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