Consider the following premises. forall x_6: P_6(x_6) or ~T_4(x_6). forall x_6: T(x_6) -> T_4(x_6). forall x_6: P_6(x_6) -> R_1(x_6). Can we infer exists x_6: ~(T(x_6) -> R_1(x_6)) from them?
 From forall x_6: P_6(x_6) or ~T_4(x_6) we can infer forall x_6: T_4(x_6) -> P_6(x_6). From forall x_6: T_4(x_6) -> P_6(x_6), forall x_6: P_6(x_6) -> R_1(x_6) we can infer forall x_6: T_4(x_6) -> R_1(x_6). Finally, from forall x_6: T(x_6) -> T_4(x_6), forall x_6: T_4(x_6) -> R_1(x_6) we can infer forall x_6: T(x_6) -> R_1(x_6), which contradicts exists x_6: ~(T(x_6) -> R_1(x_6)). Therefore, the answer is no.