This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHIL12A Section answers, 23 February 2011 Julian Jonker 1 How much do you know? 1. The following questions are adapted form exercises 5.15.6. 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....
View
Full
Document
This note was uploaded on 09/19/2011 for the course PHILOS 12A taught by Professor Fitelson during the Spring '08 term at Berkeley.
 Spring '08
 FITELSON

Click to edit the document details