Translate the following inference to logic notation: For all x, if y sleeps, then y is working. There is at least one x for which y will not work. For all x, y will sleep if and only if y will work. Therefore there is at least one x for which y will not sleep.
forall x: P(x) -> Q(x). exists x: ~Q(x). forall x: P(x) <-> Q(x). Therefore exists x: ~P(x).