Translate the following inference to logic notation: For all x, if x_5 were a janitor, then x_5 is a police officer. For all x, x_5 is fast or x_5 is a janitor. For all x, if x_5 sleeps, then x_5 is a police officer. There is at least one x for which if x_5 works, then x_5 is not a police officer. For all x, if x_5 were fast, then x_5 is sleeping. Therefore there is at least one x for which x_5 will not work.
forall x: P(x) -> Q(x). forall x: R(x) or P(x). forall x: S(x) -> Q(x). exists x: T(x) -> ~Q(x). forall x: R(x) -> S(x). Therefore exists x: ~T(x).