Consider the following premises. forall x_11: Q_11(x_11) -> R(x_11). forall x_11: R(x_11) <-> P_11(x_11). forall x_11: P_11(x_11) <-> Q_11(x_11). Can we infer exists x_11: ~(R(x_11) <-> P_11(x_11)) from them?
 From forall x_11: P_11(x_11) <-> Q_11(x_11) we can infer forall x_11: P_11(x_11) -> Q_11(x_11). From forall x_11: R(x_11) <-> P_11(x_11) we can infer forall x_11: R(x_11) -> P_11(x_11). From forall x_11: P_11(x_11) -> Q_11(x_11), forall x_11: Q_11(x_11) -> R(x_11) we can infer forall x_11: P_11(x_11) -> R(x_11). Finally, from forall x_11: R(x_11) -> P_11(x_11), forall x_11: P_11(x_11) -> R(x_11) we can infer forall x_11: R(x_11) <-> P_11(x_11), which contradicts exists x_11: ~(R(x_11) <-> P_11(x_11)). Therefore, the answer is no.