Translate the following inference to logic notation: For all x, if x_5 sleeps, then x_5 is a janitor. There is at least one x for which x_5 will sleep. For all x, x_5 is not a janitor. Therefore there is at least one x for which x_5 is poor.
forall x: P(x) -> Q(x). exists x: P(x). forall x: ~Q(x). Therefore exists x: R(x).