Translate the following inference to logic notation: There is at least one x for which x is a janitor if and only if x will sleep. For all x, if x were a janitor, then x is sleeping. Therefore there is at least one x for which if x sleeps, then x is a janitor.
exists x: P(x) <-> Q(x). forall x: P(x) -> Q(x). Therefore exists x: Q(x) -> P(x).