Translate the following inference to logic notation: For all x, z is curious. For all x, if z plays tennis, then z is working. There is at least one x for which z will drink water. For all x, if z were curious, then z is playing tennis. For all x, if z works, then z is a carpenter. Therefore there is at least one x for which z is a carpenter and z will drink water.
forall x: P(x). forall x: Q(x) -> R(x). exists x: S(x). forall x: P(x) -> Q(x). forall x: R(x) -> T(x). Therefore exists x: T(x) and S(x).