Translate the following inference to logic notation: For all x, x_9 is a janitor if and only if x_9 is a lawyer. For all x, x_9 is a janitor. There is at least one x for which x_9 is not a lawyer. Therefore there is at least one x for which x_9 is bored.
forall x: P(x) <-> Q(x). forall x: P(x). exists x: ~Q(x). Therefore exists x: R(x).