Some Solutions to the Practice Midterm Exam (1) The argument with premises A1,...,An and conclusion B is valid if: every truth assignment that makes all of A1,...,An true also makes B true. (2) A sentence B is inconsistent if there is no truth assignment relative to which it is true. i.e. in every row of its truth table, B is assigned the value F. (3.a) F & E (3.b) F -> H (3.c) (F -> E) v -M also equivalent to: M -> (F -> E) also equivalent to (M & F) -> E [In other words, this is saying that Edmund's happiness follows from Fanny's loving Mr Crawford, and Miss Crawford loving Edmund.] (4.a) 1 (1) -P A 2 (2) -Q A 3 (3) P v Q A 4 (4) P A 1,4 (5) P & -P 4,1 &I 1,4 (6) -(P v Q) 3,5 RAA 7 (7) Q A 2,7 (8) Q & -Q 7,2 &I 2,7 (9) -(P v Q) 3,8 RAA 1,2,3 (10) -(P v Q) 3,4,6,7,9 vE 1,2,3 (11) (P v Q) & -(P v Q) 3,10 &I 1,2 (12) -(P v Q) 3,11 RAA (4.b) 1 (1) (P -> Q) v (P -> R) A 2 (2) P A 3 (3) P -> Q A 2,3 (4) Q 3,2 MPP 2,3 (5) Q v R 4 vI 6 (6) P -> R A 2,6 (7) R 6,2 MPP 2,6 (8) Q v R 7 vI 1,2 (9) Q v R 1,3,5,6,8 vE 1 (10) P -> (Q v R) 2,9 CP (5) 1 (1) -(Q v -Q) A 2 (2) Q A 2 (3) Q v -Q 2 vI 1,2 (4) (Q v -Q) & -(Q v -Q) 3,1 &I 1 (5) -Q 2,4 RAA 1 (6) Q v -Q 5 vI 1 (7) (Q v -Q) & -(Q v -Q) 6,1 &I - (8) --(Q v -Q) 1,7 RAA - (9) Q v -Q 8 DN 10 (10) P A 10 (11) P & (Q v -Q) 10,9 &I - (12) P -> (P & (Q v -Q)) 10,11 CP 13 (13) P & (Q v -Q) A 13 (14) P 13 &E - (15) (P & (Q v -Q)) -> P 13,14 CP - (16) (P -> (P & (Q v -Q))) & ((P & (Q v -Q)) -> P) 12,15 &I - (17) P <-> (P & (Q v -Q)) 16 Df.<-> (6) It is a tautology. This sentence is a conditional. Its antecedent is equivalent to --P v P, which is equivalent to P. If P is true, then the conditional T -> P is true, and so is the conditional S -> (T -> P), etc.. So, whenever the antecedent is true, so is the consequent. (7) False. If A is a contingency, then it is false under some truth assignment. Under this same truth assignment, A -> B is true. Hence A -> B is not an inconsistency. (8) Yes. Suppose that the conclusion is false. Then both disjuncts, P -> R and P -> S, are false. In this case, P is true and R,S are false. So, P v Q is true and R v S is false, and the premise (P v Q) -> (R v S) is false. Therefore, whenever the conclusion is false, so is the premise. (9) No. Consult lecture notes from Thursday, March 9. (10) P & -P (11) True. Line 1 is legitimate since it is an assumption. For line n, we perform a truth table test on the sentences on lines 1 and n. We find that whenever #1 is true, then so is #n. That is, #1 logically implies #n. The Completeness Theorem then tells us that if A logically implies B, then there is a correctly written proof with A on line 1, and B on line n with dependency number 1, as such: 1 (1) A 1 (n) B