Translate the following inference to logic notation: For all x, x_1 will not play squash. There is at least one x for which if x_1 were a police officer, then x_1 is playing squash. For all x, if x_1 were a carpenter, then x_1 is a police officer. Therefore there is at least one x for which x_1 is not a carpenter.
forall x: ~P(x). exists x: Q(x) -> P(x). forall x: R(x) -> Q(x). Therefore exists x: ~R(x).