1. **Stating the problem:** We are given two logical statements: $A: (\forall x, P(x)) \to (\exists x, Q(x))$ and $B: (\exists x, P(x)) \to Q(x)$. We want to determine if $A \to B$ is a tautology and if $B \to A$ is a tautology. 2. **Recall definitions:** - A tautology is a statement that is true in every possible interpretation. - $A \to B$ means "if $A$ is true, then $B$ is true". 3. **Analyze $A \to B$:** - $A$ says: "If for all $x$, $P(x)$ holds, then there exists some $x$ such that $Q(x)$ holds." - $B$ says: "If there exists some $x$ such that $P(x)$ holds, then $Q(x)$ holds" (note $Q(x)$ here is not quantified, so it is ambiguous; assuming $Q(x)$ means $Q$ holds for that same $x$). 4. **Check if $A \to B$ is always true:** - Suppose $A$ is true: whenever $P$ holds for all $x$, there is some $x$ with $Q(x)$. - Does this imply $B$? $B$ requires that if there exists an $x$ with $P(x)$, then $Q(x)$ holds for that $x$. - This is stronger than $A$ because $A$ only guarantees existence of some $x$ with $Q(x)$ when $P$ holds for all $x$, but $B$ requires $Q(x)$ for the same $x$ where $P(x)$ holds. - Therefore, $A \to B$ is **not** a tautology. 5. **Analyze $B \to A$:** - Suppose $B$ is true: if there exists $x$ with $P(x)$, then $Q(x)$ holds for that $x$. - Does this imply $A$? $A$ says if $P$ holds for all $x$, then there exists $x$ with $Q(x)$. - If $P$ holds for all $x$, then certainly there exists $x$ with $P(x)$, so by $B$, $Q(x)$ holds for that $x$. - Hence, there exists $x$ with $Q(x)$, so $A$ is true. - Therefore, $B \to A$ is a tautology. **Final answers:** - $A \to B$ is **not** a tautology. - $B \to A$ **is** a tautology.