Consider the following premises. exists z: ~Q_4(z). forall z: S(z) -> Q_4(z). forall z: Q_4(z) -> S(z). forall z: T(z) <-> S(z). Can we infer forall z: T(z) from them?
 From forall z: S(z) -> Q_4(z), forall z: Q_4(z) -> S(z) we can infer forall z: S(z) <-> Q_4(z). From forall z: S(z) <-> Q_4(z), exists z: ~Q_4(z) we can infer exists z: ~S(z). Finally, from forall z: T(z) <-> S(z), exists z: ~S(z) we can infer exists z: ~T(z), which contradicts forall z: T(z). Therefore, the answer is no.