Question

我正在努力寻找看似简单的引理，它涉及2个定点定义。以下两个是CoLoR库的轴向定义：

From Coq Require Import Vector Program.
Import VectorNotations.

Program Fixpoint Vnth {A:Type} {n} (v : t A n) : forall i, i < n -> A :=
  match v with
  | nil _ => fun i ip => !
  | cons _ x _ v' => fun i =>
                      match i with
                      | 0 => fun _ => x
                      | S j => fun H => Vnth v' j _
                      end
  end.
Admit Obligations.

Fixpoint Vmap {A B : Type} (f: A -> B) n (v : t A n) : t B n :=
  match v with
  | nil _ => nil _
  | cons _ a _ v' => cons _ (f a) _ (Vmap f _ v')
  end.

实际问题：

Fixpoint Ind (n:nat) {A:Type} (f:A -> A -> A)
         (initial: A) (v: A) {struct n} : t A n
  :=
    match n with
    | O => []
    | S p => cons _ initial _ (Vmap (fun x => f x v) _ (Ind p f initial v))
    end.

Lemma Foo {A: Type} (n : nat) (f : A -> A -> A) (initial v : A)
      (b : nat) (bc : S b < n) (bc1 : b < n)
  : Vnth (Ind n f initial v) _ bc = f (Vnth (Ind n f initial v) _ bc1) v.
Proof.
Qed.

通常我会在这里n进行归纳，但这并没有让我更进一步。我觉得我在这里错过了一些东西。我也在这里试过program induction。

Answer 1

您需要简化Vnth_vmap和广义归纳以实现此目的：

From Coq Require Import Vector Program.
Import VectorNotations.

Program Fixpoint Vnth {A:Type} {n} (v : t A n) : forall i, i < n -> A :=
  match v with
  | nil _ => fun i ip => !
  | cons _ x _ v' => fun i =>
                  match i with
                  | 0 => fun _ => x
                  | S j => fun H => Vnth v' j _
                  end
  end.
Admit Obligations.

Fixpoint Vmap {A B : Type} (f: A -> B) n (v : t A n) : t B n :=
  match v with
  | nil _ => nil _
  | cons _ a _ v' => cons _ (f a) _ (Vmap f _ v')
  end.

Lemma Vnth_vmap {A B i n p} (v : t A n) f : Vnth (Vmap (B:=B) f n v) i p = f (Vnth v i p).
Proof.
  induction i in n, p, v |- *. destruct v. inversion p.
  simpl. reflexivity.
  destruct v. simpl. bang.
  simpl.
  rewrite IHi. f_equal. f_equal.
  (* Applies proof-irrelevance, might also be directly provable when giving the proofs in Vnth *) pi.
Qed.

Fixpoint Ind (n:nat) {A:Type} (f:A -> A -> A)
     (initial: A) (v: A) {struct n} : t A n
  :=
match n with
| O => []
| S p => cons _ initial _ (Vmap (fun x => f x v) _ (Ind p f initial v))
end.

Lemma Foo {A: Type} (n : nat) (f : A -> A -> A) (initial v : A)
  (b : nat) (bc : S b < n) (bc1 : b < n)
: Vnth (Ind n f initial v) _ bc = f (Vnth (Ind n f initial v) _ bc1) v.
Proof.
induction n in b, bc, bc1 |- *; simpl.
- bang.
- rewrite Vnth_vmap. f_equal.
  destruct b.
  + destruct n. simpl. bang. simpl. reflexivity.
  + rewrite Vnth_vmap. apply IHn.
Qed.

通过归纳证明两个固定点功能

1 个答案: