Translate the following inference to logic notation: For all x, if x_9 were a mechanic, then x_9 is rich. There is at least one x for which x_9 will not take a plane. For all x, x_9 will take a plane if and only if x_9 is rich. For all x, if x_9 drinks water, then x_9 is a mechanic. Therefore there is at least one x for which x_9 will not drink water.
forall x: P(x) -> Q(x). exists x: ~R(x). forall x: R(x) <-> Q(x). forall x: S(x) -> P(x). Therefore exists x: ~S(x).