Applying laws of propositional logic question: $\neg \big( p \lor ( \neg p \land q ) \big) \equiv \neg p \land \neg q$

I need to prove that $$\neg \big( p \lor ( \neg p \land q ) \big)$$ is logically equivalent to $$\neg p \land \neg q$$ using the laws of propositional logic instead of a truth table. I can't figure out which rules work to transform it to $$\neg p \land \neg q$$. I am taking a logic course and I want to practice my logical thinking so that I can be ready on test day. This is one I am practicing with and I tried to do De Morgan's, then associativity then idempotence laws, but I don't think this is correct. I also tried to do absorption and then De Morgan's, but I don't think that works either because of the negations and because I need it to be equivalent to $$\neg p \land \neg q$$.

• Please include all the intermediate results so that we can better locate where you ran into trouble, and provide better answers. Sep 7, 2021 at 22:57
• Use in first step distribution and then De Morgan. Add done steps to question if/when you stack and you'll get help. Sep 7, 2021 at 23:35

Using distibution we have: $$p \lor ( \neg p \land q ) \equiv \big(p\lor (\neg p)\big)\land \big(p\lor q\big) \equiv 1 \land \big(p\lor q\big) \equiv p\lor q$$ Then De Morgan gives answer
• @Skytendo I suppose that $u$ means $\vee$ and $n$ means $\wedge$. Sep 8, 2021 at 0:10