Consider the following premises. forall x_15: S_4(x_15) -> Q_7(x_15). forall x_15: U(x_15) or ~Q_12(x_15). forall x_15: U(x_15) -> Q_15(x_15). forall x_15: Q_15(x_15) -> S_4(x_15). exists x_15: Q_7(x_15) -> Q_12(x_15). Can we infer forall x_15: ~(U(x_15) <-> S_4(x_15)) from them? If possible, name the inference rules being used at each step.
 From forall x_15: U(x_15) -> Q_15(x_15), forall x_15: Q_15(x_15) -> S_4(x_15) we can infer forall x_15: U(x_15) -> S_4(x_15) via universal transitivity. From forall x_15: U(x_15) or ~Q_12(x_15) we can infer forall x_15: Q_12(x_15) -> U(x_15) via universal rewriting. From forall x_15: S_4(x_15) -> Q_7(x_15), exists x_15: Q_7(x_15) -> Q_12(x_15) we can infer exists x_15: S_4(x_15) -> Q_12(x_15) via existential transitivity. From exists x_15: S_4(x_15) -> Q_12(x_15), forall x_15: Q_12(x_15) -> U(x_15) we can infer exists x_15: S_4(x_15) -> U(x_15) via existential transitivity. Finally, from forall x_15: U(x_15) -> S_4(x_15), exists x_15: S_4(x_15) -> U(x_15) we can infer exists x_15: U(x_15) <-> S_4(x_15) via existential biconditional introduction, which contradicts forall x_15: ~(U(x_15) <-> S_4(x_15)). Therefore, the answer is no.