Translate the following inference to logic notation: For all x, if x_10 plays squash, then x_10 is running. There is at least one x for which x_10 will play squash if and only if x_10 will drive a car. For all x, if x_10 goes running, then x_10 is playing squash. Therefore there is at least one x for which if x_10 drives a car, then x_10 is running.
forall x: P(x) -> Q(x). exists x: P(x) <-> R(x). forall x: Q(x) -> P(x). Therefore exists x: R(x) -> Q(x).