How do you simplify $¬ ((¬Q \land ¬P) \lor ¬P)$ Please help. I've missed some lectures, and now I'm stuck (my fault!). The lectures notes don't explain elaborately, and I can't find good tutorials online. I've somehow managed to arrive at $(Q \lor P) \land P$. If this is correct, can this be simplified further? Thanks heaps.
 A: Replacing $\lor$ by $+$ and $\land$ by $\cdot$ makes it easy to see the further possible simplification
$$(Q+P)\cdot P = Q\cdot P+P = (Q+1)\cdot P = P$$
$P\cdot P = P\land P = P$ because "$\land$" of two inputs is true when "both" its inputs are true. Saying both $P, P$ are true is same as saying $P$ is true.  
$Q+1=1$ because "$\lor$" of two inputs is true when "at least" one input is true. Since one input is already true, $Q+1=1$. 
A: Yes, it can still be simplified further, the idea is using Commutative law and Absorption law,
I suppose you need to simplify this with Logical equivalence, and the proof is simple:
$$¬((¬Q \land ¬P) \lor ¬P)\tag*{Start from}$$
$$\neg(¬Q \land ¬P)\land P\tag*{De Morgan's law}$$
$$(Q\lor P)\land P\tag*{De Morgan's law}$$
$$P\land(P\lor Q)\tag*{ Commutative law$\times2$}$$
$$P\tag*{Absorption law}$$
Therefore this simplified the expression.
A: Using de Morgan's laws and the double negation law:
$$\neg ((\neg Q \land \neg P) \lor \neg P) \equiv \neg (\neg Q \land \neg P) \land \neg \neg P$$
$$\equiv \neg (\neg Q \land \neg P) \land P$$
$$\equiv  (\neg \neg Q \lor \neg \neg P)\land P$$
$$\equiv (Q \lor P)\land P$$
Next, you could use the absorption law and commutative law to conclude that 
$$ (Q \lor P)\land P \equiv P \land (Q \lor P) \equiv P \land (P \lor Q) \equiv P$$
If you haven't proved the absorption law in class then you could directly prove the result through the same strategy in this question and here. 
