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