I have defined an inductive definition of lists (called listkind) in order make it easy
for me to prove a specific theorem by induction on listkind rather than on list.
Inductive listkind {X}: list X -> Prop :=
| l_nil : listkind []
| l_one : forall a:X, listkind [a]
| l_app : forall l, listkind l -> forall a b, listkind ([a]++l++[b]).
(With this property, to prove things about lists, I have to prove the cases where a list is [], [a], or [a]++l++[b], rather than the cases where a list is [] or a::l. In my particular theorem, those cases fit better and makes the proof simpler.)
However, to be able to use listkind in my proof, I have to prove
Lemma all_lists_are_listkind: (forall {X} (l:list X), listkind l).
Having tried various approaches, I find myself stuck at this point. I would very much appreciate seeing how to perform such a proof, preferably with minimal coq magic applied.
Here is a solution:
Require Import List Omega.
Lemma all_lists_are_listkind_size: forall {X} (n:nat) (l:list X), length l <= n -> listkind l.
Proof.
intros X.
induction n as [ | n hi]; simpl in *; intros l hl.
- destruct l as [ | hd tl]; simpl in *.
+ now constructor.
+ now inversion hl.
- destruct l as [ | hd tl]; simpl in *.
+ now constructor.
+ induction tl using rev_ind.
* now constructor.
* constructor.
apply hi.
rewrite app_length in hl; simpl in hl.
omega. (* a bit overkill but it does the arithmetic job *)
Qed.
Lemma all_lists_are_listkind: forall {X} (l:list X), listkind l.
Proof.
intros.
apply all_lists_are_listkind_size with (length l).
apply le_refl.
Qed.
The main idea is that your lists have the same size as regular list, and induction on a natural is goes more smoothly than induction on a non trivial shape of list.
Hope it helps, V.