Translate the following inference to logic notation: For all x, if x_4 were rich, then x_4 is not playing squash. There is at least one x for which x_4 is a police officer or x_4 will play squash. For all x, x_4 is not a police officer. Therefore there is at least one x for which x_4 is not rich.
forall x: P(x) -> ~Q(x). exists x: R(x) or Q(x). forall x: ~R(x). Therefore exists x: ~P(x).