1. **Stating the problem:** We want to determine if the implication $A \to B$ is a tautology, where: - $A = \exists x \forall y, P(x) \to Q(y)$ - $B = (\forall y, Q(y)) \lor (\exists x, \neg P(x))$ 2. **Recall the implication rule:** An implication $A \to B$ is a tautology if it is true for all interpretations, i.e., whenever $A$ is true, $B$ must also be true. 3. **Analyze $A$: $\exists x \forall y, P(x) \to Q(y)$** - This means there exists some element $x$ such that for every $y$, if $P(x)$ is true, then $Q(y)$ is true. - Note that $P(x) \to Q(y)$ is logically equivalent to $\neg P(x) \lor Q(y)$. - So $\forall y, P(x) \to Q(y)$ means $\forall y, \neg P(x) \lor Q(y)$. - Since $\neg P(x)$ does not depend on $y$, this is equivalent to $\neg P(x) \lor \forall y, Q(y)$. - Therefore, $A$ is equivalent to $\exists x (\neg P(x) \lor \forall y, Q(y))$. 4. **Rewrite $A$ using distributive logic:** $$ A \equiv \exists x (\neg P(x) \lor \forall y, Q(y)) $$ - This is logically equivalent to: $$ (\exists x \neg P(x)) \lor (\exists x \forall y, Q(y)) $$ - Since $\forall y, Q(y)$ does not depend on $x$, $\exists x \forall y, Q(y)$ is equivalent to $\forall y, Q(y)$. - So: $$ A \equiv (\exists x \neg P(x)) \lor (\forall y, Q(y)) $$ 5. **Compare $A$ and $B$:** - $B = (\forall y, Q(y)) \lor (\exists x, \neg P(x))$ - $A$ and $B$ are logically equivalent. 6. **Conclusion:** - Since $A$ and $B$ are logically equivalent, the implication $A \to B$ is always true. - Therefore, $A \to B$ is a tautology. **Final answer:** $$ \boxed{\text{Yes, } A \to B \text{ is a tautology.}} $$