Translate the following inference to logic notation: For all x, if z plays squash, then z is working. There is at least one x for which z will work or z will not play squash. For all x, if z works, then z is playing squash. Therefore there is at least one x for which if z plays squash, then z is working.
forall x: P(x) -> Q(x). exists x: Q(x) or ~P(x). forall x: Q(x) -> P(x). Therefore exists x: P(x) -> Q(x).