Finding simpler formula I have to find simpler formula to this one:
$$\lnot(p \land q) \lor (\lnot p \land q)$$
I started by using De Morgan's law to get: $$(\lnot p \lor \lnot \lnot q) \lor (\lnot p \land q)$$ then used the Double negation law to get: $$(\lnot p \lor q) \lor (\lnot p \land q)$$
Probably I am missing something because right now it seems it cannot go further. Thanks.

You have four different cases: $p$ true and $q$ true, $p$ true and $q$ false, $p$ false and $q$ true, $p$ and $q$ both false. Then
\begin{align}
& p \mbox{ true and }q \mbox{ true implies } \lnot p \mbox{ or } \lnot q \mbox{ false, } \lnot p \mbox{ and } q \mbox{ false } \\
& p \mbox{ true and }q \mbox{ false implies } \lnot p \mbox{ or } \lnot q \mbox{ true, } \lnot p \mbox{ and } q \mbox{ false } \\
& p \mbox{ false and }q \mbox{ true implies } \lnot p \mbox{ or } \lnot q \mbox{ true, } \lnot p \mbox{ and } q \mbox{ true } \\
&p \mbox{ false and }q \mbox{ false implies } \lnot p \mbox{ or } \lnot q \mbox{ true, } \lnot p \mbox{ and } q \mbox{ false.} 
\end{align}
Hence $\lnot (p \mbox{ and } q) \mbox{ or } (\lnot p \mbox{ and } q)$ is false when both $p$ and $q$ are true, true in the other cases. Thus is equivalent to $q \rightarrow \lnot p$, which is false when both $p$ and $q$ are true and true in the other cases. You can also rewrite it as $\lnot q \mbox{ or } \lnot p$, which is $\lnot (p \mbox{ and } q)$.
