Translate the following inference to logic notation: For all x, if z were a soldier, then z is a carpenter. For all x, if z were a soldier, then z is sleeping. For all x, if z sleeps, then z is a soldier. There is at least one x for which if z were a carpenter, then z is a soldier. Therefore there is at least one x for which if z sleeps, then z is a carpenter.
forall x: P(x) -> Q(x). forall x: P(x) -> R(x). forall x: R(x) -> P(x). exists x: Q(x) -> P(x). Therefore exists x: R(x) -> Q(x).