Translate the following inference to logic notation: For all x, if z_1 sleeps, then z_1 is driving a car. There is at least one x for which if z_1 drives a car, then z_1 is playing squash. For all x, if z_1 plays squash, then z_1 is sleeping. Therefore there is at least one x for which if z_1 plays squash, then z_1 is sleeping.
forall x: P(x) -> Q(x). exists x: Q(x) -> R(x). forall x: R(x) -> P(x). Therefore exists x: R(x) -> P(x).