Consider the following premises. exists x_6: U_2(x_6) -> Q_6(x_6). forall x_6: ~U_3(x_6). forall x_6: R_6(x_6) -> U_3(x_6). forall x_6: Q_6(x_6) -> R_6(x_6). Can we infer forall x_6: U_2(x_6) from them?
 From exists x_6: U_2(x_6) -> Q_6(x_6), forall x_6: Q_6(x_6) -> R_6(x_6) we can infer exists x_6: U_2(x_6) -> R_6(x_6). From exists x_6: U_2(x_6) -> R_6(x_6), forall x_6: R_6(x_6) -> U_3(x_6) we can infer exists x_6: U_2(x_6) -> U_3(x_6). Finally, from exists x_6: U_2(x_6) -> U_3(x_6), forall x_6: ~U_3(x_6) we can infer exists x_6: ~U_2(x_6), which contradicts forall x_6: U_2(x_6). Therefore, the answer is no.