Consider the following premises. forall x_7: S_2(x_7) -> Q_1(x_7). forall x_7: Q_1(x_7) -> T(x_7). exists x_7: S_2(x_7) or ~T(x_7). Can we infer forall x_7: ~(S_2(x_7) <-> T(x_7)) from them?
 From exists x_7: S_2(x_7) or ~T(x_7) we can infer exists x_7: T(x_7) -> S_2(x_7). From forall x_7: S_2(x_7) -> Q_1(x_7), forall x_7: Q_1(x_7) -> T(x_7) we can infer forall x_7: S_2(x_7) -> T(x_7). Finally, from forall x_7: S_2(x_7) -> T(x_7), exists x_7: T(x_7) -> S_2(x_7) we can infer exists x_7: S_2(x_7) <-> T(x_7), which contradicts forall x_7: ~(S_2(x_7) <-> T(x_7)). Therefore, the answer is no.