Homework assignment 1: Solutions

Transcription

1 Math 240: Discrete Structures I Due 4:30pm Friday 29 September McGill University, Fall 2017 Hand in to the mailbox at Burnside Homework assignment 1: Solutions Discussing the assignment with other students is allowed (and encouraged!) but you should hand in your own write-up. Only one name per assignment. Please cite the sources you used, including the books and websites you consulted, and the names of the people you collaborated with or who helped you. Late assignments are not accepted. Only your best four out of five homework assignments count. You can earn up to 5 bonus points for submiting your homework assignment typed up in L A TEX. Total: 60 points Problem 1 (Truth tables). Use truth tables to decide whether each of the following sentences is a tautology, a contradiction, or a contigency. [5 points] a) (P Q) ( P Q) b) (DeMorgan s Law) ( P Q) P Q c) (P Q) ((P Q) ( P Q)) d) (X Y ) (Y X) e) ( P Q) (R Q) a) P Q P P Q P Q (P Q) ( P Q) T T F T T T T F F F F T F T T T T T F F T T T T Tautology. b) P Q P Q P Q ( P Q) P Q ( P Q) P Q T T F F F T T T T F F T F T T T F T T F F T T T F F T T T F F T Tautology

2 c) P Q P Q P Q P Q P Q (P Q) ( P Q) T T F F T T F T T F F T F F F F F T T F F F F F F F T T T F T T (P Q) (P Q) ( P Q) T T T T Tautology. d) X Y X Y (X Y ) Y X (X Y ) (Y X) T T T F T T T F F T T T F T T F F T F F T F T T Tautology. e) P Q R P Q P Q R Q ( P Q) (R Q) T T T F F F T T T T F F F F T T T F T F T F F T T F F F T F T T F T T T F F T T F T F T F F T T F F T T T T F F F F F T T T T T Contingency. Problem 2 (Logical identities). Use the logical identities given in class to prove the equivalences listed below. The logical identities can be found on our Piazza site or at [4 points] a) (X Y ) (X Z) (X (Y Z)) b) (P Q) ((P Q) ( P Q)) a) (X Y ) (X Z) ( X Y ) ( X Z) X Y Z X (Y Z)

3 b) P Q (P Q) ( P Q) (P ( P Q)) (Q ( P Q)) (P P ) (P Q) (Q P ) (Q Q) (P Q) (Q P ) ( P Q) (Q P ) (P Q) ( P Q) Problem 3 (Order of precedence). For each pair of sentences given below, show that they are not logically equivalent. [4 points] a) (P Q) R; P (Q R) b) (X Y ) Z; X (Y Z) The following truth assignments show that the sentence pairs are not equivalent. a) b) P Q R F T F X Y Z F T T Problem 4 (Symbolization). Symbolize the following English sentences in logic, using the abbreviation scheme provided. [18 points] a) I only leave my door open when I m in my office. [D: I leave my door open; M: I m in my office] b) Neither Peter nor Jake has his driver s licence. [P : Peter has his driver s licence; J: Jake has his driver s licence] c) If only kryptonite kills me, them I am superman. [ k: kryptonite; K(x): x kills me; S: I am superman ] d) Alice is not drinking beer underage. [B: Alice is drinking beer; Q: Alice is at least 18] (Note: legal drinking age is 18.) e) = 4, even if my heart would break. [A: = 4; B: my heart would break]

4 f) You can t jump over buildings unless you re a superhero. [ B: you can jump over buildings; H: you are a superhero. ] g) Only a knight can marry my daughter. [ K(x): x is a knight; M(x, y): x can marry y; d: my daughter ] h) There are at least four distinct people. [ P (x) : x is a person ] i) The square-root of 5 is irrational. [ 5: the number five; x y: the product of x and y; I(x): x is a nonzero integer] a) D M b) P J c) x (K(x) x = k) S ( We will also accept (K(k) x (K(x) x = k)) S. The reason is that there is more than one way to interpret the sentence (A) Only kryptonite kills me. These are: (1) Nothing kills me, except perhaps kryptonite. (And maybe nothing kills me). (2) Kryptonite kills me, but nothing else does. We have learned in class to interpret sentence (A) as (1). But I think when most people listen to (A), they hear (2). This is interesting: applying the same reasoning, the sentence (see part (g)) (B) Only a knight can marry my daughter. would have two analogous possible interpretations: (1) No one can marry my daughter, except perhaps a knight. (And maybe no one can marry her.) (2) There is a knight who can marry my daughter, but no non-knight can marry my daughter. With sentence (B), I think most people hear (1). These two sentences appear to be of a similar form, but why do we hear (2) when we listen to sentence (A), and (1) when we listen to sentence (B)?

