Consider the following premises. forall x: ~T_1(x). forall x: T_1(x) -> T_2(x). forall x: T_1(x) <-> S(x). forall x: T_2(x) -> S(x). exists x: R(x) <-> S(x). Can we infer forall x: R(x) from them?
 From forall x: T_1(x) <-> S(x) we can infer forall x: S(x) -> T_1(x). From forall x: T_1(x) -> T_2(x), forall x: T_2(x) -> S(x) we can infer forall x: T_1(x) -> S(x). From forall x: S(x) -> T_1(x), forall x: T_1(x) -> S(x) we can infer forall x: S(x) <-> T_1(x). From forall x: S(x) <-> T_1(x), forall x: ~T_1(x) we can infer forall x: ~S(x). Finally, from exists x: R(x) <-> S(x), forall x: ~S(x) we can infer exists x: ~R(x), which contradicts forall x: R(x). Therefore, the answer is no.