Proving tautology without using truth tables

Question

I have a statement (P∧Q∧(R∧P⇒~Q))⇒~R that I need to prove tautology without using truth tables. I understand I'll be using inference rules.

Here's what I've tried so far, step by step:

(P∧Q∧(~(R∧P)V~Q))⇒~R
((P∧Q∧~(R∧P)) V P∧Q∧~Q))⇒~R
((P∧Q∧~(R∧P)) V P∧False))⇒~R
((P∧Q∧~(R∧P)) V False))⇒~R
(P∧Q∧~(R∧P))⇒~R

I'm stuck at this point, unsure of what the next step is.

Answer 1

$\begin{align} (P∧Q∧(\neg(R∧P)\vee\neg Q))\to\neg R \\ ((P∧Q∧\neg(R∧P)) \vee P∧Q∧\neg Q))\to\neg R \\ ((P∧Q∧\neg(R∧P)) \vee P∧\bot))\to\neg R \\ ((P∧Q∧\neg(R∧P)) \vee \bot))\to\neg R \\ (P∧Q∧\neg (R∧P))\to\neg R \end{align}$

It's all good so far. The key steps to completion is to apply DeMorgan's Negation, Distribution, and Implication Equivalence.

$$\neg (X\wedge Y) \iff \neg X\vee \neg Y \\ X\wedge(Y\vee Z) \iff (X\wedge Y)\vee(X\wedge Z) \\ X\to Y \iff \neg X\vee Y$$

Extra Hint: $X\wedge(Y\vee \neg X) \iff X\wedge Y$

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That however is using boolean calculus. Using the rules of inference would be

$\begin{array}{|ll} (P∧Q∧(R∧P\to \neg Q))\to \neg R \\ (P\wedge \neg(\neg Q)\wedge (R\wedge P\to (\neg Q)))\to \neg R & \text{Double Negation} \\ (P\wedge \neg(R\wedge P))\to \neg R & \text{Modus Tollens} \\ P\to( \neg(R\wedge P)\to \neg R) & \text{Exportation} \\ P\to( \neg\neg R\to\neg\neg(R\wedge P)) & \text{Contraposition} \\ P\to( R\to\neg\neg(R\wedge P)) & \text{Double Negation Elimination} \\ P\to( R\to(R\wedge P)) & \text{Double Negation Elimination} \\ (P\wedge R)\to (R\wedge P) & \text{Importation} \\ (P\wedge R)\to (P\wedge R) & \text{Association} \\ \top & \text{Tautology} \end{array}$