Translate the following inference to logic notation: For all x, if x_4 makes tea, then x_4 is rich. For all x, if x_4 were rich, then x_4 is working. For all x, x_4 is poor if and only if x_4 will make tea. Therefore for all x, if x_4 were poor, then x_4 is working.
forall x: P(x) -> Q(x). forall x: Q(x) -> R(x). forall x: S(x) <-> P(x). Therefore forall x: S(x) -> R(x).