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