Translate the following inference to logic notation: There is at least one x for which if x_3 plays a game, then x_3 is making tea. For all x, x_3 is not a mechanic. For all x, x_3 is a mechanic if and only if x_3 will make tea. Therefore there is at least one x for which x_3 will not play a game.
exists x: P(x) -> Q(x). forall x: ~R(x). forall x: R(x) <-> Q(x). Therefore exists x: ~P(x).