Translate the following inference to logic notation: For all x, if y_3 were a janitor, then y_3 is drinking water. For all x, y_3 will not drink water. There is at least one x for which if y_3 drinks water, then y_3 is a janitor. Therefore there is at least one x for which y_3 is not a janitor.
forall x: P(x) -> Q(x). forall x: ~Q(x). exists x: Q(x) -> P(x). Therefore exists x: ~P(x).