Translate the following inference to logic notation: For all x, if x_7 plays squash, then x_7 is climbing a mountain. For all x, x_7 will play squash and x_7 is a doctor. There is at least one x for which if x_7 were an office clerk, then x_7 is not climbing a mountain. Therefore there is at least one x for which x_7 is not an office clerk.
forall x: P(x) -> Q(x). forall x: P(x) and R(x). exists x: S(x) -> ~Q(x). Therefore exists x: ~S(x).