Discrete 101: Validity of proof: Finding that p→q ∨ ¬r, q→p∧r, therefore p→r is invalid.

Question

I'm sorry to bother with what apparently is a very easy Basic Logic question, but in my class'es notes there's an example that the professor probably explained in class:

Show that the following proof is invalid
p→(q∨¬r)
q→(p∧r)
therefore p→r

How would I go around showing this, and finding that something's wrong along the way? Truth tables?

Answer 1

The only way to prove that proposition $a\to b$ is false is if $a$ is true and $b$ is false. Hence, you must have $p$ true and $r$ false, by the conditions of the task you were set. If $q$ were true, then $r$ would be true (by the second hypothesis), so we must have $q$ false.

Now verify that $\{p$ true, $q,r$ false$\}$ satisfy both hypothesis and falsify the conclusion.