; q/ D exp.d; q/

exp.n;j /

80,81 T13.46c

; q/  exp.n; j /

83,84 DE A (g 79_E)

exp.m;j /

exp.m;j /

89.

71 T13.46f

/  len.exp.m; j //

exp.m;j /

q < len.c  2

exp.c  2

/ D len.d / C 1

exp.n;j /

77.

85.

54,55,62 T13.49i 57 T13.46f 67 T13.11u 68 T13.44g 55,69 DE

D 1/

/ D len.c/ C 1 ^ len.d  2

len.c  2

exp.m;j /

19,53 T13.11b cap 57 T13.46f 54,58 def 20,59 T13.11n 18,56,60 (8E) A (g 619E)

62 T13.49a

exp.m;j /

72.

exp.m;j /

; q/ D exp.2

; ;/ ^ exp.d  2 exp.n;j /

; ;/ D exp.m; j / ^ exp.2

exp.m;j /

exp.n;j /

exp.n;j /

; q/ D exp.2

; ;/

; ;/ D exp.n; j /

; q/  exp.m; j / ^ exp.d  2

exp.n;j /

87,86 DE

exp.n;j / Exercise 13.40 T13.49.l ; q/  exp.m; j / ^ exp.d 2 ; q/  exp.n; j / exp.m;j /

.8i < len.c  2 Pjk

91. 92.

Pjk Pjk

exp.m;j /

//.exp.c  2

86,65 T13.46g T13.43i

; q/  exp.n; j /

exp.m;j /

exp.c  2

90.

94.

42-46 (8I) 39,41,47 9I

l < len.m/ len.pS q/ D 1 len.pS q  exp.m; l// D 1 C len.exp.m; l// len.exp.m; l// < len.exp.m; j // len.exp.m; l//  x Plk Tsubseq.c; d; exp.m; l/; v; s; exp.n; l// len.c/  len.exp.m; l// .8i < len.c//.exp.c; i /  exp.m; l/ ^ exp.d; i /  exp.n; l//

65.

93.

; i /  exp.n; j //

79,80-89 _E exp.n;j /

; i /  exp.m; j / ^ exp.d  2

; i /  exp.n; j //

77-90 (8I) 66,73,76,91 9I 61,62-92 9E 52,53-93 (9E)

974

ANSWERS FOR CHAPTER 13

95.

M.m; n; j /

A (g 23_E)

96.

Pjk

similarly

97.

N.m; n; j /

A (g 23_E)

98.

Pjk

similarly

99. 100. 101. 102. 103. 104. 105. 106.

Pjk

23,24-98 _E

len.exp.m; j //  S x !

Pjk

.8k < len.m//Œlen.exp.m; k//  S x ! P  .8k < len.m//Œlen.exp.m; k//  x ! P  ! .8k < len.m//Œlen.exp.m; k//  S x ! P  8x.8k < len.m//Œlen.exp.m; k//  x ! P  len.m/ > ; : len.m/ 1 < len.m/ Pk : len.m/ 1

20-99 !I 19-100 (8I) 18-101 !I 17,102 IN 7 T13.44j 104 T13.21i 103,105 (8E)

107. 108. 109.

Tsubseq.a; b; t; v; s; u/ len.a/  len.t / .8i < len.a//.exp.a; i /  t ^ exp.b; i /  u

A (g 106,4,6 9E)

110. 111. 112. 113. 114. 115. 116. 117. 118. 119. 120.

len.a/ D len.b/ Œpi.len.t //t len.t /  val.a; len.a// Œpi.len.t //u len.t /  val.b; len.b// Termseq.a; t / a>1 len.a/ > ; len.b/ > ; b>1 a  Œpi.len.t //t len.t / ^ b  Œpi.len.t //u len.t / .9x  Xt /.9y  Yt;u /Tsubseq.x; y; t; v; s; u/ Termsub.t; v; s; u/

107 T13.49a 108,109 T13.45o 110,108,109 T13.45o 107 T13.49a 113 T13.47a 114 T13.44j 115,110 DE 116 T13.44g 111,112,114,117 T13.45n 107,118 (9I) 119 T13.49b

121.

Termsub.t; v; s; u/

106,107-120 9E

122. Tsubseq.m; n; t; v; s; u/ ! Termsub.t; v; s; u/

1-121 !I

T13.49.m. PA ` ŒTerm.t /^Term.s/ ! 9uŒTermsub.t; v; s; u/^len.u/  len.t / len.s/ ^ .8k < len.u//exp.u; k/  t C s Let P .m; i; v; s/ = 9x9y9uŒTsubseq.x; y; exp.m; i /; v; s; u/^len.u/  len.exp.m; i // len.s/ ^ .8k < len.u//exp.u; k/  exp.m; i / C s

Exercise 13.40 T13.49.m

ANSWERS FOR CHAPTER 13

975

1.

Term.t / ^ Term.s/

A (g !I)

2. 3. 4. 5. 6. 7. 8.

Termseq.m; t / : exp.m; len.m/ 1/ D t ^ m > 1 ^ .8k < len.m//ŒA.m; k/ _ B.m; k/ _ C.m; k/ _ D.m; k/ len.m/ > ; : len.m/ 1  len.m/ s>1 len.s/ > ; ; < len.m/

1 T13.47b, 9E 2 T13.47a 3 T13.44j 4 T13.21i 1 T13.47f 6 T13.44j A (g !I)

9. 10. 11. 12. 13. 14. 15.

Term.exp.m; ;// exp.m; ;/ > 1 len.exp.m; ;// > ; A.m; ;/ _ B.m; ;/ _ C.m; ;/ _ D.m; ;/ A.m; ;/ exp.m; ;/ D p;q _ Var.exp.m; ;// exp.m; ;/ D p;q

16.

len.p;q/ D 1

17. 18. 19. 20.

Tsubseq.2 ;2 ; exp.m; ;/; v; s; p;q/ len.p;q/  len.p;q/  len.s/ len.p;q/  len.exp.m; ;//  len.s/ k < len.p;q/

p;q

13 abv A (g 14_E) cap

p;q

exp.p;q; k/  p;q p;q  p;q C s exp.p;q; k/  exp.m; ;/ C s

21. 22. 23.

2,8 T13.47n T13.47f T13.44j 3,8 (8E) A (g 12_E)

15 T13.49f 7 T13.11z 15,18 DE A (g (8I)) T13.43g T13.11u 15,21,22 T13.11a

24. 25.

.8k < len.p;q//exp.p;q; k/  exp.m; ;/ C s P;i

20-23 (8I) 17,19,24 9I

26.

Var.exp.m; ;//

A (g 14_E)

27. 28.

exp.m; ;/ D v _ exp.m; ;/ ¤ v exp.m; ;/ D v

T3.1 A (g 27_E)

exp.m;;/

exp.m;;/

29. 30. 31. 32.

Tsubseq.2 ;2 ; exp.m; ;/; v; s; s/ len.exp.m; ;// > ; len.s/  len.exp.m; ;//  len.s/ k < len.s/

26 T13.49h 26 T13.47d 30 T13.11z A (g (8I))

33. 34. 35. 36. 37.

exp.s; k/  s s  exp.m; ;/ C s exp.s; k/  exp.m; ;/ C s .8k < len.s//exp.s; k/  exp.m; ;/ C s P;i

T13.43g T13.11u 33,34 T13.11a 32-35 (8I) 29,31,36 9I

38.

exp.m; ;/ ¤ v

A (g 27_E)

39. 40. 41.

Tsubseq.2 ;2 ; exp.m; ;/; v; s; exp.m; ;// len.exp.m; ;//  len.exp.m; ;//  len.s/ k < len.exp.m; ;//

exp.m;;/

exp.exp.m; ;/; k/  exp.m; ;/ exp.m; ;/  exp.m; ;/ C s exp.exp.m; ;/; k/  exp.m; ;/ C s

42. 43. 44.

.8k < len.exp.m; ;///exp.m; ;/; k/  exp.m; ;/ C s P;i

45. 46. 47.

exp.m;;/

P;i

38 T13.49g 7 T13.11z A (g (8I)) T13.43g T13.11u 42,43 T13.11a 41-44 (8I) 39,40,45 9I 27,28-46_E

48.

P;i

14,15-47 _E

49.

B.m; ;/ _ C.m; ;/ _ D.m; ;/

A (g 12_E)

50.

P;i

51. 52.

trivial

P;i ; < len.m/ ! P;i

12,13-50 _E 8-51 !I

Exercise 13.40 T13.49.m

976

ANSWERS FOR CHAPTER 13

53.

.8z  i /.z < len.m/ ! Pzi /

A (g !I)

54.

S i < len.m/

A (g !I)

55. 56.

A.m; S i / _ B.m; S i / _ C.m; S i / _ D.m; S i / A.m; S i /

3,54 (8E) A (g 55_E)

57.

PSi i

as above

58.

B.m; S i /

A (g 55_E)

59. 60. 61.

