Translate the following inference to logic notation: For all x, if x_3 were a mechanic, then x_3 is rich. There is at least one x for which x_3 is not rich. For all x, if x_3 were a mechanic, then x_3 is a police officer. For all x, if x_3 were a police officer, then x_3 is a mechanic. Therefore there is at least one x for which x_3 is not a police officer.
forall x: P(x) -> Q(x). exists x: ~Q(x). forall x: P(x) -> R(x). forall x: R(x) -> P(x). Therefore exists x: ~R(x).