I want to define a dependent type as eqType
.
For example, suppose that we have defined the following dependent type Tn
:
From mathcomp Require Import all_ssreflect.
Variable T: nat -> eqType.
Inductive Tn: Type := BuildT: forall n, T n -> Tn.
To make this of eqType
, I defined an equality function Tn_eq
for 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)).
Then I tried to prove the equality axiom for Tn_eq
but it fails.
Lemma Tn_eqP: Equality.axiom Tn_eq.
Proof.
case=>n1 t1; case=>n2 t2//=.
case_eq(n1==n2).
I had an error here:
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".
How should I prove this lemma?
Here we go:
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.