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