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