Consider the following premises. forall x_6: R(x_6) -> Q_6(x_6). exists x_6: Q_6(x_6) -> T(x_6). forall x_6: T(x_6) -> Q_1(x_6). forall x_6: Q_1(x_6) -> R_1(x_6). Can we infer forall x_6: ~(R(x_6) -> R_1(x_6)) from them?
That contradicts the premises. Therefore the answer is no.