Show that (p∨q)∧(¬p∨r)→(q∨r)is a tautology.

Question

How to do this without using a truth table?

Answer 1

The key observation is the antecedent "(p or q) and ((not p) or r)" is equivalent to "q or r", which makes the whole thing equivalent to "(q or r) implies (q or r)" which is obviously a tautology.

You can see why the observation is true in a few different ways. The simplest is to use a truth table, but you can also see this by the following argument: if p is true, then r has to be true, and if p is false then q has to be true. Therefore, since p is either true or false, one of r or q has to be true.