Sorry for the basic question, but I'm getting slightly tripped up. I know that if $S\subset D$, then we have the logical equivalence of the statements $$(\forall x\in S) P(x)\equiv (\forall x\in D)(x\in S\implies P(x)),$$ and $$(\exists x\in S) P(x)\equiv (\exists x\in D)((x\in S)\wedge P(x)).$$ I was wondering about what happens when we have nested quantifiers. For example, I'm especially interested in if it's true that if $S_1\subset D_1$, then $$(\exists x\in S_1)(\forall y\in D_2) P(x,y)\equiv (\exists x\in D_1)(\forall y\in D_2)((x\in S_1)\wedge P(x,y)),$$ or possibly
$$(\exists x\in S_1)(\forall y\in D_2) P(x,y)\equiv (\exists x\in D_1)(\forall y\in D_2)((x\in S_1)\implies P(x,y)).$$ I feel like it should be the former, if I consider an example like $$(\exists x\in\mathbb{R}^+)(\forall y\in \mathbb{N})(xy=y),$$ then this seems to be the same as saying $$(\exists x\in\mathbb{R})(\forall y\in \mathbb{N})(x>0\wedge xy=y).$$ I don't think it's the implication because it looks like the statement could be true, even if $x\leq 0$.
From the second equivalence rule, for $P'(x) = (\forall y\in D_2)P(x,y)$, we get
\begin{align} &(\exists x\in S_1)(\forall y\in D_2)P(x,y)\\ \equiv~&(\exists x\in S_1) P'(x)\\ \equiv~&(\exists x\in D_1)(x\in S_1 \land P'(x))\\ \equiv~&(\exists x\in D_1)(x\in S_1 \land (\forall y\in D_2)P(x,y))\\ \equiv~&(\exists x\in D_1)(\forall y\in D_2)(x\in S_1 \land P(x,y) \end{align}