Consider the following premises. forall z: R(z) or Q_1(z). forall z: T_1(z) <-> R(z). exists z: Q_1(z) -> T_1(z). Can we infer forall z: ~T_1(z) from them?
 From forall z: T_1(z) <-> R(z) we can infer forall z: R(z) -> T_1(z). Finally, from forall z: R(z) -> T_1(z), exists z: Q_1(z) -> T_1(z), forall z: R(z) or Q_1(z) we can infer exists z: T_1(z), which contradicts forall z: ~T_1(z). Therefore, the answer is no.