show this statement is a tautology using conditional statements

Question

Have been doing some homework questions and got stuck on this question

$$(p ∨ q) ∧ (p → r) ∧ (q → r) → r $$

I got it simplified to

$$((¬p ∧ ¬q) ∨ ¬r ∧ (p ∨ q)) ∨ r$$

but now i'm not sure what to do now. Any help would be appreciated.

Answer 1

The Implication Equivalence is typically known as $p \to q = \neg p \lor q$, but note that that means that $\neg (p \to q) = \neg(\neg p \lor q) = \neg \neg p \land \neg q = p \land \neg q$

So, let's use the $\neg (p \to q) =p \land \neg q$ principle (which some texbooks actually have as an implicit part of the Implication equivalence).

Then:

$$[(p ∨ q) ∧ (p → r) ∧ (q → r)] → r =$$

$$\neg \neg ([(p \lor q) \land (\neg p \lor r) \land (\neg q \lor r)] \to r) =$$

$$\neg ((p \lor q) \land (\neg p \lor r) \land (\neg q \lor r) \land \neg r) $$

OK, now use:

Reduction

$p \land (\neg p \lor q) = p \land q$ (in the context of $p$, the $\neg p \lor q$ term 'reduces to just $q$)

Applied to where we were:

$$\neg ((p \lor q) \land (\neg p \lor r) \land (\neg q \lor r) \land \neg r) =$$

$$\neg ((p \lor q) \land \neg p \land \neg q \land \neg r) =$$

$$\neg (q \land \neg p \land \neg q \land \neg r) =$$

$$\neg (\bot) = \top$$