Port a Coq lemma over Z to a similar lemma over nat
Question
I have a lemma that is proved for Z. All the variables are bounded to be greater that or equal to zero.
Q: How can one as easily and generally as possible "port" that lemma to nat, i.e. use that lemma to prove a similar lemma for nat by using the lemma for Z?
Example:
Require Import ZArith. Open Scope Z.Admitted. Close Scope Z.forall n n0 n1 n2 n3 n4 n5 n6 : Z, n >= 0 -> n0 >= 0 -> n1 >= 0 -> n2 >= 0 -> n3 >= 0 -> n4 >= 0 -> n5 >= 0 -> n6 >= 0 -> n5 + n4 = n6 + n3 -> n1 + n0 = n2 + n -> n5 * n1 + n6 * n2 + n3 * n0 + n * n4 = n5 * n2 + n1 * n6 + n3 * n + n0 * n4(* prove this using `Z_lemma` *)forall n n0 n1 n2 n3 n4 n5 n6 : nat, n5 + n4 = n6 + n3 -> n1 + n0 = n2 + n -> n5 * n1 + n6 * n2 + n3 * n0 + n * n4 = n5 * n2 + n1 * n6 + n3 * n + n0 * n4
Answer
You can do it rather generically for all the lemmas which have this shape by defining a tactic exploiting the fact that Z.of_nat is injective and distributes over ( + ) and ( * ):
Ltac solve_using_Z_and lemma :=
(* Apply Z.of_nat to both sides of the equation *)
apply Nat2Z.inj;
(* Push Z.of_nat through multiplications and additions *)
repeat (rewrite Nat2Z.inj_mul || rewrite Nat2Z.inj_add);
(* Apply the lemma passed as an argument*)
apply lemma;
(* Discharge all the goals with the shape Z.of_nat m >= 0 *)
try (apply Z.le_ge, Nat2Z.is_nonneg);
(* Push the multiplications and additions back through Z.of_nat *)
repeat (rewrite <- Nat2Z.inj_mul || rewrite <- Nat2Z.inj_add);
(* Peal off Z.of_nat on each side of the equation *)
f_equal;
(* Look up the assumption in the environment*)
assumption.
The proof of nat_lemma now simply becomes:
forall n n0 n1 n2 n3 n4 n5 n6 : nat, n5 + n4 = n6 + n3 -> n1 + n0 = n2 + n -> n5 * n1 + n6 * n2 + n3 * n0 + n * n4 = n5 * n2 + n1 * n6 + n3 * n + n0 * n4forall n n0 n1 n2 n3 n4 n5 n6 : nat, n5 + n4 = n6 + n3 -> n1 + n0 = n2 + n -> n5 * n1 + n6 * n2 + n3 * n0 + n * n4 = n5 * n2 + n1 * n6 + n3 * n + n0 * n4solve_using_Z_and Z_lemma. Qed.n, n0, n1, n2, n3, n4, n5, n6: nat
H: n5 + n4 = n6 + n3
H0: n1 + n0 = n2 + nn5 * n1 + n6 * n2 + n3 * n0 + n * n4 = n5 * n2 + n1 * n6 + n3 * n + n0 * n4