Translate the following inference to logic notation: There is at least one x for which if z_4 works, then z_4 is sleeping. For all x, if z_4 were rich, then z_4 is sleeping. For all x, z_4 is rich or z_4 will work. For all x, z_4 will sleep if and only if z_4 is curious. Therefore there is at least one x for which z_4 is curious.
exists x: P(x) -> Q(x). forall x: R(x) -> Q(x). forall x: R(x) or P(x). forall x: Q(x) <-> S(x). Therefore exists x: S(x).