Consider the following premises. forall x_9: S(x_9) -> U(x_9). forall x_9: U(x_9) -> Q_5(x_9). forall x_9: Q_5(x_9) -> R_5(x_9). forall x_9: U(x_9) -> S(x_9). Can we infer exists x_9: ~(S(x_9) -> R_5(x_9)) from them?
 From forall x_9: S(x_9) -> U(x_9), forall x_9: U(x_9) -> S(x_9) we can infer forall x_9: S(x_9) <-> U(x_9). From forall x_9: U(x_9) -> Q_5(x_9), forall x_9: Q_5(x_9) -> R_5(x_9) we can infer forall x_9: U(x_9) -> R_5(x_9). From forall x_9: S(x_9) <-> U(x_9) we can infer forall x_9: S(x_9) -> U(x_9). Finally, from forall x_9: S(x_9) -> U(x_9), forall x_9: U(x_9) -> R_5(x_9) we can infer forall x_9: S(x_9) -> R_5(x_9), which contradicts exists x_9: ~(S(x_9) -> R_5(x_9)). Therefore, the answer is no.