Coq：多个构造函数的单一表示法 [英] Coq: a single notation for multiple constructors

问题描述

是否可以在Coq中为多个构造函数定义一个符号？如果构造函数的参数类型不同，则可能无法从中推断出它们。最小（非）工作示例：

Is it possible to define a single notation for multiple constructors in Coq? If the constructors differ by their argument types, they might be inferrable from them. A minimal (non-)working example:

Inductive A : Set := a | b | c: C -> A | d: D -> A
with C: Set := c1 | c2
with D: Set := d1 | d2.

Notation "' x" := (_ x) (at level 19).
Check 'c1. (*?6 c1 : ?8*)

在这种情况下，构造函数推断无效。也许还有另一种方法可以将构造函数指定为变量？

In this case, constructor inference doesn't work. Maybe there's another way to specify a constructor as a variable?

推荐答案

您可以创建一个以构造函数为实例的类型类，实例解析机制推断构造函数要为您调用：

You can create a typeclass with the constructors as instances and let the instance resolution mechanism infer the constructor to call for you:

Class A_const (X:Type) : Type :=
  a_const : X -> A.
Instance A_const_c : A_const C := c.
Instance A_const_d : A_const D := d.

Check a_const c1.
Check a_const d2.

顺便说一句，使用Coq 8.5，如果您真的想要一种符号' x 导致将精确的构造函数应用于 x ，而不是例如 @a_const C A_const_c c1 ，则可以使用ltac术语来实现：

By the way, with Coq 8.5, if you really want a notation ' x to result in the exact constructor applied to x, rather than e.g. @a_const C A_const_c c1, then you can use ltac-terms to accomplish that:

Notation "' x" := ltac:(match constr:(a_const x) with
                        | @a_const _ ?f _ =>
                          let unfolded := (eval unfold f in f) in
                          exact (unfolded x)
                        end) (at level 0).
Check 'c1. (* c c1 : A *)
Check 'd2. (* d d2 : A *)

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