Translate the following inference to logic notation: For all x, y is a carpenter or y will not play tennis. There is at least one x for which y will drive a car if and only if y will play tennis. For all x, if y were a carpenter, then y is playing tennis. Therefore there is at least one x for which if y were a carpenter, then y is driving a car.
forall x: P(x) or ~Q(x). exists x: R(x) <-> Q(x). forall x: P(x) -> Q(x). Therefore exists x: P(x) -> R(x).