Axiom
Z_nat_axiom :
forall x y : Z,
(0 <= x)%Z ->
(0 <= y)%Z -> (y < x)%Z -> convert_to_nat y < convert_to_nat x.
(*  What I want to prove. Lemma Z_nat_business2: (x:Z)  `0<=x` -> `0<x` -> (lt (0) (convert_to_nat `x`)).  Lemma Z_nat_business3: (x,y:Z)  `0<=x` -> `0<=y` -> `y<x` -> (lt (convert y) (convert x)).  Lemma Z_nat_business4: (x,y:positive)  (lt (convert y) (convert x)) -> (lt (convert y) (convert x)).  *)
(* use Theorem bij1 : (m:nat) (convert (anti_convert m)) = (S m). *)

Lemma no_copy_succ_decreases :
relation_decreases legal_config no_copy_succ no_copy_measure.