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Extra Exercises for Chapter 3 Use Tarski’s World for all exercises on this page. H 3.1 (Building a world) Open the file Sentences Cstern 0301. 1. ¬Cube(a) 6. ¬(d ≠ b 2. ¬SameShape(a,b) 7. ¬SameCol(a,e) 3. ¬RightOf(b,c) 8. ¬¬SameCol(c,e) 4. ¬¬Tet(b) 9. ¬LeftOf(a,c) 5. c ≠ d 10. ¬¬SameShape(c,d) Start a new world file and build a world where all these sentences are true. As you modify the world to make the later sentences true, make sure you have not accidentally falsified any of the earlier sentences. Save your world as World Cstern 0301. Submit both your world and your sentences to the GradeGrinder. H 3.2

(Building a world) Open the file Sentences Cstern 0302. 1. ¬Dodec(a) 6. b≠a 2. ¬Tet(a) 7. ¬(b ≠ c) 3. ¬Tet(b) 8. ¬¬SameSize(b,d) 4. ¬FrontOf(a,b) 9. ¬SameShape(d,c) 5. ¬¬SameShape(b,e) 10. ¬¬SameRow(c,d) Start a new world file and build a world where all these sentences are true. As you modify the world to make the later sentences true, make sure you have not accidentally falsified any of the earlier sentences. Save your world as World Cstern 0302. Submit both your world and your sentences to the GradeGrinder.

H 3.3

Open Sentences Cstern 0303. Show that the sentence ¬ Small(a) ∧ ¬Small(b) is not a conseqence of ¬(Small(a) ∧ Small(b)) Do this by building a counterexample world in which the second sentence is true but the first is false. Save as World Cstern 0303 and submit it to the GradeGrinder.

H 3.4

Open Sentences Cstern 0304. Show that the sentence ¬Small(a) ∧ Cube(a) is not a conseqence of ¬(Small(a) ∧ Cube(a)) Do this by building a counterexample world in which the second sentence is true but the first is false. Save as World Cstern 0304 and submit it to the GradeGrinder.

H 3.5

Open Sentences Cstern 0305. Show that the sentence (Small(a) ∧ Cube(a)) ∨ (Small(b) ∧ Cube(b)) is not a conseqence of (Small(a) ∨ Small(b)) ∧ (Cube(a) ∨ Cube(b)) Do this by building a counterexample world in which the second sentence is true but the first is false. Save as World Cstern 0305 and submit it to the GradeGrinder.

H 3.6

Open Sentences Cstern 0306. Show that the sentence ¬(Large(a) ∨ Tet(a)) is not a conseqence of ¬Large(a) ∨ ¬Tet(a) Do this by building a counterexample world in which the second sentence is true but the first is false. Save as World Cstern 0306 and submit it to the GradeGrinder. 1

Extra Exercises for Chapter 4 Use Boole to construct truth tables and to indicate whether each sentence is TT-possible and whether each is a tautology. Build your own reference columns and fill them in yourself. (Do not have Boole do these steps for you.) Open file Table Cstern 0401 for H 4.1, etc. H 4.1

(A ∧ B) ∨ (¬A ∧ ¬B)

H 4.2

(A ∧ B) ∨ (¬A ∨ ¬B)

H 4.3

¬(¬(A ∨ B) ∧ A)

H 4.4

¬((¬A ∧ ¬B) ∧ (A ∨ B))

H 4.5

¬(A ∨ B) ∧ ¬ (¬A ∧ ¬B)

H 4.6

¬(¬A ∨ ¬B) ∧ ¬(A ∧ B))

H 4.7

(A ∨ B) ∨ (¬A ∧ ¬B)

H 4.8

¬(A ∨ B) ∨ (A ∧ B)

Use Boole to construct truth tables and to indicate whether each pair of sentences is tautologically equivalent. Build your own reference columns and fill them in yourself. (Do not have Boole do these steps for you.) Open Table Cstern 0409 for exercise H 4.9, and Table Cstern 0410 for exercise H 4.10, etc. H 4.9 ¬(A ∧ B) H 4.10 ¬(A ∧ B) H 4.11 ¬(A ∨ B) ¬A ∧ ¬B ¬A ∨ ¬B ¬A ∨ ¬B Use the truth table method to determine whether the conclusion of each argument is a tautological consequence of the premises. Build your own reference columns and fill them in yourself. (Do not have Boole do these steps for you.) Open file Table Cstern 0412 for exercise H 4.12, etc. H 4.12

Tet(a) ∨ Small(a) Tet(a) ¬Small(a)

H 4.13

Tet(a) ∧ Small(a) (Tet(a) ∨ Cube(a)) ∧ (Small(a) ∨ Large(a))

H 4.14

A ∧ ¬B B∨C A∧C

H 4.15

¬ (A ∧ B) B∨C ¬(¬A ∧ C)

H 4.16

¬ (¬A ∨ B) B∨C A∧C

H 4.17

¬A ∨ B B∨C ¬(A ∧ C)

H 4.18

A ∧ (B ∨ C) ¬(A ∧ E) ¬(B ∧ C) ¬E ∧ (¬A ∨ D)

H 4.19

A ∨ (B ∧ C) ¬(C ∨ (B ∧ D)) ¬C ∧ A

H 4.20

(A ∧ B) ∨ (C ∧ D) ¬(A ∧ E) ¬(B ∧ C) ¬E ∧ (¬A ∨ D)

H 4.21

(A ∨ B) ∧ (C ∨ D) ¬(A ∧ (C ∨ E)) ¬(¬E ∧ ¬D) B ∨ (A ∧ (¬E ∧ D))

2

Extra Problems for Chapter 7 For each pair of sentences, use Boole to determine whether (a) the two sentences are tautologically equivalent, (b) the second sentence is a tautological consequence of the first, and (c) the first sentence is a tautological consequence of the second. You may have Boole build the reference columns and fill them out for you. Open file Table Cstern 0701 for the pair in H 7.1, etc., and Table Cstern 0710 for the pair in H 7.10, etc. H 7.1

AB BA

H 7.2

A↔B B↔A

H 7.3

AB ¬A  ¬B

H 7.4

A  ¬B ¬(A  B)

H 7.5

A↔B ¬A  ¬B

H 7.6

A↔B (A ∧ B) ∨ (¬A ∧ ¬B)

H 7.7

(A ∧ B)  C (A  C) ∧ (B  C)

H 7.8

(A ∧ B)  C (A  C) ∨ (B  C)

H 7.9

(A ∨ B)  C (A  C) ∧ (B  C)

H 7.10

(A ∨ B)  C (A  C) ∨ (B  C)

H 7.11

A  (B ∧ C) (A  B) ∧ (A  C)

H 7.12

A  (B ∨ C) (A  B) ∨ (A  C)

Use the truth table method (Boole) to determine whether each of the following arguments is tautologically valid -- that is, whether the conclusion is a tautological consequence of the premises. Open file Table Cstern 0713 for H 7.13, etc. H 7.13

AB A  ¬C B∨C ¬A

H 7.14

AB (A ∧ B)  C AC

H 7.15

(A ∨ B) ↔ (A ∧ B) A↔B

H 7.16

(A ∨ B)  C ¬A ↔ B C

H 7.17

A ↔ (B ∨ C) B  (A ↔ ¬C) B  ¬C

H 7.18

¬ (A  B) C  (¬B  ¬A) ¬C

H 7.19

A  (B ↔ (C ∧ D)) C  (¬B ∧ A) ¬D  ¬C

H 7.20

¬A  ¬(B ∧ C) (C ∨ D)  B ¬(¬A ∧ D)

H 7.21

(A ∧ B)  (C ∨ D) ¬(A ∧ C) A ↔ (D  B) A  (B  D)

H 7.22

A  ¬(B ∨ C) ¬D  (A ↔ B) ¬(C  D) ¬A ∧ ¬D

3

Proof Rules = Intro i

= Elim

… a=a

i k

=Intro

∧ Intro

∧ Elim

… i A … k B … A ∧ B ∧Intro: i,k

… i A∧B … A ∧Elim: i

∨ Intro … i A … A∨B

P(a) … a…= b P(b)

=Elim,: i,k

OR:

∨ Elim OR: ∨Intro, i

… i A … B ∨ A ∨Intro: i

… i A∧B … B ∧Elim,: i … A∨B … k A … m …C n B … p C C ∨Elim: i,k-m,n-p i

⊥ Intro … i A … k ¬A … ⊥ ⊥Intro: i,k

⊥ Elim … i ⊥ … C ⊥Elim: i

¬ Intro … i A … k ⊥ ¬A ¬Intro: i -k

Intro

Elim

↔Intro

… … A i AB … … k B k A AB Intro:i -k … B Elim:i,k i

i k l n

4

¬ Elim … i ¬¬A … A ¬Elim: i

↔Elim … … i A↔B … A k A … … B ↔Elim: i,k B OR: … B i A↔B … … A k B … A↔B ↔Intro:i-k,l-n A ↔Elim: i,k

PROOFS: Some very easy problems to start on Problems 0.1, 0.1, and 8.1 - 8.14 do not require subderivations. For 8.15 - 8.23, subderivations may be needed. Use Fitch to open Proof CStern 080x for H 8.1 - 8.9, and Proof CStern 08xx for H 8.10 -8.24. Submit assigned solutions to the GradeGrinder. Do not use TautCon. Apply AnaCon only to literals (atomic sentences and their negations). H 0.1

H 0.2 H 8.1 H 8.2 H 8.3

The starter must be malfunctioning. The car won't start, but the lights are working. If the lights work, the battery must be charged. If the car won't start when the battery is charged, there must be a problem with the starter. (S = The starter is functioning properly; C = The car starts; L = The lights are working; B = The battery is charged) If Mia moved out of her parents' house, she must be able to afford to. She can afford it only if she found both a roommate and a job. Since she moved out, she must have found a job. {Dodec(a), Dodec(a) → Medium(a)} /∴ Dodec(a) ∧ Medium(a) {Tet(a), Tet(a) → Tet(b), (Tet(a) ∧ Tet(b)) → Larger(a,b)}/∴ Tet(b) ∧ ¬SameSize(a,b) {SameCol(b,c) → (Dodec(b) ∧ Cube(c)), SameCol(b,c), Cube(c) → c=d} /∴ Cube(d)

H 8.6

Small(a) Cube(a) → (Large(c) → Cube(c)) (Small(a) ∧ Medium(b)) → Large(c)) Cube(a) ∧ Medium(b)) SameShape(a,c)

H 8.7

(Tet(b) ∨ Dodec(b)) → Small(b) Tet(b) ↔ Medium(c) Medium(c) Larger(c,b)

H 8.8 H 8.9 H 8.10 H 8.11 H 8.12 H 8.13 H 8.14 H 8.15.

{ (K ∧ L) ↔ (B ∧ C), C ∧ L, B} /∴ ((B ∧ C) ∧ K) ∧ L { [(A ∧ B) ∧ C] → D, A → (R → C), A ∧ R, (C ∧ A) → B} /∴ D { (J ∨ S) ↔ L, S } /∴ L { A, (C ∨ A) → (T ∧ M) } /∴ T ∧ M { B ∧ E, (E ∨ (F ∧ C))  K } /∴ M ∨ K { (Q ∧ M) ↔ (P ∨ R), P } /∴ M {(G ∨ B)  L, (S ∧ L)  H, S ∧ B} /∴ L ∧ H {P, ¬(P ∧ Q)} /∴ ¬Q H 8.17 ¬Small(b) H 8.18 Tet(a) ∨ Tet(b) H 8.16 Cube(a)  Tet(b) b=c  Small(b) ¬(Tet (a) ∧ Tet(b)) Tet(b)  Dodec(c) Small(b) ↔ b=c ¬Tet (a) ↔ Tet (b) Cube(a)  Dodec(c)

H 8.19

H 8.21 H 8.22 H 8.23 H 8.24

RightOf(a,b) ∨ SameRow(a,b) RightOf(a,b)  (BackOf(a,b) ∧ Larger(a,b)) SameRow(a,b) ↔ Larger(a,b) Larger(a,b)

H 8.20

Cube(a) ∨ Cube(b) ∨ Cube(c) Cube(a)  Cube(b) Cube(c) ↔ Cube(b) Cube(c)

{ (J ∧ K)  L, J ∧ M } /∴ ¬L  ¬K { A  B, (A ∧ B)  C, (C ∨ D)  A } /∴ A ↔ C { L  (F ∨ G), F  (J ↔ K), K ∧ L, (G ∧ L)  M } /∴ J ∨ M { (P ∧ Q) ↔ (¬R ∨ ¬S) } /∴ (P ∧ S)  (Q  ¬R) 5

SAMPLE ANSWERS TO SELECTED PRACTICE PROBLEMS H 0.1 {¬C ∧ L, L → B, (¬C ∧ B) → ¬S} /∴ ¬S 1 ¬C ∧ L P 2 L→B P 3 (¬C ∧ B) → ¬S P 4 ¬C ∧Elim: 1 5 L ∧Elim: 1 6 B →Elim: 2,5 7 ¬C ∧ B ∧Intro: 4,6 8 ¬S →Elim: 7,3

H 8.21 1 (J ∧ K) → L P 2 J∧M P 3 ¬L 4 K 5 J ∧Elim: 2 6 J∧K ∧Intro:4,5 7 L →Elim: 6,1 8 ⊥ ⊥Intro: 3,7 9 ¬K ¬Intro: 4-(7,8) 10 ¬L → ¬K →Intro: 3-9

2. { M → A, A → (R ∧ J), M } /∴ J (The second premise could be symbolized instead as ¬ (R ∧ J) → ¬A The derivation would then require a subderivation.) 1 M→A P 2 A → (R ∧ J) P 3 M P 4 A → Elim: 1,3 5 R∧J →Elim: 2,4 6 J ∧Elim: 5

H 8.22 1 A→B 2 (A ∧ B) ↔ C 3 (C ∨ D) → A 4 A 5 B 6 A∧B 7 C

P P P →Elim: 1,4 ∧Intro: 4,5 ↔Elim: 2,6

8 C 9 C∨D ∨Intro: 8 10 A →Elim: 3.9 11 A ↔ C ↔Intro: 4-7,8-10

H 8.11 1 A P 2 (C ∨ A) → (T ∧ M) P 3 C∨A ∨Intro: 1 4 T∧M →Elim: 2,3

H 8.23 1 2 3 4 5 6 7 8 9 10 11

H 8.12 1 B∧E P 2 [E ∨ (F ∧ C)] → K P 3 E ∧Elim: 1 4 E ∨ (F ∧ C) ∨Intro: 3 5 K →Elim: 2,4 6 M∨K ∨Intro: 5 H 8.13 1 (Q ∧ M) ↔ (P ∨ R) P 2 P P 3 P∨R ∨Intro: 2 4 Q∧M ↔Elim: 1,3 5 M ∧Elim: 4

L → (F ∨ G) F → (J ↔ K) K∧L (G ∧ L) → M L F∨G F J↔K K J J∨M

P P P P ∧Elim: 3 →Elim: 1,5 →Elim: 2,7 ∧Elim: 3 ↔Elim: 8,9 ∨Intro: 10

12 G 13 G∧L ∧Intro: 12,5 14 M →Elim: 4,13 15 J∨M ∨Intro: 14 16 J ∨ M ∨Elim: 6,7-11,12-15

STRATEGIES FOR PROOFS 1.

Try to pull goal sentence out of a sentence you already have that contains the goal sentence as a component. For example, suppose your goal is '(J ∧ K)' and one of your earlier sentences (either a premise or something you have already derived) is '((J ∧ K) ↔ L)'. In that case, aim for 'L', and then use it with '((J ∧ K) ↔ L)' to get '(J ∧ K)' by ↔Elim.

2.

If goal sentence cannot be extracted as a whole from any sentence you already have, base your strategy on the structure of the goal sentence. ∧ conjunction Aim for each conjunct separately, then apply ∧Intro. → conditional Plan to use →Intro. To do this, start a subderivation with the antecedent as provisional assumption. Aim for the consequent in the subderivation. ↔ biconditional Plan to use ↔Intro. Start one subderivation with the left side of the biconditional and aim for the right in this subderivation. Set up a second subderivation going from the right side of the biconditional to the left. ¬ negation If goal has ¬ as its main operator, try reaching it by ¬Intro. To do this, start a subderivation with the statement to be negated (but without the ¬) as provisional assumption. Within this subderivation, get ⊥ by aiming for any contradiction you can get. ∨ disjunction a) If one disjunct is obviously easy to get, get that one. Then use ∨Intro to reach goal. b) If neither disjunct is obviously easy to get, look for an earlier disjunction. Try ∨Elim on earlier disjunction c) If neither of these works, assume the negation of the goal sentence. You will need to use ⊥Intro followed by ¬Intro and then ¬¬Elim to reach the goal.

3.

If you have a disjunction already, and you can't use it with one of our easy rules, you will probably have to use the ∨Elim rule. If you will have to use ∨Elim, set up for it early. (A disjunction can be used with an easy rule if that disjunction forms the antecedent of a conditional or forms one side of a biconditional that you have on another line.

4.

When aiming for ⊥, look for a negation you already have to use as one member of the contradictory pair of sentences.

5.

If you have no idea what to do, you try applying any easy rules you can. Perhaps the results of this process will give you some ideas for other things to do.

6.

When all else fails, assume the opposite of what you want, and aim for a contradiction.

Example using strategies 1 & 2: {(A Ÿ B) Æ (C ´ D), (A Ÿ B) ´ (F ⁄ G), G Ÿ H} /\ C ´ D

Our first strategy hint tells us to try to extract our goal sentence from a more complex sentence in which it occurs as a whole. Here we can get the goal sentence from step 1 by ‡Elim if we have ‘A Ÿ B’, so we aim for that.

1 (A Ÿ B) Æ (C ´ D) P 2 (A Ÿ B) ´ (F ⁄ G) P 3 GŸH P C´D

The same hint tells us to look for a complex sentence in which ‘A Ÿ B’ appears, from which we can pull out this sentence as a whole. We could get this from step 2 if we had the other side of the biconditional, ‘F ⁄ G’, as a whole, so now we aim for that.

1 (A Ÿ B) Æ (C ´ D) P 2 (A Ÿ B) ´ (F ⁄ G) P 3 GŸH P AŸB C´D

1 (A Ÿ B) Æ (C ´ D) 2 (A Ÿ B) ´ (F ⁄ G) 3 GŸH F⁄G AŸB C´D

Æ Elim: 1,?

P P P

Élim: 2,? ÆElim: 1,?

Our new goal, ‘F ⁄ G’, does not appear in its entirety as a part of any more complex sentence except in step 2. To get it from there, we would have to have ‘A Ÿ B’ already, but the whole point of getting ‘F ⁄ G’ is that we don’t have ‘A Ÿ B’ yet, and we need ‘F ⁄ G’ before ‘A Ÿ B’ to help us get ‘A Ÿ B’. So we won’t be able to get ‘F ⁄ G’ by pulling it out of line 2. Our second strategy hint suggests building up our goal sentence. The main operator in ‘F ⁄ G’ is the ‘v’, so we ask whether one disjunct is obviously very easy to get from what we already have above. If so we, we will do that, then use ⁄ Intro to get the disjunction we want. It’s easy to get ‘G’ from step 3, so we follow this strategy.

All that remains to be done is numbering the rest of the lines, and using these line numbers to complete our justifications.

1 2 3 4 5 6 7

(A Ÿ B) Æ(C ´ D) P (A Ÿ B) ´ (F ⁄ G) P GŸH P G Ÿ Elim: 3 F⁄G ⁄ Intro: 4 AŸB Élim: 2,5 C´D ÆElim: 1,6 8

1 (A Ÿ B) Æ (C ´ D) P 2 (A Ÿ B) ´ (F ⁄ G) P 3 GŸH P G ŸElim: 3 F⁄G ⁄ Intro AŸB Élim: 2,? C´D ÆElim: 1,?

1 C´D 2 C Ÿ ¬(D Ÿ B)

P P

STRATEGY: {C ´ D, C Ÿ ¬(D Ÿ B)} /\ ¬B

¬B 1 C´D 2 C Ÿ ¬(D Ÿ B) B

P P

? ¬? ^ ¬B

^ Intro ¬ Intro

Notice that we don’t have ‘¬B’ in the premises, so we can’t expect to get it directly using rules like Ÿ Elim and Élim. Thus we assume ‘B’ and try to show that it leads to a contradiction. We don’t yet know what the contradiction will be, but we know we need some sentence and its negation.

1 C´D 2 C Ÿ ¬(D Ÿ B) 3 B

We notice that we can get a negation, ‘¬(D Ÿ B)’, easily from 2 by ŸElim., so we pick that as the negation to aim for. Then we try to get ‘D Ÿ B’ as the other member of our contradictory pair of sentences. fi When aiming for a conjunction, try to get each conjunct separately. We already have ‘B’ on line 3, so we just need ‘D’.

Notice that we can get ‘C’ from 2 by Elim. With 1, that will let us get ‘D’ by using the ´ Elim rule.

1 C´D 2 C Ÿ ¬(D Ÿ B) 3 B D DŸB ¬(D B) ^ ¬B

DŸB ¬(D Ÿ B) ^ ¬B

P P

ŸElim: 2 ^ Intro ¬ Intro

P P

Ÿ Intro Ÿ Elim: 2 ^ Intro ¬ Intro

1 C´D P 2 C Ÿ ¬(D Ÿ B) P 3 B C Ÿ Elim: 2 D ´ Elim DŸB Ÿ Intro ¬(D Ÿ B) Ÿ Elim: 2 ^ ^ Intro ¬B ¬ Intro

No other steps are needed. We fill in the line numbers and complete the justifications. fi

9

1 C´D 2 C Ÿ ¬(D Ÿ B) 3 B 4 C 5 D 6 DŸB 7 ¬ (D Ÿ B) 8 ^ 9 ¬B

P P Ÿ Elim: 2 Élim: 1,4 Ÿ Intro: 3,5 Ÿ Elim: 2 ^ Intro: 6,7 ¬ Intro: 3-8

STRATEGY: {¬(Tet(a) Ÿ Small(a)), RightOf(b,c) Æ Small(b), LeftOf(c,b), a=b} /\ ¬Tet(a) ‹ 1 2 3 4 5

¬(Tet(a) Ÿ Small(a)) RightOf(b,c) Æ Small(b) LeftOf(c,b) a=b Tet(a)

The meaning relationship fl between ‘RightOf’ and ‘LeftOf’, is programmed into Fitch, so it lets us derive ‘RightOf(b,c)’ from ‘LeftOf(c,b). (Without that, we need a premise, not stated here, specifying that each implies the other. See variant on next page.)

Since goal is a negation, assume its opposite, planning for ¬Intro.

^ ¬Tet(a) fl

fl 1 ¬(Tet(a) Ÿ Small(a)) 2 RightOf(b,c) Æ Small(b) 3 LeftOf(c,b) 4 a=b 5 Tet(a)

The only negation we have is on line 1, so we aim for ‘Tet(a) Ÿ Small(a)’ to contradict line 1.

fi 1 2 3 4 5 6 7

Tet(a) Ÿ Small(a) ^ ¬Tet(a) 1 2 3 4 5

Small(a) Tet(a) Ÿ Small(a) ^

¬(Tet(a) Ÿ Small(a)) RightOf(b,c) Æ Small(b) LeftOf(c,b) a=b Tet(a)

¬(Tet(a) Ÿ Small(a)) RightOf(b,c) Æ Small(b) LeftOf(c,b) a=b Tet(a) RightOf(b,c) Small(b)

ÆElim

Small(a) Tet(a) Ÿ Small(a) ŸIntro ^ Întro ¬Tet(a) ¬Intro

ŸIntro Întro ¬Tet(a)

‹ We already have the first conjunct, but still need the second, ‘Small(a)’.

Small(a) Tet(a) Ÿ Small(a) ŸIntro ^ Întro ¬Tet(a) 1 2 3 4 5

¬(Tet(a) Ÿ Small(a)) P RightOf(b,c) Æ Small(b) P LeftOf(c,b) P a=b P Tet(a) RightOf(b,c) AnaCon:3 Small(b) ÆElim

fl fl fl

‹

Given line 4, if we could get ‘Small(b)’ out of line 2, we should eventually be able to get ‘Small(a)’. Aim for the antecedent of line 2 to get ‘Small(b)’ 10

1 2 3 4 5 6 7 8 9 10 11 12 13

If we apply=Elim strictly, allowing term to the right of ‘=’ to replace that on left but not vice versa, we need ‘b = a’ to be able to derive ‘Small(a)’ from ‘Small(b)’. (Fitch is not as strict, letting us omit this reversal. See variant on next page.) ¬(Tet(a) Ÿ Small(a)) P RightOf(b,c) Æ Small(b) P LeftOf(c,b) P a=b P Tet(a) RightOf(b,c) AnaCon:3 Small(b) ÆElim: 2,6 a=a =Intro b=a =Elim: 4,8 Small(a) =Elim: 7,9 Tet(a) Ÿ Small(a) ŸIntro:5,10 ^ Întro:1,11 ¬Tet(a) ¬Intro: 5-12

The derivation on the previous page uses AnaCon, and applies =Elim strictly: 1 2 3 4 5 6 7 8 9 10 11 12 13

¬(Tet(a) Ÿ Small(a)) P RightOf(b,c) Æ Small(b) P LeftOf(c,b) P a=b P Tet(a) RightOf(b,c) AnaCon:3 Small(b) ÆElim: 2,6 a=a =Intro b=a =Elim: 4,8 Small(a) =Elim: 7,9 Tet(a) Ÿ Small(a) ŸIntro: 5,10 ^ Întro:1,11 ¬Tet(a) ¬Intro: 5-12

Remember that AnaCon is not really a derivation rule in F, but a shortcut permitted by the software, Fitch. In most systems of logic, the set of rules does not have any rules for predicates, except the very special ‘=’, representing (numerical) identity. Like any standard set of rules of logic, F does not include AnaCon, since each related pair of predicates in the language would require its own rule. The system of rules would then be infinite, making it impractical. To compensate for lack of AnaCon in a standard set of rules, in any argument where we use AnaCon, we make explicit an originally unstated premise specifying the relevant relationship between the specific predicates used in the argument (as in step 5 below). It is common for a system of rules of logic to state their versions of the =Elim to allow the replacement of either term in an identity sentence for the other. Fitch, not being as strict as F, lets us do this, too. Taking these two differences into account, a more standard variation on the derivation above would look like this: 1 2 3 4 5 6 7 8 9 10 11 12 13

¬(Tet(a) Ÿ Small(a)) P RightOf(b,c) Æ Small(b) P LeftOf(c,b) P a=b P RightOf(b,c) ´ LeftOf(c,b) P Tet(a) Tet(b) =Elim: 4,6 RightOf(b,c) Élim: 3,5 Small(b) ÆElim: 2,8 Tet(b) Ÿ Small(b) ŸIntro: 7,9 ¬(Tet(b) Ÿ Small(b) =Elim: 1,10 ^ Întro: 10,11 ¬Tet(a) ¬Intro: 6-12

11

1 J 2 ¬H Æ (J Æ K) 3 (K Ÿ J) ´ ¬H

P P P

STRATEGY: {J, ¬H Æ (J Æ K), (K Ÿ J) ´ ¬H} /\ K ´ ¬H

K ´ ¬H 1 J To get a biconditional (a sentence with ‘´’ as its main 2 ¬H Æ (J Æ K) operator), use the ´ Intro rule. This requires 2 subproofs, 3 (K Ÿ J) ´ ¬H one starting from the sentence to the left of the ‘´’ and 4 K going to the right, and the other in the opposite direction. Our first step is to set up these subproofs. ¬H

P P P

¬H 1 J 2 ¬H Æ (J Æ K) 3 (K Ÿ J) ´ ¬H 4 K KŸJ ¬H

P P P

Ÿ Intro: 1,3 ´ Elim

¬H K K ´ ¬H

´ Intro

We now try to get from ‘K’ to ‘¬H’. We could use line 3 to get ‘¬H’ if we only had ‘K Ÿ J’, so we aim for that. We just need the two conjuncts, which we have already on lines 1 and 4.

Having completed the first subproof, we try to fill in the second. To get ‘K’, 2 looks like the most promising premise, although it will not give us ‘K’ directly. With line 7, though, line 2 will let us get ‘J Æ K’. If we can also get ‘J’, we will be able to get ‘K’. 1 J P 2 ¬H Æ (J Æ K) P 3 (K Ÿ J) ´ ¬H P 4 K 5 KŸJ Ÿ Intro: 1,4 6 ¬H ´ Elim: 3,5 7 ¬H 8 JÆK 9 K 10 K´ ¬H

K K´ ¬H

Íntro

1 J P 2 ¬H Æ (J Æ K) P 3 (K Ÿ J) ´ ¬H P 4 K 5 KŸJ Ÿ Intro: 1,4 6 ¬H Élim: 3,5 7 8

¬H JÆK

ÆElim: 2,7

K K ´ ¬H

Íntro

Looking back at the premises, we see that we already have ‘J’. So we just need to fill in the justifications to complete this proof.

Æ Elim: 2,7 Æ Elim: 1,8 ´ Intro: 4-6,7-9

12

1 2 3

C⁄L LÆM (C Ÿ J) Æ M

P P P

STRATEGY for: { C ⁄ L, L Æ M, (C Ÿ J) Æ M}/\ J Æ M

J‡M

When aiming for a conditional, set up for Æ Intro by assuming the antecedent and aiming for the consequent.

1 2 3 4

C⁄L LÆM (C Ÿ J) Æ M J M JÆM

P P P

Æ Intro

Line 1 says ‘C’ is true or ‘L’ is. We will now show that ‘M’ follows either way. We first assume ‘C’ and show that it leads to M. Then we give up ‘C’, assume ‘L’ instead, and show that ‘L’ leads to ‘M’. fi fi fi 1 2 3 4 5

C⁄L LÆM (C Ÿ J) Æ M J C

P P P

M L M M JÆM

P P P

⁄ Elim ÆIntro

‹

M L M M JÆM

fl 1 C⁄L 2 LÆM 3 (C Ÿ J) Æ M 4 J 5 C

ÆElim ⁄ Elim ÆIntro

We notice that ‘L’ leads directly to ‘M’, but ‘C’ does not. This shows us how to justify ‘M’ in the subproof that begins with L.

Now we need to get from ‘C’ to ‘M’. We see that we could do that by ÆElim with the help of line 3, if only we could get the conjunction ‘C Ÿ J’. And to get that, all we need to do is join 4 and 5 by Ÿ Intro. Once we’ve done this, we just need to complete the justifications to finish the proof. fi

13

1 2 3 4 5 6 7

C⁄L P LÆM P (C Ÿ J) Æ M P J C CŸJ Ÿ Intro: 4,5 M Æ Elim: 3,6

8 L 9 M 10 M 11 J Æ M

Æ Elim: 2,8 ⁄ Elim: 1,5-7,8-9 Æ Intro: 4-10

1 G 2 ¬F Æ (G ´ H)

P P

Example using strategy 2.c: {G, ¬F Æ (G ´ H)} /\ F ⁄ H

F⁄H

When aiming for a disjunction: A. If one disjunct is easy to get, do that, then use ⁄I. B. If neither disjunct is easy to get, see if you have or can easily get another disjunction. If so, maybe each disjunct in that one will lead to one disjunct in the goal disjunction. So apply ⁄Elim to the one you have, using ⁄Intro in each subproof to get the disjunction you want. C. If neither (A) nor (B) works, assume the opposite of your goal. Then use ¬I or ¬E to get the disjunction you want. Assume one disjunct as PA. From this, get the disjunction, contradicting the PA. Apply ¬Intro. In this case, ‘F’ was chosen as PA because its negation can be used with 2 to apply the ÆElim rule. 1 G P 2 ¬F Æ (G ´ H) P 3 ¬(F ⁄ H) 4 F 5 F ⁄ H ⁄Intro: 4 6 ^ ^ Intro: 3,5 7 ¬F ¬Intro: 4-6 H F⁄H ^ F⁄H

We see that by applying ÆElim to 2 and 6 we get ‘G ´ H’. Using this and line 1, we reach our goal of ‘H’ by Élim.

‹ Next try to get the OTHER disjunct in our target disjunction. Then we can use ⁄I to get that disjunction, again contradicting line 2.

⁄ Intro

1 G P 2 ¬F Æ (G ´ H) P 3 ¬(F ⁄ H) F⁄H

¬E

1 G P 2 ¬F Æ (G ´ H) P 3 ¬(F ⁄ H) 4 F 5 F⁄H ⁄ Intro: 4 6 ^ ^ Intro: 3,5 7 ¬F ¬Intro: 4-6 F⁄H ^ F⁄H

⁄ Intro ^ Intro: 3,? ¬ Elim

Stages 2 and 3 of the process above use a common GENERAL STRATEGY:

¬ Elim 1 G P 2 ¬F Æ (G ´ H) P 3 ¬(F ⁄ H) 4 F 5 F⁄H ⁄Intro: 4 6 ^ ^ Intro: 3,5 7 ¬F ¬Intro: 4-6 8 G´H ÆElim: 2,7 9 H Élim: 1,8 10 F⁄H ⁄ Intro: 9 11 ^ ^ Intro: 3,10 12 F ⁄ H ¬Elim: 3-11 14

¬(P ⁄ Q) P P⁄Q ^ ¬P Q P⁄Q ^ P⁄Q

⁄ Intro ^ Intro ¬ Intro

⁄ Intro ^ Intro ¬ Elim

EXAMPLES USING THE ⁄ELIM RULE 1 A v P 2 (A ‡ S) Ÿ (S ‡ H)

P P

3 H ‡ ¬W 4 (P ‡ ¬E) Ÿ (¬E ‡ ¬W)

P P

5

6 7 8 9 10

Teresa has either appendicitis or a severe case of food poisoning. If she has appendicitis, she'll need surgery, which would keep her in the hospital tomorrow. Of course, she can't be at work tomorrow if she's in the hospital. On the other hand, if she has severe food poisoning, she won't be up to eating anything for at least another day, in which case she won't be able to come to work tomorrow. Consider the possibility that Teresa has appendicitis. (Line 1 tells us she has appendicitis or food poisoning (or both) but doesn't say which. We are temporarily assuming she has appendicitis.) If she has appendicitis, she'll need surgery. So on this assumption, she'll need surgery.. If she needs surgery, she'll still be in the hospital tomorrow. So on this assumption, she'll be in the hospital tomorrow. So on this assumption, she will not be at work tomorrow.

A

A‡S S S‡ H H ¬W

ŸElim: 2 ‡ Elim: 5,6 Ÿ Elim: 2 ‡ Elim: 7,8 ‡ Elim: 3,9

11 P 12 P ‡ ¬E Ÿ Elim: 4 13 ¬E ‡ ¬W Ÿ Elim: 4 14 ¬E ‡Elim 11,12 15 ¬W ‡Elim: 13,14 16 ¬W v Elim:1,5-10,11-15

1 2 3 4 5 6 7 8 9 10 11 12 13 14

L(n) v P(n) L(n) ‡ S(n) S(n) ‡ ¬V(n) C(n) ´ P(n) ¬(C(n) Ÿ V(n)) L(n) S(n) ¬V(n)

On the other hand, suppose she has food poisoning. If she has food poisoning, she won't eat for at least another day. If she can't eat for another day, she can't work tomorrow. So ON THIS ASSUMPTION, she won't eat for another day. So ON THIS ASSUMPTION, she won't work tomorrow. So Teresa will not work tomorrow. Our conclusion doesn't depend on the specific diagnosis because we would reach the same conclusion regardless of which of the two possible problems Teresa has. We were told she has (at least) one of two problems, and we saw that either one would keep her from being at work. So we can be sure she won't be at work without figuring out which of these two things is wrong with her. P P P P P

‡Elim: 2 ‡Elim: 3,7

P(n) C(n) ´ Elim: 4,9 V(n) C(n) Ÿ V(n) ŸIntro:10,11 ^ ^ Intro: 12,5 ¬V(n) 11-13,¬ Intro

15 ¬V(n)

v Elim:1,6-8,9-14

Nancy will lose her job or get a promotion. If she loses her job, she'll limit her spending to necessities. If she's only spending on necessities, she won't take a vacation. She'll buy a new car if, but only if, she gets a promotion. She can't afford both a car and a vacation (even with a promotion). First CONSIDER THE POSSIBILITY THAT SHE LOSES HER JOB. IN THAT CASE, she will limit her spend to necessities. IN THAT CASE, she won't take a vacation. On the other hand, SUPPOSE SHE GETS THE PROMOTION. IN THAT CASE, she'll buy a new car. Suppose also that she goes on vacation. THEN she would be both buying a car and taking a vacation, contradicting the premise that she CAN'T do BOTH of those. So, on the assumption that she gets the promotion, she would not take a vacation. So Nancy will not be taking a vacation. We don’t need to find out whether Nancy is losing her job or getting a promotion to be sure that she won’t take a vacation. Step 1 says at least one of those two things will happen. We showed that either way, she wouldn't take a vacation. So she definitely won't take a vacation.

15

Example using strategy 3: {J Æ (K ⁄ L), M ´ (J K), L Æ (N Ÿ P), J Ÿ ¬S} /\ N ⁄ M

1 2 3 4

J Æ (K ⁄ L) M ´ (J Ÿ K) L Æ (N Ÿ P) J Ÿ ¬S

P P P P

N⁄M

1 2 3 4 5 6 7

J Æ (K ⁄ L) P M ´ (J Ÿ K) P L Æ (N Ÿ P) P J ¬S P J ŸElim: 4 K ⁄ L ÆElim: 1,5 K N⁄M

Our goal sentence is not in any of our premises. Neither of its disjuncts is obviously the easy one to get. We do not have a disjunction, but we can get one easily. Our strategy hints tell us to get that disjunction, then set up the subderivations we would need to apply ⁄ Elim to that disjunction.

1 2 3 4 5 6 7 8 9 10

J Æ (K ⁄ L) M ´ (J Ÿ K) L Æ(N Ÿ P) J Ÿ ¬S J K⁄L K JŸK M N⁄M

J Æ (K ⁄ L) P M ´ (J Ÿ K) P L Æ (N Ÿ P) P J Ÿ ¬S P J ŸElim: 4 K⁄L ÆElim: 1,5 K L N⁄M

COMMON STRATEGY: Often when we have one disjunction and are aiming for another, we get the disjunction we are aiming for inside each subderivation. Then we pull it out by ⁄Elim.

L N⁄M N⁄M

1 2 3 4 5 6 7

⁄ Elim

P P P P ŸElim: 4 ÆElim: 1,5 ŸIntro: 5,7 Élim: 2,8 ⁄ Intro: 9

11 L 12 N Ÿ P ÆElim: 3,11 13 N Ÿ Elim: 12 14 N ⁄ M ⁄ Intro: 13 15 N ⁄ M ⁄Elim:6,7-10,11-14

In the first subderivation, we see if we can get one disjunct of the desired disjunction easily. Here it’s easy to get ‘M’. So we do that, and then apply⁄Intro to get the disjunction.

The second subderivation uses the same strategy in the first. In this case it is easy to get ‘N’. So we get ‘N’, then apply ⁄Intro to get ‘N ⁄ M’ inside the subderivation. Once we have ‘N ⁄ M’ in both subderivations, we can justify ‘N ⁄ M’ outside the subderivations by ⁄Elim. 16

1 2 3 4 5 6 7 8 9 10

⁄Elim

J Æ (K ⁄ L) P M ´ (J Ÿ K) P L Æ (N Ÿ P) P J Ÿ ¬S P J ŸElim: 4 K⁄L ÆElim: 1,5 K JŸK ŸIntro: 5,7 M Élim: 2,8 N⁄M ⁄ Intro: 9 L N⁄M N⁄M

⁄Elim

COMMON STRATEGY to get one disjunction from another: Use ⁄Elim on 1st A⁄B disjunction to get the second. Inside the 2 A subderivations, get the 2 disjuncts P in the P ⁄ Q ⁄Intro goal disjunction, B then reach goal by D ⁄Intro P ⁄ Q ⁄Intro in each. P⁄Q ⁄Elim Finally, apply ⁄Elim.

1 A Ÿ (B ⁄ C) 2 B Æ (¬D ´ A) 3 (A Ÿ C) Æ E

P P P

STRATEGY: {A Ÿ (B ⁄ C), B Æ (¬D ´ A), (A Ÿ C) Æ E} /\ ¬D ⁄ E

¬D ⁄ E When aiming for a disjunction, we first see if one disjunct is obviously very simple to get. If so, get it, then use ⁄ Intro. Since neither disjunct is clearly easy, we look for a different disjunction we have or can easily get. Here, it is easy to get ‘B ⁄ C’ from line 1. 1 2 3 4

1 2 3 4

A Ÿ (B ⁄ C) B Æ (¬D ´ A) (A Ÿ C) Æ E B⁄C ¬D ⁄ E

A Ÿ (B ⁄ C) P We set up subproofs to apply B Æ (¬D ´ A) P (A Ÿ C) Æ E P ⁄Elim to 'B ⁄ C', to reach the disjunction we B⁄C Elim: 1 want. We aim for our goal disjunction within B each subproof, then use the ⁄Elim rule. ¬D ⁄ E

1 2 3 4

C ¬D ⁄ E ¬D ⁄ E

⁄ Elim

A Ÿ (B ⁄ C) P B Æ (¬D ´ A) P (A Ÿ C) Æ E P B⁄C ŸElim: 1 B ¬D ¬D ⁄ E

In each subproof, we try to get one disjunct, then use ⁄Intro to build the disjunction we want. Since line 2 connects ‘B’ and ‘¬D’, but no sentence connects ‘B’ and ‘E’, we aim for ‘¬D’.

10

A Ÿ (B v C) P B Æ (¬D ´ A) P (A Ÿ C) Æ E P B⁄C Ÿ Elim: 1 A Ÿ Elim: 1 B ¬DÁ ÆElim: 2,6 ¬D Élim: 6,7 ¬D ⁄ E ⁄Intro: 8 C ¬D ⁄ E ¬D ⁄ E

⁄ Elim ⁄ Elim

⁄ Intro

C ¬D ⁄ E ¬D ⁄ E

1 2 3 4 5 6 7 8 9

P P P

⁄ Elim

We saw we’ll use line 2. Now we realize we also need ‘A’. We can get it by Ÿ Elim. We put ‘A’ before the first subproof because we see we'll use it in the second subproof, too. (If we put it inside the subproof, we'll need it in both of them.)

fi fi fi fi fi fi fi fi fi fi To finish, we get ‘¬D ⁄ E’ by ⁄Intro through the other disjunct, ‘E’, in the second subproof. 17

1 2 3 4 5 6 7 8 9

A Ÿ (B ⁄ C) P B Æ (¬D ´ A) P (A Ÿ C) Æ E P B⁄C Ÿ Elim: 1 A Ÿ Elim: 1 B ¬D ´ A ÆElim: 2,6 ¬D Élim: 6,7 ¬D ⁄ E ⁄Intro: 8

10 C 11 A Ÿ C ŸIntro: 5,10 12 E ÆElim: 3,11 13 ¬D ⁄ E ⁄ Intro: 12 14 ¬D ⁄ E ⁄Elim:4,6-9,10-13

To show a statement is truth-functionally true, start proof with no premises and end with the statement:

STRATEGY to show that (F ∨ G) → (¬F → G) is a tautology (is truth-functionally true).

(F ∨ G) → (¬F → G) Same strategy as in step 2:

We are aiming for a conditional, so we set up a subproof to get it by → Intro. 1 2 (F ∨ G) ¬F → G (F ∨ G) → (¬F → G)

1 2 3

(F ∨ G) ¬F G ¬F → G (F ∨ G) → (¬F → G)

Intro

1 2 3 4

Now what we are aiming for is a single letter, so we look to see how we can get it from earlier steps. A. Can we get it by an easy rule? (∧ Elim, → Elim, ↔ Elim) B. Can we get it from any other complex sentence containing ‘G’, or if there are non with that, with ‘¬G’? Here we find a ‘G’ in line 1, which is a disjunction. The only thing we can do with that is use ∨Elim, so we set that up.

F∨G ¬F

F

G G ¬F → G (F ∨ G) → (¬F → G)

1 2 3 4 5 6

We see that it is easy to get from ‘G’ to ‘G’, so the only work left is to get from ‘F’ to ‘G’. To get ‘G’ under ‘F’, we notice that we already have '¬F' and 'F' available in this subproof. So we can use ⊥ Intro and then ⊥ Elim ‘G’.

→Intro

∨Elim →Intro →Intro

F∨G ¬F

F ⊥ G

⊥ Intro: 3,4 ⊥ Elim: 5

7 G 8 G ∨ Elim: 2,3-6,7-7 9 ¬F → G →Intro: 3-8 10 (F v G) → (¬F  G) →Intro: 2-9

18

| :. | ¬(A ∨ B) ↔ (¬A ∧ ¬B)

To show 2 sentences are tautologically equivalent: From null (empty) set of premises, derive biconditional.

To get (¬A ∧ ¬B), aim for each conjunct.

Start 2 subproofs to reach goal by ↔I:

¬(A ∨ B)

¬(A ∨ B)

¬A

¬A ∧ ¬B

¬B ¬A ∧ ¬B ¬A ∧ ¬B

¬A ∧ ¬B

¬(A v B) ¬(A ∨ B) ↔ (¬A ∧ ¬B)

¬(A ∨ B) ¬(A ∨ B) ↔ (¬A ∧ ¬B)

STRATEGY to show ¬(A ∨ B) and (¬A ∧ ¬B) are tautologically equivalent. 1 2 3 4 5 6 7 8 9 10 11

¬(A ∨ B) A A∨B ∨ Intro: 2 ⊥ ⊥ Intro: 1,3 ¬A ¬Intro: 2-4 B A∨B ∨ Intro: 7 ⊥ ⊥ Intro: 1,8 ¬B ¬Intro: 7-9 ¬A ∧ ¬B ∧Intro: 6,10

12 ¬A ∧ ¬B .. .. To get ¬A, assume A, then use ¬Intro. We already have a ¬(A ∨ B) negation, ‘¬(A ∨ B)', on line 1, so that will be one member of our ¬(A ∨ B) ↔ (¬A ∧ ¬B) contradictory pair. Get '(A ∨ B)' directly from A by ∨Intro for the desired contradiction. The same strategy gives us ¬B:

Our goal in the second subproof is a negation. We can't get it directly by easy rules like ∧Elim or Elim. So we assume '(A ∨ B)'. 1 2 ¬(A ∨ B) 3 A 4 A ∨ B ∨ Intro, 3 5 ⊥ ⊥ Intro: 2,4 6 ¬A ¬Intro: 3-5 7 B 8 A ∨ B ∨ Intro: 7 9 ⊥ ⊥ Intro: 2,8 10 ¬B ¬Intro: 7-9 11 ¬A ∧ ¬B ∧In: 6,10 12 13

¬A ∧ ¬B A∨B : : : : : : ¬(A ∨ B) ¬(A ∨ B) ↔ (¬A ∧ ¬B)

We must use ∨Elim. We need a sentence and its negation. Take negated conjunct from 11. Get unnegated statement by ∨Elim. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

¬(A ∨ B) A A ∨ B ∨ Intro, 3 ⊥ ⊥ Intro: 2,4 ¬A ¬Intro: 3-5 B A ∨ B ∨ Intro: 7 ⊥ ⊥ Intro: 2,8 ¬B ¬Intro: 7-9 ¬A ∧ ¬B ∧In: 6,10 ¬A ∧ ¬B A∨B A : : : :

B

A A ¬A ⊥ ¬(A ∨ B) ¬(A ∨ B) ↔ (¬A ∧ ¬B) 19

We already have a contradiction. Use it for ⊥Intro. Then use ⊥Elim to get A again so we can pull it out by ∨Elim for a contradiction under 'A ∨ B' 1 2 ¬(A ∨ B) 3 A 4 A∨B ∨ Intro, 3 5 ⊥ ⊥ Intro: 2-4 6 ¬A ¬Intro: 3-5 7 B 8 A∨B ∨ Intro: 7 9 ⊥ ⊥ Intro: 2,8 10 ¬B ¬Intro: 7-9 11 ¬A ∧ ¬B ∧ Intro: 6,10 12 ¬A ∧ ¬B 13 A∨B 14 A 15 B 16 ¬B ∧Elim: 12 17 ⊥ ⊥ Intro: 15,16 18 A ⊥ Elim: 17 19 A ∨Elim: 13,14-14,15-18 20 ¬A ∧ Elim: 12 21 ⊥ ⊥ Intro: 19,20 22 ¬(A ∨ B) ¬ Intro: 13-21 23 ¬(A ∨ B) ↔ (¬A ∧ ¬B) ↔Intro: 2-11,12-22

STRATEGY Show that (¬A ∨ B) and A → B are tautologically equivalent.

1 2 3

¬A ∨ B A

⇐

¬A

1

Derive biconditional from the null (empty) set of premises. As usual, plan to build biconditional by ↔Intro.

¬A ∨ B A→ B A→B ¬A ∨ B (¬A ∨ B) ↔ (A→ B)

In the first subproof, build « the usual way, by ↔Intro. Within this subproof, get ‘B’ by ∨Elim.

B B A→ B A→B ¬A ∨ B (¬A ∨ B) ↔ (A→ B)

⇐ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

¬A ∨ B A ¬A ⊥ ⊥ Intro: 2,3 B ⊥ Elim: 4 B B ∨Elim:1,3-5,6-6 A→B →Intro:2-7 A→B ¬(¬A ∨ B) ¬A ¬A ∨ B ∨Intro: 11 ⊥ ⊥Intro:10,12 ¬¬A ¬Intro:11-13 A ¬Elim: 14 ¬A ∨ B (¬A ∨ B) ↔ (A → B)

⇒ ‘B’ is already in the second subproof. In the first, use the standard strategy we apply when we already have or can easily get a contradiction: use ⊥Intro and ⊥Elim. Our goal is a disjunction. Neither disjunct is easy to get. We have no earlier disjunction to which we can apply ∨Elim. So assume the opposite. Use the common strategy for getting a contradiction when one of the things we have to work from is the negation of a disjunction. Here we assume ‘¬A’ because its opposite will be useful in combination with line 9.

⇒ In keeping with that standard strategy, get the other disjunct, and build the disjunction again. Fill in the justifications to complete the proof. 20

1 2 3 4 5 6 7 8

¬A ∨ B A ¬A ⊥ ⊥ Intro: 2,3 B ⊥ Elim: 4 B B ∨ Elim:1,3-5,6-6 A→B →Intro: 2-7 A→B ¬A ∨ B (¬A ∨ B) ↔ (A → B)

1 2 ¬A ∨ B 3 A 4 ¬A 5 ^ ⊥ Intro: 3,4 6 B ⊥ Elim: 5 7 B 8 B ∨ Elim: 2,4-6,7-7 9 A→B →Intro: 3-8 10 A → B 11 ¬(¬A ∨ B) 12 ¬A 13 ¬A ∨ B ∨ Intro: 12 14 ⊥ ⊥ Intro: 11,13 15 ¬¬A ¬ Intro: 12-14 16 A ¬ Elim: 15 17 B →Elim: 10,16 18 ¬A ∨ B ∨ Intro: 17 19 ⊥ ⊥ Intro: 11,18 20 ¬¬(¬A v B) ¬Intro: 11-19 21 ¬A v B ¬¬Elim: 20 22 (¬A ∨ B) ↔ (A → B) ↔ Intro: 2-9,10-21

To show that a set is tautologically inconsistent, derive a contradiction from the members of the set.

1 2 3 4

A ∧ (B ↔ C) A→B (A ∧ B) → D A → (C → ¬D)

STRATEGY Show that this set is tautologically inconsistent: {A ∧ (B ↔ C), A → B, (A ∧ B) → D, A → (C → ¬D)}

P P P P

1 2 3 4

The only ‘¬’ anywhere in this set is in ‘¬D’ inside line 4. So we’ll aim for ‘¬D’ and for ‘D’, then apply ⊥ Intro.

A ∧ (B ↔ C) A→B (A ∧ B) → D A → (C → ¬D) ¬D

⊥ 1 2 3 4 5 6

A ∧ (B ↔ C) P A→B P (A ∧ B) → D P A → (C → ¬D) P A ∧ Elim: 1 C → ¬D →Elim: 4,5

⇐

We just saw that ‘¬D’ comes from line 4. To start breaking line 4 apart, we need ‘A’. We notice that we can get it from 1 by ∧Elim.

¬D D ⊥

⇒ We also need ‘C’. That, too will have to come from breaking apart 1, but not directly. First we pull out 'B ↔ C'.

⊥ Intro

D ⊥ 1 2 3 4 5 6 7

A ∧ (B ↔ C) P A→B P (A ∧ B) → D P A → (C → ¬D) P A ∧ Elim: 1 C → ¬D →Elim: 4,5 B↔C ∧ Elim: 1 B →Elim: 2,5 C ↔ Elim: 7,8 ¬D →Elim: 6,9 D ⊥

⊥ Intro

⇒ Now we have all we need to justify ‘¬D’. Next we work on getting ‘D’. It must come from line 3. To get it, we must first have ‘A ∧ B’, the antecedent of the conditional. We already have both conjuncts separately, so we can build ‘A ∧ B’ by ∧Intro. That gives us all the steps we need, so all that’s left is completing the justifications.

21

⊥ Intro

A ∧ (B ↔ C) P A→B P (A ∧ B) → D P A → (C → ¬D) P A ∧ Elim: 1 C → ¬D →Elim: 4,5 B↔C ∧ Elim: 1 C ¬D D ⊥

All that is missing to get ‘C’ is ‘B’. We can get that from 2 and 5 by →Elim. 1 2 3 4 5 6 7 8 9 10

P P P P

1 2 3 4 5 6 7 8 9 10 11 12 13

⊥ Intro

A ∧ (B ↔ C) P A→B P (A ∧ B) → D P A → (C → ¬D) P A ∧ Elim: 1 C → ¬D →Elim: 4,5 B↔C ∧ Elim: 1 B →Elim: 2,5 C ↔ Elim: 7,8 ¬D →Elim: 6,9 A∧B ∧ Intro: 5,8 D →Elim: 3,11 ⊥ ⊥ Intro: 10,12

To show that a set of sentences is tautologically inconsistent, derive a contradiction from the members of the set. 1 2 3 4

J∨K J → (L ∧ ¬N) K ↔ (N ∧¬L) ¬(L ∨ N)

P P P P

⇒ Look for a negated sentence you already have or can easily get as one member of the contradictory pair. Here, if we try for ‘L ∨ N’ to contradict the sentence on line 4, we use the common strategy of applying ∨Elim to one disjunction to get another.

⊥

1 2 3 4 5 6

J∨K P J → (L ∧ ¬N) P K ↔ (N ∧ ¬L) P ¬(L ∨ N) P J L ∧ ¬N →Elim:2,5 L∨N

Show that {J ∨ K, J (L ∧ ¬N), K ↔ (N ∧ ¬L), ¬(L ∨ N)} is tautologically inconsistent

For the first subderivation, we notice that ‘J’ is the antecedent in line 2, and the first disjunct in ‘L ∨ N’ is in the consequent, so we apply →Elim to line 2.

1 2 3 4 5 6 7 8 ⇒

To get ‘L ∨ N’, we just need to separate ‘L’ from 6 by ∧Elim.

1 2 3 4 5 6 7 8

J∨K J → (L ∧ ¬N) K ↔ (N ∧ ¬L) ¬(L ∨ N) J L∨N K L∨N L∨N ⊥

⇐

K L∨N L∨N ⊥

1 2 3 4

J∨K P J → (L ∧ ¬N) P K ↔ (N ∧ ¬L) P ¬(L ∨ N) P J L ∧ ¬N → Elim:2,5 L ∧ Elim: 6 L∨N ∨ Elim: 7

J∨K P J → (L ∧ ¬N) P K ↔ (N ∧ ¬L) P ¬(L ∨ N) P J L ∧ ¬N →Elim:2,5 L ∧ Elim: 6 L∨N ∨ Elim: 7 K L∨N L∨N ⊥

⇐ Completion of the second subderivation is similar. Apply ↔Elim to 3, then separate one of the desired disjuncts by ∧Elim.

9 K 10 N ∧ ¬L ↔Elim: 3,9 11 N ∧Elim: 10 12 L∨N ∨Intro: 11 13 L ∨ N ∨Elim: 1,5-8,9-12 14 ⊥ ⊥Intro: 4,13 22

P P P P

To show that a set of Strategy: sentences is tautologically Show that this set is tautologically inconsistent: inconsistent, derive a {¬(P ∧ Q), R → (P ↔ S), S ∧ R, S → (¬Q → ¬R)} contradiction from the set’s members. ¬(P ∧ Q) 1 ¬(P ∧ Q) ⇒ R → (P ↔ S) 2 R → (P ↔ S) Look for a negation S∧R 3 S∧R we already have as one member of the S → (¬Q → ¬R) 4 S → (¬Q → ¬R) contradictory pair. Here, try for ‘P ∧ Q’ to contradict the sentence on line 1. P∧Q ⊥ ⊥

1 2 3 4

1 2 3 4

:

¬(P ∧ Q) R → (P ↔ S) S∧R S → (¬Q → ¬R) P↔S P

⇐

To get ‘P ∧ Q’, we need each conjunct. First get ‘P’. ‘P’ occurs only in lines 1 and 2. We can’t break apart line 1, so we must use line 2. We need to separate out its consequent. ⇒

Q P∧Q ⊥

To get ‘P’ from ‘P ↔ S’, we need ‘S’, which we can also get from 3. 1 2 3 4 5 6 7 8

¬(P ∧ Q) R → (P ↔ S) S∧R S → (¬Q → ¬R) R ∧ Elim: 3 P ↔ S → Elim: 2,5 S ∧ Elim: 3 P ↔Elim: 6,7 Q P∧Q ⊥

1 2 3 4 5 6

To get ‘P ↔ S’ from 2, we must first get the antecedent, ‘R’. Get ‘R’ from 3 by ∧E. ‘Q’ appears above only in lines 1 & 4. We can’t get ‘Q’ directly, so plan to get a contradiction. 'R' and '¬R' look promising. Assume '¬Q' to help us get '¬R' and ‘⊥’. 1 2 3 4 5 6 7 8 9

¬(P ∧ Q) R → (P ↔ S) S∧R S → (¬Q → ¬R) R ∧ Elim: 3 P↔S → Elim: 2,5 S ∧ Elim: 3 P ↔ Elim: 6,7 ¬Q ¬R ⊥ ¬¬Q Q P∧Q ⊥

23

¬(P ∧ Q) R → (P ↔ S) S∧R S → (¬Q → ¬R) R ∧Elim: 3 P ↔ S → Elim: 2,5 P Q P∧Q ⊥

‘¬Q’ leads easily to ‘¬R’. With ‘R’ (line 5), we have a contradiction. This completes the proof. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

¬(P ∧ Q) R → (P ↔ S) S∧R S → (¬Q → ¬R) R ∧ Elim: 3 P↔S → Elim: 2,5 S ∧ Elim: 3 P ↔ Elim: 6,7 ¬Q ¬Q → ¬R →Elim: 4,7 ¬R →Elim: 9,10 ⊥ ⊥ Intro: 5,11 ¬¬Q ¬ Intro: 9-12 Q ¬¬Elim: 13 P∧Q ∧Intro: 8,14 ⊥ ⊥ Intro: 1,15

Truth-functional compounds containing quantifiers Domain limited to people in this room 1. If anyone cheats, someone will be punished. 2. If anyone cheats, he or she will be punished. 3. Only those who cheat will be punished. 4. If everyone studies, no one will fail. 5. No one who studies will fail. (Anyone who studies will not fail.) 6. I'll be disappointed in anyone who cheats. 7. If anyone cheats, I'll be disappointed. 8. Everyone will be punished unless someone confesses. Symbolization with multiple quantifiers Domain = people, L(x,y) = x likes y $x"yL(x,y) = There is someone who likes everyone. $x"yL(y,x) = There is someone whom everyone likes. "x$yL(x,y) = Everyone likes someone (or other). "x$yL(y,x) = Everyone is liked by someone (or other). (For everyone, there is someone who likes that person.) Domain not limited:

F(x,y) = x is a flavor of y i = ice cream

P(x) = x is a person L(x,y)= x likes y

1. Everyone likes all flavors of ice cream. 2. Some people like all flavors of ice cream. 3. Nobody likes every flavor of ice cream. 4. For every flavor of ice cream, there is someone who likes it. 5. Everyone likes at least one flavor of ice cream. 6. There's a flavor of ice cream everyone likes.

24

Identity and numbers

Dictionary:

Domain = states in the US L(x,y) = x is larger than y

a = Alaska c = California t = Texas

1. Alaska is larger than any other state. 2. Alaska is the only state that is larger than Texas. 3. No state is larger than Alaska. 4. At least one state is larger than Texas. 5. At least two states are larger than California. 6. At most one state is larger than Texas. 7. At most two states are larger than California. 8. Exactly one state is larger than Texas. 9. Exactly two states are larger than California. 10. Alaska and Texas are the only states that are larger than California.

Dictionary: f = Felix g = Garfield t = Tina 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Dictionary: C(x) = x is a cat h = Herman H(x,y) = x hates y P(x) = x is a person j = Jason L(x,y) = x likes y F(x,y) = x is more famous than y Domain = people L(x,y) = x likes y Everyone likes at least one cat. 13. No one hates himself. There's no cat that everyone likes. 14. Everyone likes himself. Felix is the only cat Tina likes. 15. Nobody hates everybody. There is only one cat that Tina likes. 16. Nobody hates anybody. Tina likes at most one cat. 17. Nobody hates anybody who likes him or her. Tina likes at least two cats. 18. Everyone who likes anyone likes himself or Tina likes exactly two cats. herself. Tina likes at most two cats. 19. If Jason likes anyone, he likes Herman. Tina likes at least three cats. 20. Herman hates everyone except Jason. If any cat is more famous than 21. Jason and Herman do not like any of the Felix, Garfield is. same people. Felix is more famous than any 22. There is at least one person who likes, and is other cat. liked by, Jason. Felix is more famous than any 23. There are exactly two people whom both other cat except Garfield. Jason and Herman like. 24. Jason hates at most three people.

25

Answers to Exercises 1. "x(P(x) ‡ $y[C(y) Ÿ L(x,y)]) 2. ¬$x(C(x) Ÿ "y[P(y) Æ L(y,x)]) "x(C(x) ‡ $y[P(y) Ÿ ¬H(y,x)]) 3. (C(f) Ÿ L(t,f)) Ÿ "x([C(x) Ÿ L(t,x)] ‡ x=f) (C(f) Ÿ L(t,f)) Ÿ ¬$x([C(x) Ÿ L(t,x)] Ÿ x≠f) 4. $x([C(x) Ÿ L(t,x)] Ÿ "y([C(y) Ÿ L(t,y)] ‡ y=x)) $x([C(x) Ÿ L(t,x)] Ÿ ¬$y([C(y) Ÿ L(t,y)) Ÿ y≠x)) 5. ¬$x$y([(C(x) Ÿ C(y)) Ÿ (L(t,x) Ÿ L(t,y))] Ÿ x≠y) 6. $x$y([(C(x) Ÿ C(y)) Ÿ (L(t,x) Ÿ L(t,y))] Ÿ x≠y) 7. $x$y[([(C(x) Ÿ C(y)) Ÿ (L(t,x) Ÿ L(t,y))] Ÿ x≠y) Ÿ ¬$z([C(z) Ÿ L(t,z)] Ÿ (z≠x Ÿ z≠y))] $x$y[([(C(x) Ÿ C(y)) Ÿ (L(t,x) Ÿ L(t,y))] Ÿ x≠y) Ÿ "z([C(z) Ÿ L(t,z)] ‡ (z=x v z=y))] $x$y[([(C(x) Ÿ C(y)) Ÿ (L(t,x) Ÿ L(t,y))] Ÿ x≠y) Ÿ "z[¬(z=x v z=y) ‡ ¬(C(z) Ÿ L(t,z))]] $x$y[([(C(x) Ÿ C(y)) Ÿ (L(t,x) Ÿ L(t,y))] Ÿ x≠y) Ÿ "z[(z≠x Ÿ z≠y) ‡ ¬(C(z) Ÿ L(t,z))]] 8. ¬$x$y$z[([(C(x) Ÿ C(y)) Ÿ C(z)] Ÿ [(L(t,x) Ÿ L(t,y)) Ÿ L(t,z)]) Ÿ [(x≠y Ÿ x≠z) Ÿ y≠z]] 9. $x$y$z[([(C(x) Ÿ C(y))Ÿ C(z)] Ÿ [(L(t,x) Ÿ L(t,y)) Ÿ L(t,z)]) Ÿ [(x≠y Ÿ x≠z) Ÿ y≠z]] 10. $x[C(x) Ÿ F(x,f)] Æ F(g,f) "x([C(x) Ÿ F(x,f)] Æ F(g,f)) 11. C(f) Ÿ "x[(C(x) Ÿ x≠f) Æ F(f,x)] C(f) Ÿ "x[C(x) Æ (x≠f Æ F(f,x))] 12. [(C(f) Ÿ C(g)) Ÿ ¬F(f,g)] Ÿ "x[(C(x) Ÿ (x≠f Ÿ x≠g)) Æ F(f,x)] [(C(f) Ÿ C(g)) Ÿ ¬F(f,g)] Ÿ "x[(C(x) Ÿ ¬(x=f v x=g)) Æ F(f,x)] [(C(f) Ÿ C(g)) Ÿ ¬F(f,g)] Ÿ "x[(C(x) Ÿ ¬F(f,x)) Æ (x=f v x=g)] [(C(f) Ÿ C(g)) Ÿ ¬F(f,g)] Ÿ ¬$x[(C(x) Ÿ (x≠f Ÿ x≠g)) Ÿ ¬F(f,x)] 13. ¬$xH(x,x) (preferred) "x¬H(x,x) 14. "xL(x,x) 15. ¬$x"yH(x,y) (preferred) "x$y¬H(x,y) 16. ¬$x$yH(x,y) (preferred) "x"y¬H(x,y) 17. ¬$x$y(L(y,x) Ÿ (x,y)) or ¬$x$y(L(x,y) Ÿ H(y,x)) "x"y(L(y,x) Æ ¬H(x,y)) or "x"y(L(x,y) Æ ¬H(y,x)) 18. "x($yL(x,y) Æ L(x,x)) (preferred) "x(¬L(x,x) Æ ¬--$yL(x,y)) 19. $xL(j,x) Æ L(j,h) (preferred) "x[L(j,x) Æ L(j,h)] 20. ¬H(h,j) Ÿ "x(x≠j Æ H(h,x)) (preferred) ¬H(h,j) Ÿ "x(x=j ⁄ H(h,x)) ¬H(h,j) Ÿ ¬$x(x≠j Ÿ ¬H(h,x)) 21. ¬$x[L(j,x) Ÿ L(h,x)] (preferred) "x¬[L(j,x) Ÿ L(h,x)] "x[¬L(j,x) ⁄ ¬L(h,x)] 22. $x(L(x,j) Ÿ L(j,x)) (preferred) ¬"x(L(x,j) Æ ¬L(j,x)) 23. $x$y[([(L(j,x) Ÿ L(h,x)) Ÿ (L(j,y) Ÿ L(h,y))] Ÿ x≠y) Ÿ ¬$z[(L(j,z) Ÿ L(h,z)) Ÿ (z≠x Ÿ z≠y)]] $x$y[([(L(j,x) Ÿ L(h,x)) Ÿ (L(j,y) Ÿ L(h,y))] Ÿ x≠y) Ÿ "z([L(j,z) Ÿ L(h,z)] Æ (z=x ⁄ z=y))] $x$y[([(L(j,x) Ÿ L(h,x)) Ÿ (L(j,y) Ÿ L(h,y))] Ÿ x≠y) Ÿ "z[¬(z=x ⁄ z=y) Æ ¬(L(j,z) Ÿ L(h,z))]] $x$y[([(L(j,x) Ÿ L(h,x)) Ÿ (L(j,y) Ÿ L(h,y))] Ÿ x≠y) Ÿ "z[(z≠x Ÿ z≠y) Æ ¬(L(j,z) Ÿ L(h,z))]] 24. ¬$x1$x2$x3$x4([([H(j,x1) Ÿ H(j,x2)] Ÿ H(j,x3)) Ÿ H(j,x4)] Ÿ ([(x1≠x2 Ÿ x1≠x3) Ÿ x1≠x4] Ÿ (x2≠x3 Ÿ x2≠x4) Ÿ x3≠x4)]

26

PROOF RULES FOR QUANTIFIERS constants: a - v

variables: w, x, y, z (with subscripts if needed)

The following rules apply only where the quantifier put in or taken out has the entire rest of the line as its scope (i.e, is the main operator). (So, for example, the "Elim rule cannot be applied to '¬"xP(x)' because its main operator is the '¬'.) Strictly speaking, this restriction means that a series of quantifiers at the beginning of a sentence must be removed or applied one at a time. However, for a series of quantifiers of the same type (multiple universal quantifiers or multiple existential quantifiers) we will allow ourselves the shortcut of removing or introducing more than one at a time. However, when a statement begins with mixed quantifiers (for example: "x$yL(x,y)), the quantifiers must be taken out one at a time, starting at the left. To put in more than one, add one at a time, always adding at the left end. Remember that one quantifier cannot fall within the scope of another for the same variable. " Elim -- Universal Elimination (examples 1-4) When you have a universal statement (one in which the main operator is a universal quantifier), you may take out the quantifier and replace the variable it was binding throughout the rest of the line by any constant. 1. Every occurrence of the variable bound by the universal quantifier must be replaced, and all of them must be replaced by the same constant. 2. You may apply " Elim to the same line as many times you choose, replacing the variable with whatever constant you wish on each application of the rule. $ Intro -- Existential Introduction (examples 5-9) You may substitute a variable for one or more occurrences of a single constant in a sentence you already have, putting an existential quantifier for that variable at the beginning of the line (with the rest of the line in its scope). " Intro -- Universal Introduction (examples 10-14) Start a subproof with a boxed constant not found outside that subproof. Replace all occurrences of the constant with the same variable (for either form of " Intro). No other quantifier for the same variable can occur within the scope of the new quantifier. FOR SIMPLICITY, we will use only the basic Universal Introduction rule, although the text also uses a variation they call General Conditional Proof.

UNIVERSAL INTRODUCTION i

c

n

P(c) "xP(x)

"Intro: i-n

$ Elim -- Existential Elimination (examples 15-19) Apply this rule to a line with an existential sentence. Begin a subproof with a boxed constant that does not occur outside the subproof. To form the first line of the subproof, drop the existential quantifier from the sentence to which you plan to apply this rule; in the rest of the line, replace every occurrence of the variable for that quantifier with the constant introduced in the box. The $E rule takes a line from this subproof back out of it (one column to the left). The line that moves to the left cannot contain the (boxed) constant introduced in the first line of the subderivation.

27

EXAMPLES USING QUANTIFIER RULES

1 2 3 4 5 6 7 8 9

"x(H(x) ‡ G(x)) P H(a) Ÿ H(b) P H(a) ‡ G(a) "E:1 H(a) ŸE:2 G(a) ‡E: 3,4 H(b) ‡ G(b) "E:1 H(b) ŸE: 2 G(b) ‡ E: 6,7 G(a) Ÿ G(b) ŸI:5,8

1 2 3 4 5 6 7

1 N(a) Ÿ R(a) P 2 $x(N(x) Ÿ R(x)) $I:1

1 N(a) Ÿ R(a) P 2 $y(N(a) Ÿ R(y)) $I:1 3 $x$y(N(x) Ÿ R(y)) $I:2

1 2 3 4 5 6 7 8 9 10 11

B(a) P "x(B(x) ´ C(x)) P "x(C(x) ‡ D(x)) P B(a) ´ C(a) "E:2 C(a) É:1,4 C(a) ‡ D(a) "E:3 D(a) ‡E: 5,6

1 2 3 4 5 6

N(a) Ÿ R(b) N(a) $xN(x) R(b) $xR(x) $xN(x) Ÿ$xR(x)

1 2 3 "x(J(x) ‡K(x)) P 4 "x[(J(x) Ÿ K(x)) ‡ L(x)] P 5 6 J(a) 7 J(a) ‡ K(a) "E:1 8 K(a) ‡E:3,4 9 J(a) Ÿ K(a) ŸI:3,5 10 (J(a) Ÿ K(a) ‡L(a) "E:2 L(a) ‡E:6,7 J(a) ‡ K(a) ‡I: 4-9 "x(J(x) ‡ L(x)) "I: 3-10

1 2 3 4

"x(S(x ) ´ N(x)) P S(a) P S(a) ´ N(x) "E:1 N(x) É: 2,3

1 $ xC(x,a) 2 $x$yC(x,y)

P(a) P $xPx ‡ "x[Q(x) ´ R(x)] P "xR(x) P $xP(x) $I:1 "x(Q(x) ´ R(x)] ‡E: 2,4 "E:3 "E:5 É:7,8 "I:6-9

P $I:1

1 "x(G(x) ‡ H(x)) P 2 G(a) P 3 4 G(a) ‡ H(a) "E:1 5 H(a) ‡E:2,4 6 "xH(x) "I:3-5 1 "xN(x) 2 "xS(x ) 3 4 Na 5 Sb 6 7 Na Ÿ Sb 8 "x(Nx Ÿ Sx)

P P "E:1 "E:2 ŸI:4,6 "I:5-7

1 $ x[L(x) Ÿ M(x)] P 2 L(a) Ÿ M(a) 3 M(a) ŸE:2 4 M(a) $E :1,2-3 5 $xM(x) $I:4

1 $ x(C(x) Ÿ D(x)) P 2 C(a) Ÿ D(a) 3 D(a) ŸE:2 4 $xD(x) $I:3 5 $xD(x) $E:1,2-4

1 "x(R(x) ´ "yS(y)) P 2 $ xR(x) P 3 R(b) 4 R(b) ´ "y S(y) "E:1 5 "yS(y) É:3,4 6 S(a) "E:5 7 S(a) $E: 2,3-6

"x(P(x) ‡ Q(x)) P P(a) Ÿ R(b) P P(a) ‡ Q(b) "E:1 P(a) ŸE:2 Q(b) ‡E:3,4

1 N(a) Ÿ R(b) P 2 $x(N(x) Ÿ R(x)) $I:1

P 1,ŸE $I:2 ŸE:1 $I:4 ŸI:3,5

R(b) Q(b) ´ R(b) Q(b) "xQx

1 2 3 4 5

1 2 3 4 5 6 7 8 9

"x[A(x) ´ D(x)] P $xA(x) P $ xD(x) ´ "xD(x) P A(b) A(b) ´ D(b) "E:1 D(b) É:4,5 $xD(x) $I:6 "xD(x) É: 3,7 "xDx $E: 2,4-8

28

1 J(a) ´ K(a) P 2 $ x[K(x) Ÿ L(x]) P 3 K(a) Ÿ L(a) 4 K(a) ŸE:3 5 J(a) É:1,4 6 $xJ(x) $I:5 7 $xJ(x) $E: 2,3-6

Rules for Derivations with Quantifiers Ÿ Intro Using Ÿ, join any sentences you already have. Ÿ Elim From a conjunction, take any conjunct. ^ Intro From a sentence and its negation, get ^. ^ Elim From ^, get any sentence.

¬ Elim … i ¬¬A … A ¬Elim: i ¬ Intro … i A … j ^ … ¬A ¬Intro: i-j

Æ Elim From a conditional and its antecedent, get its consequent. Æ Intro … i A … j B … AÆB

k l

C

B … C

⁄Elim: h,i-j,k-l

´ Elim From a biconditional and the sentence on one side of its ´, get the sentence on the other side. ´ Intro … i A … j B k B … l A A´B Íntro: i-j, k-l

ÆIntro: i-j

"Elim From a sentence with '"x' as its main operator, drop '"x'; replace all occurrences of 'x' uniformly with any constant.

⁄ Intro Using ⁄, join any sentence you'd like with any sentence you already have. ⁄ Elim … h A⁄B … i A … j C

$ Intro In any sentence A you already have, replace 0 or more occurrences of a constant with 'x' (where 'x' is any variable not in A) and apply '$x' to the resulting formula.

" Intro … j a … j P "xP(…x…) "Intro: i-j where 'a' is a constant that does not occur outside the subproof, 'P' is any sentence, 'x' is any variable not occurring in 'P', and 'P(…x…)' results from substituting an 'x' for each occurrence of 'a' in 'P'.

$ Elim

… i $xP(…x…) … j a P(…a…) … k Q Q $ Elim: i, j-k where 'a' is a constant that does not occur outside the subproof, 'P(…a…)' results from substituting 'a' for every 'x' in 'P(…x…)', and 'Q' does not contain 'a'. 29

STRATEGIES FOR PROOFS 1.

Try to extract goal statement from a statement you already have, in which the goal statement is a subformula. Instances of a quantified sentence are subformulas of that quantified sentence. The "Elim rule allows you to extract an instance. As before, ŸElim, ÆElim, and Élim allow you to extract immediate components of formulas. (So if your goal is 'J(a)' and you have ' "x(J(x) Ÿ K(x))', apply "Elim and then ŸElim.)

2.

If goal sentence cannot be extracted as a whole from any statement you already have, base your strategy on the structure of the goal statement. Ÿ conjunction Aim for each conjunct separately, then apply ŸIntro. Æ conditional Plan to use ÆIntro. To do this, start a subproof with the antecedent as proivisional assumption. Aim for the consequent in the subproof. ´ biconditional Plan to use Íntro. Start one subproof with the left side of the biconditional and aim for the right in this subproof. Set up a second subproof going from the right side of the biconditional to the left. ¬ negation If goal has as its main operator, try reaching it by ¬Intro. To do this, start a subproof with the statement to be negated (but without the ¬) as provisional assumption. Within this subproof, aim for any contradiction you can get. ⁄ disjunction a) If one disjunct is obviously easy to get, get it, then use ⁄Intro. b) If neither disjunct is obviously easy to get, look for an earlier disjunction. Try ⁄Elim on earlier disjunction c) If neither of these works, assume the negation of the goal statement. You will then need to use ¬Intro to reach goal. " universal Start a subproof, introducing a constant that does not occur outside the subproof. Aim for an instance of the universal with that constant. Build the universal outside the subproof by "Intro. $ existential a) If one instance is obviously easy to get, get it. Then use $Intro. b) If no instance is obviously easy to get, look for an earlier existential statement. Try $Elim on earlier existential. c) If neither of these works, assume the negation of the goal statement. You will then need to use ¬Intro to reach goal.

3.

a) If you have an existential sentence already and you can't use it with any of our easy rules, you will probably have to use $Elim. If you will have to use $Elim, set up for it early. (An existential sentence can be used with an easy rule if the existential forms the antecedent of a conditional or one side of a biconditional on another line.) b) If you have a disjunction already, and you can't use it with an easy rule, you’ll probably have to use ⁄Elim rule. If you’ll have to use ⁄Elim, set up for it early.

4.

When aiming for ^, look for a negation you already have to use as one member of the contradictory pair of sentences.

5.

If you have no idea what to do, try applying any easy rules you can. Perhaps the results of this process will give you some ideas for other things to do.

6.

When all else fails, assume the opposite of what you want. 30

EXAMPLES USING THE $ Elim RULE 1 $x(S(x) Ÿ L(x)) 2 "x(S(x) Æ P(x)) 3 a S(a) Ÿ L(a) 4 5 6 7 8 9

S(a) Ÿ Elim: 3 S(a) Æ P(a) " Elim: 2 P(a) Æ Elim: 4,5 L(a) Ÿ Elim: 3 P(a) Ÿ L(a) ŸIntro: 6,7 $x(P(x) Ÿ L(x)) $ Intro: 8

10 $x(P(x) Ÿ L(x)) $Elim: 1,3-9

1 "x(P(x) Æ ¬H(x)) 2 $x(P(x) Ÿ H(x)) 3 a P(a) Ÿ H(a) 4 P(a) ‡ ¬H(a) "Elim: 1 5 P(a) ŸElim: 3 6 ¬H(a) ÆElim:4,5 7 H(a) ŸElim: 3 8 ^ ^ Intro: 6,7 9 ^ $ Elim: 2,3-8

10 ¬$x(P(x) Ÿ H(x))

¬Intro: 2-9

At least one (perhaps former) Senator has lied under oath.. Every Senator is a politician. Let's suppose a is a Senator who has lied under oath. (Line 1 says there's at least one, but doesn't identify one. We are temporarily assuming a is such a Senator.) On this supposition, a is a Senator. If a is a Senator, a is a politician. So a is a politician. Also, a has lied under oath. So a both is a politician and has lied under oath. ON OUR ASSUMPTION, at least one politician has lied under oath. So at least one politician has lied under oath. (This conclusion doesn't depend on the truth of our assumption about a, because we could have reached the same conclusion no matter what constant had been used in step 3.) All politicians are dishonest. Suppose there is a politician who is honest. Let's assume for now a is such a person. THEN, if a is a politician, a is dishonest. Also, a is a politician, so a is dishonest. But (we were already told) a is honest. Contradiction: a both is and is not honest. The contradiction follows from the claim that there's an honest politician, regardless of who might be one. (It doesn't depend on a in particular being such a person. Anyone we might pick as an honest politician could be shown, as a was, both to be honest and not to be honest.) So it's not true that there's an honest politician (because the assumption that there is one leads to a contradiction). In other words, no politicians are honest.

1 "x(P(x) Æ Q(x,a)) 2 b $x(P(x) Ÿ ¬Q(x,a)) 3 P(b) Ÿ ¬Q(b,a) 4 P(b) Æ Q(b,a) "Elim: 1 5 P(b) Ÿ Elim: 3 6 Q(b,a) Æ Elim: 4,5 7 ¬Q(b,a) Ÿ Elim: 3 8 ^ ^ Intro: 6,7 9 ^ $ Elim: 2,3-8 10 ¬$x(P(x) Ÿ ¬Q(x,a)) ¬ Intro: 2-9

31

EXTRA PROBLEMS FOR CHAPTER 13, GROUP 1 If the argument is FO-valid, use Fitch to open Proof Cstern 130x for H 13.x, and give a proof. Use AnaCon only for literals (atomic sentences and their negations). Do not use TautCon. If the argument is FO-invalid, use Tarski’s World to open Sentences Cstern 130x for H 13.x, and give a counterexample. Save worlds as World Cstern 130x, corresponding to Sentences Cstern 130x. H 13.1

"x (Small(x) ‡ Cube(x)) "x Dodec(x) "x¬Small(x)

H 13.2

"x((Small(x) ⁄ Large(x)) ‡ Tet(x)) "x (Small(x) ⁄ Medium(x) ⁄ Large(x)) "x Cube(x) "x Medium(x)

H 13.3

"x Dodec(x) ‡ "x Small(x) "x (Dodec(x) ‡ Small(x))

H 13.4

$x (Larger(x,a) Ÿ Cube(x)) $x Larger(x,a) Ÿ $x Cube(x)

H 13.5

$x Larger(x,a) $x Cube(x) $x (Larger(x,a) Ÿ Cube(x))

H 13.6

"x (Dodec(x) ‡ Medium(x)) $x Large(x) ‡ $x ¬Dodec(x)

H 13.7

"x(Small(x) ⁄ Large(x)) "x(Small(x) ´ Tet(x)) "x(Dodec(x) ⁄ ¬Large(x)) "x(Tet(x) ⁄ Dodec(x))

H 13.8

"x(Dodec(x) ‡ Large(x)) ¬$x(Small(x) Ÿ Cube(x)) "x((Cube(x) ⁄ Dodec(x)) ‡ ¬Small(x))

32

1 ∀x(Larger(x,a) →Tet(x)) 2 rm(a) = b 3 Large(b) ∧ Small(a)

STRATEGY for a derivation using =Elim and AnaCon

Tet(rm(a)) ∧ rm(a)≠a We plan to build the goal conjunction by ∧Intro. So we aim for the 2 conjuncts.

1 ∀x(Larger(x,a) → Tet(x)) 2 rm(a) = b 3 Large(b) ∧ Small(a) Tet(rm(a)) rm(a) ≠ a Tet(rm(a)) ∧ rm(a)≠a

1 ∀x(Larger(x,a) →Tet(x)) 2 rm(a) = b 3 Large(b) ∧ Small(a)

To help get ‘Tet(rm(a))’, we notice ‘Tet’ in line 1. So we apply ∀Elim. We aim the antecedent of that conditional so that we will be able to apply →Elim

Because we have ‘rm(a) = b’ on line 2, we can substitute either term for the other. We could do that to reach this new goal if we could use the information in line 3 to get ‘Larger(b,a)’.

∧ Intro

Larger(rm(a),a) Larger(rm(a),a) →Tet(rm(a)) Tet(rm(a)) rm(a) ≠ a Tet(rm(a)) ∧ rm(a)≠a

∀Elim →Elim ∧ Intro

1 ∀x(Larger(x,a) → Tet(x)) 2 rm(a) = b 3 Large(b) ∧ Small(a) Larger(b,a) Larger(rm(a),a) Larger(rm(a),a) → Tet(rm(a)) Tet(rm(a)) rm(a) ≠ a Tet(rm(a)) ∧ rm(a)≠a

1 2 3 4 5 6 7 8 We also notice that we now have all we 9 need to justify line ‘rm(a)≠a’, because 10 nothing is larger than itself. 11 ‘Larger(b,a)’ follows from the meanings of ‘Large’, ‘Small’, and ‘Larger’. Since we are applying AnaCon only to literals (atomic sentences and their negations), we must break apart line 3.

∀Elim →Elim ∧ Intro

∀x(Larger(x,a) → Tet(x)) rm(a) = b Large(b) ∧ Small(a) Large(b) ∧ Elim: 3 Small(a) ∧ Elim: 3 Larger(b,a) AnaCon: 4,5 Larger(rm(a),a) =Elim: 2,6 Larger(rm(a),a) →Tet(rm(a)) ∀Elim: 1 Tet(rm(a)) →Elim: 7,8 rm(a) ≠ a AnaCon:7 Tet(rm(a)) ∧ rm(a)≠a ∧ Intro: 9,10

If we did not have AnaCon, we would need two additional premises. What general premises would express the relevant meaning relations? 33

Our goal is a negation, so we assume the sentence it negates, and plan to use ¬Intro. To do this, we need ⊥ in the subproof.

1 ∃xB(x) 2 ¬∃x(B(x) ↔ C(x)) 3 ∀xC(x) ⊥ ¬∀xC(x)

We need to apply ∃Elim to use the first premise. So we start a subproof with a boxed constant that does not occur outside the subproof, using that constant in an instance of line 1. Our goal in that subproof, too, will be ⊥, so we can move it to the first subproof by ∃Elim.

P P

STRATEGY for: {∃xB(x), ¬∃x(B(x) ↔ C(x))} /∴ ¬∀xC(x)

¬Intro

1 ∃xB(x) P 2 ¬∃x(B(x) ↔ C(x)) P 3 ∀xC(x) 4 B(a) 1 ∃xB(x) 2 ¬∃x(B(x) ↔ C(x)) 3 ∀xC(x) 4 B(a)

⊥ ⊥ ¬∀xC(x)

¬Intro ⇒

We need a sentence and its negation to get ⊥. We already have a negation on line 2, so aim for the sentence that it negates. 1 ∃xB(x) 2 ¬∃x(B(x) ↔ C(x)) 3 ∀xC(x) 4 B(a)

P P

∃x(B(x) ↔ C(x)) ⊥ ⊥Intro ⊥ ∃Elim ¬∀xC(x) ¬Intro

To get ‘∃x(B(x) ↔ C(x))’ inside this subproof, aim for an instance. ‘B(a) ↔ C(a)’ should be the easiest one to get because we already have ‘B(a)’.

⇐

B(a) ↔ C(a) ∃x(B(x) ↔ C(x)) ∃Intro ⊥ ⊥Intro ⊥ ∃Elim ¬∀xC(x) ¬Intro

1 ∃xB(x) 2 ¬∃x(B(x) ↔ C(x)) 3 ∀xC(x) 4 B(a) 5 B(a)

C(a)

P P

⇒

∀Elim: 3

7 C(a) 8 B(a) Reit: 4 9 B(a) ↔ C(a) ↔Intro: 5-6,7-8 10 ∃x(B(x) ↔ C(x)) ∃Intro: 9 11 ⊥ ⊥Intro: 9,2 12 ⊥ ∃Elim:1,4-10 13 ¬∀xC(x) ¬Intro: 3-12

P P

C(a)

Start 2 subproofs to get 'B(a) ↔ C(a)’ by ↔Intro. 1 ∃xB(x) 2 ¬∃x(B(x) ↔ C(x)) 3 ∀xC(x) 4 B(a) 5 B(a) 6 C(a)

P P

⇐

34

B(a) B(a) ↔ C(a) ↔Intro ∃x(B(x) ↔ C(x)) ∃Intro ⊥ ⊥Intro ⊥ ∃Elim ¬∀xC(x) ¬Intro

We already have all we need to justify all the steps. So complete the proof by filling in justifications.

STRATEGY: {∀x(C(x) ↔ D(x)), C(a), ∃xD(x)  ∀xM(x), ∀x(M(x)  L(x))} /∴ ∀xL(x)

1 2 3 4

∀x(C(x) ↔ D(x)) C(a) ∃xD(x) → ∀xM(x) ∀x(M(x) → L(x))

We can't reach our goal by easy rules like ∧, ∃, or ∀Elim. Next best strategy: build it by ∀Intro. Start a subproof with a boxed constant that's not outside the subproof; get an instance of our goal with the constant in the box. So aim for 'L(b)' in a subproof starting with .

P P P P

∀xL(x)

The only premise with 'L' is 4, so use the instance of 4 with 'b'.

