Translate the following inference to logic notation: There is at least one x for which if x_7 were an office clerk, then x_7 is a janitor. For all x, x_7 is not bored. For all x, x_7 is a janitor if and only if x_7 is bored. Therefore there is at least one x for which x_7 is not an office clerk.
exists x: P(x) -> Q(x). forall x: ~R(x). forall x: Q(x) <-> R(x). Therefore exists x: ~P(x).