Prove that $\neg (\neg P \lor (P \land \neg Q)) \equiv P \land Q$ without using truth tables
Using de Morgan's laws:
$$\neg(\neg P\lor (P\land \neg Q))\equiv \neg\neg P\land \neg(P\land \neg Q)$$
$$\equiv P\land (\neg P\lor \neg\neg Q)$$
$$\equiv P\land (\neg P\lor Q)$$
$$\equiv (P\land \neg P)\lor(P\land Q)$$
$$\equiv P\land Q.$$
Hint : a repeated application of DeMorgan's is all you need.
To start, $$\neg (\neg P \lor (P \land \neg Q)) = P \land (\neg{(P \land \neg Q))} $$
Can you proceed?
$$\neg(\neg P \lor (P\land\neg Q)) \leftrightarrow P\land Q$$ $$ P \land \neg(P\land\neg Q)\leftrightarrow P\land Q$$ $$ P \land (\neg P\lor Q)\leftrightarrow P\land Q$$ $$ (P \land \neg P )\lor( P\land Q)\leftrightarrow P\land Q$$ $$ P\land Q\leftrightarrow P\land Q$$