Translate the following inference to logic notation: There is at least one x for which if z were a carpenter, then z is sleeping. For all x, z will not sleep. For all x, z is a carpenter if and only if z is happy. Therefore there is at least one x for which z is not happy.
exists x: P(x) -> Q(x). forall x: ~Q(x). forall x: P(x) <-> R(x). Therefore exists x: ~R(x).