How do I prove that these are equal using laws of logic?

Question

Prove that $\lnot(p\lor(\lnot p\land q)$ is equivalent to $\lnot p\land\lnot q$

I always end up at $\lnot p$, here's my solution: \begin{equation} \lnot(p\lor(\lnot p\land q))=\lnot p\land\lnot q \end{equation} let $\lnot p\land q$ be $r$ \begin{gather} \lnot(p\lor r)=\lnot p\land\lnot q\\ \lnot p\land\lnot r=\lnot p\land\lnot q\\ \lnot p\land\lnot(\lnot p\land q)=\lnot p\land\lnot q\\ \lnot p\land(p\land\lnot q)=\lnot p\land\lnot q\\ \lnot p=\lnot p\land\lnot q \end{gather} not equal.

Answer 1

$\neg P \wedge \neg ( \neg P \wedge Q)$

= $\neg P \wedge (P \vee \neg Q) \qquad \text{(De Morgan's law)}$

= $(\neg P \wedge P) \vee (\neg P \wedge \neg Q) \qquad \text{(ditributivity)}$

= $\neg P \wedge \neg Q$

The conclusion is justified by the fact that $\neg P \wedge P$ is paradoxical, hence $(\neg P \wedge P) \vee (\neg P \wedge \neg Q)$ is true if and only if $\neg P \wedge \neg Q$ is.

Answer 2

Using an equational style (Boolean algebra identities at work)

$\neg (p \vee (\neg p \wedge q)) = \neg ((p \vee \neg p) \wedge (p \vee q))$ -- Distributivity

$\neg (p \vee (\neg p \wedge q)) = \neg (\top \wedge (p \vee q))$

$\neg (p \vee (\neg p \wedge q)) = \neg (p \vee q)$

$\neg (p \vee (\neg p \wedge q)) = \neg p \wedge \neg q$ -- De Morgan