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