Prove logical equivalence with negations and implications

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Leanna
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I'm stuck on this proof question:
(¬(Q⇒¬P) ∧ ¬((Q∧¬R)⇒¬P )) ⇔ ¬(R ∨ (P ⇒¬Q))

I've tried to get rid of the negation and implications but I keep going in circles and I'm getting nowhere near to the equivalence required. I would appreciative if anyone can help me solve this because it's really been doing my head in :/
 

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If you see question marks that's the negation
 
Hi Leanna,

I'm not sure what kind of proof structure is required, but you can obtain the results if you use the following:
$\lnot (Q \implies \lnot P) \iff (Q \land P)$, De Morgan's, and the fact that $Q \land Q \land P \iff Q \land P$.
 
Leanna said:
I'm stuck on this proof question:
(¬(Q⇒¬P) ∧ ¬((Q∧¬R)⇒¬P )) ⇔ ¬(R ∨ (P ⇒¬Q))
What exactly is the question? What you have written is a formula.