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.

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.
