$(\lnot p \lor q) \land \lnot[(p\lor \lnot q)\Rightarrow (p\lor q)]$ is equivalent to $\lnot (p\lor q)$
Ok so I'm supposed to simplify the expression to get the other one, but I'm stuck. Currently I'm with this: $$(\lnot p \lor q) \land \lnot[(\lnot p\land q)\lor (p\lor q)]$$
I don't know how to solve what's between $[{}\cdot{}]$. If I had 2 $\land$ it would be easy. Can someone explain how I can surpass this? Thank you
Hint:
The attempts are correct so far, now see if we can find somewhere to apply Absorption law.
Answer:
\begin{align} &(\lnot p \lor q) \land \lnot[(\lnot p\land q)\lor (p\lor q)]\\ \equiv&(\lnot p \lor q) \land \lnot[((\lnot p\land q)\lor q) \lor p]\tag*{Reordering}\\ \equiv&(\lnot p \lor q) \land \lnot(q \lor p)\tag*{Absorption law}\\ \equiv&(\lnot p \lor q) \land (\lnot q \land \lnot p)\tag*{De Morgan's law}\\ \equiv&((\lnot p \lor q) \land \lnot p) \land \lnot q)\tag*{Reordering}\\ \equiv&\lnot p \land \lnot q\tag*{Absorption law}\\ \equiv&\lnot (p \lor q)\tag*{De Morgan's law}\\ \end{align}