5 I don t know how to explain the difference. I m a mathematician, and this is psychology/linguistics/philosophy. The difference in how we interpret these sentences may be connected to the fact that (1) mentions a specific thing (kryptonite) which may satisfy the predicate, whereas (2) does not. I would be interested to hear your thoughts in a piazza discussion. ) d) A Q e) A f) B H g) x (M(x, d) = K(x)) h) w x y z (P (w) P (x) P (y) P (z) w x w y w z x y x z y z) i) x (x x = 5 q (I(q) I(q x)) Problem 5 (Converse and contrapositive). For each of the following implications, write both the converse and the contrapositive in English. [6 points] a) If Susan is a sophomore, then she needs to take Math 240. b) The only way to succeed is to keep trying. c) You can only graduate if you have paid your library fines. a) Converse: If Susan needs to take Math 240, then she is a sophomore. Contrapositive: If Susan does not need to take Math 240, then she is not a sophomore. b) Converse: If you keep trying, then you will succeed. Contrapositive: If you don t keep trying, then you won t succeed. c) Converse: If you have paid your library fines, then you can graduate. Contrapositive: If you have not paid your library fines, then you cannot graduate. Problem 6 (Knights and knaves). You encounter three inhabitants A, B, and C on the island of knights and knaves. A says, B is a knave. B says, A and C are of the same type. What is C? (I.e. a knight, or a knave?) Use truth tables, the logic identities, or a combination of both to justify your answer. [6 points]

6 We define the following propositions: P : A is a knight. Q : B is a knight. R : C is a knight. A has said Q, and B has said P R. So P Q (1) and Q (P R) (2) Note that (1) is equivalent to P Q, and putting this together with (2) tells us P (P R) (3) must be true. Now consider the following truth table, which has (3) in the last column. P R P P R P (P R) T T F T F T F F F T F T T F F F F T T T This table shows that the only way to make (3) true is to make R false. In other words, C must be a knave. Problem 7 (Negation). Match up each sentence on the lefthand side with its negation on the righthand side, and justify your answers with the logic identities given in class, together with the rules for negation with quantifiers. [8 points] A) x(p (x) [( yp (y)) ( yr(x, y)) ( yp (y))]) B) x y( R(x, y) P (x)) C) x(p (x) yr(x, y)) D) x y(p (x) R(x, y) R(y, x)) 1) x y(p (x) R(x, y) R(y, x)) 2) x([p (x) yr(x, y)] [ P (x) y R(x, y)]) 3) x y(p (x) R(x, y)) 4) x P (x) A4: x (P (x) [( yp (y)) ( yr(x, y)) ( yp (y))]) x ( P (x) [( yp (y)) ( yr(x, y)) ( yp (y))]) x ( P (x) [( yp (y)) ( yr(x, y)) ( yp (y))]) x ( P (x) F ) x P (x),

7 where the third comes from noticing that the sentence yp (y) occurs together with its negation in a conjuction. B3: C2: D1: x y( R(x, y) P (x)) x y (R(x, y) P (x)) x (P (x) y R(x, y)) x ((P (x) y R(x, y)) ( P (x) y R(x, y))) x ((P (x) y R(x, y)) ( P (x) y R(x, y))) x ([P (x) y R(x, y)] [ P (x) y R(x, y)]) x ([P (x) y R(x, y)] [ P (x) y R(x, y)]) x y(p (x) R(x, y) R(y, x)) x y(p (x) R(x, y) R(y, x)) Problem 8 (Thinking). Let us introduce a new logical connective, which we define with the following truth table. [9 points] P Q P Q T T F T F T F T T F F T The sentence P Q is logically equivalent to (((P P ) Q) ((P P ) Q)) (((P P ) Q) ((P P ) Q)). Find a sentence written only with parentheses,, and the proposition letters X and Y, which is logically equivalent to (X ( X Y )). Prove that your answer is correct. (For instance, you could provide a truth table.) You can actually rewrite any sentence in propositional logic using only the connective. To see this, note that you can rewrite the connectives and using only,, and. Using DeMorgan s law, you can rewrite in terms of and (or in terms

8 of and ). So we just need to show how to rewrite and in terms. And here s how. First note that P P P. Now, inspection of the truth table for reveals that P Q (P Q). So (P Q) (P Q) P Q. That s the proof. This proof gives us an algorithm for turning any sentence in propositional logic into one which uses only. We could apply the algorithm to get an answer for the problem, but it s easier to do some simplification first. (X ( X Y )) X ( X Y ) X X Y X Y. Now we see that a correct answer is (X Y ) (X Y ).