.9j < S i /exp.m; S i / D pS q  exp.m; j / exp.m; S i / D pSq  exp.m; j / j < Si

58 abv A (g 59 (9E))

62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78.

j i j < len.m/ Pji Tsubseq.a; b; exp.m; j /; v; s; r/ len.r/  len.exp.m; j //  len.s/ .8k < len.r//exp.r; k/  exp.m; j / C s pSqexp.m;j /

pSqr

;b  2 ; pSq  exp.m; j /; v; s; pS q  r/ Tsubseq.a  2 len.pS q/ D 1 len.pS q  r/ D 1 C len.r/ len.pS q  exp.m; j // D 1 C len.exp.m; j // 1 C len.r/  1 C len.exp.m; j //  len.s/ len.pS q  r/  1 C len.exp.m; j //  len.s/ 1 C len.exp.m; j //  len.s/  len.s/ C len.exp.m; j //  len.s/ len.pS q  r/  len.s/ C len.exp.m; j //  len.s/ len.pS q  r/  Œ1 C len.exp.m; j //  len.s/ len.pS q  r/  len.exp.m; S i //  len.s/ k < len.pSq  r/ k < len.pSq/ C len.r/ k < len.pSq/ _ k  len.pS q/ k < len.pS q/

79. 80. 81. 82. 83. 84. 85. 86. 87. 88.

exp.pSq  r; k/ D exp.pSq; k/ exp.pSq; k/  pS q exp.pSq  r; k/  pS q pS q  pSq  exp.m; j / pS q  exp.m; j /  pS q  exp.m; j / C s pS q  exp.m; j /  exp.m; S i / C s exp.pSq  r; k/  exp.m; S i / C s

89.

k  len.pS q/

: k D len.pSq/ C .k len.pS q// : exp.pSq  r; k/ D exp.r; k len.pS q// : : k len.pS q/ < .len.pS q/ C len.r// len.pS q/ : k len.pS q/ < len.r/ : exp.r; k len.pS q//  exp.m; j / C s exp.pSq  r; k/  exp.m; j / C s exp.m; j /  pS q  exp.m; j / exp.m; j /  exp.m; S i / exp.pSq  r; k/  exp.m; S i / C s

90. 91. 92. 93. 94. 95. 96. 97. 98.

exp.pS q  r; k/  exp.m; S i / C s

99. 100. 101. 102. 103.

.8k < len.pS q  r//exp.pS q  r; k/  exp.m; S i / C s PSi i PSi i

61 T13.11n 54,61 T13.11b 53,62,63 (8E) A (g 64 9E)

65 T13.49i cap 69 T13.46f 69 T13.46f 66 T13.11v 72,70 DE 7 T13.11v 73,74 T13.11a 75 T6.65 60,71,76 DE A (g (8I)) 78 T13.46f T13.11q A (g 80_E) 81 T13.46c T13.43g 82,83 DE T13.46n T13.11u 60,85 DE 84,85,87 T13.11a A (g 80_E) 89 T13.21a 90 T13.46g 89,79 T13.21e 92 T13.21l,b 93,67 (8E) 91,94 DE T13.46o 60,96 DE 95,97 T13.11v 80,81-98 _E 78-99 (8I) 68,77,100 9I 64,65-101 9E

PSi i

59,60-102 (9E)

Exercise 13.40 T13.49.m

ANSWERS FOR CHAPTER 13

977

104.

C.m; S i /

A (g 55_E)

105.

PSi i

similarly

106.

D.m; S i /

A (g 55_E)

107.

PSi i

similarly

108. 109. 110. 111. 112. 113. 114. 115. 116.

PSi i

S i < len.m/ ! PSi i Œ.8z  i /.z < len.m/ ! Pzi / ! ŒS i < len.m/ ! PSi i  8i Œi < len.m/ ! P  9x9y9uŒTsubseq.x; y; t; v; s; u/ ^ len.u/  len.t /  len.s/ ^ .8k < len.u//exp.u; k/  t C s Tsubseq.x; y; t; v; s; u/ ^ len.u/  len.t /  len.s/ ^ .8k < len.u//exp.u; k/  t C s Termsub.t; v; s; u/ ^ len.u/  len.t /  len.s/ ^ .8k < len.u//exp.u; k/  t C s 9uŒTermsub.t; v; s; u/ ^ len.u/  len.t /  len.s/ ^ .8k < len.u//exp.u; k/  t C s 9uŒTermsub.t; v; s; u/ ^ len.u/  len.t /  len.s/ ^ .8k < len.u//exp.u; k/  t C s

117. ŒTerm.t / ^ Term.s/ ! 9uŒTermsub.t; v; s; u/ ^ len.u/  len.t /  len.s/ ^ .8k < len.u//exp.u; k/  t C s

T13.49.n. PA ` ŒAtomic.p/ ^ Term.s/ ! 9qŒAtomsub.p; v; s; q/ ^ len.q/  len.p/  len.s/ ^ .8k < len.q//exp.q; k/  p C s

Exercise 13.40 T13.49.n

55,56-107 _E 54-108 !I 53-109 !I 52,110 T13.11ag 3,5,111 8E A (g 112 9E) 113 T13.49l 114 9I 112,113-115 9E 1-116 !I

ANSWERS FOR CHAPTER 13

978

1.

Atomic.p/ ^ Term.s/

A (g !I)

2. 3. 4. 5. 6. 7.

s>1 len.s/ > ; .9x  p/.9y  p/ŒTerm.x/ ^ Term.y/ ^ p D pDq  x  y Term.a/ ^ Term.b/ p D pDq  a  b a p^b p

1 T13.47f 2 T13.44j 1 T13.48c A (g 4 (9E))

8. 9. 10. 11. 12.

9a0 ŒTermsub.a; v; s; a0 / ^ len.a0 /  len.a/  len.s/ ^ .8k < len.a0 //exp.a0 ; k/  a C s 9b 0 ŒTermsub.b; v; s; b 0 / ^ len.b 0 /  len.b/  len.s/ ^ .8k < len.b 0 //exp.b 0 ; k/  b C s Termsub.a; v; s; a0 / len.a0 /  len.a/  len.s/ .8k < len.a0 //exp.a0 ; k/  a C s

1,5 T13.49m 1,5 T13.49m A (g 89E)

13. 14. 15.

Termsub.b; v; s; b 0 / len.b 0 /  len.b/  len.s/ .8k < len.b 0 //exp.b 0 ; k/  b C s

A (g 99E)

16. 17. 18. 19. 20. 21.

b 0  pDq  a0  b 0 a0  pDq  a0 pDq  a0  pDq  a0  b 0 a0  pDq  a0  b 0 a0  pDq  a0  b 0 ^ b 0  pDq  a0  b 0 Term.a/ ^ Term.b/ ^ p D pDq  a  b ^ Termsub.a; v; s; a0 / ^ Termsub.b; v; s; b 0 / ^ pDq  a0  b 0 D pDq  a0  b 0 Atomsub.p; v; s; pDq  a0  b 0 / len.pDq/  len.pDq/  len.s/ len.pDq  a  b/ D len.pDq/ C len.a/ C len.b/ len.pDq  a0  b 0 / D len.pDq/ C len.a0 / C len.b 0 / len.pDq/ C len.a0 / C len.b 0 /  len.pDq/  len.s/ C len.a/  len.s/ C len.b/  len.s/ len.pDq/  len.s/ C len.a/  len.s/ C len.b/  len.s/ D Œlen.pDq/ C len.a/ C len.b/  len.s/ len.pDq/ C len.a0 / C len.b 0 /  Œlen.pDq/ C len.a/ C len.b/  len.s/ len.pDq  a0  b 0 /  len.pDq  a  b/  len.s/ len.pDq  a0  b 0 /  len.p/  len.s/ j < len.pDq  a0  b 0 /

T13.46o T13.46o T13.46n 17,18 T13.11a 19,16 ^I

22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39.

j < len.pDq/ _ j  len.pDq/ j < len.pDq/ exp.pDq  a0  b 0 ; j / D exp.pDq; j / exp.pDq; j /  pDq pDq  pDq  a  b pDq  a  b  pDq  a  b C s exp.pDq; j /  pDq  a  b C s exp.pDq  a0  b 0 ; j /  pDq  a  b C s

Exercise 13.40 T13.49.n

5,6,10,13 ^I 7,20,21 (9I) 3 T13.11z T13.46f T13.46f 23,11,14 T13.11v T6.65 26,27 DE 28,25,24 DE 29,6 DE A (g (8I)) T13.11q A (g 32_E) 33 T13.46c T13.43g T13.46n T13.11u 35,36,37 T13.11a 34,38 DE

ANSWERS FOR CHAPTER 13

979

40.

j  len.pDq/

A (g 32_E)

41. 42. 43. 44. 45.

len.pDq/ D le : j D le C .j le / : exp.pDq  a0  b 0 ; j / D exp.a0  b 0 ; j le / : : j le < len.a0 / _ j le  len.a0 / : j le < len.a0 / : : exp.a0  b 0 ; j le / D exp.a0 ; j le / : 0 0 0 exp.pDq  a  b ; j / D exp.a ; j le / : exp.a0 ; j le /  a C s exp.pDq  a0  b 0 ; j /  a C s a  pDq  a pDq  a  pDq  a  b a  pDq  a  b exp.pDq  a0  b 0 ; j /  pDq  a  b C s : j le  len.a0 / : .j le / C le  le C len.a0 / j  le C len.a0 / le C len.a0 / D la : j D la C .j la / 0 len.pDq  a / D la : exp.pDq  a0  b 0 ; j / D exp.b 0 ; j la / len.pDq  a0  b 0 / D la C len.b 0 / j < la C len.b 0 / : la C .j la / < la C len.b 0 / : j la < len.b 0 / : exp.b 0 ; j la /  b C s b  pDq  a  b b C s  pDq  a  b C s : exp.b 0 ; j la /  pDq  a  b C s exp.pDq  a0  b 0 ; j /  pDq  a  b C s

def 40 T13.21a 42 T13.46g T13.11q A (g 44 _E)

