I came across this question and I'm not sure if I fully understand how to do it. The question asks to show that ∀x(P(x) → q) is logically equivalent to ∃xP(x) → q. The way that I was taught to do this is to show that both sides have the same truth value when q is true and q is false (at least I think this is the correct approach).
When q is true, you know that P(x) → q is true, so the left side of the equation reduces to ∀x(T), which I'd imagine reduces to T. The right side of the equation, I think, reduces to T as well, but I'm not sure about that.
When q is false, P(x) → q depends on P(x), so I think that the left side reduces to ∀x(P(x)). Again, I'm not sure about the right side.
Can anybody tell me if I'm approaching this correctly? I thought I understood this stuff but I'm second guessing myself and getting very confused.
You may want to look in to natural deduction. A verbose / natrual language proof along those rules goes as follows:
Assume $\forall x(P(x)\to q)$. We want to show $\exists x P(x)\to q$. So assume $\exists x P(x)$, say $P(a)$. By specialization of $\forall x(P(x)\to q)$ we get $P(a)\to q$. So together with $P(a)$ we get $q$. All in all we have shown (under the assumption of $\forall x(P(x)\to q)$) that $\exists x P(x)$ implies $q$, so $\exists x P(x)\to q$ as desired.
Now assume $\exists x P(x)\to q$. We want to show that $\forall x(P(x)\to q)$. So let $x$ be arbitrary. Then either $P(x)$ or $\neg P(x)$. In the second case $P(x)\to q$ is true. In the first, $P(x)$ implies $\exists x P(x)$, hence from the given $\exists x P(x)\to q$ we find $q$ and hence also $P(x)\to q$. So in both cases $P(x)\to q$ follows. As $x$ was arbitrary, we arrive at $\forall x(P(x)\to q)$.