I'm asked to verify the following formula is valid:
$$\exists x (p(x)\implies q(x)) \iff (\forall x [p(x)\implies\exists x \ q(x)])$$
For the first direction, I want to assume that $p(x)\implies q(x)$ is true for some $a$, so $p(a)\implies q(a)$. I don't understand how this implies the RHS however. Take, for instance, $p(x)$ to be false when $x=a$, but true for all other $x$, and $q(x)$ is always false. Then, the LHS would be true, but the RHS would be false. Am I thinking about the problem incorrectly?