46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77.

exp.pDq  a0  b 0 ; j /  pDq  a  b C s exp.pDq  a0  b 0 ; j /  pDq  a  b C s exp.pDq  a0  b 0 ; j /  p C s .8k < len.pDq  a0  b 0 //exp.pDq  a0  b 0 ; k/  p C s 9qŒAtomsub.p; v; s; q/ ^ len.q/  len.p/  len.s/ ^ .8k < len.q//exp.q; k/  p C s 9qŒAtomsub.p; v; s; q/ ^ len.q/  len.p/  len.s/ ^ .8k < len.q//exp.q; k/  p C s 9qŒAtomsub.p; v; s; q/ ^ len.q/  len.p/  len.s/ ^ .8k < len.q//exp.q; k/  p C s 9qŒAtomsub.p; v; s; q/ ^ len.q/  len.p/  len.s/ ^ .8k < len.q//exp.q; k/  p C s

78. ŒAtomic.p/ ^ Term.s/ ! 9qŒAtomsub.p; v; s; q/ ^ len.q/  len.p/  len.s/ ^ .8k < len.q//exp.q; k/  p C s

E13.42. Work the case marked “similarly” on line 115 of T13.51a and the D case from T13.51f. Hard core: show each of the results from T13.51. T13.51.a.

Exercise 13.42 T13.51.a

45 T13.46c 43,46 DE 12,45 (8E) 47,48 DE T13.46o T13.46n 50,51 T13.11a 49,52 T13.11v A (g 44_E) 54 T13.11v 42,55 DE def 56 T13.21a T13.46f 58,59 T13.46g T13.46f 31,61 DE 58,62DE 63 T13.11w 15,64 (8E) T13.46o 66 T13.11v 65,67 T13.11a 60,68 DE 44,45-69 _E 32,33-70 _E 71,6 DE 31-72 (8I) 22,30,73 9I 9,13-74 9E 8,10-75 9E 4,5-76 (9E) 1-77 !I

980

ANSWERS FOR CHAPTER 13

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65.

8uŒ.P .u/ ^ len.u/  x/ ! .8k < len.u/P .val.u; k/ : : : : val.c; j /  val.a; j l1 /  val.c1 ; j l2 /  val.b; j l3 /  val.c2 ; j l4 / D c  d  c1  e  c2 P .a/ ^ P .b/ ^ P .d / ^ P .e/ 8v.P .v/ ! v > 1/ len.c/ D 1 ^ c1 > ; ^ c2 > ; ^ len.c1 /  1 ^ len.c2 /  1 j < l ^ Sx  l

P P P P P P

a >1^b >1^d >1^e >1 val.a; len.a// D a ^ val.b; len.b// D b ^ val.d; len.d // D d ^ val.e; len.e// D e c>; val.c; len.c// D c ^ val.c1 ; len.c1 // D c1 ^ val.c2 ; len.c2 // D c2 l D S.len.a/ C len.c1 / C len.b/ C len.c2 // S x  S.len.a/ C len.c1 / C len.b/ C len.c2 // x  len.a/ C len.c1 / C len.b/ C len.c2 / len.a/  x ^ len.b/  x : : : : len.val.c; j /  val.a; j l1 /  val.c1 ; j l2 /  val.b; j l3 /  val.c2 ; j l4 // D : : : : len.val.c; j // C len.val.a; j l1 // C len.val.c1 ; j l2 // C len.val.b; j l3 // C len.val.c2 ; j l4 // : : : : len.val.c; j // C len.val.a; j l1 // C len.val.c1 ; j l2 // C len.val.b; j l3 / C len.val.c2 ; j l4 //  l len.c  d  c1  e  c2 /  l len.c  d  c1  e  c2 / D S.len.d / C len.c1 / C len.e/ C len.c2 // S.len.d / C len.c1 / C len.e/ C len.c2 //  S x len.d / C len.c1 / C len.e/ C len.c2 /  x len.d /  x ^ len.e/  x j < l1

3,4 8E 7 T13.45n 5 T13.44g 5,9 T13.45n 5 T6.48 6,11 DE 12 T13.11j 13 T13.11u

j D; : : : : j l1 D ; ^ j l2 D ; ^ j l3 D ; ^ j l4 D ; : : : : val.c; j / D 1 ^ val.a; j l1 / D 1 ^ val.c1 ; j l2 / D 1 ^ val.b; j l3 / D 1 ^ val.c2 ; j l4 / D 1 : : : : val.c; j /  val.a; j l1 /  val.c1 ; j l2 /  val.b; j l3 /  val.c2 ; j l4 / D 1 c  d  c1  e  c2 D 1 len.c  d  c1  e  c2 /  1 c  d  c1  e  c2 > 1 1>1 11 ? j – l1 j  l1 val.c; j / D c : : : : val.a; j l1 /  val.c1 ; j l2 /  val.b; j l3 /  val.c2 ; j l4 / > ; d  c1  e  c2 > ; : : : : val.a; j l1 /  val.c1 ; j l2 /  val.b; j l3 /  val.c2 ; j l4 / D d  c1  e  c2 j < l2 : : : j l2 D ; ^ j l3 D ; ^ j l4 D ; : : : val.c1 ; j l2 / D 1 ^ val.b; j l3 / D 1 ^ val.c2 ; j l4 / D 1 : val.a; j l1 / > ; : : : : : val.a; j l1 /  val.c1 ; j l2 /  val.b; j l3 /  val.c2 ; j l4 / D val.a; j l1 / : val.a; j l1 / D d  c1  e  c2 len.d  c1  e  c2 / D len.d / C len.c1 / C len.e/ C len.c2 / len.d /  len.d  c1  e  c2 / : len.d /  len.val.a; j l1 // : : len.val.a; j l1 //  j l1 : len.d /  j l1 z < len.d / exp.d  c1  e  c2 ; z/ D exp.d; z/ : z < len.val.a; j l1 // : : : : : exp.val.a; j l1 /  val.c1 ; j l2 /  val.b; j l3 /  val.c2 ; j l4 ; z/ D exp.val.a; j l1 /; z/ : z len.a/ len.d / < len.a/

66 T13.11r T13.11u 66,67 T13.21d T13.21l 68,69 DE 7,70 T13.45n T13.11p A (c 72_E)

74.

z < len.d /

A (g (8I))

75. 76. 77. 78.

exp.d  c1  e  c2 ; z/ D exp.d; z/ : z < len.val.a; j l1 // : : : : : exp.val.a; j l1 /  val.c1 ; j l2 /  val.b; j l3 /  val.c2 ; l l4 /; z/ D exp.val.a; j l1 /; z/ exp.a; z/ D exp.d; z/

74 T13.46c 71,73,74 T13.11c 76 T13.46c 71,77,75,38 DE

79. 80. 81. 82. 83. 84.

.8z < len.d //exp.a; z/ D exp.d; z/ val.a; len.d // D val.d; len.d // val.a; len.d // D d P .val.a; len.d /// P .val.a; len.d /// ?

74-78 (8I) 79 T13.45m 80,8 DE 3,81 DE 1,3,14,73 8E 82,83 ?I

85.

len.d / > len.a/

A (c 72_E)

86.

?

similarly (72)

87.

len.d / D len.a/

88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99. 100. 101. 102. 103. 104. 105. 106. 107. 108. 109. 110. 111. 112. 113. 114.

A (c 72_E)

z < len.d /

A (g (8I))

z < len.a/ : z < len.val.a; j l1 // exp.d  c1  e  c2 ; z/ D exp.d; z/ : : : : : exp.val.a; j l1 /  val.c1 ; j l2 /  val.b; j l3 /  val.c2 ; j l4 /; z/ D exp.val.a; j l1 /; z/ : exp.val.a; j l1 /; z/ D exp.d; z/ : .8z < len.d //exp.val.a; j l1 /; z/ D exp.d; z/ : : val.val.a; j l1 /; len.val.a; j l1 /// D val.d; len.d // : val.a; j l1 / > ; : : : val.val.a; j l1 /; len.val.a; j l1 /// D val.a; j l1 / : val.a; j l1 / D d : : : val.c1 ; j l2 /  val.b; j l3 /  val.c2 ; j l4 // > ; c1  e  c2 > ; : : : val.c1 ; j l2 /  val.b; j l3 /  val.c2 ; j l4 / D c1  e  c2 j < l3 ?

?

116.

j – l4

88-93 (8I) 94,71,87 T13.45m T13.45i 96 T13.45n 95,97,8 DE T13.46c T13.46c 98,38,99,100 T13.46l A (c I) similarly (22)

j – l3 j  l3 l3  l2 : : j l2  l3 l2 : l3 l2 D len.c1 / : j l2  len.c1 / : val.c1 ; j l2 / D c1 : : val.b; j l3 /  val.c2 ; j l4 / > ; e  c2 > ; : : val.b; j l3 /  val.c2 ; j l4 / D e  c2 j < l4

115.

