First order logic equivalence proof

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I have this homework problem that I can't figure out. I have to show that the following sentences are equivalent:

$\neg \forall c\; A(c)\Rightarrow\exists d\; B(d) \land \neg C(c,d)$

$\exists c\; A(c)\land\forall d\; B(d)\Rightarrow C(c,d)$

Here's my approach. Starting with:

$\neg \forall c\; A(c)\Rightarrow\exists d\; B(d) \land \neg C(c,d)$

Eliminate the $\Rightarrow$:

$\neg \forall c\; \neg A(c)\lor\exists d\; B(d) \land \neg C(c,d)$

Move the $\neg$ inwards ($\neg\forall c \; p \equiv \exists c\; \neg p$):

$\exists c\; A(c)\land\neg\exists d\; [B(d) \land \neg C(c,d)]$

Now use the same equivalence:

$\exists c\; A(c)\land\forall d\; [\neg B(d) \lor C(c,d)]$

And I don't know where to go from there. Distributing the $\land$ doesn't seem to lead anywhere.

Any help would be appreciated.

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$\neg p\vee q$ is equivalent to $p \Rightarrow q$, so $\neg B(d) \vee C(c,c)~$ is equivalent to $B(d) \Rightarrow C(c,d)$.