简体   繁体   English

Coq中列表的布尔相等?

[英]Boolean equality of lists in Coq?

I want to be able to compare two items of type "list" in Coq and get a boolean "true" or "false" for their equivalence. 我希望能够在Coq中比较两个类型为“list”的项,并为它们的等价获得一个布尔“true”或“false”。

Right now, I'm comparing the two lists this way: 现在,我正在用这种方式比较两个列表:

Eval vm_compute in (list 1 = list 2). 

I get a Prop of the form: 我得到了表格的支柱:

= nil
   :: (2 :: 3 :: nil)
      :: (2 :: nil)
         :: (3 :: nil) :: nil =
   nil
   :: (2 :: 3 :: nil)
      :: (2 :: nil)
         :: (3 :: nil) :: nil
 : Prop

Obviously list1 = list2, so how do I get it to just return true or false? 显然list1 = list2,那么如何让它返回true或false?

I use the Mathematical Components Library boolean equality operators : 我使用数学组件库布尔等于运算符

From mathcomp Require Import all_ssreflect.

...

Eval vm_compute in list 1 == list 2

You can generate a boolean list equality function that takes as input a boolean equality over the elements automatically using Coq's commands: 您可以生成一个布尔列表相等函数,该函数使用Coq命令自动将元素的布尔相等性作为输入:

Require Import Coq.Lists.List Coq.Bool.Bool.

Import Coq.Lists.List.ListNotations.

Scheme Equality for list.

This prints: 这打印:

list_beq is defined
list_eq_dec is defined

where list_beq is a boolean equality function on lists that takes as first parameter a comparison function for the lists elements and then two lists: 其中list_beq是列表上的布尔相等函数,它将列表元素的比较函数作为第一个参数,然后是两个列表:

Print list_beq.

Gives

list_beq = 
fun (A : Type) (eq_A : A -> A -> bool) =>
fix list_eqrec (X Y : list A) {struct X} : bool :=
  match X with
  | [] => match Y with
          | [] => true
          | _ :: _ => false
          end
  | x :: x0 => match Y with
               | [] => false
               | x1 :: x2 => eq_A x x1 && list_eqrec x0 x2
               end
  end
     : forall A : Type, (A -> A -> bool) -> list A -> list A -> bool

and

Check list_eq_dec

gives

list_eq_dec
     : forall (A : Type) (eq_A : A -> A -> bool),
       (forall x y : A, eq_A x y = true -> x = y) ->
       (forall x y : A, x = y -> eq_A x y = true) -> forall x y : list A, {x =  y} + {x <> y}

showing that list equality is decidable if the underlying types equality is agrees with leibniz equality. 如果底层类型相等与leibniz相等,则表明列表相等是可判定的。

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM