Math 131 - Dr. Miller - HW: Tarski's World - SOLUTIONS, Sept. 18 ...

Math 131 - Dr. Miller - HW: Tarski’s World - SOLUTIONS, Sept. 18, 2013 Use the following Tarski World to answer the questions that follow; unempty boxes contain first the letter name of the object, and then a verbal description since I can’t print in color. For instance, the top left box contains a blue triangle whose name is A.

A blue 4

C blue D gray

B red 4

J blue 4 F gray ♥ G gray 4

I gray ♥

E gray 4

K blue H red 4

L gray ♥

The “rules” are based on those from our handout of last week plus these: SameRow(x,y) means “x is in the same horizontal row as y.” SameColumn(x,y) is defined similarly. Below(x,y) means “x is below y” (but possibly in different rows). Classify each statement as true or false, then justify your claim with a sentence of explanation. 1. ∃x, Blue(x) True - C is blue, so there does exist an object that is blue. 2. ∀x, Blue(x) False - D is not blue, so it is not true that all objects are blue. 3. ∃x, Blue(x) ∧ SameRow(x, B) False - all objects in the same row as B are colors other than blue. or, False - all blue objects are in different rows from B. 4. ∀x, Blue(x) =⇒ (SameRow(x, A) ∨ SameColumn(x, J ) True - the statement is equivalent to the implication “If x is blue, then x is in the same row as A or in the same column as J . When we check each blue object (those making the hypothesis true), we find that only A, C, J , and K are blue. Of those, A, C, and J are certainly in the same row as A while K is in the same column as J , and so the conclusion is true whenever the hypothesis is. That makes the implication true. 5. ∀x, Blue(x) ∨ Gray(x) False - B is not blue and also not gray, so it’s untrue to say that all objects are blue or gray. 6. ∃x, Blue(x) ∧ Gray(x) False - Each object is just a single color. There are none that are simultaneously blue AND gray.

7. ∃x ∧ ∃y, SameRow(x, y) ∧ Red(x)∧ ∼ Red(y). True - the statement says there exist two objects in the same row as each other where one of them is red and the other isn’t. B and G meet these conditions. 8. ∀x, RightOf(x, J ) =⇒ Red(x) True - We discussed in class how it’s best to merge the universal quantifier into the implication since they’re interchangeable. So the statement claims that “if an object is to the right of J , then the object is red.” This implication’s hypothesis is ALWAYS false - there are no object to the right of J - making the entire implication true. 9. OMIT ∃x, ∀y 6= x, Below(x, y) =⇒ Red(x) We omitted this problem. 10. ∀x, Above(A, x) False - A isn’t above C: they’re level with each other. Thus, it’s not true to say that A is above all objects. 11. ∀x, ∀y, (Blue(x) ∨ Blue(y)) =⇒∼ SameRow(y, D) False - again, we should merge the universal phrase into the implication. The statement now says the “if one object is blue or another is blue, then the second object is not in the same row as D. Take C as the first object - it’s blue, so the “or” hypothesis is true and take F as the second object. Unfortunately, the second object IS in the same row as D, so the conclusion is false. True hypothesis mixed with false conclusion means false implication overall. 12. ∃x, ∀y, (Gray(y) ∧ Heart(y)) =⇒ Above(y, x) False - this time, let’s merge the implication into the ∀ portion of the statement. The entire set-up claims that there is an object where all gray hearts are above this object. L isn’t above anything, so that makes the claim false. 13. ∀y, (Gray(y) ∧ Heart(y)) =⇒ (∃x, Above(x, y)) True - Here, let’s merge into an implication again - see how handy it’s been to be able to trade back and forth? We wonder about the statement “if an object is a gray heart, then there’s an object above the gray heart.” It’s true because gray heart F has C above it (remember, different columns are okay), gray heart I has C above it, and gray heart L also has C above it. I didn’t have to use the same item, C, every time, but I decided to because it was easy to see.

Also complete p. 93 #5-8 and p. 95 #46-54. I won’t publish answers for textbook problems.