Translate the following inference to logic notation: For all x, if x_3 sleeps, then x_3 is a police officer. There is at least one x for which if x_3 were a janitor, then x_3 is sleeping. For all x, if x_3 were a police officer, then x_3 is a janitor. For all x, x_3 is not a police officer. Therefore there is at least one x for which x_3 is not a janitor.
forall x: P(x) -> Q(x). exists x: R(x) -> P(x). forall x: Q(x) -> R(x). forall x: ~Q(x). Therefore exists x: ~R(x).