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