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