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