Consider the following premises. forall x_4: S(x_4) -> T_3(x_4). exists x_4: T_3(x_4) -> S(x_4). forall x_4: T_3(x_4) or ~S(x_4). Can we infer forall x_4: ~(S(x_4) <-> T_3(x_4)) from them? If possible, name the inference rules being used at each step.
 From forall x_4: S(x_4) -> T_3(x_4), exists x_4: T_3(x_4) -> S(x_4) we can infer exists x_4: S(x_4) <-> T_3(x_4) via existential biconditional introduction. From exists x_4: S(x_4) <-> T_3(x_4) we can infer exists x_4: T_3(x_4) -> S(x_4) via existential biconditional elimination. From forall x_4: T_3(x_4) or ~S(x_4) we can infer forall x_4: S(x_4) -> T_3(x_4) via universal rewriting. Finally, from forall x_4: S(x_4) -> T_3(x_4), exists x_4: T_3(x_4) -> S(x_4) we can infer exists x_4: S(x_4) <-> T_3(x_4) via existential biconditional introduction, which contradicts forall x_4: ~(S(x_4) <-> T_3(x_4)). Therefore, the answer is no.