Translate the following inference to logic notation: For all x, z_4 is not bored. For all x, z_4 is rich if and only if z_4 will sleep. There is at least one x for which z_4 is rich if and only if z_4 is bored. Therefore there is at least one x for which z_4 will not sleep.
forall x: ~P(x). forall x: Q(x) <-> R(x). exists x: Q(x) <-> P(x). Therefore exists x: ~R(x).