Can't seem to show that (P∧R)⇒¬(Q∧¬(P∧R)) is a tautology

Question

I'm trying to show that (P∧R)⇒¬(Q∧¬(P∧R)) is a tautology, which according to Wolfram Alpha it is. However, I'm getting stuck with my calculations, I have some ideas but can't figure out how to apply the rules to achieve my answer.

Show with a calculation that (P∧R)⇒¬(Q∧¬(P∧R)) is a tautology. I have tried:

{Implication}

¬(P∧R)∨¬(Q∧¬(P∧R))

{De Morgan, 2x}

(¬P∨¬R)∨(¬Q∨¬¬(P∧R))

{Double negation}

(¬P∨¬R)∨(¬Q∨(P∧R))

{Associativity}

¬Q∨(¬P∨¬R)∨(P∧R)

{Distributivity}

¬Q∨(P∧(¬P∨¬R))∨(R∧(¬P∨¬R))

{Distributivity, 2x}

¬Q∨(P∧¬P)∨(P∧¬R)∨(R∧¬P)∨(R∧¬R)

I could apply {Contradiction} twice, however don't see where this will help me. I don't really know how to go on from here, so some explanation would be greatly appreciated.

Answer 1

From $\neg Q\lor(\neg P\lor\neg R)\lor(P\land R)$ you can get $\neg Q\lor\big(\neg(P\land R)\lor(P\land R)\big)$, and from there $\neg Q\lor\top$ and then $\top$. (Thus, your first application of De Morgan is unnecessary.)

Answer 2

I think it will simplify things to replace "$P\wedge R$" by "$A$".

Then you're trying to show that $$(*)\quad A\implies \neg (Q\wedge \neg A)$$ is a tautology. Following the steps you've laid out, it's enough to show that $$(**)\quad \neg Q\vee \neg A\vee A$$ is a tautology (I'm slightly rewriting the line after "Associativity" in your OP). But $\neg A\vee A$ is clearly a tautology, and we can always deduce $S\vee T$ from $T$. So do you see how to finish this?

Answer 3

Establish that $\lnot$(Q $\land$ $\lnot$Z) == (Q $\to$ Z).

Substitute each instance of Z in the above with (P$\land$R).

Then we have ((P$\land$R)$\to$(Q$\to$(P$\land$R))).

Forget about that for a moment and check that 1. (A $\to$ (Q $\to$ A)) is a tautology.

Finally, substitute A with (P$\land$R) in all places in 1. where A appears.

This solution uses one equivalence, one tautology checked, and two substitutions.