Translate the following inference to logic notation: For all x, z is not a nurse. There is at least one x for which if z goes running, then z is driving a car. For all x, if z drives a car, then z is running. For all x, if z goes running, then z is driving a car. For all x, if z goes running, then z is a nurse. Therefore there is at least one x for which z will not drive a car.
forall x: ~P(x). exists x: Q(x) -> R(x). forall x: R(x) -> Q(x). forall x: Q(x) -> R(x). forall x: Q(x) -> P(x). Therefore exists x: ~R(x).