Consider the following premises. forall x_15: Q(x_15) -> Q_15(x_15). exists x_15: P_15(x_15) <-> Q_15(x_15). forall x_15: P_15(x_15) -> R_1(x_15). forall x_15: Q(x_15). Can we infer forall x_15: ~R_1(x_15) from them?
 From exists x_15: P_15(x_15) <-> Q_15(x_15) we can infer exists x_15: Q_15(x_15) -> P_15(x_15). From forall x_15: Q(x_15) -> Q_15(x_15), exists x_15: Q_15(x_15) -> P_15(x_15) we can infer exists x_15: Q(x_15) -> P_15(x_15). From exists x_15: Q(x_15) -> P_15(x_15), forall x_15: P_15(x_15) -> R_1(x_15) we can infer exists x_15: Q(x_15) -> R_1(x_15). Finally, from exists x_15: Q(x_15) -> R_1(x_15), forall x_15: Q(x_15) we can infer exists x_15: R_1(x_15), which contradicts forall x_15: ~R_1(x_15). Therefore, the answer is no.