We make a truth table which contains all the lines of the argument form:
| \(P\) | \(Q\) | \(P\imp Q\) | \(\neg P\) | \(\neg P \imp Q\) |
| T | T | T | F | T |
| T | F | F | F | T |
| F | T | T | T | T |
| F | F | T | T | F |
(we include a column for \(\neg P\) just as a step to help getting the column for \(\neg P \imp Q\)).
Now look at all the rows for which both \(P \imp Q\) and \(\neg P \imp Q\) are true. This happens only in rows 1 and 3. Hey! In those rows \(Q\) is true as well, so the argument form is valid (it is a valid deduction rule).