PHIL12A Section Answers, 23 February 2011

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PHIL12A Section Answers, 23 February 2011 PHIL12A Section answers, 23 February 2011 Julian Jonker 1 How much do you know? 1. The following questions are adapted form exercises 5.1-5.6. Decide whether each pattern of inference is valid. If it is, show that it is using truth tables. If it is not, give example sentences that show how the conclusion can be false though the premises are true. (a) From P and :Q, infer P ^ Q. This is invalid, as the following sentences exemplify: 1 P = Logic is fun. True 2 :Q = Logic is not easy. True 3 P ^ Q = Logic is fun and easy. False (b) From :P _:Q and :P , infer :Q. This is invalid, as the following sentences exemplify: 1 :P _:Q = Either soft drinks are unhealthy or water is unhealthy. True 2 :P = Soft drinks are unhealthy. True 3 :Q = Water is unhealthy. False (c) (Ex 5.6) From P ^ Q and :P , infer R. This is valid, as a truth table will show. You need to show that every row of the truth table which makes both P ^ Q and :P true also makes R true. However, because P ^ Q and :P are contradictory, there is no row of the truth table which makes both of them true (try it!). So the definition for TT validity is (vacuously) fulfilled. 1 2. Write informal proofs for the following arguments, using proof by cases. Be as explicit as possible about each step in your proof. (a) Suppose you know that (Cube(a)^Tet(b))_(Cube(c)^Tet(b)). Show that Tet(b) follows. There are two cases to consider. First, suppose that (Cube(a)^Tet(b)) is true. Since both of the conjuncts of a true conjunction are true, we can conclude that Tet(b) is true. Second, consider that (Cube(c)^Tet(b)) is true. Again, both the conjuncts of a true conjunction are true, so we conclude that Tet(b) is true. Since Tet(b) follows in each case, we conclude that Tet(b) is true. (b) Suppose you know that (Cube(a)^Small(b))_(Cube(c)^Small(c)). Show that Small(b)_Small(c) follows. There are two cases to consider. First, suppose that Cube(a)^Small(b) is true. Since both con- juncts of a conjunction are true, we know that Small(b) is true. But then we can conclude that Small(b)_Small(c) is true, since a disjunction is true whenever one of the disjuncts is true. Second, suppose that Cube(c)^Small(c) is true. Since both conjuncts of a conjunction are true, we know that Small(c) is true. But then we can conclude that Small(b)_Small(c) is true, since a disjunction is true whenever one of the disjuncts is true. It follows that Small(b)_Small(c) is true, since it is true in every case. (c) (Ex 5.7) 1 Home(max) _ Home(claire) 2 :Home(max) _ Happy(carl) 3 :Home(claire _ Happy(carl) 4 Happy(carl) We want to show that if we make the premises of the argument true, the conclusion must be true. In holding premise 1 true, there are two cases we can consider: either Max is home or Claire is home. First, suppose that Max is home. By the second premise, Carl is happy, since at least one of the disjuncts of the disjunction must be true. This also makes the third premise true. Second, suppose that Claire is home. By the third premise, Carl is happy, since at least one of the disjuncts of the disjunction must be true. This also makes the second premise true. It follows that Carl is happy, since we can conclude this in every possible case. 2 3. Write informal proofs to show that the following arguments are valid. (a) (Ex 5.16) 1 Max or Claire is at home but either Scruffy or Carl is unhappy. 2 Either Max is not home or Carl is happy. 3 Either Claire is not home or Scruffy is unhappy. 