1 2 3 4 5

∀x(C(x) ↔ D(x)) C(a) ∃xD(x) → ∀xM(x) ∀x(M(x) → L(x)) M(b) → L(b) L(b) ∀xL(x)

'M' is in only two premises. In 4, it's in the antecedent of the conditional to which '∀x' applies, so that’s no help. But in 3 it's in the consequent. Also, that consequent is universal, so it would yield 'M(b)' by ∀Elim. So use 3 to get ‘∀xM(x)’. Our goal is an existential. Since we can’t pull it out of anything, build it by ∃Intro. So we need an instance. Except for 3, the only premise with ‘D’ is line 1, so we’ll use an instance of that to get an instance of ‘∃xD(x)’. We need a constant we can get after ‘C’. Line 2 is ‘C(a)’, so we use ‘a’.

1 2 3 4 5

P P P P

1 2 To use this 3 to reach 4 'L(b)', we 5 also need 'M(b)', so aim for that.

∀Elim:4 ∀Intro


P P P P

∀xM(x) →Elim M(b) ∀Elim M(b) → L(b) ∀Elim:4 L(b) →Elim ∀xL(x) ∀Intro

1 2 3 4 5


P P P P

D(a) ∃xD(x) ∀xM(x) →Elim M(b) ∀Elim M(b) → L(b) ∀Elim:4 L(b) →Elim ∀xL(x) ∀Intro 35

To get ‘∀xM(x)’ from 3, first aim for ‘∃xD(x)’.

Now all we need to do is fill in the justifications.

1 2 3 4

∀x(C(x) ↔ D(x)) C(a) ∃xD(x) → ∀xM(x) ∀x(M(x) → L(x)) L(b) ∀xL(x)

P P P P

∀Intro


P P P P

M(b) M(b) → L(b) ∀Elim: 4 L(b) →Elim ∀xL(x) ∀Intro 1 2 3 4 5

1 2 3 4 5 6 7 8 9 10 11 12 13


P P P P

∃xD(x) ∀xM(x) →Elim M(b) ∀Elim M(b) → L(b) ∀Elim:4 L(b) →Elim ∀xL(x) ∀Intro


P P P P

C(a) ↔ D(a) ∀Elim: 1 D(a) ↔Elim: 2,5 ∃xD(x) ∃Intro: 7 ∀xM(x) →Elim: 3,8 M(b) ∀Elim: 9 M(b) → L(b) ∀Elim: 4 L(b) →Elim:10,11 ∀xL(x) ∀Intro: 5-12

Our conclusion is a conjunction. So we try to get each conjunct alone, then join them by ∧Intro.

1 ∃x(J(x) ∧ Kx) P 2 ¬∀x(K(x) → J(x)) P : : ∃xJ(x) : : ∃x¬J(x) ∃xJ(x) ∧ ∃x¬J(x)

STRATEGY for {∃x(J(x) ∧ K(x)), ¬∀x(K(x)  J(x))} /∴ ∃xJ(x) ∧ ∃x¬J(x)

We should be able to use ⇒ ⇓ the first premise to get ‘∃xJ(x)’. We begin a Within the subproof, we subproof, planning to need something without ‘a’ apply the ∃Elim rule. that we can pull out of the subproof by the ∃Elim rule. 1 ∃x(J(x) ∧ Kx) P We want ’∃xJ(x)’, so we’ll 2 ¬∀x(K(x) → J(x)) P try to get that. We need an 3 J(a) ∧ K(a) : : instance of ‘∃xJ(x)’, which : ∃xJ(x) we can get from 3 by ∧Elim. : : ∃x¬J(x) To get ‘∃x¬J(x)’, we must use ∃xJ(x) ∧ ∃x¬J(x) line 2. The only way to use it is in a contradictory pair. So assume negation of desired statement, and aim for a contradiction with 2.

1 ∃x(J(x) ∧ Kx) P 2 ¬∀x(K(x) → J(x)) P 3 J(a) ∧ K(a) 4 Ja ∧Elim: 3 5 ∃xJ(x) ∃Intro: 4 6 ∃xJ(x) ∃Elim: 1,3-5 7

Plan to use "Intro. Start a ⇓ subderivation with a new boxed constant constant, and aim for the desired conditional with that constant. Set up subderivation for → Intro. 1 ∃x(J(x) ∧ Kx) P 2 ¬∀x(K(x) → J(x)) P 3 J(a) ∧ K(a) 4 J(a) ∧Elim: 3 5 ∃xJ(x) ∃Intro: 4 6 ∃xJ(x) ∃Elim: 1,3-5 7 ¬∃x¬J(x) 8 : : : K(b) → J(b) ∀x(K(x) → J(x)) ∀Intro ⊥ ⊥ Intro ∃x¬J(x) ¬Intro ∃xJ(x) ∧ ∃x¬J(x) ∧Intro

1 ∃x(J(x) ∧ Kx) P 2 ¬∀x(K(x) → J(x)) P 3 J(a) ∧ K(a) 4 Ja ∧Elim: 3 5 ∃xJ(x) ∃Intro: 4 6 ∃xJ(x) ∃Elim: 1,3-5 : ∃x¬J(x) ∃xJ(x) ∧ ∃x¬J(x)

: :

¬∃x¬J(x) ∀x(K(x) → J(x)) ⊥ ⊥Intro ∃x¬J(x) ¬Intro ∃xJ(x) ∧ ∃x¬J(x) ∧Intro

⇓ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

∃x(J(x) ∧ Kx) ¬∀x(K(x) → J(x)) J(a) ∧ K(a) J(a) ∃xJ(x) ∃xJ(x) ¬∃x¬J(x) K(b) ¬J(b) ∃x¬J(b) ⊥ J(b) K(b) → J(b) ∀x(K(x) → J(x)) ⊥ ∃x¬J(x) ∃xJ(x) ∧ ∃x¬J(x) 36

P P ∧Elim: 3 ∃ Intro: 4 ∃Elim: 1,3-5

∃ Intro: 10 ⊥ Intro: 1017 ¬intro: 10-12 →Intro: 9-13 ∀Intro: 8-14 ⊥ Intro: 2,15 ¬Intro: 7-16 ∧Intro: 6,17

⇐

We have no way to get ‘J(b)’ by use of easy rules. Also, the only way we can use line 7 is in a contradiction. So we assume ‘¬J(b)’. That justifies ‘∃x¬J(x)’, contradicting line 7.

1 ∀x[(F(x) v G(x)) → H(x)] 2 ∀x(F(x) ↔ ¬ G(x))

P P

STRATEGY for: {∀x[(F(x) v G(x)) → H(x)], ∀x(F(x) ↔ ¬G(x))} /∴ ∀xH(x)

∀xH(x) 1 ∀x[(F(x) v G(x)) → H(x)] P 2 ∀x(F(x) ↔ ¬ G(x)) P 3 H(a) ∀xH(x)

The goal is a universal. For ∀Intro, start a subproof with a boxed constant that does not occur outside the subproof. There are no constants outside the subproof, so we can use any constant. In the subproof, aim for an instance of the universal with the constant introduced in the box.

∀Intro

To get our instance, weʼll be plugging the constant from our new goal into our universal premises. Since line 1 has the predicate ʻHʼ that we need, begin there.

1 ∀x[(F(x) v G(x))  H(x)] P 2 ∀x(F(x) ↔ ¬ G(x)) P 3 (F(a) v G(a))  4 H(a) ∀Elim:1 Elim ∀ Intro (F(a) v G(a)) H(a) ∀xH(x)

⇓

1 ∀x[(F(x) v G(x)) → H(x)] 2 ∀x(F(x) ↔ ¬ G(x)) 3 4 (F(a) v G(a)) → H(a) H(a) ∀xH(x)

We can get ʻH(a)ʼ with the help of line 3 if we can get ʻF(a) v G(a)ʼ.

⇒ ⇒ ⇒ Neither disjunct is obviously the easy one to get, so we assume the negation of the disjunction.

P P

∀Intro

1 ∀x[(F(x) v G(x))  H(x)] P 2 ∀x(F(x) ↔ ¬ G(x)) P 3 4 (F(a) v G(a))  H(a) ∀Elim 5 ¬F(a) v G(a)) (F(a) v G(a)) H(a) ∀xH(x)

Elim ∀ Intro

Apply common strategy for getting a contradiction under the negation of a disjunction.. Assume ʻG(a)ʼ because we have an easy way to use ʻ¬G(a)ʼ (with line 2), but we have no easy way to use ʻ¬F(a)ʼ. We will use ʻ¬G(a)ʼ to get ʻF(a)ʼ, then ʻF(a) v G(a)ʼ.

1 ∀x[(F(x) v G(x)) → H(x)] P 2 ∀x(F(x) ↔ ¬G(x)) P 3 4 (F(a) v G(a))  H(a) ∀Elim:1 5 ¬(F(a) v G(a)) 6 G(a) 7 F(a) v G(a) vIntro: 6 8 ⊥ ⊥Intro: 5,7 9 ¬G(a) ¬Intro: 5-8 F(a) F(a) v G(a) ⊥ ¬ ¬ (F(a) v G(a)) F(a) v G(a) H(a) ∀xH(x)

vIntro ⊥Intro ¬Intro ¬¬Elim Elim ∀ Intro

⇒ ⇒ ⇒ ⇒ To get ʻF(a)ʼ, use line 2. When we get ʻF(a) v G(a)ʼ again, we can use ⊥ Intro and ¬ Intro to get ʻ¬¬(F(a) v G(a))ʼ, then drop the double negation to get ʻF(a) v G(a)ʼ where we really want it, outside of any subproof.

37

1 ∀x[(F(x) v G(x)) → H(x)] P 2 ∀x(F(x) ↔ ¬G(x)) P 3 4 (F(a) v G(a)) → H(a) ∀Elim:1 5 ¬(F(a) v G(a)) 6 G(a) 7 F(a) v G(a) vIntro: 6 8 ⊥ ⊥Intro: 5,7 9 ¬G(a) ¬Intro: 5-8 10 F(a) ↔ ¬G(a) ∀Elim: 2 11 F(a) ↔Elim: 9,10 12 F(a) v G(a) vIntro: 11 13 ⊥ ⊥Intro: 12,5 14 ¬ ¬ (F(a) v G(a)) ¬Intro: 5-13 15 F(a) v G(a) ¬¬Elim: 14 16 H(a) →Elim: 4,15 17 ∀xH(x) ∀Intro: 3-16

1 ∃xR(x,x) → ∀x(Q(x) ∨ S(x)) 2 ∀x(R(a,x) ∧ ¬Q(x)) ∀xS(x)

STRATEGY for: {∃xR(x,x) → ∀x(Q(x) ∨ S(x)), ∀x(R(a,x) ∧ ¬Q(x))} /∴ ∀xS(x)

The only ‘S’ in the premises is in 1. So aim for the antecedent. Get the consequent by →Elim, then try to get ‘∀xS(x)’ from that. ⇒ ⇒ ⇒ 1 2 3 4 5 6

∃xR(x,x) → ∀x(Q(x) ∨ S(x)) ∀x(R(a,x) ∧ ¬Q(x)) R(a,a) ∧ ¬Q(a) ∀Elim: 2 R(a,a) ∧Elim: 3 ∃xR(x,x) ∃ Intro: 4 ∀x(Q(x) ∨ S(x)) →Elim: 1,5

1 ∃xR(x,x) → ∀x(Q(x) ∨ S(x)) 2 ∀x(R(a,x) ∧ ¬Q(x)) ∃xR(x,x) ∀x(Q(x) ∨ S(x)) ∀xS(x)

⇐

To get ‘∃xR(x,x)’, we’ll get an instance, then apply ∃ Intro. Use line 2 to get the instance ‘R(a,a)’.

∀xS(x) ‘∀xS(x)’ is not in any statement we have yet. Plan to build it by "Intro. Begin a subproof with a boxed constant that is not in any assumption, and aim for an instance of ‘∀xS(x)’ with that constant. Since ‘a’ is in a premise, we must use a different constant. To get ‘S(b)’, begin by applying ∀Elim to 6:

1 2 3 4 5 6 7 8

∃xR(x,x) → ∀x(Q(x) ∨ S(x)) P ∀x(R(a,x) ∧ ¬Q(x)) P R(a,a) ∧ ¬Q(a) ∀Elim: 2 R(a,a) ∧Elim: 3 ∃xR(x,x) ∃ Intro: 4 ∀x(Q(x) v S(x)) →Elim: 1,5

Set up ⇓ for ∨ Elim: 1 ∃xR(x,x) → ∀x(Q(x) ∨ S(x)) P 2 ∀x(R(ax) ∧ ¬Q(x)) P 3 R(a,a) ∧ ¬Q(a) ∀Elim: 2 4 R(a,a) ∧Elim: 3 5 ∃xR(x,x) ∃ Intro: 4 6 ∀x(Q(x) ∨ S(x)) →Elim: 1,5 7 8 Q(b) ∨ S(b) ∀Elim: 6 9 Q(b)

S(b) ∀xS(x)

∃xR(x,x) → ∀x(Q(x) ∨ S(x)) P ∀x(R(a,x) ∧ ¬Q(x)) P R(a,a) ∧ ¬Q(a) ∀Elim: 2 R(a,a) ∧Elim: 3 ∃xR(x,x) ∃ Intro: 4 ∀x(Q(x) ∨ S(x)) →Elim: 1,5 S(b) ∀xS(x)

Q(b) v S(b) S(b) ∀xS(x)

S(b)

⇒ 1 2 3 4 5 6 7

⇒ ⇒ Note that we can get ‘¬Q (b)’ by applying ∀Elim to line 2 again. This yields a contradiction, which we can put to work for us by applying ⊥Elim to get ‘S(b)’. 38

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

∃xR(x,x) → ∀x(Q(x) ∨ S(x)) P ∀x(R(a,x) ∧ ¬Q(x)) P R(a,a) ∧ ¬Q(a) ∀Elim: 2 R(a,a) ∧Elim: 3 ∃xR(x,x) ∃ Intro: 4 ∀x(Q(x) ∨ S(x)) →Elim: 1,5 Q(b) ∨ S(b) ∀Elim: 6 Q(b) R(a,b) ∧ ¬Q(b) ∀Elim: 2 ¬Q(b) ∧Elim: 10 ⊥ ⊥ Intro: 9,11 S(b) ⊥ Elim: 12 S(b) S(b) ∨Elim: 8,9-13,14-14 ∀xS(x) ∀Intro: 7-15

STRATEGY: {Cube(a), ¬∃x∃y(x≠y ∧ SameShape(x,y))} /∴ ∀x(Cube(x) → x = a) 1 Cube(a) 2 ¬∃x∃y(x≠y ∧ SameShape(x,y)) ∀x(Cube(x) → x = a)

We aim for an instance of the existential. We have one conjunct, so we need the other, which we can get right away by AnaCon.

We set up a subderivation to reach our goal by ∀Intro. 1 Cube(a) 2 ¬∃x∃y(x≠y ∧ SameShape(x,y)) 3 b

1 Cube(a) 2 ¬∃x∃y(x≠y ∧ SameShape(x,y)) 3 b 4 Cube(b) 5 b≠a 6 SameShape(b,a) 7 b≠a ∧ SameShape(b,a)

Cube(b) → b = a ∀x(Cube(x) → x = a)

∃x∃y(x≠y ∧ SameShape(x,y)) ⊥ b=a Cube(b) → b = a ∀x(Cube(x) → x = a)

We start another subderivation to build the conditional we need. 1 Cube(a) 2 ¬∃x∃y(x≠y ∧ SameShape(x,y)) 3 b 4 Cube(b) b=a Cube(b) → b = a ∀x(Cube(x) → x = a) The only way to use line 2 is in a contradiction. So we assume the negation of our current goal, and try to get a contradiction with line 2. 1 Cube(a) 2 ¬∃x∃y(x≠y ∧ SameShape(x,y)) 3 b 4 Cube(b) 5 b≠a ∃x∃y(x≠y ∧ SameShape(x,y)) ⊥ b=a Cube(b) → b = a ∀x(Cube(x) → x = a)

⇓

We just need to build up the existential sentence, and complete the justifications.

1 Cube(a) P 2 ¬∃x∃y(x≠y ∧ SameShape(x,y)) P 3 b 4 Cube(b) 5 b≠a 6 SameShape(b,a) AnaCon: 1,4 7 b≠a ∧ SameShape(b,a) ∧Intro: 5,6 8 ∃y(b=y ∧ SameShape(b,y)) ∃Intro: 7 9 ∃x∃y(x≠y ∧ SameShape(x,y)) ∃Intro: 8 10 ⊥ ⊥Intro: 2,9 11 ¬b ≠ a ¬Intro: 5-10 12 b=a ¬Elim: 11 13 Cube(b) → b = a 14 ∀x(Cube(x) → x = a)

→Intro: 4-13 ∀Intro: 3-13

Without AnaCon, we’d need a third premise. You should be able to complete the modified derivation: 1 Cube(a) 2 ¬∃x∃y(x≠y ∧ SameShape(x,y)) 3 ∀x∀y((Cube(x) ∧ Cube(y)) → SameShape(x,y)) ∀x(Cube(x) → x = a) 39

Strategy: {∃x∀y(F(x) → G(x,y)))} /∴ ∀xF(x) → ∃xG(x,x)

1 ∃x∀y(F(x) → G(x,y)) P : : ∀xF(x) → ∃xG(x,x)

Since conclusion is a conditional, set up a subproof to build it by →Intro. 1 ∃x∀y(F(x) → G(x,y)) P 2 ∀xF(x) : : : : ∃xG(x,x) ∀xF(x) → ∃xG(x,x) →Intro

Now we are aiming for an existential statement. When we have one existential statement and are aiming for another one, a good way to try to reach that goal is by ∃Elim. So set up a subproof for later use of the ∃Elim rule. Aim for the desired existential statement at the end of the subproof. Then ∃Elim will let us move it to the left, out of the subproof.

To get ‘∃xG(x,x)’, aim for an instance, then use the ∃Intro rule. So you want ‘G’, followed by two occurrences of the same constant. Since you have a ‘G’ with an ‘a’ in line 3, try for ‘G(a,a)’.

