Show that $∀xP(x) ∨ ∀xQ(x)$ and $∀x∀y(P(x) ∨ Q(y))$, where all quantifiers have the same nonempty domain, are logically equivalent. (The new variable y is used to combine the quantification's correctly.)
I'm not sure what the y is there for, and how to go about solving this. I know that I can prove they're logically equivalent by proving that they imply each other but how do I do that?
Note that $P(x)$ is not bounded by $\forall y$. So $P(x)$ can be moved out of $∀y$, i.e. $$ ∀y(P(x) ∨ Q(y))\iff P(x) ∨∀yQ(y) $$ And $Q(y)$ is not bounded by $\forall x$. So $∀yQ(y)$ can be moved out of $∀x$, i.e. $$ ∀x(P(x) ∨∀y Q(y))\iff ∀xP(x) ∨∀y Q(y)\iff ∀xP(x) ∨∀x Q(x) $$ So there is $$ ∀x∀y(P(x) ∨ Q(y))\iff∀x(P(x) ∨∀y Q(y))\iff ∀xP(x)∨∀x Q(x) $$