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