Translate the following inference to logic notation: For all x, if x_15 were excited, then x_15 is working. There is at least one x for which if x_15 were an office clerk, then x_15 is a scientist. For all x, if x_15 works, then x_15 is climbing a mountain. For all x, if x_15 climbs a mountain, then x_15 is an office clerk. For all x, x_15 is excited or x_15 is not a scientist. Therefore there is at least one x for which x_15 is excited if and only if x_15 will climb a mountain.
forall x: P(x) -> Q(x). exists x: R(x) -> S(x). forall x: Q(x) -> T(x). forall x: T(x) -> R(x). forall x: P(x) or ~S(x). Therefore exists x: P(x) <-> T(x).