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