Translate the following inference to logic notation: For all x, if x_16 drives a car, then x_16 is sad. For all x, x_16 is curious if and only if x_16 will drive a car. There is at least one x for which if x_16 were sad, then x_16 is playing tennis. For all x, x_16 will play tennis if and only if x_16 will play squash. Therefore there is at least one x for which if x_16 were curious, then x_16 is playing squash.
forall x: P(x) -> Q(x). forall x: R(x) <-> P(x). exists x: Q(x) -> S(x). forall x: S(x) <-> T(x). Therefore exists x: R(x) -> T(x).