What are examples of predicates $P(x)$ and $Q(x)$ and domains where the above two statements are equivalent?
My stab at the problem:
Let the domain of discourse be all positive whole numbers, $P(x)$ be defined as "$x \geq 0$, and $Q(x)$ be defined as $x + 1 \geq 0$. In this case, both the first statement and the second would evaluate to true... however, I don't know if this proves equivalency, because I don't think I can say that two statements are equivalent just because they both happen to evaluate to true.
Any help is much appreciated. Thanks in advance!
EDIT: initially had $\land$'s when I should have had $\lor$'s...
The two statements using $\land$, as in your original question, are equivalent, so for every model (a domain, and interpretations of $P$ and $Q$ in that domain), both are true or both are false. This is because either statement is true if and only if the interpretations of both $P$ and $Q$ equal the entire domain.
Your "Let the domain of discourse..." paragraph doesn't prove anything in general, it just gives an example of two predicates $P,Q$ whose interpretations are the same in that model.
Now you've changed the question to ask about $\lor$, and the answer is very different: the two statements are not equivalent. $$ \forall x \, P(x) \lor \forall x \, Q(x) \tag{1} $$ is true when (and only when) either $P$ is interpreted as the entire domain, or $Q$ is. However, $$ \forall x \, (P(x) \lor Q(x)) \tag{2} $$ is true in a model $M$ iff for every element $a$ of the universe of $M$ (the domain, in your usage), either $a\in P^M$ or $a\in Q^M$, where the superscripted predicate symbols are the interpretations in $M$ of those symbols. (1) implies (2), but (2) does not imply (1). A simple example: let $M$ be the integers, and interpret $P$ and $Q$ as the even and the odd integers respectively. Obviously (2) holds — every integer is either even or odd — but just as obviously (1) is false: it's not true that every integer is even, or every integer is odd.