Translate the following inference to logic notation: For all x, x_3 will not play a game. For all x, if x_3 drives a car, then x_3 is playing a game. For all x, if x_3 plays a game, then x_3 is rich. There is at least one x for which if x_3 were rich, then x_3 is driving a car. Therefore there is at least one x for which x_3 is not rich.
forall x: ~P(x). forall x: Q(x) -> P(x). forall x: P(x) -> R(x). exists x: R(x) -> Q(x). Therefore exists x: ~R(x).