88,87 DE 71,89 DE 88 T13.46c 90 T13.46c 92,91,38 DE

102-103 I 104 T13.11r T13.11u 105,106 T13.21d T13.21l 107,108 DE 5,105 T13.45n T13.46c T13.46c 101,110,111,112 T13.46l A (c I) similarly (39) 114-115 I

Exercise 13.42 T13.51.a

982

ANSWERS FOR CHAPTER 13

117. 118. 119. 120. 121. 122. 123. 124.

j  l4 l4  l3 : : j l3  l4 l3 : l4 l3 D len.b/ : j l3  len.b/ : val.b; j l3 / D b len.e/ < len.b/ _ len.e/ D len.b/ _ len.e/ > len.b/ len.e/ < len.b/

116 T13.11r T13.11u 117,118 T13.21d T13.21l 119,120 DE 121,7 T13.45n A (c 123_E)

125.

?

similarly (72)

126.

len.e/ > len.b/

A (c 123_E)

127.

?

similarly (72)

128.

len.e/ D len.b/ : val.b; j l3 / D e : val.c2 ; j l4 / > ; : val.c2 ; j l4 / D c2 : len.val.c2 ; j l4 // D len.c2 / : : len.val.c2 ; j l4 //  j l4 : j l4  len.c2 / : .j l4 / C l4  len.c2 / C l4 : j D l4 C .j l4 / j l j –l ?

A (c 123_E)

129. 130. 131. 132. 133. 134. 135. 136. 137. 138. 139. 140.

?

141. ?

similarly T13.45i 113,130,5 T13.46l 131 DE T13.45j 133,134 DE 134 T13.11v 117 T13.21a 135,136 DE 137 T13.11r 138,6 ?I 123,124-139 _E 72,73-140 _E

T13.51.f. PA ` Term.t / ! .8k < len.t //Term.val.t; k//

Exercise 13.42 T13.51.f

983

ANSWERS FOR CHAPTER 13

1. 2.

Term.t / ^ len.t /  ;

A (g !I)

k < len.t /

A (g (8I))

3.

Term.val.t; k//

A (c I)

4. 5. 6.

k; j D;

T13.11d,m A (g 16_E)

18. 19. 20.

val.a; j / D 1 val.a; j /  1 Term.val.a; j //

17 def 18 T13.11m,r 19 T13.47f

21.

j >;

A (g 16_E)

22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34.

: j D S.j 1/ : S.j 1/ < len.a/ Termseq.m; a/ : exp.m; len.m/ 1/ D a m>1 .8k < len.m//ŒA.m; k/ _ B.m; k/ _ C.m; k/ _ D.m; k/ len.m/ > ; : len.m/ 1 < len.m/ : : : : A.m; len.m/ 1/ _ B.m; len.m/ 1/ _ C.m; len.m/ 1/ _ D.m; len.m/ 1/ Term.val.a; j // : A.m; len.m/ 1/ a D p;q _ Var.a/ a D p;q

35.

len.a/ D 1

36.

Var.a/

37.

.9x  a/a D 2 23C2x

39.

len.a/ D 1

45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57.

32,25 abv A (g 33_E)

A (g 33_E)

aD2

41. 42. 43. 44.

A (c 30 _E)

34 cap 23C2x

38.

40.

21 T13.21j 15,22 DE 13 T13.47b 24 T13.47a 24 T13.47a 24 T13.47a 26 T13.44j 28 T13.21i 27,29 (8E) A (c I)

len.a/ D 1 len.a/ D 1 j ; exp.m; l/ > ; : val.pSq  exp.m; l/; j / D val.pS q; j /  val.exp.m; l/; j 1/ Exercise 13.42 T13.51.f val.pSq; j / D pS q : val.a; j / D pSq  val.exp.m; l/; j 1/ : 9rŒpS q  val.exp.m; l/; j 1/ D pS q  r ^ Term.r/

36 def A (g 37 (9E)) 38 T13.44k 37,38-39 (9E) 33,34-35,36-40 _E 15,41 T13.11n 42 T13.11d,r 21,43 ?I A (c 30 _E) 45,25 abv A (c 46 (9E)) 29,47 T13.11b 24,49 T13.47n cap 51 T13.44g 49,24 T13.47e 51,52,53 T13.46m 21,51 T13.45n 54,48,55 DE 31,56 T13.51c

ANSWERS FOR CHAPTER 13 : pS q  val.exp.m; l/; j 1/ D pSq  r Term.r/ : val.exp.m; l/; j 1/ > ; r >; : val.exp.m; l/; j 1/ D r : Term.val.exp.m; l/; j 1// len.a/ D len.pS q/ C len.exp.m; l// len.a/ D 1 C len.exp.m; l// len.a/ D S len.exp.m; l// S len.exp.m; l//  S x len.exp.m; l//  x : S.j 1/ < Slen.exp.m; l// : j 1 < len.exp.m; l// : Term.val.exp.m; l/; j 1// ?

58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73.

? ?

75.

C.m; len.m/

76. 77. 78. 79.

: : .9i < len.m/ 1/.9j < len.m/ 1/a D pCq  exp.m; i /  exp.m; j / : k < len.m/ 1 : l < len.m/ 1 a D pCq  exp.m; k/  exp.m; l/

89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99. 100.

1/

A (c 30 _E) 75 abv A (c 76(9E))

?

91,92-99 9E

102.

D.m; len.m/

103.

?

107.

46,47-73 (9E) :

29,77 T13.11b 29,78 T13.11b 24,80 T13.47n 24,81 T13.47n cap 84 T13.44g 77,24 T13.47e 78,24 T13.47e

?

106.

T13.45i 59 T13.47f 58,60,61 T13.46l 59,62 DE 48 T13.46f 64,51 DE 65 T6.48 14,66 DE 67 T13.11j 66,23 DE 69 T13.11k 11,50,68,70 (8E) 63,71 ?I

k < len.m/ l < len.m/ Term.exp.m; k// Term.exp.m; l// len.pCq/ D 1 pCq > ; exp.m; k/ > ; exp.m; l/ > ; val.pCq  exp.m; k/  exp.m; l/; j / D : : val.pCq; j /  val.exp.m; k/; j 1/  val.exp.m; l/; j .1 C len.exp.m; k//// val.pCq; j / D pCq : : val.a; j / D pCq  val.exp.m; k/; j 1/  val.exp.m; l/; j .1 C len.exp.m; k//// : : 9r9sŒpCq  val.exp.m; k/; j 1/  val.exp.m; l/; j .1 C len.exp.m; k//// D pCq  r  s ^ Term.r/ ^ Term.s/ : : pCq  val.exp.m; k/; j 1/  val.exp.m; l/; j .1 C len.exp.m; k//// D pCq  r  s Term.r/ Term.s/ : : val.pCq; j /  val.exp.m; k/; j 1/  val.exp.m; l/; j .1 C len.exp.m; k//// D pCq  r  s Term.exp.m; k// ^ Term.exp.m; l// ^ Term.r/ ^ Term.s/ 8v.Term.v/ ! v > 1/ len.a/  S x ^ j < len.a/ ?

101.

104. 105.

A (c 579E)

57,58-72 9E

74.

80. 81. 82. 83. 84. 85. 86. 87. 88.

984

84,85,86,87 T13.46m 21,84 T13.45n 88,79,89 DE 31,90 T13.51d A (c 919E)

92,89 DE 82,83,93,94 ^I T13.47f 14,15 ^I 11,95,96,97,84,98 T13.51a

76,77-100 (9E) :

1/

? Term.val.a; j // Term.val.a; j // .8k < len.a//Term.val.a; k//

A (c 30 _E) similarly 30,32-103 _E 31-104 I 16,17-20,21-105 _E 15-106 (8I)

.Term.a/ ^ len.a/  S x/ ! .8k < len.a//Term.val.a; k// 8t Œ.Term.t / ^ len.t /  S x/ ! .8k < len.t //Term.val.t; k//

12-108 !I 108 8I

110. 8t Œ.Term.t / ^ len.t /  x/ ! .8k < len.t //Term.val.t; k// ! Exercise T13.51.f 8t Œ.Term.t / ^ len.t /  S x/ !13.42 .8k < len.t //Term.val.t; k// 111. 8t Œ.Term.t / ^ len.t /  x/ ! .8k < len.t //Term.val.t; k//

11-109 !I 10,110 IN

108. 109.

ANSWERS FOR CHAPTER 13

985

E13.43. Show (g) including at least the A case, and (k) from T13.52. Hard core: show each of the results from T13.52. T13.52.h. PA ` ŒWff .p//^Wff .q/^Wff .a/^Wff .b/ ! Œcnd.p; q/ D cnd.a; b/ ! .p D a ^ q D b/ 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Wff .p// ^ Wff .q/ ^ Wff .a/ ^ Wff .b/

A (g !I)

cnd.p; q/ D cnd.a; b/

A (g !I)

p.q  p  p!q  q  p/q D p.q  a  p!q  b  p/q p >1^q >1^a >1^b >1 p  p!q  q  p/q > 1 ^ a  p!q  b  p/q > 1 p  p!q  q  p/q D p  p!q  q  p/q p  p!q  q > 1 ^ a  p!q  b > 1 p  p!q  q D a  p!q  b len.p/ < len.a/ _ len.p/ D len.a/ _ len.p/ > len.a/ len.p/ < len.a/

2 def 1 T13.48e 4 T13.46n 3,5 T13.46l 4 T13.46n 6,7 T13.46k T13.11p A (g I)

11.

i < len.p/

A g (8I))

12. 13. 14.

i < len.a/ exp.p  p!q  q; i / D exp.p; i / ^ exp.a  p!q  b; i / D exp.a; i / exp.p; i / D exp.a; i /

11,10 T13.11b 11,12 T13.46c 8,13 DE

15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33.

