Proof using.
induction l1; simpl in *; intuition eauto.
Qed.

Lemma collate_ls_split_eq :
forall l1 l2 (f : A -> A -> list B) h m from to,
h <> from ->
collate_ls A_eq_dec (l1 ++ h :: l2) f to m from to =
collate_ls A_eq_dec  (l1 ++ l2) f to m from to.