Prove that $\neg (\neg P \lor (P \land \neg Q)) \equiv P \land Q$ without using truth tables 
Prove that
$$\neg (\neg P \lor (P \land \neg Q)) \equiv P \land Q$$
without using truth tables. Instead, use various logic properties like De Morgan's, etc.

 A: Using de Morgan's laws:
$$\neg(\neg P\lor (P\land \neg Q))\equiv \neg\neg P\land \neg(P\land \neg Q)$$
$$\equiv P\land (\neg P\lor \neg\neg Q)$$
$$\equiv P\land (\neg P\lor  Q)$$
$$\equiv (P\land \neg P)\lor(P\land  Q)$$
$$\equiv P\land Q.$$
A: 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? 
A: $$\neg(\neg P \lor (P\land\neg Q)) \leftrightarrow P\land Q$$
$$  P \land \neg(P\land\neg Q)\leftrightarrow P\land Q$$
$$  P \land (\neg P\lor  Q)\leftrightarrow P\land Q$$
$$  (P \land \neg P )\lor( P\land  Q)\leftrightarrow P\land Q$$
$$   P\land  Q\leftrightarrow P\land Q$$
A: I haven't seen this question until now, but I would like to propose another approach (essentialy same as any other):
$$\neg (\neg P\lor (P\wedge \neg Q))\equiv\neg((\neg P\lor P)\wedge (\neg P\lor \neg Q))\equiv\neg(1\wedge (\neg P\lor \neg Q))\equiv P\wedge Q.$$
You can see it immediately, but I wrote the most just in case. Whenever you see something like $\neg P\lor P$, you should consider it when you have some disjunction. It might be useful.
