Consider the following premises. forall y_2: Q(y_2) <-> P_2(y_2). forall y_2: Q(y_2) -> P_2(y_2). forall y_2: Q(y_2) -> P_2(y_2). Can we infer exists y_2: ~(Q(y_2) -> P_2(y_2)) from them?
 From forall y_2: Q(y_2) <-> P_2(y_2) we can infer forall y_2: P_2(y_2) -> Q(y_2). From forall y_2: Q(y_2) -> P_2(y_2), forall y_2: P_2(y_2) -> Q(y_2) we can infer forall y_2: Q(y_2) <-> P_2(y_2). From forall y_2: Q(y_2) <-> P_2(y_2) we can infer forall y_2: P_2(y_2) -> Q(y_2). From forall y_2: P_2(y_2) -> Q(y_2), forall y_2: Q(y_2) -> P_2(y_2) we can infer forall y_2: P_2(y_2) <-> Q(y_2). Finally, from forall y_2: P_2(y_2) <-> Q(y_2) we can infer forall y_2: Q(y_2) -> P_2(y_2), which contradicts exists y_2: ~(Q(y_2) -> P_2(y_2)). Therefore, the answer is no.