Logical Equivalences - (P and not Q) or((P and(not R)) and Q)

Question

I have been studying logical equivalences recently and came across this question; s = P and(R implies( not(Q and P))) $$s= P\land(R\to (\lnot(Q\land P)))$$

Show that (P and not Q) or((P and(not R)) and Q) is equivalent to s using logical equivalence.$$(P\land\lnot Q)\lor((P\land(\lnot R)) \land Q)$$

I have used distributive law to multiple the brackets but have not managed to get much further. Help would be greatly appreciated.

No truth table allowed

Thanks

Answer 1

Here is a step-by-step way without using a truth table: $$P \ \wedge (R \to \neg\left(Q \ \wedge P\right) ) $$

$$P \ \wedge (R \to (\neg Q \ \vee \neg P) ) $$

$$P \ \wedge (\neg R \vee(\neg Q \ \vee \neg P) $$

$$P \ \wedge (\neg R \vee \neg Q \ \vee \neg P ) $$

$$P \ \wedge (\neg R \vee \neg Q \ ) $$

$$P \ \wedge ((\neg R \vee \neg Q ) \wedge (\neg Q \vee Q) ) $$

$$P \ \wedge ((\neg Q \vee (\neg R \wedge Q) ) $$

$$(P \ \wedge\neg Q) \vee (P \wedge (\neg R \wedge Q) ) $$

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In the third line, we use the fact that $p \to q \equiv \neg p \vee q $

In the fifth line, the $\neg P$ is removed because it is unnecessary; $p \wedge \neg p$ is a contradiction and will never be true

In the sixth line, we can then add in $\neg Q \vee Q$, as in general, $p \equiv p \wedge (\neg q \vee q)$