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