.8i < len.p//exp.p; i / D exp.a; i / val.p; len.p// D val.a; len.p// p D val.a; len.p// Wff .val.a; len.p/// Wff .val.a; len.p/// ? len.p/ – len.a/ len.p/ > len.a/

11-14 (8I) 15 T13.45m 16,4 T13.45n 17,1 DE 1,10 T13.51g 18,19 ?I 10-20 I A (g I)

?

similarly

len.p/  len.a/ len.p/ D len.a/ p!q  q > 1 ^ p!q  b > 1 p!q  q D p!q  b qDb p  p!q > 1 ^ a  p!q > 1 p  p!q D a  p!q pDa p Da^q Db cnd.p; q/ D cnd.a; b/ ! .p D a ^ q D b/

34. ŒWff .p// ^ Wff .q/ ^ Wff .a/ ^ Wff .b/ ! Œcnd.p; q/ D cnd.a; b/ ! .p D a ^ q D b/

22-23 I 9,21,24 DS 4 T13.46o 8,25,26 T13.46l 27,4 T13.46l 4 T13.46n 8,28,29 T13.46k 30,4 T13.46k 31,28 ^I 2-32 !I 1-33 !I

T13.52.j. PA ` Axiompa.p/ ! Wff .p/ The cases for axioms of Q are immediate by capture. The following should be sufficient to see how other cases will go. PA ` Axiomad6.n/ ! Wff .n/

Exercise 13.43 T13.52.j

986

ANSWERS FOR CHAPTER 13

1.

Axiomad6.n/

A (g !I)

2. 3. 4.

.9v  n/ŒVar.v/ ^ n D pDq  v  v vn Var.v/ ^ n D pDq  v  v

1 T13.39e A (g 2 (9E))

5. 6. 7. 8. 9. 10.

Term.v/ Term.v/ ^ Term.v/ ^ n D pDq  v  v .9x  n/.9y  n/ŒTerm.x/ ^ Term.y/ ^ n D pDq  x  y Atomic.n/ Wff .n/ Wff .n/

4 T13.47p 5,4 ^I 6,3 (9I) 7 T13.48c 8 T13.48m 2,3-9 (9E)

11. Axiomad6.n/ ! Wff .n/

1-10 !I

PA ` Axiompa7.n/ ! Wff .n/ 1.

Axiompa7.p/

A (g !I)

2.

.9p  n/.9v  n/ŒWff .p/ ^ Var.v/ ^ n D cnd.neg.cnd.formsub.p; v; p;q/; neg.unv.v; cnd.p; formsub.p; v; pS q  v//////; unv.v; p// Wff .p/ ^ Var.v/ n D cnd.neg.cnd.formsub.p; v; p;q/; neg.unv.v; cnd.p; formsub.p; v; pS q  v//////; unv.v; p//

1 T13.39e A (g 2 (9E))

3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

Term.p;q/ ^ Term.pS q  v/ Wff .formsub.p; v; p;q// Wff .formsub.p; v; pS q  v// Wff .cnd.p; formsub.p; v; pSq  v/// Wff .unv.v; cnd.p; formsub.p; v; pSq  v//// Wff .neg.unv.v; cnd.p; formsub.p; v; pS q  v///// Wff .cnd.formsub.p; v; p;q/; neg.unv.v; cnd.p; formsub.p; v; pSq  v////// Wff .neg.cnd.formsub.p; v; p;q/; neg.unv.v; cnd.p; formsub.p; v; pS q  v/////// Wff .unv.v; p// Wff .cnd.neg.cnd.formsub.p; v; p;q/; neg.unv.v; cnd.p; formsub.p; v; pS q  v//////; unv.v; p/// Wff .n/ Wff .n/

3 T13.47o,p 3,5 T13.50m 3,5 T13.50m 3,7 T13.48o 3,8 T13.48p 9 T13.48c 6,10 T13.48o 11 T13.48c 3 T13.48p 12,13 T13.48o 4,14 DE 2,3-15 (9E)

17. Axiompa7.n/ ! Wff .n/

1-16 !I

E13.44. As a start to a complete demonstration of T13.53, provide a demonstration through part (C) that does not skip any steps. PA ` Prvt.cnd.p; q// ! .Prvt.p/ ! Prvt.q//. (a)

Exercise 13.44 T13.53

987

ANSWERS FOR CHAPTER 13

1.

Prvt.cnd.p; q//

A (g !I)

2. 3.

Wff .cnd.p; q// Prvt.p/

1 T13.52k A (g !I)

4. 5. 6. 7. 8. 9. 10. 11.

Wff .p/ Wff .q/ Mp.cnd.p; q/; p; q/ Mp.cnd.p; q/; p; q/ _ .cnd.p; q/ D p ^ Gen.p; q// Icon.cnd.p; q/; p; q/ 9vPrft.v; cnd.p; q// 9vPrft.v; p/ Prft.j; cnd.p; q//

12.

Prft.k; p/

13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

l =j k2 : exp.j; len.j / 1/ D cnd.p; q/ : exp.k; len.k/ 1/ D p len.j  k/ D len.j / C len.k/ q>; q len.2 / D 1 q .8i < 1/Œexp.l; i C len.j  k// D exp.2 ; i / ; ; : exp.j; len.j / 1/ > ; : len.j / 1 < len.j / : : exp.l; len.j / 1/ D exp.j; len.j / 1/ p>; : exp.k; len.k/ 1/ > ; : len.k/ 1 < len.k/ : : exp.l; len.j / C len.k/ 1/ D exp.k; len.k/ 1/ : : IconŒexp.l; len.j / 1/; exp.l; len.j / C len.k/ 1/; exp.l; len.j / C len.k//

Exercise 13.44 T13.53

29-34 (8I) 2 T13.48e 14,36 DE 37 T13.44h 26,38 (8E) 4 T13.48e 40,15 DE 41 T13.44h 35,42 (8E) 25,39,43 DE

ANSWERS FOR CHAPTER 13

988

(c1) 45. 46. 47. 48. 49.

.8i < len.j //ŒAxiomt.exp.j; i // _ .9m < i /.9n < i /Icon.exp.j; m/; exp.j; n/; exp.j; i // a < len.j / Axiomt.exp.j; a// _ .9m < a/.9n < a/Icon.exp.j; m/; exp.j; n/; exp.j; a// exp.l; a/ D exp.j; a/ Axiomt.exp.j; a//

T13.39g A (g (8I)) 45,46 (8E) 26,46 (8E) A (g 47_E)

50. 51.

Axiomt.exp.l; a// Axiomt.exp.l; a// _ .9m < a/.9n < a/Icon.exp.l; m/; exp.l; n/; exp.l; a//

49,48 DE 50 _I

52.

.9m < a/.9n < a/Icon.exp.j; m/; exp.j; n/; exp.j; a//

A (g 47_E)

53. 54. 55.

Icon.exp.j; m0 /; exp.j; n0 /; exp.j; a// m0 < a n0 < a

A (g 529E)

56. 57. 58. 59. 60. 61.

m0 < len.j / n0 < len.j / exp.l; m0 / D exp.j; m0 / exp.l; n0 / D exp.j; n0 / Icon.exp.l; m0 /; exp.l; n0 /; exp.l; a// .9m < a/.9n < a/Icon.exp.l; m/; exp.l; n/; exp.l; a//

46,54 T13.11b 46,55 T13.11b 26,56 (8E) 26,57 (8E) 53,58,59,48 DE 60,54,55 (9I)

62. 63. 64. 65.

.9m < a/.9n < a/Icon.exp.l; m/; exp.l; n/; exp.l; a// Axiomt.exp.l; a// _ .9m < a/.9n < a/Icon.exp.l; m/; exp.l; n/; exp.l; a// Axiomt.exp.l; a// _ .9m < a/.9n < a/Icon.exp.l; m/; exp.l; n/; exp.l; a// .8i < len.j //ŒAxiom.exp.l; i // _ .9m < i /.9n < i /Icon.exp.l; m/; exp.l; n/; exp.l; i //

52,53-61 (9E) 62 _I 47,49-51,52-63 _E 46-64 (8I)

(c2) The argument is similar for, .8i 1 n  Œpi.len.p//p len.p/ n  Bp : exp.n; len.n/ 1/ D p a < len.n/

1 T13.48b 2 T13.48a 2 T13.48a 4 T13.44j 5 T13.21i 2,3,6 T13.48j 7 T13.45o 7 T13.48a 8,9 T13.45n 10 T13.56b 7 T13.48a A (g (8I))

E.n; a/ _ F .n; a/ _ G.n; a/ _ H.p; n; a/ E.n; a/

7,13 T13.48a A (g 14_E)

16. 17. 18.

