Inductive sublist (A : Type) : (list A) -> (list A) -> Prop :=
| sublist_refl : forall l, sublist l l
| sublist_skip : forall l1 h t, sublist l1 t -> sublist l1 (h :: t)
| sublist_cons : forall h t1 t2, sublist t1 t2 -> sublist (h :: t1) (h :: t2).Inductive sublist (A : Type) : (list A) -> (list A) -> Prop :=
| sublist_refl : forall l, sublist l l
| sublist_skip : forall l1 h t, sublist l1 t -> sublist l1 (h :: t)
| sublist_cons : forall h t1 t2, sublist t1 t2 -> sublist (h :: t1) (h :: t2).Inductive sublist (A : Type) : (list A) -> (list A) -> Prop :=
| sublist_refl : forall l, sublist l l
| sublist_skip : forall l1 h t, sublist l1 t -> sublist l1 (h :: t)
| sublist_cons : forall h t1 t2, sublist t1 t2 -> sublist (h :: t1) (h :: t2).Inductive sublist (A : Type) : (list A) -> (list A) -> Prop :=
| sublist_refl : forall l, sublist l l
| sublist_skip : forall l1 h t, sublist l1 t -> sublist l1 (h :: t)
| sublist_cons : forall h t1 t2, sublist t1 t2 -> sublist (h :: t1) (h :: t2).

Lemma In_In_list_decompose {A} (x y : A) l : In x l -> In y l -> x = y \/ exists l1 l2 l3, (l = l1 ++ x :: l2 ++ y :: l3) \/ (l = l1 ++ y :: l2 ++ x :: l3).