Translate the following inference to logic notation: For all x, z_4 is not a lawyer. There is at least one x for which if z_4 were a lawyer, then z_4 is an office clerk. For all x, if z_4 were an office clerk, then z_4 is a lawyer. Therefore there is at least one x for which z_4 is not an office clerk.
forall x: ~P(x). exists x: P(x) -> Q(x). forall x: Q(x) -> P(x). Therefore exists x: ~Q(x).