Translate the following inference to logic notation: For all x, if z_1 sleeps, then z_1 is a police officer. There is at least one x for which if z_1 were a police officer, then z_1 is sleeping. For all x, z_1 will not sleep. Therefore there is at least one x for which z_1 is not a police officer.
forall x: P(x) -> Q(x). exists x: Q(x) -> P(x). forall x: ~P(x). Therefore exists x: ~Q(x).