Translate the following inference to logic notation: There is at least one x for which if x_9 were fast, then x_9 is bored. For all x, if x_9 were a police officer, then x_9 is fast. For all x, if x_9 were bored, then x_9 is fast. For all x, x_9 is not bored. Therefore there is at least one x for which x_9 is not a police officer.
exists x: P(x) -> Q(x). forall x: R(x) -> P(x). forall x: Q(x) -> P(x). forall x: ~Q(x). Therefore exists x: ~R(x).