proposition: Theorem sum_inversion (f: nat -> nat -> X A B) i i' k k': sum i k (fun u => (sum i' k' (f u))) == sum i' k' (fun u'=> (sum i k (fun u => f u u'))).
proof: Proof. induction k'; simpl_sum_r. apply sum_zero; reflexivity. rewrite sum_cut_fun. rewrite IHk'; reflexivity. Qed.
imports: ATBR(Common) ATBR(Classes) ATBR(Graph)