I have $¬(¬(P ∧ ¬Q) ∨ (¬R ∨ ¬Q)),$ I know I can distribute to get:
$¬((¬P ∨ Q) ∨ (¬R ∨ ¬Q))$ ... my question is a little basic but I don't know whether my next step will be:
$(P ∧ ¬Q) ∨ (R ∧ Q)$ or $(P ∧ ¬Q) ∧ (R ∧ Q)$
I'm inclined to believe the second of the two is correct, as my truth table is giving false for everything and the second gives $¬Q ∧ Q$ which will be false.
By DeMorgan's laws, $^\dagger$ starting with$$¬((¬P ∨ Q) ∨ (¬R ∨ ¬Q))$$ we first get $$\lnot (\lnot P\lor Q) \land \lnot (\lnot R \lor \lnot Q)$$
Then, using DeMorgan's laws, again, twice, we get $$(P \land \lnot Q) \land (R \land Q)$$
And you are correct, that gives us $$P\land \lnot Q \land R \land Q$$ (The second of the two options you though might be correct), which gives us $$P \land R \land (\lnot Q \land Q) \equiv P \land R \land False \equiv False.$$
$\dagger$ DeMorgan's laws: $\lnot (a \lor b) \equiv (\lnot a \land \lnot b)$, and $\lnot (a\land b) \equiv (\lnot a \lor \lnot b)$.
