Question 26

Geetha's conjecture is $\phi: \forall x (P(x) \implies \exists y Q(x, y))$.

1. Option (B) $\forall x \forall y Q(x, y)$: If $Q(x, y)$ is true for all $x$ and all $y$, then for any given $x$, choosing any $y$ satisfies $Q(x, y)$, so $\exists y Q(x, y)$ is universally true. Thus, the implication $P(x) \implies \exists y Q(x, y)$ is trivially true for all $x$. Therefore, (B) implies $\phi$.
2. Option (C) $\exists y \forall x (P(x) \implies Q(x, y))$: Suppose there exists some $y_0$ such that for all $x$, $P(x) \implies Q(x, y_0)$. Then for any $x$, if $P(x)$ holds, $Q(x, y_0)$ holds, which proves that there exists a $y$ (namely $y_0$) such that $Q(x, y)$ holds. Thus $\forall x (P(x) \implies \exists y Q(x, y))$ is satisfied. Therefore, (C) implies $\phi$.
3. Options (A) and (D) only assert existential quantification $\exists x P(x)$, which provides no guarantee for all $x$ required by the universal quantifier in Geetha's conjecture.