Equivalence $\displaystyle((\lnot p \lor q) \land (q \lor r)) \land (p \land \lnot q) ≡ c$

Question

I am trying to prove this. Using only laws. I am getting nowhere. The 3 variables on the left hand side keep tripping me up. I've tried DeMorgans, Distributive, Identity and Negation as a starting point but hit dead ends.

Answer 1

Distributing, $$((\lnot p \lor q)\land (q\lor r)) \iff (((\lnot p \lor q)\land q ) \lor ((\lnot p \lor q)\land r)) \iff q\,\lor ((\lnot p \lor q)\land r)$$ $$\iff (\lnot p \lor q)\land (q\lor r) \iff \lnot(p \land \lnot q)\land (q\lor r)$$ Then the original expression is $$\lnot(p \land \lnot q)\land (q\lor r) \land (p\land \lnot q)$$ Which is false.

Answer 2

$$((\neg p\vee q)\wedge (q\vee r)) \wedge (p\wedge \neg q)$$

Use commutation, association, and deMorgan's to obtain:

$$(q\vee r)\wedge (\neg (p\wedge \neg q)\wedge (p\wedge \neg q))$$

Then you should be able to take it from there.