Does the order of nested quantifiers matter for $\exists x \forall y P(x) \vee Q(y)$? I fail to see difference between $\exists x \forall y P(x) \vee Q(y)$ and $\forall y\exists x  P(x) \vee Q(y)$.
after all, there is no relation between x & y, and all we need to do is to check if one of the statements is true for the overall statement to be true since the or relation $\vee$.
like,
for every $x$ there is at least a $y$ that makes the statement $P(x) \vee Q(y)$ true
for every $y$ there is at least an $x$ that makes the statement $P(x) \vee Q(y)$ true.
 A: If $P(x)$ has a single value that makes it true,  both statements are true due to the fact that it is an or.
If $P(x)$ is a contradiction,  then the truth value of the or reduces to $Q(y)$, so it will only be true if $Q$ is a tautology.  So yes, they are equivalent in this case.
A: Working with the formulae as given:

*

*$$\exists x \forall y P(x) \vee Q(y)$$ is first-order equivalent to
$$\exists x P(x) \vee Q(y),$$ which is first-order equivalent to
$$\forall y\exists x  P(x) \vee Q(y).$$

Working with the given formulae but with brackets inserted (in case they had accidentally been omitted):

*

*$$\exists x \forall y [P(x) \vee Q(y)]$$ is first-order equivalent to
$$\exists x [P(x) \vee \forall y Q(y)],$$ which is first-order
equivalent to $$\exists x P(x) \vee \exists x \forall y Q(y),$$ which
is first-order equivalent to $$\exists x P(x) \vee \forall y Q(y),$$ which is first-order equivalent to $$\forall y [\exists x P(x) \vee Q(y)],$$ which is first-order equivalent to $$\forall y [\exists x P(x) \vee \exists x Q(y)],$$ which is first-order equivalent to $$\forall y \exists x [P(x) \vee Q(y)].$$

Similarly, these two sentences happen to be first-order equivalent:

*

*$$\exists x \forall y [P(x) \rightarrow Q(y)]$$

*$$\forall y \exists x [P(x) \rightarrow Q(y)].$$