1 ∃x∀y(F(x) → G(x,y)) P 2 ∀xF(x) 3 ∀y(F(a) → G(a,y)) : : : : : : ∃xG(x,x) ∃xG(x,x) ∃Elim ∀xF(x) → ∃xG(x,x) →Intro

1 ∃x∀y(F(x) → G(x,y)) 2 ∀xF(x) 3 ∀y(F(a) → G(a,y)) : : : : : : G(a,a) ∃xG(x,x) ∃xG(x,x) ∀xF(x) → ∃xG(x,x)

Notice that if you plug in ‘a’ in place of ‘y’ using the ∀Elim rule on line 3, you will get a conditional that has your current goal as it’s consequent. That is a promising strategy. In this case, it is particularly promising because its antecedent is easy to get by applying ∀Elim to line 2.

P

∃Elim →Intro

1 ∃x∀y(F(x) → G(x,y)) P 2 ∀xF(x) 3 ∀y(F(a) → G(a,y)) 4 F(a) → G(a,a) ∀Elim: 3 5 F(a) ∀Elim: 2 6 G(a,a) →Elim: 4,5 7 ∃xG(x,x) ∃ Intro: 6 8 ∃xG(x,x) ∃Elim: 1,3-7 9 ∀xF(x) → ∃xG(x,x) →Intro: 2-8 40

STRATEGY: {∀x(Gxa ↔ x = b), ¬∃x(Hx ∧ ∃yG(b,y)), Hb} /∴ ¬∃xG(x,a)

1 ∀x(Gxa ↔ x = b) 2 ¬∃x(Hx ∧ ∃yG(b,y)) 3 H(b)

We have an existential sentence at the top, and are aiming for another. So we start a subderivation to apply ∃Elim to the first one to get the second.

¬∃xG(x,a)

Our goal is a negation, so we plan for ¬Intro and assume the opposite. The contradiction will be with the only negation in the premises.

1 ∀x(Gxa ↔ x = b) 2 ¬∃x(Hx ∧ ∃yG(b,y)) 3 H(b) 4 ∃xG(x,a) 5 c G(c,a)

1 ∀x(Gxa ↔ x = b) 2 ¬∃x(Hx ∧ ∃yG(b,y)) 3 H(b) 4 ∃xG(x,a)

∃yG(b,y) ∃yG(b,y) H(b) ∧ ∃yG(b,y) ∃x(H(x) ∧ ∃yG(x,y)) ⊥ ¬∃xG(x,a)

∃x(H(x) ∧ ∃yG(x,y)) ⊥ ¬∃xG(x,a) ¬Intro

To build the existential that is now our goal by ∃Intro, we need an instance. We could get ‘G(b,a)’ as our instance if we could replace the ‘c’ in line 4 with ‘b’. We can do that with the help of line 1.

To build the existential that is now our goal by ∃Intro, we need an instance. If we use ‘b’, we already have one conjunct on line 3, so we just need to aim for the other conjunct, ‘∃yG(b,y)’. 1 2 3 4 5 6 7 8 9 10 11 12 13 14

1 ∀x(Gxa ↔ x = b) 2 ¬∃x(Hx ∧ ∃yG(b,y)) 3 H(b) 4 ∃xG(x,a) ∃yG(b,y) H(b) ∧ ∃yG(b,y) ∃x(H(x) ∧ ∃yG(x,y)) ⊥ ¬∃xG(x,a)

∧Intro ∃Intro ⊥Intro ¬Intro

41

∀x(Gxa ↔ x = b) ¬∃x(Hx ∧ ∃yG(b,y)) H(b) ∃xG(x,a) c G(c,a) G(c,a) ↔ c = b c=b G(b,a) ∃yG(b,y) ∃yG(b,y) H(b) ∧ ∃yG(b,y) ∃x(H(x) ∧ ∃yG(x,y)) ⊥ ¬∃xG(x,a)

∀Elim: 1 ↔Elim: 5,6 =Elim: 5,7 ∃Intro: 8 ∃Elim: 4,5-9 ∧Intro: 3,10 ∃Intro: 11 ⊥Intro: 2,12 ¬Intro: 4-13

EXTRA EXERCISES FOR CHAPTER 13, GROUP 2 If the argument is FO-valid, use Fitch to give a proof. Start by opening Proof CStern 130x, where 1≤x≤9, or Proof CStern 13x where x≥10. Use AnaCon only for literals (atomic sentences and their negations). Do not use TautCon. If the argument is FO-invalid, use Tarski’s World to give a counterexample. Start by opening Sentences CStern 130x for H 13.x, where 1≤x≤9, or Sentences CStern 13x where x≥10. Save worlds as World CStern 130x or 13x, corresponding to Sentences CStern 130x H 13.9

∃x(Tet(x) ∧ ∀y[Cube(y)  Adjoins(y,x)] ∀x(Cube(x)  ∃y(Tet(y) ∧ Adjoins(x,y)))

H 13.10

∃xCube(x) ∀x(Cube(x)  ∃y(Dodec(y) ∧ Adjoins(x,y))) ∃x(Dodec(x) ∧ ∀y[Cube(y)  Adjoins(y,x)])

H 13.11

∀x∀y([Cube(x) ∧ Tet(y)]  SameSize(x,y)) ∃x∃y (Cube(x) ∧ Tet(y)) ∃x[Dodec(x) ∧ ∀y(Tet(y)  Smaller(x,y))] ∃x∃y (Dodec(x) ∧ Cube(y) ∧ Smaller(x,y))

H 13.12

∀x∀y([Cube(x) ∧ Tet(y)]  SameSize(x,y)) ∃x∃y(Cube(x) ∧ Tet(y)) ∃x[Dodec(x) ∧ ∀y(Tet(y)  Smaller(x,y))] ∀x∀y[(Dodec(x) ∧ Cube(y))  Larger(y,x)]

H 13.13

∃x(Tet(x) ∧ ∀y[(Tet(y) ∧ y≠x)  Larger(x,y)] ∃x∃y(Tet(x) ∧ Tet(y) ∧ y≠x) ∀x∀y[(Tet(x) ∧ SameCol(x,y))  SameSize(x,y)] ∀x∃y(SameCol(x,y) ∧ x≠y) ¬∃x∃y(Tet(x) ∧ Tet(y) ∧ x≠y ∧ SameCol(x,y))

H 13.14

∃x∃y(Dodec(x) ∧ Tet(y) ∧ Larger(x,y)) ¬∃x(Tet(x) ∧ ∀y(Cube(y)  Smaller(x,y))) ∃x(Large(x) ∧ Cube(x))  ∀x(Large(x)  Cube(x))

H 13.15

∃x(Dodec(x) ∧ ∀y(Cube(y)  Larger(x,y)) ∃x(Cube(x) ∧ ∀y(Tet(y)  Larger(x,y))) ∀x((Cube(x) ∨ Tet(x))  ∃y(Dodec(y)  Larger(y,x)))

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Derive conclusion from premises: 1 ∀x(Q(x) → ∃yS(x,y)) 2 ¬∃xQ(x) → ∃x∃yR(x,y))

STRATEGY: {∀x(Q(x) → ∃yS(x,y)), ¬∃xQ(x) → ∃x∃yR(x,y))} /∴ ∃x∃y(R(x,y) ∨ S(x,y))

∃x∃y(R(x,y) v S(x,y)) Our goal is an existential sentence. We have no constant, so no particular instance is obviously easy to get to use for ∃Intro. We have no existential premise, which would lead us to reach our goal by ∃Elim. So as a last resort, assume the negation of the goal sentence. A contradiction will yield the double negation of the goal sentence. Then drop the double negation.

The only negated sentence we have to work from is line 3, so our contradictory pair will probably consist of 3 and the sentence it negates.

∃x∃y(R(x,y) ∨ S(x,y)) ⊥ ¬¬∃x∃y(R(x,y) ∨ S(x,y)) ∃x∃y(R(x,y) ∨ S(x,y))

¬∃xQ(x) ∃x∃yR(x,y) ∃x∃y(R(x,y) ∨ S(x,y)) ⊥ ¬¬∃x∃y(R(x,y) ∨ S(x,y)) ∃x∃y(R(x,y) ∨ S(x,y))

⇐ ¬∃xQ(x) ∃x∃yR(x,y) ⊥ Intro ¬Intro ¬Elim 43

⊥ Intro ¬Intro ¬Elim

1 ∀x(Q(x) → ∃yS(x,y)) 2 ¬∃xQ(x) → ∃x∃yR(x,y)) 3 ¬∃x∃y(R(x,y) ∨ S(x,y)) 4 ∃xQ(x)

∀x(Q(x) → ∃yS(x,y)) ¬∃xQ(x) → ∃x∃yR(x,y)) ¬∃x∃y(R(x,y) ∨ S(x,y)) ∃xQ(x) a Q(a)

∃x∃y(R(x,y) ∨ S(x,y)) ⊥ ¬¬∃x∃y(R(x,y) ∨ S(x,y)) ∃x∃y(R(x,y) ∨ S(x,y))

⊥ ⊥ Intro ¬¬∃x∃y(R(x,y) ∨ S(x,y)) ¬Intro ∃x∃y(R(x,y) ∨ S(x,y)) ¬Elim

1 ∀x(Q(x) → ∃yS(x,y)) 2 ¬∃xQ(x) → ∃x∃yR(x,y)) 3 ¬∃x∃y(R(x,y) ∨ S(x,y))

Further strategizing requires us to think about what provisional assumption might be useful. Assuming '∃xQ(x)' will enable us to use the first premise. Getting a contradiction under that PA would yield '¬∃xQ(x)', which we could use with line 2. So this is promising for our next stage.

1 2 3 4 5

1 ∀x(Q(x) → ∃yS(x,y)) 2 ¬∃xQ(x) → ∃x∃yR(x,y)) 3 ¬∃x∃y(R(x,y) ∨ S(x,y))


The only way we will be able to use line 4 is by applying the ∃ Elim rule to it. So we start a subproof for that, using a boxed constant that does not occur outside this subproof.

⇒ ⇒ CONTINUE

⇒ ⇒

( CONTINUED) 1 ∀x(Q(x) → ∃yS(x,y)) 2 ¬∃xQ(x) → ∃x∃yR(x,y)) 3 ¬∃x∃y(R(x,y) ∨ S(x,y)) To get ⊥ ∃xQ(x) under line 3, 4 5 a Q(a) aim for ⊥ under line 4, ∃x∃y(R(x,y) ∨ S(x,y)) then apply ⊥ ¬Intro. Get ⊥ ⊥ under line 4 ¬∃xQ(x) by ∃Elim. ∃x∃yR(x,y) To get ⊥ ∃x∃y(R(x,y) ∨ S(x,y)) under line 5, ⊥ ⊥ Intro aim for a ¬¬∃x∃y(R(x,y) ∨ S(x,y)) ¬Intro contradiction ∃x∃y(R(x,y) ∨ S(x,y)) ¬Eelim with line 3. ⇒

From '∃yS(a,y)', it should be easy to get '∃x∃y(R(x,y) ∨ S(x,y))' by ∃Elim: 1 ∀x(Q(x) → ∃yS(x,y)) 2 ¬∃xQ(x) → ∃x∃yR(x,y)) 3 ¬∃x∃y(R(x,y) ∨ S(x,y)) 4 ∃xQ(x) 5 a Q(a) 6 Q(a) → ∃yS(a,y) 7 ∃yS(a,y) 8 S(a,b) 9 R(a,b) ∨ S(a,b) 10 ∃y(R(a,y) ∨ S(a,y)) 11 ∃x∃y(R(x,y) ∨ S(x,y)) 12 ∃x∃y(R(x,y) ∨ S(x,y)) 13 ∃x∃y(R(x,y) ∨ S(x,y)) 14 ⊥ 15 ¬∃xQ(x) 16 ∃x∃yR(x,y) ∃x∃y(R(x,y) ∨ S(x,y)) ⊥ ¬¬∃x∃y(R(x,y) ∨ S(x,y)) ∃x∃y(R(x,y) ∨ S(x,y)) ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ Similarly, it should be easy to get '∃x∃y(R(x,y) ∨ S(x,y))' from '∃x∃yR(x,y)' by ∃Elim.

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 44

1 2 3 4 5 6 7

To make use of 'Q(a)', apply ∀E rule to line 1. That gives us '∃yS(a,y)' by →Elim. ∀x(Q(x) → ∃yS(x,y)) ¬∃xQ(x) → ∃x∃yR(x,y)) ¬∃x∃y(R(x,y) ∨ S(x,y)) ∃xQ(x) a Q(a) Q(a) → ∃yS(a,y) ∀Elim:1 ∃yS(a,y) →Elim:5,6 ∃x∃y(R(x,y) ∨ S(x,y)) ⊥ ⊥ Intro ⊥ ∃Elim ¬∃xQ(x) ∃x∃yR(x,y) ∃x∃y(R(x,y) ∨ S(x,y)) ⊥ ¬¬∃x∃y(R(x,y) ∨ S(x,y)) ∃x∃y(R(x,y) ∨ S(x,y))


∀x(Q(x) → ∃yS(x,y)) ¬∃xQ(x) → ∃x∃yR(x,y)) ¬∃x∃y(R(x,y) ∨ S(x,y)) ∃xQ(x) a Q(a) Q(a) → ∃yS(a,y) ∀Elim: 1 ∃yS(a,y) → Elim: 5,6 b S(a,b) R(a,b) ∨ S(a,b) ∨ Intro: 8 ∃y(R(a,y) ∨ S(a,y)) ∃ Intro: 9 ∃x∃y(R(x,y) ∨ S(x,y)) ∃ Intro: 10 ∃x∃y(R(x,y) ∨ S(x,y)) ∃Elim:7,8-11 ∃x∃y(R(x,y) ∨ S(x,y)) ∃Elim: 4,5-12 ⊥ ⊥ Intro: 3,13 ¬∃xQ(x) ¬ Intro: 4-13 ∃x∃yR(x,y) → Elim: 2,15 c ∃yR(c,y) d R(c,d) R(c,d) ∨ S(c,d) ∨ Intro: 18 ∃y(R(c,y) ∨ S(c,y)) ∃ Intro: 19 ∃x∃y(R(x,y) ∨ S(x,y)) ∃ Intro: 20 ∃x∃y(R(x,y) ∨ S(x,y)) ∃Elim:17,18-21 ∃x∃y(R(x,y) ∨ S(x,y)) ∃Elim:16,17-22 ⊥ ⊥ Intro: 3,23 ¬¬∃x∃y(R(x,y) ∨ S(x,y)) ¬Intro: 3-24 ∃x∃y(R(x,y) ∨ S(x,y)) ¬Elim: 25

To show that a sentence is FO-valid, derive ¬∃xF(x) → ∀x(F(x) → G(x)) it from the null set of premises. 1 2

Since the main operator in this sentence is the arrow, we try to build it by →Intro. So we assume the antecedent as PA and aim for the consequent. 1 2 3 4

To get ‘∀x(F(x) → G(x))’, we begin a subproof for ∀Intro with a new boxed constant an easy choice since no constant has appeared yet. We need an instance of the universal with that constant. Plan to use →Intro, then ∀Intro.

STRATEGY to show: ¬∃xF(x) → ∀x(F(x) → G(x)) is first-order valid.

¬∃xFx ∀x(F(x) → G(x)) ¬∃xF(x) → ∀x(F(x) →G(x)) →Intro

¬∃xF(x) a F(a) G(a) F(a) → G(a) ∀x(F(x) → G(x)) ¬∃xF(x) → ∀x(F(x) → G(x))) 1 2 3 4

There is no way we can use line 2 or line 3 directly right now. The only way we can use ‘¬∃xF(x)’ is as a member of a contradictory pair. Since ‘¬∃xF(x)’ will be one member of our contradictory pair, the other will be ‘∃xF(x)’.

→Intro ∀Intro →Intro

¬∃xF(x) a F(a) ∃xF(x) ⊥ G(a) F(a) → G(a) ∀x(F(x) → G(x)) ¬∃xF(x) → ∀x(F(x) → G(x))

We already have all we need to justify 'G(a)’. All that is missing is the rules and the line numbers.

1 2 ¬∃xF(x) 3 a 4 F(a) 5 ∃xF(x) 6 ⊥ 7 G(a) 8 F(a) → G(a) 9 ∀x(F(x) → G(x)) 10 ¬∃xF(x) → ∀x(F(x) → G(x))

45

→Intro ∀Intro →Intro

∃ Intro: 4 ⊥Intro: 2,5 ⊥Elim: 6 →Intro: 4-7 ∀Intro: 3-8 →Intro: 2-9

STRATEGY to show that ∀x∀y((J(x,y) → x≠y) → ¬∃xJ(x,x) is first-order valid

Derive the sentence from the null (empty) set of premises: 1

Removing the universal quantifiers from line 1 will give us a conditional whose antecedent we already have on line 4:

∀x∀y((J(x,y) → x≠y) → ¬∃xJ(x,x)

1 2 3 4 5 6

Goal sentence is a conditional, so start a subderivation for →Intro: 1 2

∀x∀y((J(x,y) → x≠y)

∀x∀y((J(x,y) → x≠y) ∃xJ(x,x) a J(a,a) ∀y((J(a,y) → a≠y) J(a,a) → a ≠ a ⊥ ⊥ ¬∃xJ(x,x) ∀x∀y((J(x,y) → x≠y) → ¬∃xJ(x,x)

¬∃xJ(x,x) ∀x∀y((J(x,y) → x≠y) → ¬∃xJ(x,x)

Our current goal is a negation, so we start a subderivation for ¬Intro: 1 2 3

∀x∀y((J(x,y) → x≠y) ∃xJ(x,x)

⇓

⊥ ¬∃xJ(x,x) ∀x∀y((J(x,y) → x≠y) → ¬∃xJ(x,x)

We have an existential sentence to work with now, so we start a subderivation for ∃Elim, aiming for our current goal of ⊥. 1 2 3 4

∀x∀y((J(x,y) → x≠y) ∃xJ(x,x) a J(a,a) ⊥ ⊥ ¬∃xJ(x,x) ∀x∀y((J(x,y) → x≠y) → ¬∃xJ(x,x)

Applying →Elim to lines 4 and 6 gives us ‘a≠a’. We can get a contradiction immediately by applying =Intro:

1 2 ∀x∀y((J(x,y) → x≠y) 3 ∃xJ(x,x) 4 a J(a,a) 5 ∀y(J(a,y) → a≠y) 6 J(a,a) → a ≠ a 7 a≠a 8 a=a 9 ⊥ 10 ⊥ 11 ¬∃xJ(x,x) 12 ∀x∀y((J(x,y) → x≠y) → ¬∃xJ(x,x)

46

∀Elim:1 ∀Elim: 5 →Elim:4,6 =Intro ⊥Intro: 7,8 ∃Elim: 3,4-9 ¬Intro: 3-10 →Intro: 3-11

STRATEGY to show: ∃x∀y(Q(y) → R(x,y)) → (∀xQ(x) → ∃xR(x,x)) is first-order valid.

