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