Consider the following premises. forall x_6: P_6(x_6) -> Q_6(x_6). R_6(a_5) -> P_6(a_5). forall x_6: Q_6(x_6) -> R_6(x_6). Can we infer ~(R_6(a_5) <-> P_6(a_5)) from them?
 From forall x_6: P_6(x_6) -> Q_6(x_6), forall x_6: Q_6(x_6) -> R_6(x_6) we can infer forall x_6: P_6(x_6) -> R_6(x_6). From forall x_6: P_6(x_6) -> R_6(x_6) we can infer P_6(a_5) -> R_6(a_5). Finally, from R_6(a_5) -> P_6(a_5), P_6(a_5) -> R_6(a_5) we can infer R_6(a_5) <-> P_6(a_5), which contradicts ~(R_6(a_5) <-> P_6(a_5)). Therefore, the answer is no.