Let p, q, r and z be four primitive statements. Consider the following arguments:
P: [(p v q) ^ (p -> r) ^ (q -> r)] -> r
Q: [(p -> q) ^ (p -> ~q)] -> ~p
R: [(p -> q) ^ ~p] -> ~q
S: [(p -> q) -> r] -> (q -> r)
Which of the above arguments are valid?
A. P and Q only
B. P and R only
C. P and S only
D. P, Q, R and S
GATE 2004 · Discrete Mathematics · Propositional Logic · medium
Answer: P and S only (Option C)
Test Argument P: proof by cases: Premises: (p v q), (p -> r), (q -> r). Suppose r = False. Then p -> r forces p = False, and q -> r forces q = False. But then p v q = False, contradicting the first premise. No counterexample exists. P is VALID.
Test Argument Q: contradiction / reductio: Premises: (p -> q), (p -> ~q). Suppose ~p = False, i.e., p = True. Then p -> q gives q = True and p -> ~q gives q = False — contradiction. So p must be False, meaning ~p = True. No counterexample. Q is VALID.
Test Argument R: denying the antecedent (counterexample): Try p = False, q = True, r arbitrary. Premises: p -> q = (F -> T) = True; ~p = True. Conclusion ~q = False. Both premises true, conclusion false. R is INVALID.
Test Argument S: tautology verification: Premise: (p -> q) -> r. Conclusion: q -> r. Suppose S is false: premise true, conclusion false. Then q = True, r = False. For premise to be true with r = False, need (p -> q) = False, i.e., p = True and q = False. But q = True (assumed) — contradiction. No counterexample. S is VALID.