Consider the following premises. forall z_1: Q_3(z_1) <-> Q(z_1). forall z_1: ~Q(z_1). exists z_1: Q_3(z_1) -> T(z_1). forall z_1: T(z_1) -> Q_3(z_1). Can we infer forall z_1: T(z_1) from them?
 From forall z_1: Q_3(z_1) <-> Q(z_1), forall z_1: ~Q(z_1) we can infer forall z_1: ~Q_3(z_1). From exists z_1: Q_3(z_1) -> T(z_1), forall z_1: T(z_1) -> Q_3(z_1) we can infer exists z_1: Q_3(z_1) <-> T(z_1). Finally, from exists z_1: Q_3(z_1) <-> T(z_1), forall z_1: ~Q_3(z_1) we can infer exists z_1: ~T(z_1), which contradicts forall z_1: T(z_1). Therefore, the answer is no.