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

@larsr's answer inspired this alternative answer.

First of all, let's define fib_v2:

Require Import Coq.Arith.Arith. 

Definition fib_v2 n := visit_fib_v2 n 0 1.

Then, we are going to need a lemma, which is the same as fib_v2_lemma in @larsr's answer. I'm including it here for consistency and to show an alternative proof.

Lemma visit_fib_v2_main_property n: forall a0 a1,
  visit_fib_v2 (S (S n)) a0 a1 =
  visit_fib_v2 (S n) a0 a1 + visit_fib_v2 n a0 a1.
Proof.
  induction n; intros a0 a1; auto with arith.
  change (visit_fib_v2 (S (S (S n))) a0 a1) with
         (visit_fib_v2 (S (S n)) a1 (a0 + a1)).
  apply IHn.
Qed.

As suggested in the comments by larsr, the visit_fib_v2_main_property lemma can be also proved by the following impressive one-liner:

now induction n; firstorder.

Because of the nature of the numbers in the Fibonacci series it's very convenient to define an alternative induction principle:

Lemma pair_induction (P : nat -> Prop) :
  P 0 ->
  P 1 ->
  (forall n, P n -> P (S n) -> P (S (S n))) ->
  forall n, P n.
Proof.
  intros H0 H1 Hstep n.
  enough (P n /\ P (S n)) by tauto.
  induction n; intuition.
Qed.

The pair_induction principle basically says that if we can prove some property P for 0 and 1 and if for every natural number k > 1, we can prove P k holds under the assumption that P (k - 1) and P (k - 2) hold, then we can prove forall n, P n.

Using our custom induction principle, we get the proof as follows:

Lemma fib_v1_eq_fib2 n :
  fib_v1 n = fib_v2 n.
Proof.
  induction n using pair_induction.
  - reflexivity.
  - reflexivity.
  - unfold fib_v2.
    rewrite visit_fib_v2_main_property.
    simpl; auto.
Qed.