Coq 我们如何为依赖类型定义“eqType”?
我想将依赖类型定义为Coq 我们如何为依赖类型定义“eqType”?,coq,ssreflect,Coq,Ssreflect,我想将依赖类型定义为eqType。 例如,假设我们定义了以下依赖类型Tn: From mathcomp Require Import all_ssreflect. Variable T: nat -> eqType. Inductive Tn: Type := BuildT: forall n, T n -> Tn. Definition Tn_eq: rel Tn := fun '(BuildT n1 t1) '(BuildT n2 t2) => (if n
eqType
。
例如,假设我们定义了以下依赖类型Tn
:
From mathcomp Require Import all_ssreflect.
Variable T: nat -> eqType.
Inductive Tn: Type := BuildT: forall n, T n -> Tn.
Definition Tn_eq: rel Tn :=
fun '(BuildT n1 t1) '(BuildT n2 t2) =>
(if n1 == n2 as b return (n1==n2) = b -> bool
then fun E => t1 == eq_rect_r T t2 (elimTF eqP E)
else fun _ => false) (erefl (n1 == n2)).
为了实现eqType
,我为Tn
定义了一个等式函数Tn_eq
:
From mathcomp Require Import all_ssreflect.
Variable T: nat -> eqType.
Inductive Tn: Type := BuildT: forall n, T n -> Tn.
Definition Tn_eq: rel Tn :=
fun '(BuildT n1 t1) '(BuildT n2 t2) =>
(if n1 == n2 as b return (n1==n2) = b -> bool
then fun E => t1 == eq_rect_r T t2 (elimTF eqP E)
else fun _ => false) (erefl (n1 == n2)).
然后我试图证明Tn_eq
的等式公理,但失败了
Lemma Tn_eqP: Equality.axiom Tn_eq.
Proof.
case=>n1 t1; case=>n2 t2//=.
case_eq(n1==n2).
我这里有个错误:
Illegal application:
The term "elimTF" of type
"forall (P : Prop) (b c : bool), reflect P b -> b = c -> if c then P else ~ P"
cannot be applied to the terms
"n1 = n2" : "Prop"
"b" : "bool"
"true" : "bool"
"eqP" : "reflect (n1 = n2) (n1 == n2)"
"E" : "b = true"
The 4th term has type "reflect (n1 = n2) (n1 == n2)" which should be coercible to
"reflect (n1 = n2) b".
我应该如何证明这个引理呢?现在我们开始:
From Coq Require Import EqdepFacts.
From mathcomp Require Import ssreflect ssrfun ssrbool eqtype ssrnat.
Variable T: nat -> eqType.
Inductive Tn: Type := BuildT: forall n, T n -> Tn.
Definition Tn_eq: rel Tn :=
fun '(BuildT n1 t1) '(BuildT n2 t2) =>
(if n1 == n2 as b return (n1==n2) = b -> bool
then fun E => t1 == eq_rect_r T t2 (elimTF eqP E)
else fun _ => false) (erefl (n1 == n2)).
Lemma Tn_eqP: Equality.axiom Tn_eq.
Proof.
case=> n1 t1; case=> n2 t2 /=.
case: eqP => [eq1 | neq1]; last by constructor; case.
case: eqP.
- move=> ->; constructor; move: t2; rewrite [elimTF _ _]eq_irrelevance.
by case: _ / eq1.
move=> neq2; constructor.
case=> _ exT; move: (eq_sigT_snd exT) => Cast; apply: neq2.
rewrite -Cast.
rewrite [eq_sigT_fst _]eq_irrelevance [elimTF _ _]eq_irrelevance.
by case: _ / eq1.
Qed.
在mathcomp中实现这一点的最佳当前方法是使用(部分)双射,而不是手动定义等式及其正确性证明:
Definition encode x := let: BuildT _ tn := x in Tagged T tn.
Definition decode (x : sigT T) := BuildT (tagged x).
Lemma encodeK : cancel encode decode. Proof. by case. Qed.
Definition Tn_eqMixin := CanEqMixin encodeK.
Canonical Tn_eqType := EqType Tn Tn_eqMixin.
自动化方法包括:
Lemma Tn_eqE : Tn_eq =2 eq_op.
Proof.
case=> [n tn] [m tm]; rewrite [RHS]/eq_op/= -tag_eqE/= /tag_eq/= /tagged_as/=.
by case: eqP => //= p; rewrite [elimTF _ _](eq_irrelevance _ p).
Qed.
(有时,证明程序相等性确实比从头开始证明正确性更容易。)