Translate the following inference to logic notation: For all x, if y sleeps, then y is an office clerk. There is at least one x for which y will not sleep. For all x, if y were an office clerk, then y is sleeping. Therefore there is at least one x for which y is not an office clerk.
forall x: P(x) -> Q(x). exists x: ~P(x). forall x: Q(x) -> P(x). Therefore exists x: ~Q(x).