Translate the following inference to logic notation: For all x, if x_10 were rich, then x_10 is drinking water. There is at least one x for which if x_10 drinks water, then x_10 is rich. For all x, if x_10 were poor, then x_10 is drinking water. For all x, if x_10 were rich, then x_10 is poor. Therefore there is at least one x for which if x_10 drinks water, then x_10 is rich.
forall x: P(x) -> Q(x). exists x: Q(x) -> P(x). forall x: R(x) -> Q(x). forall x: P(x) -> R(x). Therefore exists x: Q(x) -> P(x).