Consider the following premises. forall x_9: U(x_9) -> U_1(x_9). forall x_9: U_1(x_9) -> U(x_9). forall x_9: U(x_9). exists x_9: R(x_9). Can we infer forall x_9: ~(R(x_9) and U_1(x_9)) from them?
 From forall x_9: U(x_9) -> U_1(x_9), forall x_9: U_1(x_9) -> U(x_9) we can infer forall x_9: U(x_9) <-> U_1(x_9). From forall x_9: U(x_9) <-> U_1(x_9) we can infer forall x_9: U(x_9) -> U_1(x_9). From forall x_9: U(x_9) -> U_1(x_9), forall x_9: U(x_9) we can infer forall x_9: U_1(x_9). Finally, from exists x_9: R(x_9), forall x_9: U_1(x_9) we can infer exists x_9: R(x_9) and U_1(x_9), which contradicts forall x_9: ~(R(x_9) and U_1(x_9)). Therefore, the answer is no.