Translate the following inference to logic notation: There is at least one x for which if z plays tennis, then z is a carpenter. For all x, if z plays tennis, then z is a carpenter. For all x, if z were a carpenter, then z is playing tennis. Therefore there is at least one x for which z is a carpenter if and only if z will play tennis.
exists x: P(x) -> Q(x). forall x: P(x) -> Q(x). forall x: Q(x) -> P(x). Therefore exists x: Q(x) <-> P(x).