Consider the following premises. forall x_4: R(x_4) -> Q_4(x_4). ~R(a_2). forall x_4: R_4(x_4) -> R(x_4). forall x_4: Q_4(x_4) -> R_4(x_4). Can we infer R_4(a_2) from them?
 From forall x_4: R(x_4) -> Q_4(x_4), forall x_4: Q_4(x_4) -> R_4(x_4) we can infer forall x_4: R(x_4) -> R_4(x_4). From forall x_4: R(x_4) -> R_4(x_4), forall x_4: R_4(x_4) -> R(x_4) we can infer forall x_4: R(x_4) <-> R_4(x_4). From forall x_4: R(x_4) <-> R_4(x_4) we can infer R(a_2) <-> R_4(a_2). Finally, from R(a_2) <-> R_4(a_2), ~R(a_2) we can infer ~R_4(a_2), which contradicts R_4(a_2). Therefore, the answer is no.