Sample Solutions on HW1 (29 questions in total) Sec. 1.1 2, 6(d, e), 14( e), 22, 28(b), 32(f), 48 2. a) This is not a proposition; its a command. b) This is not a proposition; its a question. c) This is a proposition that is false. (This is not a common knowledge to our students, so a full credit will be given as long as a proposition is been determined.) d) This is not a proposition; its truth value depends on the value of x. e) This is a proposition that is false. f) This is not a proposition; its truth value depends on the value of n. 6. d) False, because the hypothesis of this conditional statement is true and the conclusion is false. e) False, because the first part of this biconditional statement is false and the second part is true. 14(e) (p q) r 22 a) If you get promoted, then you wash the bosss car. b) If the winds are from the south, then there will be a spring thaw. c) If you bought the computer less than a year ago, then the warranty is good. d) If Willy cheats, then he gets caught. e) If you can access the website, then you must pay a subscription fee. f) If you know the right people, then you will be elected. g) If Carol is on a boat, then she gets seasick. 28(b) Converse: Whenever I go to the beach, it is a sunny summer day. Contrapositive: Whenever I do not go to the beach, it is not a sunny summer day. Inverse: Whenever it is not a sunny day, I do not go to the beach. 32(f) p q q p q p q (p q) (p q) T T F T F T T F T F T T F T F F T T F F T T F T 48 This cannot be a proposition, because it cannot have a truth value. Indeed, if it were true, then it would be truly asserting that it is false, a contradiction; on the other hand if it were false, then its assertion that it is false must be false, so that it would be true again a contradiction. Thus this string of letters, while appearing to be a proposition, is in fact meaningless. Sec. 1.2 4, 10, 18 4 The condition stated here is that if you use the network, then either you pay the fee or you are a subscriber. Therefore the proposition in symbols is w d s. 10 We write these symbolically: u a, a s, s u. Note that we can make all the conclusion true by making a false, s true, and u false. Therefore if the users cannot access the file system, they can save new files, and the system is not being upgraded, then all the conditional statements are true. Thus the system is consistent. 18 We will translate these conditions into statements in symbolic logic, using j, s, and k for the propositions that Jasmine, Samir, and Kanti attend, respectively. The first statement is j s. The second statement is s k, . The last statement is k j, , because “unless” means “or.” (We could also translate this as k j. From the comments following Definition 5 in the text, we know that jp q is equivalent to “q unless p. In this case p is j and q is k .) First, suppose that s is true. Then the second statement tells us that k is also true, and then the last statement forces j to be true. But now the first statement forces s to be false. So we conclude that s must be false; Samir cannot attend. On the other hand, if s is false, then the first two statements are automatically true, not matter what the truth values of k and j are. If we look at the last statement, we see that it will be true as long as it is not the case that k is true and j is false. So the only combinations of friends that make everybody happy are Jasmine and Kanti, or Jasmine alone (or no one!). Sec. 1.3 6, 8(a,b), 12(b), 30, 32, 40, 51 6 We see that the fourth and seventh columns are identical. 8 a) Kwame will not take a job in industry and will not go to graduate school. b) Yoshiko does not know Java or does not know calculus. 12(b) )()()(rprqqp )()()(rprqqp )()()(rprqqp rprqqp)()( )()(rrqpqp )()()()(rrrqpqpp )()(rqpq p q p q (p q) p q p q T T T F F F F T F F T F T T F T F T T F T F F F T T T T rpqq)( TrpT 30 Solution 1: )()()(rqrpqp )()()(rqrpqp )()()(rqrpqp )()(rrpqqp Trqpprpqp)()( Solution 2: Besides using the equivalences to show this is a tautology, we can argue directly as follows: The conclusion q r will be true in every case except when q and r are both false. But if q and r are both false, then one of p q or p r is false, because one of p or p is false. Thus in this case the hypothesis (p q) ( p r) is false. A conditional statement in which the conclusion is true or the hypothesis is false is true, and that completes the argument. 32 We just need to find an assignment of truth values that makes one of these propositions true and the other false. We can let p be true and the other two variables be false. Then the first statement will be F F, which is true, but the second will be F T, which is false. 40 Following the hint, we see that the answer i