Translate the following inference to logic notation: For all x, if x_4 takes a plane, then x_4 is excited. There is at least one x for which if x_4 climbs a mountain, then x_4 is writing a letter. For all x, if x_4 sleeps, then x_4 is taking a plane. For all x, if x_4 writes a letter, then x_4 is sleeping. Therefore there is at least one x for which if x_4 climbs a mountain, then x_4 is excited.
forall x: P(x) -> Q(x). exists x: R(x) -> S(x). forall x: T(x) -> P(x). forall x: S(x) -> T(x). Therefore exists x: R(x) -> Q(x).