Translate the following inference to logic notation: For all x, x_6 will play a game or x_6 will listen to a song. For all x, if x_6 plays a game, then x_6 is not driving a car. For all x, x_6 will drive a car. There is at least one x for which x_6 is rich. Therefore there is at least one x for which x_6 will listen to a song and x_6 is rich.
forall x: P(x) or Q(x). forall x: P(x) -> ~R(x). forall x: R(x). exists x: S(x). Therefore exists x: Q(x) and S(x).