Atomic.exp.n; a// T .n; a/ T .n; a/ _ U.n; a/ _ V .n; a/ _ W .p; v; n; a/ _ X.p; v; v; n; a/

15 abv 16 abv 17_I

19.

F .n; a/ _ G.n; a/

A (g 14_E)

20.

T .n; a/ _ U.n; a/ _ V .n; a/ _ W .p; v; n; a/ _ X.p; v; v; n; a/

similarly

21.

H.p; n; a/

A (g 14_E)

22. 23. 24.

.9i < a/.9j < p/ŒVar.j / ^ exp.n; a/ D unv.j; exp.n; i // l ; numseq.x/ > ; pi.S x/num.S x/ > ; 8j Œexp.numseq.S x/; j / D exp.numseq.x/; j / C exp.pi.S x/num.S x/ ; j / exp.pi.S x/num.S x/ ; S x/ D num.S x/ exp.pi.S x/num.S x/ ; S x/ > ; exp.numseq.S x/; S x/  exp.pi.S x/num.S x/ ; S x/ exp.numseq.S x/; S x/ > ; len.numseq.S x// > S x len.numseq.S x//  S S x k > Sx exp.numseq.x/; k/ D ; exp.pi.S x/num.S x/ ; k/ D ; exp.numseq.S x/; k/ D ; .8k > S x/exp.numseq.S x/; k/ D ; len.numseq.S x//  S S x len.numseq.S x// D S S x

24. len.numseq.x// D S x ! len.numseq.S x// D S S x 25. len.numseq.x// D S x

T13.56.t. PA ` ŒWff .p/ ^ Var.v/ ! Freefor.num.x/; v; p/

Exercise 13.48 T13.56.t

def T13.56l T13.56m T13.42h 6,8,9 T13.43m T13.43i 7,11 DE 10 T13.11u 12,13 T13.11c 14 T13.44h 15 T13.11l A (g (8I)) 5,17 T13.44l 17 T13.43j 10,18,19 DE 17-20 (8I) 21 T13.44i 16,22 T13.20 5-23 !I 4,24 IN

998

ANSWERS FOR CHAPTER 13

Wff .p/ ^ Var.v/

A (g !I)

2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Formseq.m; p/ : exp.m; len.m/ 1/ D p Formseq.n; p/ .8i < len.n//exp.n; i /  p ^ len.n/  len.p/ Œpi.len.p//p len.p/  val.n; len.n// n>1 val.n; len.n// D n n  Bp : exp.n; len.n/ 1/ D p a < len.n/

1 T13.48b 2 T13.48a 2,3 T13.48j 2,3 T13.48j 5 T13.45o 4 T13.48a 7 T13.45n 6,8 T13.56b 4 T13.48a A (g (8I))

12. 13.

E.n; a/ _ F .n; a/ _ G.n; a/ _ H.p; n; a/ E.n; a/

4,11 T13.48a A (g 12_E)

1.

14. 15. 16.

Atomic.exp.n; a// T .n; a/ T .n; a/ _ U.n; a/ _ V .n; a/ _ W .p; v; n; a/ _ X.p; v; num.x/; n; a/

13 abv 14 abv 15 _I

17.

F .n; a/ _ G.n; a/

A (g 12_E)

18.

T .n; a/ _ U.n; a/ _ V .n; a/ _ W .p; v; n; a/ _ X.p; v; num.x/; n; a/

similarly

19.

H.p; n; a/

A (g 12_E)

20. 21. 22. 23.

.9i < a/.9j < p/ŒVar.j / ^ exp.n; a/ D unv.j; exp.n; i // i 1 len.m/ > ; ^ len.m0 / > ; : : exp.n; len.n/ 1/ D t ^ exp.n0 ; len.n0 / 1/ D t 0 len.m/ D len.n/ ^ len.m0 / D len.n0 / .8k < len.m//ŒI.m; n; k/ _ J.v; m; n; k/ _ K.v; num.y/; m; n; k/ _ L.m; n; k/ _ M.m; n; k/ _ N.m; n; k/ .8k < len.m0 //ŒI.m0 ; n0 ; k/ _ J.w; m0 ; n0 ; k/ _ K.w; num.z/; m0 ; n0 ; k/ _ L.m0 ; n0 ; k/ _ M.m0 ; n0 ; k/ _ N.m0 ; n0 ; k/ l < len.m/ ^ l 0 < len.m0 / len.exp.m; l//  ;

4 T13.49b A (g 5 (9E)) 6 T13.49a 7 T13.47a 7 T13.47a 9 T13.44j 6 T13.49a 6 T13.49a 6 T13.49a 6 T13.49a A (g (8I)) A (g !I)

0

17.

k;k Œexp.m; l/ D exp.m0 ; l 0 / ! Pl;l 0 

A (c E)

18. 19. 20.

exp.m; l/ > 1 exp.m; l/  1 ?

7,15 T13.47f 16 T13.44j 18,19 ?I

21. 22. 23. 24. 25.

k;k exp.m; l/ D exp.m0 ; l 0 / ! Pl;l 0

0

17-20 E 0

k;k len.exp.m; l//  ; ! .exp.m; l/ D exp.m0 ; l 0 / ! Pl;l 0 /

.8k < len.m//.8k 0 < len.m0 //Œlen.exp.m; k//  ; ! .exp.m; k/ D exp.m0 ; k 0 / ! P / .8k < len.m//.8k 0 < len.m0 //Œlen.exp.m; k//  x ! .exp.m; k/ D exp.m0 ; k 0 / ! P / l < len.m/ ^ l 0 < len.m0 /

16-21 !I 15-22 (8I) A (g !I) A (g (8I))

26.

len.exp.m; l//  S x ^ exp.m; l/ D exp.m0 ; l 0 /

A (g !I)

27. 28. 29.

I.m; n; l/ _ J.v; m; n; l/ _ K.v; num.y/; m; n; l/ _ L.m; n; l/ _ M.m; n; l/ _ N.m; n; l/ I.m0 ; n0 ; l 0 / _ J.w; m0 ; n0 ; l 0 / _ K.w; num.z/; m0 ; n0 ; l 0 / _ L.m0 ; n0 ; l 0 / _ M.m0 ; n0 ; l 0 / _ N.m0 ; n0 ; l 0 / I.m; n; l/

13,25 (8E) 14,25 (8E) A (g 27_E)

30. 31.

exp.m; l/ D p;q ^ exp.n; l/ D p;q I.m0 ; n0 ; l 0 /

32.

exp.m0 ; l 0 / D p;q ^ exp.n0 ; l 0 / D p;q

33.

Tsubseq.2

34.

Tsubseq.2

35.

k;k Pl;l 0

36.

J.w; m0 ; n0 ; l 0 / _ K.w; num.z/; m0 ; n0 ; l 0 / _ L.m0 ; n0 ; l 0 / _ M.m0 ; n0 ; l 0 / _ N.m0 ; n0 ; l 0 /

p;q p;q

p;q

;2

p;q

;2

30 T13.49f

; exp.n0 ; l 0 /; v; num.y/; p;q/

32 T13.49f

0

33,34 9I 0

k;k Pl;l 0

38. 39. 40.

exp.m0 ; l 0 / D p;q ŒJ.w; m0 ; n0 ; l 0 / _ K.w; num.z/; m0 ; n0 ; l 0 / _ L.m0 ; n0 ; l 0 / _ M.m0 ; n0 ; l 0 / _ N.m0 ; n0 ; l 0 / ? k;k Pl;l 0

31 abv

; exp.n; l/; w; num.z/; p;q/

37.

41.

29 abv A (g 28_E)

A (g 28_E) A (c E)

0

30,26 DE 38 T13.54c 36,39 ?I 37-40 E

Exercise 13.49 T13.57.b

1000

ANSWERS FOR CHAPTER 13

42.

J.v; m; n; l/

A (g 27 _E)

43. 44.

Var.exp.m; l// ^ exp.m; l/ ¤ v ^ exp.n; l/ D exp.m; l/ J.w; m0 ; n0 ; l 0 /

42 abv A (g 28_E)

45. 46.

Var.exp.m0 ; l 0 // ^ exp.m0 ; l 0 / ¤ w ^ exp.n0 ; l 0 / D exp.m0 ; l 0 / exp.n; l/ ¤ w ^ exp.n0 ; l 0 / ¤ v ^ exp.n; l/ D exp.n0 ; l 0 / exp.n;l/

;2

exp.n;l/

; exp.n; l/; w; num.z/; exp.n; l//

43 abv 26,43,45 DE

47.

Tsubseq.2

48.

Tsubseq.2

49.

k;k Pl;l 0

50.

K.w; num.z/; m0 ; n0 ; l 0 /

A (g 28_E)

51. 52.

Var.exp.m0 ; l 0 // ^ exp.m0 ; l 0 / D w ^ exp.n0 ; l 0 / D num.z/ Var.exp.n; l// ^ exp.n; l/ D w

50 abv 26,43,51 DE

53. 54. 55. 56.