To show a sentence is first-order valid (quantificationally true), derive it from the null set of premises. Since our goal sentence is a conditional, try to build it by →I. Start a subproof with the antecedent, and aim for the consequent. Our new goal is another conditional. Set up subproof for another →Intro.

1 2

∀xQ(x) → ∃xR(x,x) ∃x∀y(Q(y) → R(x,y)) → (∀xQ(x) → ∃xR(x,x))

1 2 3

∃x∀y(Q(y) → R(x,y)) ∀xQ(x) ∃xR(x,x) ∀xQ(x) → ∃xR(x,x) ∃x∀y(Q(y) → R(x,y)) → (∀xQ(x) → ∃xR(x,x))

To use line 2, we must apply the ∃Elim rule. As soon as we recognize that, we begin the subproof this rule requires. We need a sentence without the boxed constant introduced in the first line of the subproof. We can move that sentence out of the subproof by ∃Elim. So we aim for our next goal, ‘∃xR(x,x)’. Substituting ‘a’ for the variable in both universal sentences gives us what we need to get ‘R(a,a)’. That in turn lets us reach the goal of ‘∃xR(x,x)’. All we need to add is line numbers and justifications.

∃x∀y(Q(y) → R(x,y))

1 2 3 4

∃x∀y(Q(y) → R(x,y)) ∀xQ(x) a ∀y(Q(y) → R(x,y)) ∃xR(x,x) ∃xR(x,x) ∀xQ(x) → ∃xR(x,x) ∃x∀y(Q(y) → R(x,y)) → (∀xQ(x) → ∃xR(x,x))

1 2 ∃x∀y(Q(y) → R(x,y)) 3 ∀xQ(x) 4 a ∀y(Q(y) → R(x,y)) 5 Q(a) 6 Q(a) → R(a,a) 7 R(a,a) 8 ∃xR(x,x) 9 ∃xR(x,x) 10 ∀xQ(x) → ∃xR(x,x) 11 ∃x∀y(Q(y) → R(x,y)) → (∀xQ(x) → ∃xR(x,x))

47

∀Elim: 3 ∀Elim: 4 →Elim: 5,6 ∃Intro: 7 ∃Elim: 2,4-8 →Intro: 3-9 →Intro: 1-10

To show 2 sentences are first-order equivalent, derive biconditional from the null (empty) set of premises. Use standard strategy: build biconditional by ↔Intro. 1 2

STRATEGY: Show that ∀x(B(x) ↔ ¬C(x)) is first-order equivalent to ¬∃x[(B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))]

∀x(B(x) ↔ ¬C(x)) ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) ∀x(B(x) ↔ ¬C(x)) ∀x(B(x) ↔ ¬C(x)) ↔ ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x)))

Since the goal in the first subproof is a negation, we'll probably get it by ¬Intro. So we assume the sentence that it negates.

1 2 3

∀x(B(x) ↔ ¬C(x)) ∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x)))

To use line 3, plan on ∃Elim. Step 4 begins a subproof for ∃Elim. We need a contradiction under 3, so we aim for ⊥. We will move it left by ∃Elim to get ⊥ under line 3, where we really want it.

1 2 3 4 5

1 2 3 4

∀x(B(x) ↔ ¬C(x)) ∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) a (B(a) ∧ C(a)) ∨ (¬B(a) ∧ ¬C(a)) ⊥ ⊥ ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x)))

We will need to use line 2, so substitute the boxed constant from line 4 for the x's in line 2. 1 ∀x(B(x) ↔ ¬C(x)) 2 ∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) 3 a (B(a) ∧ C(a)) ∨ (¬B(a) ∧ ¬C(a)) 4 B(a) ↔ ¬C(a) 5 6 ⊥ ⊥ ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) ⇒

∀x(B(x) ↔ ¬C(x)) ∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) a (B(a) ∧ C(a)) ∨ (¬B(a) ∧ ¬C(a)) B(a) ↔ ¬C(a) B(a) ∧ C(a) ¬B(a) ∧ ¬C(a) ⊥ ⊥ ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x)))

To use line 4, start two subderivations for ∨Elim.

⇒ ⇒ CONTINUE 48

⇒ ⇒

1 2 3 4 5 6

∀x(B(x) ↔ ¬C(x)) ∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) a (B(a) ∧ C(a)) ∨ (¬B(a) ∧ ¬C(a)) B(a) ↔ ¬C(a) 2,∀ Elim B(a) ∧ C(a) ⊥ ¬B(a) ∧ ¬C(a) ⊥

⊥

⊥Intro ∨ Elim ∃ Elim ¬Intro

⊥ ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x)))

1 2 3 4 5 6 7 8 9 10

∀x(B(x) ↔ ¬C(x)) ∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) a (B(a) ∧ C(a)) ∨ (¬B(a) ∧ ¬C(a)) B(a) ↔ ¬C(a) 2,∀ Elim B(a) ∧ C(a) B(a) 6,∧ Elim C(a) 6,∧ Elim ¬C(a) 5,7,↔ Elim ⊥ 8,9,⊥Intro

11 12 13 14 15 16 17 18

¬B(a) ∧ ¬C(a) ¬C(a) B(a) ¬B(a) ⊥

11,∧ Elim 5,12,↔ Elim 11,∧ Elim 13,14,⊥Intro ⊥ 4,6-10, 11-15,∨ Elim ⊥ 3,4-16, ∃ Elim ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) 2-16,¬Intro

19

¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x)))

We need a contradiction under line 6. That will allow us to apply the ¬Intro rule to line 6. We can get contradictions in both subproofs for ∨ Elim. So we'll bring '⊥' out of both subproofs in our ∨Elim step.

In the first subproof, get 'B(a)' by ∧ Elim. To get'¬B(a)', notice that we can easily get both 'C(a)' (by ∧ Elim) and '¬C(a)' (by ↔ Elim). To put this contradiction to work for us, apply ⊥Intro. Getting 'B(a)' and '¬B(a)' in the second subproof is easy, giving us ⊥ again.. Moving ⊥ out of the 2 subproofs by ∨ Elim gives us ⊥ under line 3. We can then move it to the left, under line 2, by ∃ Elim, justifying ¬Intro. This completes the first subproof.

∀x(B(x) ↔ ¬C(x)) ∀x(B(x) ↔ ¬C(x)) ↔ ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x)))

⇒ ⇒ CONTINUE

49

⇒ ⇒

(CONTINUED) We are now ready to begin the second subproof.

1

∀x(B(x) ↔ ¬C(x))

18

¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x)))

19

¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) ∀x(B(x) ↔ ¬C(x)) ∀x(B(x) ↔ ¬C(x)) ↔ ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x)))

19 20 21

¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) b B(b) ¬C(b)

To get '∀x(B(x) ↔ ¬C(x))', we plan to use ∀Intro. So we start a subproof with a boxed constant that does not occur outside the subproof, 'b'. Aim for 'B(b) ↔ ¬C(b)', by ↔ Intro.

¬C(b) B(b) B(b) ↔ ¬C(b) ∀x(B(x) ↔ ¬C(x)) To get '¬C(b)' by ¬Intro, assume 'C(b)'. Conjoin with 'B(b). Use ∨ Intro and ∃ Intro to build the sentence that is negated in line 19, yielding a contradiction:

To get 'B(b)' from '¬C(b)', use a strategy like the one for getting '¬C(b)' from 'B(b)'. This completes the proof.

27

19 20 21 22 23 24 25 26 27 28

¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) b B(b) C(b) B(b) ∧ C(b) ∧Intro: 21,22 (B(b) ∧ C(b)) ∨ (¬B(b) ∧ ¬C(b)) ∨Intro: 23 ∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) ∃Intro: 24 ⊥ ⊥Intro: 19,25 ¬C(b) ¬Intro: 22-26 ¬C(b) B(b) B(b) ↔ ¬C(b) ∀x(B(x) ↔ ¬C(x))

¬C(b)

22-26,¬Intro

28 ¬C(b) 29 ¬B(b) 30 ¬B(b) ∧ ¬C(b) ∧ Intro: 28, 29 31 (B(b) ∧ C(b)) v (¬B(b) ∧ ¬C(b)) ∨ Intro: 30 32 ∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) ,∃ Intro: 31 33 ⊥ ⊥ Intro: 19,32 34 ¬¬B(b) ¬Intro: 29-33 35 B(b) ¬ Elim: 34 36 B(b) ↔ ¬C(b) ↔Intro: 21-27, 28-35 37 ∀x(B(x) ↔ ¬C(x)) ∀Intro: 20-36 38 ∀x(B(x) ↔ ¬C(x)) ↔ ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) ↔Intro: 2-18,19-37 ⇒ ⇒ Entire proof shown on the next page ⇒ ⇒ 50

In total, our proof showing that ∀x(B(x) ↔ ¬C(x)) is first-order equivalent to ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) looks like this:

1 2 3 4 5 6 7 8 9 10

∀x(B(x) ↔ ¬C(x)) ∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) a (B(a) ∧ C(a)) ∨ (¬B(a) ∧ ¬C(a)) B(a) ↔ ¬C(a) B(a) ∧ C(a) B(a) C(a) ¬C(a) ⊥

11 12 13 14 15 16 17 18

¬B(a) ∧ ¬C(a) ¬C(a) B(a) ¬B(a) ⊥ ⊥ ⊥ ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x)))

19 20 21 22 23 24 25 26 27

¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) b B(b) C(b) B(b) ∧ C(b) B(b) ∧ C(b) ∨ (¬B(b) ∧ ¬C(b)) ¬∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) ⊥ ¬C(b)

28 ¬C(b) 29 ¬B(b) 30 ¬B(b) ∧ ¬C(b) 31 (B(b) ∧ C(b)) v (¬B(b) ∧ ¬C(b)) 32 ∃x((B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))) 33 ⊥ 34 ¬¬B(b) 35 B(b) 36 B(b) ↔ ¬C(b) 37 ∀x(B(x) ↔ ¬C(x)) 38 ∀x(B(x) ↔ ¬C(x)) ↔ ¬∃x[(B(x) ∧ C(x)) ∨ (¬B(x) ∧ ¬C(x))]

51

∀ Elim: 2 ∧ Elim: 6 ∧ Elim: 6 ↔ Elim: 5,7 ⊥ Intro: 8,9 ∧Elim: 11 ↔Elim: 5,12 11,∧Elim: 11 13,14,⊥Intro: 13,14 ∨ Elim: 4,5-10,11-15 ∃ Elim: 3,4-16 ¬ Intro: 3-17

∧ Intro: 21,22 ∨ Intro: 23 ∃ Intro: 24 ⊥ Intro: 19,25 ¬ Intro: 22-26

28,29,∧ Intro: 28,29 ∨ Intro: 30 ∃ Intro: 31 ⊥ Intro:19,32 ¬ Intro: 29-33 ¬¬ Elim: 34 ↔ Intro: 21-27,28-35 ∀Intro: 20-36 ↔Intro: 2-18,19-37

EXTRA EXERCISES FOR CHAPTER 13, GROUP 3 If the sentence is FO-valid, open Proof CStern 13x, and complete and save the proof. If it is FO-invalid, use Tarski’s World to give a counterexample. (Start by opening Sentences CStern 13x. Save your solution as World CStern 13x.) ∃x¬Dodec(x) → ∃x(Dodec (x) → Small(x)) ¬∀x Large(x) ∨ ∀x (Medium(x) → Large(x)) ∀x (Tet(x) ∨ Cube(x)) → (∃x Tet(x) ∨ ∀x Cube(x)) ∀x(Cube(x) → Small(x)) → (∃x Cube(x) → ∃x Small(x)) ∀x (Tet(x) → ¬Medium(x)) ∨ ∃x(Tet(x) ∧ Medium(x))

H13.16 - 13.24:

H 13.16 H 13.17 H 13.18 H 13.19 H 13.20 H 13.21 ∃x∃y(Cube(x) ∧ Tet(y) ∧ ¬SameCol(x,y)) → ¬∃x∃y(Cube(x) ∧ Tet(y) ∧ SameCol(x,y)) H 13.22 ∀x∃y(Small(x) → RightOf(x,y)) → (∀x¬Small(x) ∨ ∃x∃yRightOf(x,y)) H 13.23 ∃x∀y ¬Larger(x,y) → ∀y∃x ¬Larger (x,y) H 13.24 ∃x[Cube(x) ∧ ¬∀y(Tet(y) → Larger(y,x))] → ∃x[Cube(x) ∧ ∀y(Tet(y) →¬Larger(y,x))] H 13.25 - H 13.32: H 13.25 H 13.26 H 13.27 H 13.28 H 13.29 H 13.30 H 13.31 H 13.32

H 13.33 - H 13.34

H 13.33 1. 2. H 13.34 1. 2.

Open Proof CStern 13x for H 13.x. Use Fitch to give a proof showing each pair of sentences is FO-equivalent. ∃x (LeftOf(x,a) ∨ BackOf(x,a)), ∃x LeftOf(x,a) ∨ ∃x BackOf(x,a) ∀x (Small(x) ∧ Tet(x)), ∀x Small(x) ∧ ∀x Tet(x) ∃x (Dodec(x) ↔ ¬Large(x)), ¬∀x(Dodec(x) ↔ Large(x)) ∀x (FrontOf(a,x) → ¬Large(x)), ¬∃x(FrontOf(a,x) ∧ Large(x)) ∀x (Tet(x) → Medium(a)), ¬∃x Tet(x) ∨ Medium(a) ∃x Large(x) ∨ Medium(b)), ∀x ¬Large(x) → Medium(b) ∃xDodec(x) → Small(a), ∀x (Dodec(x) → Small(a)) ∀x∃y (S(x) → R(x,y)), ∀x (S(x) → ∃y R(x,y))

Open Sentences CStern 13x. To show each pair of sentences is not FO-equivalent build a world where the first is true but the second is false. Save as World CStern 13x.

∀x∀y (SameCol(x,y) → SameSize(x,y)) ∀x∀y (SameSize(x,y) → SameCol (x,y)) ∀x∃y ¬Larger(x,y) ∃y∀x ¬Larger(x,y)

52

H 13.35 - 13.38

Open Proof CStern 13x.Use Fitch to give a proof showing that each set of sentences is FO-inconsistent.

H 13.35 {∀x (J(a,x) ↔ (K(x) ∨ L(x)), ∃x (J(a,x) ∧ ¬K(x)), ¬∃x L(x)} H 13.36 { ∃x Cube(x), ∃x∀y SameCol(x,y), ∀x (∃y SameCol(y,x) → ¬Cube(x)) H 13.37 { ∀x(∃y¬SameShape(x,y) → ∀ySameSize(x,y)), ¬∃x∃y(SameSize(x,y) ∧ Adjoins(x,y)), ∀x∀y(SameShape(x,y) ∨ Adjoins(x,y)), ∃x∃y¬(SameShape(x,y) ∨ Dodec(x))} H 13.38 {∀x(B(x) → ∃y(C(x,y)), ∀x∀y(C(x,y)) ↔ ¬D(y,x)), ∀x∀y(D(x,y) ∨ G(x,y)), ∃x(B(x) ∧ ¬∃y G(y,x))} H 13.39 - H 13.41 Use Tarski’s World to show that the set with the listed sentences as its members is FO-consistent: H 13.39

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

¬∃x(Tet(x) ∧ ∀y(Cube(y) → Adjoins(x,y))) ∀y (Cube(y) → ∃x (Tet(x) ∧ Adjoins (x,y))) ∀x∀y ((Tet(x) ∧ Tet(y)) → SameSize (x,y)) ∃x∃y (Dodec(x) ∧ Dodec(y) ∧ x≠y ∧ SameSize(x,y)) ¬∀x∀y ((Dodec(x) ∧ Dodec(y)) → SameSize(x,y)) ∃x∃y (Cube(x) ∧ Cube(y) ∧ ¬SameSize(x,y))

H 13.40

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

∀x ∀y ((Cube(x) ∧ Dodec(y)) → BackOf(x,y)) ∃x∃y∃z (Dodec(x) ∧ Tet(y) ∧ Tet(z) ∧ Between(x,y,z)) ∃x (Cube(x) ∧ ∃y (Tet(y) ∧ SameRow(x,y))) ¬∀x∀y ((Cube(x) ∧ Cube(y)) → SameRow(x,y)) ¬∃x∃y (Tet(x) ∧ Tet(y) ∧ x≠y ∧ SameRow(x,y)) ∃x∃y (Dodec(x) ∧ Dodec(y) ∧ ¬(SameRow(x,y) ∨ SameCol(x,y)))

H 13.41

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Cube(a) ∧ Tet(b) ∧ Tet(c) ∧ b≠c ¬∀x(Tet(x) → (x=b ∨ x=c)) ∀x(Tet(x) → ∃y(Dodec(y) ∧ Adjoins(y,x))) ¬∃x∃y(Cube(x) ∧ Dodec(y) ∧ SameCol(x,y)) ∃x∃y(Cube(x) ∧ Cube(y) ∧ x≠y ∧ ¬∃z[Cube(z) ∧ z≠x ∧ z≠y]) ∀x∀y[¬(Tet(x) ∧ Tet(y)) ∨ SameSize(x,y)] ∀x∀y[(Cube(x) ∧ Tet(y)) → Larger(x,y)] ∀x[Cube(x) → ∃y∃z(Dodec(y) ∧ Dodec(z) ∧ Between(x,y,z))] ∀x∀y[(Cube(x) ∧ Cube(y) ∧ x≠y) → ¬SameSize(x,y)] ∃x(Dodec(x) ∧ ∀y[(Dodec(y) ∧ y≠x) → Larger(x,y)])

53