Consider the following premises. forall x: R(x) <-> Q_1(x). forall x: Q_1(x) -> T(x). forall x: T(x) <-> Q_1(x). forall x: ~R(x). Can we infer exists x: T(x) from them?
 From forall x: T(x) <-> Q_1(x) we can infer forall x: T(x) -> Q_1(x). From forall x: R(x) <-> Q_1(x), forall x: ~R(x) we can infer forall x: ~Q_1(x). From forall x: Q_1(x) -> T(x), forall x: T(x) -> Q_1(x) we can infer forall x: Q_1(x) <-> T(x). Finally, from forall x: Q_1(x) <-> T(x), forall x: ~Q_1(x) we can infer forall x: ~T(x), which contradicts exists x: T(x). Therefore, the answer is no.