Does the order of nested quantifiers matter for $\exists x \forall y P(x) \vee Q(y)$?

Question

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.

Answer 1

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.

Answer 2

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

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

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