Translate the following inference to logic notation: For all x, if z_2 were excited, then z_2 is sleeping. There is at least one x for which z_2 will sleep or z_2 is an office clerk. For all x, z_2 is not excited. For all x, if z_2 sleeps, then z_2 is excited. Therefore there is at least one x for which z_2 is an office clerk.
forall x: P(x) -> Q(x). exists x: Q(x) or R(x). forall x: ~P(x). forall x: Q(x) -> P(x). Therefore exists x: R(x).