Translate the following inference to logic notation: For all x, if x were curious, then x is running. For all x, if x goes running, then x is playing tennis. For all x, if x plays tennis, then x is excited. There is at least one x for which x is an astronaut or x is not excited. Therefore there is at least one x for which if x were curious, then x is an astronaut.
forall x: P(x) -> Q(x). forall x: Q(x) -> R(x). forall x: R(x) -> S(x). exists x: T(x) or ~S(x). Therefore exists x: P(x) -> T(x).