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