Translate the following inference to logic notation: For all x, if x_6 were excited, then x_6 is poor. For all x, if x_6 works, then x_6 is excited. For all x, if x_6 were poor, then x_6 is working. There is at least one x for which if x_6 climbs a mountain, then x_6 is excited. Therefore there is at least one x for which if x_6 climbs a mountain, then x_6 is poor.
forall x: P(x) -> Q(x). forall x: R(x) -> P(x). forall x: Q(x) -> R(x). exists x: S(x) -> P(x). Therefore exists x: S(x) -> Q(x).