在Coq中建立良好的递归 [英] Well founded recursion in Coq

问题描述

我试图在Coq中编写一个用于计算自然除法的函数，但由于它不是结构性递归，因此在定义它时遇到了一些麻烦。

I am trying to write a function for computing natural division in Coq and I am having some trouble defining it since it is not structural recursion.

我的代码是：

Inductive N : Set :=
  | O : N
  | S : N -> N.

Inductive Bool : Set :=
  | True : Bool
  | False : Bool.

Fixpoint sum (m :N) (n : N) : N :=
match m with
  | O => n
  | S x => S ( sum x n)
end.

Notation "m + n" := (sum m n) (at level 50, left associativity).

Fixpoint mult (m :N) (n : N) : N :=
match m with
  | O => O
  | S x => n + (mult x n)
end.

Notation "m * n" := (mult m n) (at level 40, left associativity).

Fixpoint pred (m : N) : N :=
match m with
  | O => S O
  | S x => x
end.

Fixpoint resta (m:N) (n:N) : N :=
match n with
  | O => m
  | S x => pred (resta m x)
end.
Notation "m - x" := (resta m x) (at level 50, left associativity).

Fixpoint leq_nat (m : N) (n : N) : Bool :=
match m with
  | O => True
  | S x => match n with 
    | O => False
    | S y => leq_nat x y
  end
end.

Notation "m <= n" := (leq_nat m n) (at level 70).

Fixpoint div (m : N) (n : N) : N :=
match n with
  | O => O
  | S x => match m <= n with
    | False => O
    | True => pred (div (m-n) n)
  end
end.

如您所见，Coq不喜欢我的函数div，它说

As you can see, Coq does not like my function div, it says

错误：无法猜测 fix 的递减参数。

如何在Coq中提供此功能的终止证明？我可以证明如果n> O ^ n <= m->（m-n）< m。

How can I supply in Coq a termination proof for this function? I can prove that if n>O ^ n<=m -> (m-n) < m.

推荐答案

在这种情况下，最简单的策略可能是使用 Program 扩展名和措施。然后，您必须提供一个证明，根据度量，递归调用中使用的参数小于顶级参数。

The simplest strategy in this case is probably to use the Program extension together with a measure. You will then have to provide a proof that the arguments used in the recursive call are smaller than the top level ones according to the measure.

Require Coq.Program.Tactics.
Require Coq.Program.Wf.

Fixpoint toNat (m : N) : nat :=
match m with O => 0 | S n => 1 + (toNat n) end.

Program Fixpoint div (m : N) (n : N) {measure (toNat m)}: N :=
match n with
  | O => O
  | S x => match m <= n with
    | False => O
    | True => pred (div (m-n) n)
  end
end.
Next Obligation.
(* your proof here *)

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