Proof by case analysis in Coq - coq

Question

I am trying to prove a Proposition about the following function:
Program Fixpoint division (m:nat) (n:nat) {measure m} : nat :=
match lt_nat 0 n with
| false => 0
| true => match leq_nat n m with
| false => 0
| true => S (division (menos m n) n)
end
end.
menos is natural subtraction.
I am trying to prove some fact involving division. I wrote down an informal proof were I first consider a case analysis in lt_nat 0 n and then in the case when lt_nat is true a further case analysis in leq_nat n m. This is in order to reduce the definition of division.
However I can not find how to express this case analysis in Coq. I tried with destruct (leq_nat n m) but it does nothing. I am expecting Coq to generate two subgoals: one where I need to prove my proposition assuming leq_nat n m = false and one assuming leq_nat n m = true.
Furthermore, I can not unfold the definition of division in my proof! When I try unfold division I get: division_func (existT (fun _ : nat => nat) m n).
How can I perfom case analysis in leq_nat n m? Why is it that I can not unfold the definition of division as I usually do with other functions?
Thank you.

Answer 1

Everything is more complicated than usual because of Program Fixpoint, which does not define your function as you would expect with a classic Fixpoint, since it needs to find a structurally recursive way of defining it. What division really is, is hidden in division_func.
Therefore, to manipulate your function, you need to prove basic lemmas, including the one stating that your function can be replaced by its body.
Lemma division_eq : forall m n, division m n = match lt_nat 0 n with
| false => 0
| true => match leq_nat n m with
| false => 0
| true => S (division (menos m n) n)
end
end.
Now, the question is how to prove this result. Here is the only solution I know, which I consider really unsatisfying.
I use the tactic fix_sub_eq located in Program.Wf, or fix_sub_eq_ext in Program.Wf.WfExtensionality.
This gives something like:
Proof.
intros.
unfold division. unfold division_func at 1.
rewrite fix_sub_eq; repeat fold division_func.
- simpl. destruct (lt_nat 0 n) eqn:H.
destruct (leq_nat n m) eqn:H0. reflexivity.
reflexivity. reflexivity.
But the second goal is quite complicated. The easy and general way of solving it is to use the axioms proof_irrelevance and functional_extensionality. It should be possible to prove this particular subgoal without any axioms, but I have not found the right way to do it. Instead of manually applying the axioms, you can use the second tactic fix_sub_eq_ext which calls them directly, leaving you a single goal.
Proof.
intros.
unfold division. unfold division_func at 1.
rewrite fix_sub_eq_ext; repeat fold division_func.
simpl. destruct (lt_nat 0 n) eqn:H.
destruct (leq_nat n m) eqn:H0. reflexivity.
reflexivity. reflexivity.
Qed.
I have not found a better way to use Program Fixpoint, that's why I prefer using Function, which has other defaults, but generates directly the equation lemma.
Require Recdef.
Function division (m:nat) (n:nat) {measure (fun n => n) m} : nat :=
match lt_nat 0 n with
| false => 0
| true => match leq_nat n m with
| false => 0
| true => S (division (menos m n) n)
end
end.
Proof.
intros m n. revert m. induction n; intros.
- discriminate teq.
- destruct m. discriminate teq0.
simpl. destruct n. destruct m; apply le_n.
transitivity m. apply IHn. reflexivity. assumption. apply le_n.
Qed.
Check division_equation.
Now you have the equation lemma, you can rewrite with it and reason as usual.
About your problem with destruct, destruct does not unfold the definitions. Therefore, if you have no occurrences of the term you're destructing in your goal or any of the hypotheses, destruct will not do anything interesting, unless you save the equation it produces. You can use destruct ... eqn:H for this purpose. I did not know case_eq but it seems to do the same thing.