Consider the following premises. forall z: U(z) -> Q(z). Q(d) -> S(d). forall z: S(z) -> U(z). Can we infer Q(d) <-> S(d) from them?
 From forall z: S(z) -> U(z), forall z: U(z) -> Q(z) we can infer forall z: S(z) -> Q(z). From forall z: S(z) -> Q(z) we can infer S(d) -> Q(d). Finally, from Q(d) -> S(d), S(d) -> Q(d) we can infer Q(d) <-> S(d). Therefore, the answer is yes.