Translate the following inference to logic notation: For all x, x_6 is not a lawyer. There is at least one x for which if x_6 were a lawyer, then x_6 is a police officer. For all x, if x_6 were a police officer, then x_6 is a lawyer. Therefore there is at least one x for which x_6 is not a police officer.
forall x: ~P(x). exists x: P(x) -> Q(x). forall x: Q(x) -> P(x). Therefore exists x: ~Q(x).