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.
 A: If A follows p and B follows q then 
${(A \cap B)}' \cup (A' \cap B) = A' \cup B' \cup (A' \cap B) = A' \cup B'$
since $A' \cap B \subset A' \cup B'$.
A: 
I have to find simpler formula to this one:

not(p and not q) or (not p and q).

I started by using De Morgan's law to get (not p or not not q) or (not p and q), then used the Double negation law to get (not p or q) or (not p and q).
  Probably I am missing something because right now it seems it cannot go further. Thanks.

So far, it is good.
Associativity gives not p or (q or (not p and q))
Commutativity gives not p or (q or (q and not p))
Absorption gives not p or q
That is all.

Absorption : A or (A and B) means either A is true or both A and B are; which certainly means A is true.  Converserly A means either A is true or both A and something else is true.  So A or (A and B) is equivalent to A.
A: 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 $\lnot (p \mbox{ and } q)$.
A: not(p and q) or (not p and q)
= (PQ)'+ P'Q
= P' + Q' + P'Q [Applying De Morgan's 2nd Law]
= P' (1 + Q) + Q' [X + 1 = 1]
= P' + Q' (Ans)
I hope I'm correct. Checked using a truth table.
