Truth table of quantifiers

Question

Is it possible to draw the truth table of ∃x,(P(x)∨Q(x))?
I thought of this as follows,let x be any arbitrary domain containing, x1,x2,...,xN,
So the above expression translates to (P(x1)∨Q(x1))∨(P(x2)∨Q(x2))...∨(P(xN)∨Q(xN)).
Using this can I construct a truth table?

Answer 1

Only if the domain of intepretation is finite.

But a predicate formula is valid if it is true in all interpretations, and thus we have to take into account also intepretation with infinite domain, like $\mathbb N$.

For example, the formula $\exists y \forall x (x \le y)$ is true in every finite domain of natural numbers, but it is clearly false in $\mathbb N$.