Consider the following premises. exists x_2: Q_3(x_2) or T_2(x_2). forall x_2: Q_3(x_2) -> ~Q_2(x_2). forall x_2: Q_2(x_2). Can we infer forall x_2: ~T_2(x_2) from them?
 From forall x_2: Q_3(x_2) -> ~Q_2(x_2), forall x_2: Q_2(x_2) we can infer forall x_2: ~Q_3(x_2). Finally, from exists x_2: Q_3(x_2) or T_2(x_2), forall x_2: ~Q_3(x_2) we can infer exists x_2: T_2(x_2), which contradicts forall x_2: ~T_2(x_2). Therefore, the answer is no.