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.
 A: 
\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}

That logical entailment in Line 3 is incorrect, as evidenced by the assignment $(P,N,D)=(0,1,1).$
Here's a correct attempt, if you don't mind applying the distributive laws: \begin{aligned} (\neg P\land N)\lor(\neg D \land P) &\equiv (¬P∨¬D)∧(¬P∨P)∧(N∨¬D)∧(N∨P)\\
&\models (N∨¬D)\\
&\equiv \neg D ∨ N. \end{aligned} Thus, $$(\neg P\land N)\lor(\neg D \land P) \to \neg D ∨ N$$ is a validity, as required.
P.S. I use $≡$ and $⊨$ to mean logically equivalent and logically implies, respectively (i.e., as metalogical symbols), while I use $\to$ merely as the material conditional (i.e., as a logical operator). As for $\implies,$ I use it just to mean implies (e.g., $x=2\implies x^2=4$) rather than as the material conditional.
A: 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.
