Translate the following inference to logic notation: There is at least one x for which z will not sleep. For all x, z will sleep or z is not a janitor. For all x, z is a janitor if and only if z will sleep. Therefore there is at least one x for which z is not a janitor.
exists x: ~P(x). forall x: P(x) or ~Q(x). forall x: Q(x) <-> P(x). Therefore exists x: ~Q(x).