Translate the following inference to logic notation: There is at least one x for which if x_3 were a carpenter, then x_3 is an office clerk. For all x, x_3 will go running if and only if x_3 is an office clerk. For all x, x_3 will not go running. Therefore there is at least one x for which x_3 is not a carpenter.
exists x: P(x) -> Q(x). forall x: R(x) <-> Q(x). forall x: ~R(x). Therefore exists x: ~P(x).