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