TITLE: Prove using properties of logical equivalence $(P\wedge (P \to (Q \to R))) \to (Q \to R)$ is a tautology
QUESTION [2 upvotes]: 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.

REPLY [2 votes]: 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:
$$\equiv (\neg P \lor \neg(g))\lor (g)$$
$$\equiv \neg P \lor (\neg(g)\lor (g))~~ \mbox{ (conmutative property) } $$
$$\equiv \neg P \lor \top~~ \mbox{ (property of } \wedge )$$
$$\equiv \top~~ \mbox{(absorption)}$$
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