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