Translate the following inference to logic notation: For all x, x_12 is rich if and only if x_12 is an office clerk. For all x, if x_12 works, then x_12 is an office clerk. There is at least one x for which x_12 is not rich. Therefore there is at least one x for which x_12 will not work.
forall x: P(x) <-> Q(x). forall x: R(x) -> Q(x). exists x: ~P(x). Therefore exists x: ~R(x).