我尝试练习subtypes
在Coq
,并使用ssreflect
简化事情。但我在重写子类型时总是遇到一些问题。例如:
Require Import Omega.
From mathcomp Require Import ssreflect ssrfun ssrbool ssrnat eqtype.
(* a type A to build X *)
Inductive A: Set :=
| mkA: nat -> A.
Definition getNat_A (a: A) :=
match a with
| mkA n => n
end.
Inductive X: Set :=
| r1 : A -> X.
(* subtype of X that satisfying some property *)
Definition Instantiated_X (x : X) : bool :=
match x with
| r1 a => (getNat_A a) > 10
end.
Definition iX : Set := {x:X | (Instantiated_X x)}.
(* rewrite constructor of X, stating the fact of elements of A, under certain condition creates element of iX *)
Program Definition r1_rewrite : A -> option iX :=
fun a: A =>
match (Instantiated_X (r1 a)) with
| true => Some (exist _ (r1 a) _)
| false => None
end.
(* try to prove r1_rewrite is surjective *)
Example r1_rewrite_surj:
forall t : iX, exists (a : A),
match (r1_rewrite a) with
| None => True
| Some e => eq t e
end.
Proof.
intros.
destruct t eqn: caseiX.
destruct x eqn: caseX.
exists a.
destruct (r1_rewrite a) eqn: r_res.
- destruct (10 < getNat_A a) eqn: guard.
destruct i0.
destruct x0.
unfold r1_rewrite in r_res.
simpl in r_res.
rewrite <- guard in r_res. (* <- stuck *)
Abort.
我不明白为什么它卡在那里。错误消息显示:
Error: Abstracting over the term "true" leads to a term: ...
which is ill-typed.
我认为 Coq 会替换所有出现的 (10 < getNat_A a)
与 true
在r_res
,这会导致类似的结果:
Some (exist (fun x : X => Instantiated_X x) (r1 a)
(r1_rewrite_obligation_1 a Heq_anonymous) =
Some (exist (fun x : X => Instantiated_X x) (r1 a0) i0)
以及proof irrelevance
和r1 injectivity
,让我的证明通过。所以,我想知道我能得到一些关于如何按摩的建议r_res
在这种情况下,它有助于重写。
编辑:删除 Eq
类型类及其实例,使示例更加简洁
最佳答案
你的证明尝试的问题在于你必须小心重写的方式。这是一个可能的解决方案。
Example r1_rewrite_surj:
forall t : iX, exists (a : A),
match (r1_rewrite a) with
| None => True
| Some e => eq t e
end.
Proof.
move=> [[a] Pa]; exists a; rewrite /r1_rewrite.
move: (erefl _); rewrite {1 3}Pa.
by move=> e; rewrite (eq_irrelevance (r1_rewrite_obligation_1 _ _) Pa).
Qed.
要了解这里发生的情况有点棘手。第一行之后,证明状态如下所示:
a : A
Pa : Instantiated_X (r1 a)
============================
match
(if Instantiated_X (r1 a) as b return b = Instantiated_X (r1 a) -> option iX
then
fun H : true = Instantiated_X (r1 a) =>
Some (exist (fun x : X => Instantiated_X x) (r1 a) (r1_rewrite_obligation_1 a H))
else fun _ : false = Instantiated_X (r1 a) => None) (erefl (Instantiated_X (r1 a)))
with
| Some e => exist (fun x : X => Instantiated_X x) (r1 a) Pa = e
| None => True
end
如果我们尝试在以下任何情况下使用 Pa
重写,我们将收到类型错误。例如:
如果我们尝试替换第一次出现的
Instantiated_X (r1 a)
,Coq 将不允许我们将if
的结果应用到(erefl (Instantiated_X (r1 a))
.我们可以通过将第一个、第二个和第六个(
erefl
上的那个)出现的Instantiated_X (r1 a)
替换为true
。这也不起作用,因为它会使r1_rewrite_obligation_1
的应用程序类型错误。
解决方案是泛化erefl
(使用调用move: (erefl _)
),导致以下证明状态:
forall e : Instantiated_X (r1 a) = Instantiated_X (r1 a),
match
(if Instantiated_X (r1 a) as b return b = Instantiated_X (r1 a) -> option iX
then
fun H : true = Instantiated_X (r1 a) =>
Some (exist (fun x : X => Instantiated_X x) (r1 a) (r1_rewrite_obligation_1 a H))
else fun _ : false = Instantiated_X (r1 a) => None) e
with
| Some e0 => exist (fun x : X => Instantiated_X x) (r1 a) Pa = e0
| None => True
end
可能不容易看出,但此时可以安全地用 Pa
重写来替换第一次和第三次出现的 Instantiated_X (r1 a)
,并允许 if
减少。然后我们可以通过证明 bool 相等性的无关性来得出结论。
不用说,以这种方式推理打字问题是一场噩梦。正如 ejgallego 指出的,在这种情况下,重用 ssreflect 的子类型机制要容易得多。例如:
(* Other definitions remain the same *)
Definition r1_rewrite a : option iX := insub (r1 a).
Example r1_rewrite_surj:
forall t : iX, exists (a : A),
match (r1_rewrite a) with
| None => True
| Some e => eq t e
end.
Proof.
by move=> [[a] Pa]; exists a; rewrite /r1_rewrite insubT.
Qed.
insub
函数是 r1_rewrite
的通用版本。它检查定义子类型的属性是否成立,如果成立,则将该对象与相应的证明配对。 insubT
引理表明,当属性成立时,insub
返回 Some
。
关于coq - Coq 的子类型,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46936688/