Translate the following inference to logic notation: For all x, if y were a mechanic, then y is running. For all x, y will play squash if and only if y will go running. For all x, if y works, then y is a mechanic. For all x, y will not play squash. For all x, if y goes running, then y is working. Therefore for all x, y is not a mechanic.
forall x: P(x) -> Q(x). forall x: R(x) <-> Q(x). forall x: S(x) -> P(x). forall x: ~R(x). forall x: Q(x) -> S(x). Therefore forall x: ~P(x).