Geetha has a conjecture about integers, which is of the form: forall z(P(z)) => exists y(Q(y, y)) where P is a statement about integers, and Q is a statement about pairs of integers. Which of the following (one or more) options imply Geetha's conjecture? A. forall p(P(p) ^ ~Q(p, p)) B. forall p(P(p) -> Q(p, p)) C. exists p(P(p) -> Q(p, p)) D. exists p(P(p) ^ ~Q(p, p))

GATE 2023 · Discrete Mathematics · First Order Logic · medium

Answer: B, C. Options B and C both logically imply Geetha's conjecture forall z P(z) => exists y Q(y,y).

  1. Option A contradicts the conjecture: If A = forall p(P(p) ^ ~Q(p,p)) is true, then forall z P(z) is true (from the P(p) part). Also, ~Q(p,p) holds for every p, so NO y satisfies Q(y,y). The conjecture's conclusion exists y Q(y,y) is false while its antecedent forall z P(z) is true. The conjecture becomes True => False = False. Option A does NOT imply the conjecture; it refutes it.
  2. Option B implies the conjecture: Assume B = forall p(P(p)->Q(p,p)) and assume the conjecture's antecedent forall z P(z). By universal instantiation, for any integer p0, P(p0) is true. By modus ponens on B applied to p0: Q(p0,p0) is true. So exists y Q(y,y) holds (choosing y = p0). The conjecture is satisfied. Option B implies the conjecture.
  3. Option C implies the conjecture: Assume C = exists p(P(p)->Q(p,p)). Let p0 be the witness: P(p0)->Q(p0,p0) holds. Now assume forall z P(z). Then P(p0) is true. By modus ponens: Q(p0,p0) is true. So exists y Q(y,y) holds (y = p0). The conjecture is satisfied. Option C implies the conjecture.
  4. Option D does not imply the conjecture: Assume D = exists p(P(p) ^ ~Q(p,p)). This says some p0 has P(p0)=true and Q(p0,p0)=false. Does this force the conjecture? No. Consider: domain = {p0, p1}. P(p0)=T, Q(p0,p0)=F, P(p1)=F, Q(p1,p1)=F. Then D is true (p0 witnesses it). Forall z P(z) is false (P(p1)=F), so the conjecture's antecedent is false, making the conjecture vacuously true. But we could also have: P(p0)=T, P(p1)=T, Q(p0,p0)=F, Q(p1,p1)=F. Then D is true (p0), forall z P(z) is true, but exists y Q(y,y) is false. Conjecture fails. So D does NOT imply the conjecture.