Tsubseq.2 ;2 ; exp.n; l/; w; num.z/; num.z// Termsub.exp.n0 ; l 0 /; v; num.y/; num.z// .9x  X /.9y  Y /Tsubseq.x; y; exp.n0 ; l 0 /; v; num.y/; num.z// Tsubseq.e; f; exp.n0 ; l 0 /; v; num.y/; num.z//

57.

exp.n0 ;l 0 /

exp.n0 ;l 0 /

;2

; exp.n0 ; l 0 /; v; num.y/; exp.n0 ; l 0 //

0

47,48,46 9I

exp.n;l/

k;k Pl;l 0

43,46 T13.49g 45,46 T13.49g

num.z/

0

52 T13.49h 51 T13.56s 54 T13.49b A (g 55 (9E)) 53,56 9I

58.

k;k 0 Pl;l 0

55,56-57 (9E)

59.

I.m0 ; n0 ; l 0 / _ L.m0 ; n0 ; l 0 / _ M.m0 ; n0 ; l 0 / _ N.m0 ; n0 ; l 0 /

A (g 28_E)

60.

Var.exp.m0 ; l 0 //

43,26 DE

61.

k;k Pl;l 0

0

A (c E)

62.

exp.m0 ; l 0 / D w

A (c I)

63. 64.

ŒI.m0 ; n0 ; l 0 / _ J.w; m0 ; n0 ; l 0 / _ L.m0 ; n0 ; l 0 / _ M.m0 ; n0 ; l 0 / _ N.m0 ; n0 ; l 0 / ?

60,62 T13.54e with 59,63

65. 66. 67. 68. 69.

exp.m0 ; l 0 / ¤ w ŒI.m0 ; n0 ; l 0 / _ K.w; num.z/; m0 ; n0 ; l 0 / _ L.m0 ; n0 ; l 0 / _ M.m0 ; n0 ; l 0 / _ N.m0 ; n0 ; l 0 / ? k;k Pl;l 0

0

62-64 I 60,65 T13.54d with 59,66 61-67 E

k;k 0 Pl;l 0

28,44-68 _E

Exercise 13.49 T13.57.b

1001

ANSWERS FOR CHAPTER 13

70.

K.v; num.y/; m; n; l/

A (g 27_E)

71. 72.

Var.exp.m; l/ ^ exp.m; l/ D v ^ exp.n; l/ D num.y/ K.w; num.z/; m0 ; n0 ; l 0 /

70 abv A (g 28_E)

73.

Var.exp.m0 ; l 0 / ^ exp.m0 ; l 0 / D w ^ exp.n0 ; l 0 / D num.z/

74.

k;k Pl;l 0

75. 76.

vDw ?

0

72 abv A (c E) 26,71,73 1,75 ?I

0

77.

k;k Pl;l 0

78.

J.w; m0 ; n0 ; l 0 /

A (g 28_E)

79. 80.

Var.exp.m0 ; l 0 // ^ exp.m0 ; l 0 / ¤ w ^ exp.n0 ; l 0 / D exp.m0 ; l 0 / Var.exp.n0 ; l 0 // ^ exp.n0 ; l 0 / D v

78 abv 26,71,79 DE

81. 82. 83.

Tsubseq.2 ;2 ; exp.n0 ; l 0 /; v; num.y/; num.y// Termsub.exp.n; l/; w; num.z/; num.y// Tsubseq.e; f; exp.n; l/; w; num.z/; num.y//

84.

k;k Pl;l 0

85.

0

I.m ; n ; l / _ L.m ; n ; l / _ M.m ; n ; l / _ N.m ; n ; l /

A (g 28_E)

86.

Var.exp.m0 ; l 0 //

26,71 DE

87.

74-76 E

exp.n0 ;l 0 /

num.y/

0

80 T13.49h 71 T13.56s 82 T13.49b 81,83 9I

0

k;k Pl;l 0

0

0

0

0

0

0

0

0

0

0

0

A (c E)

88.

exp.m0 ; l 0 / D w

A (c I)

89. 90.

ŒI.m0 ; n0 ; l 0 / _ J.w; m0 ; n0 ; l 0 / _ L.m0 ; n0 ; l 0 / _ M.m0 ; n0 ; l 0 / _ N.m0 ; n0 ; l 0 / ?

86,88 T13.54e with 85,89

91. 92. 93. 94. 95.

exp.m0 ; l 0 / ¤ w ŒI.m0 ; n0 ; l 0 / _ K.w; num.z/; m0 ; n0 ; l 0 / _ L.m0 ; n0 ; l 0 / _ M.m0 ; n0 ; l 0 / _ N.m0 ; n0 ; l 0 / ? k;k Pl;l 0

0

62-64 I 86,91 T13.54d with 85,92 87-93 E

k;k 0 Pl;l 0

28,72-94 _E

Exercise 13.49 T13.57.b

1002

ANSWERS FOR CHAPTER 13

96.

L.m; n; l/

A (g 27_E)

97. 98. 99.

.9i < l/Œexp.m; l/ D pS q  exp.m; i / ^ exp.n; l/ D pS q  exp.n; i / h len.exp.m; h// len.exp.m; h// < S x len.exp.m; h//  x

26,99,103 DE 25,98,102 T13.11b 7,105 T13.47n 106,104 T13.51b 99 T13.46f 108 T13.11l 26,109 T13.11c 110 T13.11n

112.

k;k exp.m; h/ D exp.m0 ; h0 / ! Ph;h 0

24,105,111 (8E)

9q9a9b9c9d ŒTsubseq.a; b; exp.n; h/; w; num.z/; q/ ^ Tsubseq.c; d; exp.n0 ; h0 /; v; num.y/; q/ Tsubseq.a; b; exp.n; h/; w; num.z/; q/ ^ Tsubseq.c; d; exp.n0 ; h0 /; v; num.y/; q/

112,107 !E 113 abv A (g 1149E)

0

k;k 0 Ph;h 0

113. 114. 115.

pSqexp.n;h/

Tsubseq.a  2

116.

pSqq

;b  2

pSqexp.n0 ;h0 /

117.

Tsubseq.c  2

118.

Tsubseq.a  2

119.

Tsubseq.c  2

120.

k;k Pl;l 0

exp.n;l/

exp.n0 ;l 0 /

k;k Pl;l 0

121.

0

pSqq

;b  2

;b  2

; pSq  exp.n; h/; w; num.z/; pS q  q/

pSqq

;d  2

; pS q  exp.n0 ; h0 /; v; num.y/; pS q  q/

; exp.n; l/; w; num.z/; pS q  q/

pS qq

; exp.n0 ; l 0 /; v; num.y/; pS q  q/

115 T13.49i 115 T13.49i 99,116 DE 103,117 DE 118,119 9I

0

114,115-120 9E

122.

k;k 0 Pl;l 0

101,102-121 (9E)

123.

I.m0 ; n0 ; l 0 / _ J.w; m0 ; n0 ; l 0 / _ K.w; num.z/; m0 ; n0 ; l 0 / _ M.m0 ; n0 ; l 0 / _ N.m0 ; n0 ; l 0 /

A (g 28_E)

124.

exp.m0 ; l 0 / D pS q  exp.m; h/ 0

125.

k;k Pl;l 0

126. 127.

ŒI.m0 ; n0 ; l 0 / _ J.w; m0 ; n0 ; l 0 / _ K.w; num.z/; m0 ; n0 ; l 0 / _ M.m0 ; n0 ; l 0 / _ N.m0 ; n0 ; l 0 / ?

128.

k;k Pl;l 0

0

130. 131.

M.m; n; l/ _ N.m; n; l/

132.

k;k Pl;l 0

133.

k;k Pl;l 0

124 T13.54f 123,126 ?I 125-127 E

k;k 0 Pl;l 0 k;k 0 Pl;l 0

129.

26,99 DE A (c E)

28,100-128 _E 97,98-129 (9E)

0

A (g 27_E) similarly

0

27,29-132 _E

Exercise 13.49 T13.57.b

1003

ANSWERS FOR CHAPTER 13

134.

0

k;k len.exp.m; l//  S x ! .exp.m; l/ D exp.m0 ; l 0 / ! Pl;l 0 /

26-133 !I

135.

.8k < len.m//.8k 0 < len.m0 //Œlen.exp.m; k//  S x ! .exp.m; k/ D exp.m0 ; k 0 / ! P /

25-134 (8I)

136.

24-135 !I 23,136 IN 10 T13.21i

139.

.8k < len.m//.8k 0 < len.m0 //Œlen.exp.m; k//  x ! .exp.m; k/ D exp.m0 ; k 0 / ! P / ! .8k < len.m//.8k 0 < len.m0 //Œlen.exp.m; k//  S x ! .exp.m; k/ D exp.m0 ; k 0 / ! P / 8x.8k < len.m//.8k 0 < len.m0 //Œlen.exp.m; k//  x ! .exp.m; k/ D exp.m0 ; k 0 / ! P / : : len.m/ 1 < len.m/ ^ len.m0 / 1 < len.m0 / 0 : : : len.exp.m; len.m/ 1//  len.p/ ! .exp.m; len.m/ 1/ D exp.m0 ; len.m0 / 1/ ! P k;k :

140.

P k;k

137. 138.

141. 142. 143. 144. 145. 146. 147. 148.

