proposition: Lemma oplus_cons_cons : forall a n b a' n' b', (cons a n b) o+ (cons a' n' b') = match compare a a' with | Gt => cons a n (b o+ (cons a' n' b') ) | Eq => cons a (S (n + n')%nat) (b o+ b') | Lt => cons a' n' (cons a n b o+ b') end.
proof: Proof. intros; now rewrite oplus_eqn. Qed.
imports: Coq(Arith ArithRing Lia) hydras(Prelude.More_Arith Prelude.Merge_Sort Epsilon0.T1)