Consider the following premises. forall x_7: P(x_7) -> P_7(x_7). forall x_7: Q_7(x_7) -> R(x_7). forall x_7: P_7(x_7) <-> Q_7(x_7). Can we infer exists x_7: ~(P(x_7) -> R(x_7)) from them?
 From forall x_7: P_7(x_7) <-> Q_7(x_7) we can infer forall x_7: P_7(x_7) -> Q_7(x_7). From forall x_7: P(x_7) -> P_7(x_7), forall x_7: P_7(x_7) -> Q_7(x_7) we can infer forall x_7: P(x_7) -> Q_7(x_7). Finally, from forall x_7: P(x_7) -> Q_7(x_7), forall x_7: Q_7(x_7) -> R(x_7) we can infer forall x_7: P(x_7) -> R(x_7), which contradicts exists x_7: ~(P(x_7) -> R(x_7)). Therefore, the answer is no.