Question 47

The given expression is:

Pushing the outer negation inside the existential quantifier:

Applying De Morgan's law:

Let $P = \exists y (\neg \alpha)$ and $Q = \exists z (\neg \beta)$. Then $\neg P \equiv \forall y (\alpha)$ and $\neg Q \equiv \forall z (\beta)$.
The matrix inside the quantifier is $P \lor Q$.
Using the conditional equivalence $p \lor q \equiv \neg p \rightarrow q \equiv \neg q \rightarrow p$:

• $\neg P \rightarrow Q$ gives: $\forall x (\forall y (\alpha) \rightarrow \exists z (\neg \beta))$. This matches Option (C).
• $\neg Q \rightarrow P$ gives: $\forall x (\forall z (\beta) \rightarrow \exists y (\neg \alpha))$. This matches Option (B).

Now examine options (A) and (D):

• Option (A): $\forall x (\exists z (\neg \beta) \rightarrow \forall y (\alpha)) \equiv \forall x (Q \rightarrow \neg P) \equiv \forall x (\neg Q \lor \neg P)$. This is NOT logically equivalent to $P \lor Q$.
• Option (D): $\forall x (\exists y (\neg \alpha) \rightarrow \exists z (\neg \beta)) \equiv \forall x (P \rightarrow Q) \equiv \forall x (\neg P \lor Q)$. This is ALSO NOT logically equivalent to $P \lor Q$.

Because the question asks for which statement is NOT logically equivalent, and both (A) and (D) are not equivalent, there was no single unique correct option. As a result, GATE officially awarded Marks to All.