I am required to show that $\forall(x \in A)P(x) \lor \forall(x \in A)Q(x)$ is logically equivalent to $\forall(x \in A)\forall(y \in A)(P(x) \lor Q(y))$.
I'm not quite sure how to proceed, mostly because of how the first proposition only deals with $(x \in A)$, whereas the second one deals with $(x \in A)$ and $(y \in A)$.
How should I best approach this problem?
On
What you want to think of in the second sentence is ordered pairs; specifically, ordered pairs in $A^2$ (the Cartesian product of $A$ with itself). That is, $\forall (a_1,a_2) \in A(P(a_1)\lor Q(a_2)$. To some, this might obfuscate the idea, but this isn't part of the proof, this is just a way of looking at it.
The first direction of the implication, you can assume that for any $a \in A$, $P(a)$ holds. Then you can see that for any $(a_1,a_2) \in A^2$, since the first entry is from $A$, $P(a_1)$ holds. The same idea goes for when you assume that for any $a \in A$, $Q(a)$ holds.
For the second direction of the implication, after assuming the second sentence, you let $a$ be any element of $A$. Can you think of an ordered pair in $A^2$ that forces $P(a) \lor Q(a)$?
Hint: try proving contrapositive