Translate the following inference to logic notation: For all x, if x_7 were fast, then x_7 is a scientist. For all x, x_7 is not curious. For all x, if x_7 were a carpenter, then x_7 is fast. There is at least one x for which if x_7 were a scientist, then x_7 is curious. Therefore there is at least one x for which x_7 is not a carpenter.
forall x: P(x) -> Q(x). forall x: ~R(x). forall x: S(x) -> P(x). exists x: Q(x) -> R(x). Therefore exists x: ~S(x).