Proof using A dec.
intro H0. induction l as [|a l IHl].
- contradiction H0; constructor.
- destruct (NoDup_decidable l) as [H1|H1].
+ destruct (In_decidable a l) as [H2|H2].
* destruct (in_split _ _ H2) as (l1 & l2 & ->).
now exists a, nil, l1, l2.
* now contradiction H0; constructor.
+ destruct (IHl H1) as (b & l1 & l2 & l3 & ->).
now exists b, (a::l1), l2, l3.
Qed.

Lemma NoDup_list_decidable (l:list A) : NoDup l -> forall x y:A, In x l -> In y l -> decidable (x=y).
Proof using A.