Translate the following inference to logic notation: For all x, if x_5 were rich, then x_5 is an office clerk. For all x, if x_5 were an office clerk, then x_5 is rich. There is at least one x for which x_5 is an office clerk. Therefore there is at least one x for which x_5 is rich.
forall x: P(x) -> Q(x). forall x: Q(x) -> P(x). exists x: Q(x). Therefore exists x: P(x).