Equivalent formula - how do I go from $\neg (P \wedge \neg Q) \vee (\neg P \wedge Q)$ to $\neg P \vee Q$? This is item "c" of question 11 from section 1.2 in Daniel J. Velleman's "How to Prove It - A Structured Approach" (great book).
The question asks that I find a simpler formula equivalent to $\neg (P \wedge \neg Q) \vee (\neg P \wedge Q)$, and the answer at the end of the book is $\neg P \vee Q$.  I've tried DeMorgan and the associative and commutative laws, to no avail. I'm at my wit's end. All I got was $(\neg P \vee Q) \vee (\neg P \wedge Q)$.
Any clues? Thanks in advance.
 A: Note that if $(\neg P \wedge Q)$ holds then $(\neg P \vee Q)$ holds, so $(\neg P \vee Q) \vee (\neg P \wedge Q)\iff (\neg P \vee Q)$.
A: $\neg (P \wedge \neg Q) \vee (\neg P \wedge Q)$
(de Morgan's)
$(\neg P \lor Q) \lor (\neg P \land Q)$
(distributivity; associativity)
$(\neg P \lor Q \lor \neg P) \land (\neg P \lor Q \lor Q)$
Should be easy to see from here.
A: For an alternative approach you can make the truth table of both logical expressions.
$$
\begin{array}{|c|c| c|c| c|c| c|c| c|}
\hline
P                         & Q                   & 
\neg P                    &  \neg Q             &
\neg (P \wedge \neg Q)    & (\neg P \wedge Q)   & 
\neg (P \wedge \neg Q) \vee (\neg P \wedge Q)
\\\hline
V                         & V                   & 
F                         & F                   &
V                         & F                   & 
V
\\\hline
V                         & F                   & 
F                         & V                   &
F                         & F                   & 
F
\\\hline
F                         & V                   & 
V                         & F                   &
V                         & V                   &
V 
\\\hline
F                         & F                   & 
V                         & V                   &
V                         & F                   & 
V
\\\hline
\end{array}
$$
and
$$
\begin{array}{|c|c| c|c|}
\hline
 P&Q&\neg P& \neg P \vee Q 
\\\hline
 V&V&F& V 
\\\hline
 V&F&F& F 
\\\hline
 F&V&V& V 
\\\hline
 F&F&V& V 
\\\hline
\end{array}
$$
