Proof of equivalent formulas

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I've been working on proving equivalent formulas using equivalences in predicate logic however I'm stuck trying to prove that these two formulas are equivalent:

$ \forall x (A(x) \Rightarrow \exists y B(x,y)) $

$ \forall x \exists y (A(x) \Rightarrow B(x,y)) $

The only step I've managed to get so far is by swapping the quantifies on the second formula:

$ \exists y \forall x (A(x) \Rightarrow B(x,y)) $

But I'm not sure how to proceed from here. I've checked what equivalences I can use but I don't see any that I can use here anymore.

Does anyone see a way to go further from here?

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(1) $\lnot \exists x X \equiv \forall x \lnot X$ or alternatively $\exists x \lnot X \equiv \lnot \forall x X$

(2) $ \lnot (X \implies Y) \equiv (X \cdot \lnot Y)$

$\forall x (A(x) \implies \exists y B(x, y) )$

By double negation

$\lnot \lnot \forall x (A(x) \implies \exists y B(x, y) )$

By (1) $\lnot\exists x (\lnot (A(x) \implies \exists y B(x, y) ))$

By (2) $\lnot\exists x ((A(x) \cdot \lnot \exists y B(x, y) ))$

By (1) $\lnot\exists x ((A(x) \cdot \forall y \lnot B(x, y) ))$

Since A does not depend on y,

$A(x) \equiv \forall y A(x)$

$\lnot\exists x ((\forall y A(x) \cdot \forall y \lnot B(x, y) ))$

$\lnot\exists x (\forall y (A(x) \cdot\lnot B(x, y) ))$

By (1) $\forall x \lnot(\forall y (A(x) \cdot\lnot B(x, y) ))$

By(1) $\forall x (\exists y \lnot(A(x) \cdot\lnot B(x, y) ))$

By(2) $\forall x (\exists y (A(x) \implies B(x, y) ))$