Simplifying P AND (P OR NOT Q) How can I simplify this? I've tried invoking Demorgan's Law and I get
P AND (NOT (NOT P AND Q)) but I can't seem to simplify this further.
The answer is P, but how can I prove this?
 A: $$P(P+\overline{Q})=P+P\overline{Q}=P$$
A: You can think of it as (P AND P) OR (P AND NOT Q).  
This is true if and only if P is true.  If P is false, it makes both arguments of the 'OR' false.
A: $P \land (P \lor \lnot Q) = (P \land P) \lor (P \land \lnot Q) = P \lor (P\land \lnot Q)$
If $P$ then clearly this expression holds. If $\lnot P$ then $\lnot (P\land \lnot Q)$ so we see the expression depends entirely on $P$.
Stuff below is why you were confused, be cause it is not equal to $P$.
$P \land \lnot (\lnot P \land Q) = P \land (\lnot (\lnot P) \lor \lnot Q) = P \land (P \lor \lnot Q) = (P\land P) \lor (P \land \lnot Q) = P\land \lnot Q$ 
A: Shahar's answer is the most direct.  However, sometimes, these can be a bit tricky when first starting out.  What I frequently like to do when handling these sort of problems is to use this identity:
phi(a, b, c, d, ...) = a AND phi(TRUE, b, c, d, ...) OR
                       NOT a AND phi(FALSE, b, c, d, ...)

I call this the 'factor' identity for lack of a better name, but there is probably an 'official' name out there somewhere.
Applying this to your problem, I would see that:
P AND (NOT (NOT P AND Q)) = P AND (P OR NOT Q))  /* Demorgan's */
                          = P AND (TRUE AND (TRUE OR NOT Q)) OR
                            NOT P AND (FALSE AND (FALSE OR NOT Q)) /*Factor*/
                          = P AND TRUE OR NOT P AND FALSE
                          = P

Be warned that this process is not necessarily the fastest method, but it always leads to simpler expressions.
