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