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