Translate the following inference to logic notation: There is at least one x for which if x_6 makes tea, then x_6 is climbing a mountain. For all x, if x_6 plays a game, then x_6 is playing squash. For all x, x_6 will play a game or x_6 will not climb a mountain. Therefore there is at least one x for which if x_6 makes tea, then x_6 is playing squash.
exists x: P(x) -> Q(x). forall x: R(x) -> S(x). forall x: R(x) or ~Q(x). Therefore exists x: P(x) -> S(x).