How do I prove that two Fibonacci implementations are equal in Coq?

Question

I\'ve two Fibonacci implementations, seen below, that I want to prove are functionally equivalent.

I\'ve already proved properties about natural numbers, but this ex

Answer 1

A note of warning: in what follows I'll to try to show the main idea of such a proof, so I'm not going to stick to some subset of Coq and I won't do arithmetic manually. Instead I'll use some proof automation, viz. the ring tactic. However, feel free to ask additional questions, so you could convert the proof to somewhat that would suit your purposes.

I think it's easier to start with some generalization:

Require Import Arith.    (* for `ring` tactic *)

Lemma fib_v1_eq_fib2_generalized n : forall a0 a1,
  visit_fib_v2 (S n) a0 a1 = a0 * fib_v1 n + a1 * fib_v1 (S n).
Proof.
  induction n; intros a0 a1.
  - simpl; ring.
  - change (visit_fib_v2 (S (S n)) a0 a1) with
           (visit_fib_v2 (S n) a1 (a0 + a1)).
    rewrite IHn. simpl; ring.
Qed.

If using ring doesn't suit your needs, you can perform multiple rewrite steps using the lemmas of the Arith module.

Now, let's get to our goal:

Definition fib_v2 n := visit_fib_v2 n 0 1.

Lemma fib_v1_eq_fib2 n :
  fib_v1 n = fib_v2 n.
Proof.
  destruct n.
  - reflexivity.
  - unfold fib_v2. rewrite fib_v1_eq_fib2_generalized.
    ring.
Qed.