Logical equivalence please help I've been stuck on this one problem for a couple of days now with no clue on how to complete it. I need to prove the following logical equivalence:
$(\neg P \wedge \neg R) \vee (P \wedge \neg Q \wedge \neg R)$ is equivalent to $\neg R \wedge (Q \Rightarrow \neg(P ∧ \neg R))$
If anyone could shed some light on this matter, please, I noticed that $\neg R\wedge(\neg P) \wedge(P ∧ \neg Q )$ but I don't know if i'm approaching this the right way
 A: Simplification of the second expression is probably more useful initially, to disentangle the implication statement and give sight of what to work on next.
$$\begin{align}\\
\neg R \land (Q \Rightarrow \neg(P \land \neg R)) 
&= \neg R \land (\neg Q \lor \neg(P \land \neg R))\\
&= \neg R \land (\neg Q \lor \neg P \lor R)\\
&= \neg R \land (\neg Q \lor \neg P \lor R)\\
&= (\neg R \land \neg Q) \lor (\neg R \land \neg P) \lor (\neg R \land R)\\
&= (\neg R \land \neg Q) \lor (\neg R \land \neg P)\\
\end{align}$$
... which gets us really close to the desired statement; we just need to eliminate an extra term from the first expression.
$$\begin{align}\\
(\neg P \land \neg R) \lor (P \land \neg Q \land \neg R) 
&= (\neg P \lor (P \land \neg Q)) \land \neg R\\
&= ((\neg P \lor (\neg P \land \neg Q )) \lor (P \land \neg Q)) \land \neg R\\
&= (\neg P \lor ((\neg P \land P) \lor \neg Q)) \land \neg R\\
&= (\neg P \lor  \neg Q) \land \neg R\\
&= (\neg P \land \neg R) \lor (\neg Q \land \neg R) \\
\end{align}$$
which is equivalent to the second expression as required.
