Prove using properties of logical equivalence $(P\wedge (P \to (Q \to R))) \to (Q \to R)$ is a tautology Can anyone help me with this problem using properties of logical equivalence:

Prove that $(P\wedge (P \to (Q \to R))) \to (Q \to R)$ is a tautology.

 A: As you did above, you can simplify the problem: 
$$(P\wedge(P\rightarrow (Q\rightarrow R)))\rightarrow(Q\rightarrow R)$$
$$\equiv (\neg P \lor (P \wedge (Q\wedge\neg R)))\wedge (\neg Q \lor R)$$
Now,
$$\equiv ((\neg P \lor P) \wedge (\neg P \lor (Q \wedge \neg R)))\lor (\neg Q \lor R)~~ \mbox{ (distributive  property)}$$
$$\equiv (\top \wedge (\neg P \lor (Q \wedge \neg R)))\lor (\neg Q \lor R)~~ \mbox{ (property of } \wedge )$$
$$\equiv (\neg P \lor (Q \wedge \neg R))\lor (\neg Q \lor R)~~ \mbox{(absorption)}$$
$$\equiv (\neg P \lor \neg(\neg Q \lor R))\lor (\neg Q \lor R)~~ \mbox{(negating and applying Morgan's Law)}$$
Let $g:= (\neg Q \lor R)$ for clarity.
This simplifies the expression down to:
$$\equiv (\neg P \lor \neg(g))\lor (g)$$
$$\equiv \neg P \lor (\neg(g)\lor (g))~~ \mbox{ (conmutative property) } $$
$$\equiv \neg P \lor \top~~ \mbox{ (property of } \wedge )$$
$$\equiv \top~~ \mbox{(absorption)}$$
A: In a situation like that you can always recur to the brute force approach. Ugly, but it always works.
You have $3$ variables, i.e. $P$, $Q$ and $R$. Each can be true or false. You have therefore $8$ possibilities to check, not a very huge task.
Still, maybe in your case it is not a tautology. 
\begin{array}{|c|c|c|c|}
\hline
P& Q & R &  (P\wedge (P \to (Q \to R))) \wedge (Q \to R) \\ \hline
T & T & T & T \\ \hline
T & T & F & F \\ \hline
T & F & T & T \\ \hline
T & F & F & T \\ \hline
F & T & T & F \\ \hline
F & T & F & F \\ \hline
F & F & T & F \\ \hline
F & F & F & F \\ \hline
\end{array}
