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