Consider the following premises. forall x_10: S(x_10) <-> P_10(x_10). forall x_10: S(x_10) -> P_10(x_10). forall x_10: P_10(x_10) -> S(x_10). Can we infer exists x_10: ~(P_10(x_10) -> S(x_10)) from them? If possible, name the inference rules being used at each step.
 From forall x_10: P_10(x_10) -> S(x_10), forall x_10: S(x_10) -> P_10(x_10) we can infer forall x_10: P_10(x_10) <-> S(x_10) via universal biconditional introduction. From forall x_10: S(x_10) <-> P_10(x_10) we can infer forall x_10: S(x_10) -> P_10(x_10) via universal biconditional elimination. From forall x_10: P_10(x_10) <-> S(x_10) we can infer forall x_10: P_10(x_10) -> S(x_10) via universal biconditional elimination. From forall x_10: S(x_10) -> P_10(x_10), forall x_10: P_10(x_10) -> S(x_10) we can infer forall x_10: S(x_10) <-> P_10(x_10) via universal biconditional introduction. Finally, from forall x_10: S(x_10) <-> P_10(x_10) we can infer forall x_10: P_10(x_10) -> S(x_10) via universal biconditional elimination, which contradicts exists x_10: ~(P_10(x_10) -> S(x_10)). Therefore, the answer is no.