Consider the following premises. forall x_12: P(x_12) <-> Q_12(x_12). forall x_12: Q_1(x_12) -> R_1(x_12). exists x_12: S_2(x_12) -> Q_1(x_12). forall x_12: Q_12(x_12) -> S_2(x_12). Can we infer forall x_12: ~(P(x_12) -> R_1(x_12)) from them?
That contradicts the premises. Therefore the answer is no.