Translate the following inference to logic notation: For all x, if x_1 were an office clerk, then x_1 is not running. For all x, if x_1 plays a game, then x_1 is an electrician. There is at least one x for which x_1 is an office clerk if and only if x_1 is an electrician. For all x, x_1 will go running. Therefore there is at least one x for which x_1 will not play a game.
forall x: P(x) -> ~Q(x). forall x: R(x) -> S(x). exists x: P(x) <-> S(x). forall x: Q(x). Therefore exists x: ~R(x).