Coq证明阶乘N /(阶乘k *阶乘(N-k))是整数

时间:2019-03-25 07:54:26

标签: coq factorial

我在Coq标准库中找不到import a.b.c.d.e.f as ref g1 = ref.g1 h1 = ref.h.h1 是集成的证明。这个引理的简短的自包含证明是什么?

N choose k

我看到他们在ssreflect.binomial.v中通过递归定义{{​​1}} Lemma fact_divides N k: k <= N -> Nat.divide (fact k * fact (N - k)) (fact N). 并显示choose来回避了整个问题。

但是,也可以直接证明以上内容而无需求助于pascal的三角形,这将是很好的。当我在Stack。*上搜索时,出现了许多“非正式”证明。对于有理数,它们隐式地使用了代数步长,而他们并没有显示出严格地用于choose(N,k) = choose(N-1,k) + choose(N-1,k-1)除法。

编辑: 感谢以下@Bubbler的回答(基于this math),证明只是

choose(N,k) * k! * (N-k)! = N!

1 个答案:

答案 0 :(得分:1)

我将其声明为如下:

Theorem fact_div_fact_fact : forall x y, exists e, fact (x + y) = e * (fact x * fact y).

我相信您可以从中得出自己的引理,并结合Coq标准库中有关<=-的事实。

这是使用pure algebraic approach的独立的,不太简短的证明。您可以尝试运行here online

From Coq Require Import Arith.

(* Let's prove that (n+m)! is divisible by n! * m!. *)

(* fact2 x y = (x+1) * (x+2) * .. * (x+y) *)

Fixpoint fact2 x y := match y with
  | O => 1
  | S y' => (x + y) * fact2 x y'
end.

Lemma fact2_0 : forall x, fact2 0 x = fact x.
Proof.
  induction x.
  - auto.
  - simpl. rewrite IHx. auto. Qed.

Lemma fact_fact2 : forall x y, fact x * fact2 x y = fact (x + y).
Proof.
  induction x.
  - intros. simpl. rewrite fact2_0. ring.
  - induction y.
    + simpl. replace (x + 0) with x by ring. ring.
    + simpl. replace (x + S y) with (S x + y) by ring. rewrite <- IHy. simpl. ring. Qed.

Lemma fact2_left : forall x y, fact2 x (S y) = S x * fact2 (S x) y.
Proof. intros x y. generalize dependent x. induction y.
  - intros. simpl. ring.
  - intros. unfold fact2. fold (fact2 x (S y)). fold (fact2 (S x) y).
    rewrite IHy. ring. Qed.

Lemma fact_div_fact2 : forall x y, exists e, fact2 x y = e * fact y.
Proof. intros x y. generalize dependent x. induction y.
  - intros. simpl. exists 1. auto.
  - induction x.
    + unfold fact2. fold (fact2 0 y). unfold fact. fold (fact y). destruct (IHy 0). rewrite H.
      exists x. ring.
    + unfold fact2. fold (fact2 (S x) y).
      destruct (IHy (S x)). destruct IHx. exists (x0 + x1).
      replace ((S x + S y) * fact2 (S x) y) with (S x * fact2 (S x) y + S y * fact2 (S x) y) by ring.
      rewrite <- fact2_left. rewrite H0. rewrite H.
      replace (S y * (x0 * fact y)) with (x0 * (S y * fact y)) by ring.
      unfold fact. fold (fact y). ring. Qed.

Theorem fact_div_fact_fact : forall x y, exists e, fact (x + y) = e * (fact x * fact y).
Proof. intros x y. destruct (fact_div_fact2 x y). exists x0.
  rewrite <- fact_fact2. rewrite H. ring. Qed.