Translate the following inference to logic notation: For all x, x_11 is not bored. There is at least one x for which x_11 is bored or x_11 is not a server. For all x, x_11 is a server if and only if x_11 is an office clerk. Therefore there is at least one x for which x_11 is not an office clerk.
forall x: ~P(x). exists x: P(x) or ~Q(x). forall x: Q(x) <-> R(x). Therefore exists x: ~R(x).