0

: /

len.m/ 1;len.m0 / 1

:

0

:

len.m/ 1;len.m / 1 : : 9q9a9b9c9d ŒTsubseq.a; b; exp.n; len.m/ 1/; w; num.z/; q/ ^ Tsubseq.c; d; exp.n0 ; len.m0 / 1/; v; num.y/; q/ 9q9a9b9c9d ŒTsubseq.a; b; t; w; num.z/; q/ ^ Tsubseq.c; d; t 0 ; v; num.y/; q/ Tsubseq.a; b; t; w; num.z/; q/ ^ Tsubseq.c; d; t 0 ; v; num.y/; q/

Termsub.t; w; num.z/; q/ ^ Termsub.t 0 ; v; num.y/; q/ 9q9t9t 0 ŒTermsub.p; v; num.y/; t / ^ Termsub.p; w; num.z/; t 0 / ^ Termsub.t; w; num.z/; q/ ^ Termsub.t 0 ; v; num.y/; q/ 9q9t9t 0 ŒTermsub.p; v; num.y/; t / ^ Termsub.p; w; num.z/; t 0 / ^ Termsub.t; w; num.z/; q/ ^ Termsub.t 0 ; v; num.y/; q/ 9q9t 9t 0 ŒTermsub.p; v; num.y/; t / ^ Termsub.p; w; num.z/; t 0 / ^ Termsub.t; w; num.z/; q/ ^ Termsub.t 0 ; v; num.y/; q/ 9q9t 9t 0 ŒTermsub.p; v; num.y/; t / ^ Termsub.p; w; num.z/; t 0 / ^ Termsub.t; w; num.z/; q/ ^ Termsub.t 0 ; v; num.y/; q/

149. ŒTerm.p/ ^ v ¤ w ! 9q9t9t 0 ŒTermsub.p; v; num.y/; t / ^ Termsub.p; w; num.z/; t 0 / ^ Termsub.t; w; num.z/; q/ ^ Termsub.t 0 ; v; num.y/; q/

137,138 (8E) 139,8 !E 140 abv 141,11,12 DE A (g 1429E) 143 T13.49l 4,144 9I 142,143-145 9E 5,6-146 (9E) 3,4-147 9E 1-148 !I

T13.57.d. PA ` ŒWff .p/^v ¤ w ! formsub.formsub.p; v; num.y//; w; num.z// D formsub.formsub.p; w; num.z//; v; num.y// Let P = 9q9a9b9c9d ŒFsubseq.a; b; exp.n; k/; w; num.z/; q/^Fsubseq.c; d; exp.n0 ; k 0 /; v; num.y/; q/

Exercise 13.49 T13.57.d

1004

ANSWERS FOR CHAPTER 13

1.

Wff .p/ ^ v ¤ w

A (g !I)

2. 3. 4.

Term.num.y// ^ Term.num.z// 9qFormsub.p; v; num.y/; q/ ^ 9qFormsub.p; w; num.z/; q/ Formsub.p; v; num.y/; t / ^ Formsub.p; w; num.z/; t 0 /

T13.56r 1,2 T13.50k A (g 39E)

5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

formsub.p; v; num.y// D t ^ formsub.p; w; num.z// D t 0 Wff .t / ^ Wff .t 0 / .9x  X /.9y  Y /Fsubseq.x; y; p; v; num.y/; t / ^ .9x  X /.9y  Y /Fsubseq.x; y; p; w; num.z/; t 0 / Fsubseq.m; n; p; v; num.y/; t / ^ Fsubseq.m0 ; n0 ; p; w; num.z/; t 0 / Formseq.m; p/ ^ Formseq.m0 ; p/ : : exp.m; len.m/ 1/ D p ^ exp.m0 ; len.m0 / 1/ D p m > 1 ^ m0 > 1 len.m/ > ; ^ len.m0 / > ; : : exp.n; len.n/ 1/ D t ^ exp.n0 ; len.n0 / 1/ D t 0 len.m/ D len.n/ ^ len.m0 / D len.n0 / .8k < len.m//ŒO.v; num.y/; m; n; k/ _ P .m; n; k/ _ Q.m; n; k/ _ R.v; p; m; n; k/ _ S.v; p; m; n; k/ .8k < len.m0 //ŒO.w; num.z/; m0 ; n0 ; k/ _ P .m0 ; n0 ; k/ _ Q.m0 ; n0 ; k/ _ R.w; p; m0 ; n0 ; k/ _ S.w; p; m0 ; n0 ; k/ l < len.m/ ^ l 0 < len.m0 / len.exp.m; l//  ;

1,2,4 T13.55h 4,2 T13.50d 4 T13.50b A (g 7 (9E)) 8 T13.50a 9 T13.48a 9 T13.48a 11 T13.44j 8 T13.50a 8 T13.50a 8 T13.50a 8 T13.50a A (g (8I)) A (g !I)

0

19.

k;k Œexp.m; l/ D exp.m0 ; l 0 / ! Pl;l 0 

A (c E)

20. 21. 22.

exp.m; l/ > 1 exp.m; l/  1 ?

9,17 T13.48d 18 T13.44j 20,21 ?I

23. 24. 25. 26. 27.

k;k exp.m; l/ D exp.m0 ; l 0 / ! Pl;l 0

0

19-22 E 0

k;k len.exp.m; l//  ; ! .exp.m; l/ D exp.m0 ; l 0 / ! Pl;l 0 /

18-23 !I

.8k < len.m//.8k 0 < len.m0 //Œlen.exp.m; k//  ; ! .exp.m; k/ D exp.m0 ; k 0 / ! P / .8k < len.m//.8k 0 < len.m0 //Œlen.exp.m; k//  x ! .exp.m; k/ D exp.m0 ; k 0 / ! P / l < len.m/ ^ l 0 < len.m0 /

17-24 (8I) A (g !I) A (g (8I))

28.

len.exp.m; l/  S x ^ exp.m; l/ D exp.m0 ; l 0 /

A (g !I)

29. 30. 31.

O.v; num.y/; m; n; l/ _ P .m; n; l/ _ Q.m; n; l/ _ R.v; p; m; n; l/ _ S.v; p; m; n; l/ O.w; num.z/; m0 ; n0 ; l 0 / _ P .m0 ; n0 ; l 0 / _ Q.m0 ; n0 ; l 0 / _ R.w; p; m0 ; n0 ; l 0 / _ S.w; p; m0 ; n0 ; l 0 / O.v; num.y/; m; n; l/

15,27 (8E) 16,27 (8E) A (g 29_E)

32. 33. 34. 35. 36. 37. 38. 39.

Atomic.exp.m; l// ^ Atomsub.exp.m; l/; v; num.y/; exp.n; l// O.w; num.z/; m0 ; n0 ; l 0 /

31 abv A (g 30_E)

Atomic.exp.m0 ; l 0 // ^ Atomsub.exp.m0 ; l 0 /; w; num.z/; exp.n0 ; l 0 // Atomsub.exp.m; l/; w; num.z/; exp.n0 ; l 0 // Atomic.exp.n; l// ^ Atomic.exp.n0 ; l 0 // 9q9t 9t 0 ŒAtomsub.exp.m; l/; v; num.y/; t / ^ Atomsub.exp.m; l/; w; num.z/; t 0 / ^ Atomsub.t; w; num.z/; q/ ^ Atomsub.t 0 ; v; num.y/; q/ Atomsub.exp.m; l/; v; num.y/; u/ ^ Atomsub.exp.m; l/; w; num.z/; u0 / Atomsub.u; w; num.z/; r/ ^ Atomsub.u0 ; v; num.y/; r/

40. 41.

exp.n; l/ D u ^ exp.n0 ; l 0 / D u0 Atomsub.exp.n; l/; w; num.z/; r/ ^ Atomsub.exp.n0 ; l 0 /; v; num.y/; r/

42.

Fsubseq.2

43. 44. 45.

k;k 0 Pl;l 0 k;k 0 Pl;l 0 0

exp.n;l/

exp.n0 ;l 0 /

r

; 2 ; exp.n; l/; w; num.z/; r/ ^ Fsubseq.2

r

; 2 ; exp.n0 ; l 0 /; v; num.y/; r/

0

0

0

0

0

0

0

0

0

0

0

k;k Pl;l 0

47. 48. 49.

Atomic.exp.m0 ; l 0 // ŒP .m0 ; n0 ; l 0 / _ Q.m0 ; n0 ; l 0 / _ R.w; p; m0 ; n0 ; l 0 / _ S.w; p; m0 ; n0 ; l 0 / ?

51.

k;k Pl;l 0

32,35,38, T13.54j 39,40 DE 36,41 T13.50e 37,38-43 9E

0

P .m ; n ; l / _ Q.m ; n ; l / _ R.w; p; m ; n ; l / _ S.w; p; m ; n ; l /

k;k Pl;l 0

1,32 T13.57c A (g 379E)

42 9I

46.

50.

33 abv 34,28 DE 2,32,34 T13.49e

A (g 30_E) A (c E)

0

32,28 DE 47 T13.55b 45,48 ?I 46-49 E

0

30,33-50 _E

Exercise 13.49 T13.57.d

1005

ANSWERS FOR CHAPTER 13

52.

P .m; n; l/

A (g 29_E)

53. 54. 55.

.9i < l/Œexp.m; l/ D neg.exp.m; i // ^ exp.n; l/ D neg.exp.n; i // h