Consider the following premises. forall x: Q(x) -> P_2(x). forall x: R(x) <-> Q(x). forall x: P_2(x) -> R(x). Can we infer exists x: ~(R(x) -> P_2(x)) from them?
 From forall x: R(x) <-> Q(x) we can infer forall x: R(x) -> Q(x). From forall x: R(x) -> Q(x), forall x: Q(x) -> P_2(x) we can infer forall x: R(x) -> P_2(x). From forall x: P_2(x) -> R(x), forall x: R(x) -> P_2(x) we can infer forall x: P_2(x) <-> R(x). Finally, from forall x: P_2(x) <-> R(x) we can infer forall x: R(x) -> P_2(x), which contradicts exists x: ~(R(x) -> P_2(x)). Therefore, the answer is no.