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