Translate the following inference to logic notation: For all x, if x_10 works, then x_10 is a doctor. There is at least one x for which if x_10 plays squash, then x_10 is driving a car. For all x, x_10 will play squash if and only if x_10 is a janitor. For all x, if x_10 drives a car, then x_10 is working. Therefore there is at least one x for which if x_10 were a janitor, then x_10 is a doctor.
forall x: P(x) -> Q(x). exists x: R(x) -> S(x). forall x: R(x) <-> T(x). forall x: S(x) -> P(x). Therefore exists x: T(x) -> Q(x).