4 Scruffy is unhappy Premise 1 is a conjunction, and when we hold it to be true we hold both conjuncts to be true. So we accept that Max or Claire is at home, since this is one of the conjuncts. Now we use a proof by cases to show that Scruffy is unhappy. There are two cases to consider. In the first case, we suppose that Max is at home. Then by the second premise we know that Carl is happy. But not that by premise 1, it is true that either Scruffy or Carl is unhappy. This is a disjunction, so one of the disjuncts must be true. We conclude that in this case Scruffy is unhappy. In the second case, we suppose that Claire is at home. The third premise is a disjunction, so one of the disjuncts must be true. Therefore Scruffy is unhappy in this case. It follows that Scruffy is unhappy, since this is true in every case. (b) (Ex 5.17) 1 Cube(a) _ Tet(a) _ Large(a) 2 :Cube(a) _ a=b _ Large(a) 3 :Large(a) _ a=c 4 :(c=c ^ Tet(a)) 5 a=b _ a=c In order to hold premise 1 true, at least one of the disjuncts must be true. Let’s consider each disjunct: there are three cases. In each case we wish to show that the conclusion, a=b_a=c, holds. Case 1: a is a cube. By premise 2 it follows that a and b are identical, or a is large, or both. Now we show by cases that a=b_a=c holds. Suppose first that a and b are identical. Then the disjunction a=b_a=c is true since the first disjunct is true. Suppose next that a is large. Then by premise 3 we have that a and c are identical. But then it follows that the disjunction a=b_a=c is true, since the second disjunct is true. It follows that a=b_a=c is true in case 1. Case 2: a is a tetrahedron. By premise 4 we have that either c is not identical to c or a is not a tetrahedron. In either case, we have a contradiction! But anything follows from a contradiction: including a=b_a=c. 3 Case 3: a is large. By premise 3 we have that a and c are identical. But then we know that a=b_a=c is true, since the second disjunct is true. It follows that a=b_a=c is true, since it follows in each case. Note that the our proof contained proofs by cases embedded within a proof by cases. The structure of this would have been much easier to follow if we had uses a formal proof! 4. Construct formal proofs for the following arguments. (a) (Ex 6.4) 1 (A ^ B) _ C 2 C _ B Proof: 1 (A ^ B) _ C 2 (A ^ B) 3 B ^Elim: 2 4 C _ B _Intro: 3 5 C 6 C _ B _Intro: 5 7 C _ B _Elim: 1, 2-4, 5-6 4 (b) (Ex 6.2) 1 P _ (Q ^ R) 2 (P _ Q) ^ (P _ R) Proof: 1 P _ (Q ^ R) 2 P 3 P _ Q _Intro: 2 4 P _ R _Intro: 2 5 (P _ Q) ^ (P _ R) ^Intro: 3,4 6 Q ^ R 7 Q ^Elim: 6 8 P _ Q _Intro: 7 9 R ^Elim: 6 10 P _ R _Intro: 9 11 (P _ Q) ^ (P _ R) ^Intro: 8, 10 12 (P _ Q) ^ (P _ R) _Elim: 1, 2-5, 6-11 5 (c) For the following proof, you should recall that you can nest subproofs within subproofs. 1 (P _ Q) ^ (P _ R) 2 P _ (Q ^ R) (I will use Reit steps in some of these proofs to clarify things, even though not strictly necessary.) Proof: 1 (P _ Q) ^ (P _ R) 2 P _ Q ^Elim: 1 3 P 4 P _ (Q ^ R) _Intro: 3 5 Q 6 P _ R ^Elim: 1 7 P 8 P _ (Q ^ R) _Intro: 7 9 R 10 Q Reit: 5 11 Q ^ R ^Intro: 9, 10 12 P _ (Q ^ R) _Intro: 11 13 P _ (Q ^ R) _Elim: 6, 7-8, 9-12 14 P _ (Q ^ R) _Elim: 2, 3-4, 5-13 6 (d) (Ex 6.8) 1 P ^ (Q _ ::R) 2 ::P ^ (Q _ R) 1 P ^ (Q _ ::R) 2 P ^Elim: 1 3 :P 4 P Reit: 2 5 ??Intro: 3,4 6 ::P :Intro: 3-5 7 Q _ ::R ^Elim: 1 8 Q 9 Q _ R _Intro: 8 10 ::R 11 R :Elim: 10 12 Q _ R _Intro: 11 13 Q _ R _Elim: 7, 8-9, 10-12 14 ::P ^ (Q _ R) ^Intro: 6,13 7 (e) 1 :(P _ Q) 2 :P ^ :Q Proof: 1 :(P _ Q) 2 P 3 P _ Q _Intro: 2 4 :(P _ Q) Reit: 1 5 ??Intro: 3,4 6 :P :Intro: 2-5 7 Q 8 P _ Q _Intro: 7 9 :(P _ Q) Reit: 1 10 ??Intro: 8,9 11 :Q :Intro: 10 12 :P ^ :Q ^Intro: 6,11 8 (f) (Ex 6.19) 1 A _ B 2 :B _ C 3 A _ C Here’s one proof: 1 A _ B 2 :B _ C 3 A 4 A _ C _Intro: 3 5 B 6 :B 7 ??Intro: 5,6 8 A _ C ?Elim: 7 9 C 10 A _ C _Intro: 9 11 A _ C _Elim: 2, 6-8, 9-10 12 A _ C _Elim: 1, 3-4, 5-11 (Note that you might have found other proofs for some of these exercises.
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