Translate the following inference to logic notation: For all x, x_5 is bored if and only if x_5 is a lawyer. For all x, if x_5 were rich, then x_5 is a lawyer. There is at least one x for which if x_5 were a lawyer, then x_5 is rich. For all x, x_5 is not bored. Therefore there is at least one x for which x_5 is not rich.
forall x: P(x) <-> Q(x). forall x: R(x) -> Q(x). exists x: Q(x) -> R(x). forall x: ~P(x). Therefore exists x: ~R(x).