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