Translate the following inference to logic notation: For all x, if x_5 were a writer, then x_5 is poor. For all x, if x_5 were curious, then x_5 is a writer. There is at least one x for which if x_5 were a lawyer, then x_5 is curious. For all x, if x_5 were poor, then x_5 is a lawyer. Therefore there is at least one x for which x_5 is curious if and only if x_5 is a lawyer.
forall x: P(x) -> Q(x). forall x: R(x) -> P(x). exists x: S(x) -> R(x). forall x: Q(x) -> S(x). Therefore exists x: R(x) <-> S(x).