Question

问题是，如果我知道forall x, f x ≡ g x（其中≡是一些等价关系，而f，g是函数），什么是正确的{ {1}}实例，它可以让我用一个由等价关系链接的较大术语用Proper重写f吗？

假设需要，可以使用功能扩展性-我猜这是必需的吗？

一些示例代码可以更好地演示该问题：

如果我们可以将Require Import Setoid. (** Feel free to assume FunExt *) Require Import FunctionalExtensionality. Section FOOBAR. Variable T: Type. Variable f: T -> T. Variable g: T -> T. Variable t0: T. Variable combiner: (T -> T) -> T -> T. Variable equiv: T -> T -> Prop. Infix "≡" := equiv (at level 50). Axiom equivalence_equiv: Equivalence equiv. Axiom F_EQUIV_G_EXT: forall (t: T), f t ≡ g t. (** Check that coq can resolve the Equivalence instance **) Theorem equivalence_works: t0 ≡ t0. Proof. reflexivity. Qed. Theorem rewrite_in_lambda: combiner (fun t => f t) t0 ≡ combiner (fun t => g t) t0. Proof. intros. (* I wish to replace x with y. What are the Proper rules I need for this to happen? *) rewrite F_EQUIV_G_EXT. Abort. End FOOBAR.替换为f，就证明了这一点，但是我不确定该怎么做。我的等效关系需要什么附加能量才能成功？

Answer 1

解决方案是使用来自coq stdlib的pointwise_relation：Link here

我还复制粘贴了定义，以防链接bitrots：

 Definition pointwise_relation (R : relation B) : relation (A -> B) :=
    fun f g => forall a, R (f a) (g a).

因此，我们希望表单具有适当的实例：

Axiom proper: Proper (pointwise_relation T equiv ==> equiv ==> equiv) combiner.

也就是说，如果第一个函数是逐点相等的，而第二个参数是相等的，那么结果是相等的。

以下是编译的完整代码清单：

Require Import Setoid.
Require Import Relation_Definitions.
Require Import Morphisms.

(** Feel free to assume FunExt *)
Require Import FunctionalExtensionality.
Section FOOBAR.
  Variable T: Type.
  Variable x: T -> T.
  Variable y: T -> T.

  Variable t0: T.
  Variable combiner: (T -> T) -> T -> T.

  Variable equiv: T -> T -> Prop.
  Infix "≡" := equiv (at level 50).

  Axiom equivalence_equiv: Equivalence equiv.
  Axiom proper: Proper (pointwise_relation T equiv ==> equiv ==> equiv) combiner.

  Axiom X_EQUIV_Y_EXT: forall (t: T), x t ≡ y t.

  (** Check that coq can resolve the Equivalence instance **)
  Theorem equivalence_works: t0 ≡ t0.
  Proof.
    reflexivity.
  Qed.

  Theorem rewrite_in_lambda:
    combiner (fun t => x t) t0 ≡
    combiner (fun t => y t) t0.
  Proof.
    intros.
    (* I wish to replace x with y.
    What are the Proper rules  I need for this to happen? *)
    setoid_rewrite X_EQUIV_Y_EXT.
    reflexivity.
  Qed.
End FOOBAR.

在lambda中针对等价关系进行重写

1 个答案: