Translate the following inference to logic notation: There is at least one x for which if x_8 drives a car, then x_8 is working. For all x, if x_8 plays squash, then x_8 is playing tennis. For all x, if x_8 plays tennis, then x_8 is driving a car. For all x, if x_8 works, then x_8 is playing squash. Therefore there is at least one x for which if x_8 plays tennis, then x_8 is working.
exists x: P(x) -> Q(x). forall x: R(x) -> S(x). forall x: S(x) -> P(x). forall x: Q(x) -> R(x). Therefore exists x: S(x) -> Q(x).