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
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.
