coq – 定义具有类型参数约束的归纳依赖类型

Inductive bitvector : Type := Bitvector (n: nat) (v: Vector.t bool (S n)).

但是,我不明白为什么你想要在这种特殊情况下做到这一点,因为引理是完全可证明的：它是一个相当简单的推论,你可能会在以后需要更一般的引理.

你的定义(没有改变)：

Require Import Vector.

Inductive bitvector : Type := Bitvector (n: nat) (v: Vector.t bool n).

Definition bv_neg (v : bitvector) :=
    match v with
        Bitvector n w => Bitvector n (Vector.map negb w)
    end.

Vector.map上的一些结果：

Lemma map_fusion : 
  forall {A B C} {g : B -> C} {f : A -> B} {n : nat} (v : Vector.t A n),Vector.map g (Vector.map f v) = Vector.map (fun x => g (f x)) v.
Proof.
intros A B C g f n v ; induction v.
  - reflexivity.
  - simpl; f_equal; assumption.
Qed.

Lemma map_id : 
  forall {A} {n : nat} (v : Vector.t A n),Vector.map (@id A) v = v.
Proof.
intros A n v ; induction v.
  - reflexivity.
  - simpl; f_equal; assumption.
Qed.

Lemma map_extensional : 
  forall {A B} {f1 f2 : A -> B} 
         (feq : forall a,f1 a = f2 a) {n : nat} (v : Vector.t A n),Vector.map f1 v = Vector.map f2 v.
Proof.
intros A B f1 f2 feq n v ; induction v.
  - reflexivity.
  - simpl; f_equal; [apply feq | assumption].
Qed.

最后,你的结果：

Lemma double_neg :
   forall (v : bitvector),(bv_neg (bv_neg v) = v).
Proof.
intros (n,v).
 simpl; f_equal.
 rewrite map_fusion,<- map_id.
 apply map_extensional.
 - intros []; reflexivity.
Qed.