ANALYSIS EXERCISE 1 SOLUTIONS

Transcription

1 ANALYSIS EXERCISE 1 SOLUTIONS 1. (a) Let B The main course will be beef. F The main course will be fish. P The vegetable will be peas. C The vegetable will be corn. The logical form of the argument is B F P C (F C) (B P ) The premises are B F, P C, (F C); the conclusion is (B P ). The argument is invalid. If B P is true, then the premises are true, but the conclusion is false. It might help to consider the possibilities (B P, etc.). (b) Let 2. (a) S Sales will go up. H The boss will be happy. E Expenses will go up. The logical form of the argument is (S H) (E H) (S E) The premise is (S H) (E H); the conclusion is (S E). The argument is invalid. Suppose S E H. Then the premise is valid and the conclusion false. P Q P + Q T T F T F T F T T F F F (b) (P Q) ( P Q). 3. ( P Q), for example. 4. (b), (c), (d). 5. (a) A B C C B C A (B C) T T T F T T T T F T T T T F T F F F T F F T T T F T T F T F F T F T T F F F T F F F F F F T T F 1

2 2 ANALYSIS EXERCISE 1 SOLUTIONS (b) A B A B A B A B (A B) ( A B) T T F F T F F T F F T T T T F T T F T T T F F T T F T F 6. (a) (b) (c) [(A B) (B C)] (A C). Let us prove that this is a law by constructing its truth table. (Denote by D the above propositional form.) A B C A B B C (A B) (B C) A C D T T T T T F T F F F T T F T F T F T T T F F F T F F T F T T F T F T F F T T F F T F F F [ C (A B)] [A (B C)]. Let us prove that this is a law by using some laws given in the lecture notes. First, rewrite in an equivalent form the antecedent (the left hand side) of the above implication. [ C (A B)] [C (A B)] [C ( A B)] [C A B]. Now let us simplify the right hand side. Now the result is clear. [A (B C)] [ A ( B C)] [C A B]. [(A B) ( A B)] (C B). (A B) is false only if A is true and B is false. ( A B) is false only if A is true (A is false) and B is false. Hence, [(A B) ( A B)] is false if B is false, and is true if B is true. But if B is true then (C B) is true, no matter what C is. Therefore [(A B) ( A B)] (C B) is always true. 7. No. Take F instead of A, T instead of B and F instead of C. 8. (a) If A is true then A B is also true. If A is false then A (A B) is also false. (b) If A is true then both A B and A B are true. If A is false then one of the A B and A B is also false. (c) The proof is given by means of constructing the truth table for the propositional form which is denoted by C below.

3 ANALYSIS EXERCISE 1 SOLUTIONS 3 A B A B A B A B (A B) C T T F F T F F T T F F T T F F T F T T F T F F T F F T T F T T T (d) Using (c) and the double negation law we have (A B) ( A B) ( A B) ( A B). 9. (a) (b) (c) (d) A B C B C A ( B C ) T T F F F T F T T T T F F T T F T T T T F T F T T F F T T T F F F T T A B C A C B ( A C ) T T F F F T F T T T T F F F T F T T T T F T F T T F F T T T F F F T T A B C A B A C (A B) (A C) T T T F T F F T F T F T F T F F F F F F F T F T T T F F T T T T F F F T T T A B C A B (A B) C T T F T F T F T F T T F F F T F T T F T F T F F T F F T F T F F F F T

4 4 ANALYSIS EXERCISE 1 SOLUTIONS (0.1) (0.2) (0.3) (0.4) (e) A B C B C A (B C) T T F F F T F T F F T F F F F F T T T T F T F F T F F T F T F F F F T Thus, (a),(b),(d) are equivalent, and (c),(e) are equivalent. 10. Let us introduce the following propositions (a) A { Smith was drunk}. (b) B { Johns is the murderer}. (c) C { Smith lies}. (d) D { the murder took place after midnight}. Then we know that A (B C) is true. B ( A D) is true. D (B C) is true. A C is true. From (0.2) we have that either B is true, in which case we are done, or A and D are true. Then, since D is true, by (0.3), B is true (done) or C is true. But C being true and A being true contradicts (0.4). (0.1) is not needed at all. Alternatively, if you had not spotted the above argument you could use the following truth table. In fact, this is a more general kind of argument that should always work with questions like this. A B C D B C A D A (B C) B ( A D) D (B C) A C F T T T T T T T F T F T T T T T T F T T F T T T T T T F F T F T T T T T F T T T F T F T T T F T F T F T F T T T F F T F F F F F T T F F F F F F F T T F T T T F F T T F T F T T T F F T F T T F T F F T F T T T T F F T T F F F T F T F T F T F F F F T F T T T F T F F F F F F T F T T Because of (0.1)-(0.4), we know that the propositions A (B C), B ( A D), D (B C) and A C are all true. This corresponds to lines 1-4 and in the truth table. We notice that the truth value of B in all these lines is T. Hence, proposition B is true. (Notice that nothing can be deduced about propositions A, C, D, as on lines 1-4, they have truth values both T and F.)

5 ANALYSIS EXERCISE 1 SOLUTIONS (a) Assume the premises P Q and R Q. Assume P. Then Q (as P Q). The premise R Q is equivalent to Q R by the contrapositive law. So R is true. Thus P R is true. (b) Assume the premise R (P Q). Assume P. Also assume Q. By the logical law (A B) ( A B), the premise is equivalent to the disjunction R (P Q). Using the same logical law, P Q is equivalent to P Q. But this is false, as both P and Q are true. So R is true. But then Q R is true and P (Q R) is true. (There are other proofs, for example, proof by contradiction. But more on that later.) 12. A B = {3, 4}, A B = {0, 1, 2, 3, 4, 7, 10}, A B = {7, 10}, A B = {0, 1, 2, 7, 10}. 13. (x A (A B)) (x A) (x A B) (x A) (x A) (x B) (by De Morgan s law and the double negation law) (x A) [ (x A) (x B)] [(x A) (x A)] [(x A) (x B)] (x A) (x B) since (x A) (x A) is false. 14. Let us introduce the propositions P {x A}, Q {x B}, R {x C}. Then we can rewrite what has to be proved as which is a law. 15. Let us introduce the propositions [(P Q) (Q R)] (P R), P {x A}, Q {x B}. Then the first inclusion means that P (P Q) and the second one means that P Q P. Both are laws. 16. First, prove the equality A B = (A A B) B. Since A A B and B are disjoint, we have (0.5) N(A B) = N(A A B) + N(B). Notice that A = (A A B) (A B), where A A B and A B are disjoint. Therefore N(A) = N(A A B) + N(A B),

6 6 ANALYSIS EXERCISE 1 SOLUTIONS which combined with (0.5) gives the result.