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[Pierre Boutillier <pierre.boutillier AT pps.univ-paris-diderot.fr>]

Using Case0 and CaseS in
http://coq.inria.fr/distrib/V8.4/stdlib/Coq.Vectors.VectorDef.html
are answers to your problem.

I'm sure it is tackled by CPDT (maybe later in the book).

Pierre B.

On 17/10/2012 15:09, Hendrik Tews wrote:
> Hi,
>
> I am working with length-indexed lists, as they are described in
> CPDT:
>
>   Variable A : Set.
>
>   Inductive ilist : nat -> Set :=
>     | Nil : ilist 0
>     | Cons : forall n, A -> ilist n -> ilist (S n).
>
> I have a proof where I cannot destruct some ilist, therefore I
> tried to prove the following two obvious facts:
>
>   Lemma ilist_0 : forall(l : ilist 0), l = Nil.
>
>   Lemma ilist_S : forall(n : nat)(l : ilist (S n)),
>     exists(a : A)(tl : ilist n), l = Cons n a tl.
>
> However, I can only prove these by using UIP_refl, ie. uniqueness
> of reflexive identity proofs (full sources attached below).
>
> Could somebody give me a hint on how to prove these lemmas
> without using axioms?
>
> I need this for the correctness prove of an decidable equality
> predicate on ilists:
>
>   Fixpoint ilist_eq(a_eq : A -> A -> bool)
>                    (n1 n2 : nat)(l1 : ilist n1)(l2 : ilist n2) : bool :=
>     match (l1, l2) with
>       | (Nil, Nil) => true
>       | (Cons n1 a1 l1, Cons n2 a2 l2) => 
>          a_eq a1 a2 && ilist_eq a_eq n1 n2 l1 l2
>       | _ => false
>     end.
>
>   Lemma ilist_eq_correct :
>     forall(a_eq : A -> A -> bool)(n : nat)(l1 : ilist n)(l2 : ilist n),
>       (forall(a1 a2 : A), a_eq a1 a2 = true <-> a1 = a2) ->
>         (ilist_eq a_eq n n l1 l2 = true <-> l1 = l2).
>
> If there are no axiom-free proofs for the two lemmas above, I
> would appreciate any hints on how to prove ilist_eq_correct
> without axioms.
>
> Bye,
>
> Hendrik
>