Verify that $(\neg P\land N)\lor(\neg D \land P)\rightarrow \neg D\lor N$ The answer seems to be $\neg D \lor N$, but below is what I got,no idea where it goes wrong
$$\begin{aligned} (\neg P\land N)\lor(\neg D \land P) &\equiv \neg(\neg N\lor P)\lor \neg(\neg P\lor D)\\ &\equiv\neg[(N\implies P)\land(P\implies D)]\\&\rightarrow \neg(N\implies D)\\ &\equiv\neg(\neg N\lor D) \\ &\equiv\neg D\land N \end{aligned}
$$
Appreciate for any help.

It's easy to verify the relationship using Natural Deduction:
$
\dfrac
{
((\neg p\land n) \vee (\neg d \land p))
\quad
\dfrac{\dfrac{[(\neg p \land n)]}{n}(\land E)}{(\neg d \lor n)}(\lor I)
\quad
\dfrac{\dfrac{[(\neg d \land p)]}{(\neg d)}(\land E)}{(\neg d \lor n)}(\lor I)
}
{
(\neg d \lor n)
}(\lor E)
$
PS: This is my second answer. The first one, as pointed out by
spaceisdarkgreen, was completely wrong.
