Translate the following inference to logic notation: There is at least one x for which if x_15 were a mechanic, then x_15 is making tea. For all x, x_15 is a mechanic if and only if x_15 will make tea. For all x, x_15 will make tea or x_15 will not play a game. Therefore there is at least one x for which if x_15 plays a game, then x_15 is a mechanic.
exists x: P(x) -> Q(x). forall x: P(x) <-> Q(x). forall x: Q(x) or ~R(x). Therefore exists x: R(x) -> P(x).