Coq中的证明自动化如何分解证明 我在跟随软件基金会,我在一章名为“IMP”。
作者公开了一种小型语言,如下所示:Coq中的证明自动化如何分解证明 我在跟随软件基金会,我在一章名为“IMP”。,coq,proof,Coq,Proof,作者公开了一种小型语言,如下所示: Inductive aexp : Type := | ANum : nat -> aexp | APlus : aexp -> aexp -> aexp | AMinus : aexp -> aexp -> aexp | AMult : aexp -> aexp -> aexp. 下面是计算这些表达式的函数: Fixpoint aeval (a : aexp) : nat := match a w
Inductive aexp : Type :=
| ANum : nat -> aexp
| APlus : aexp -> aexp -> aexp
| AMinus : aexp -> aexp -> aexp
| AMult : aexp -> aexp -> aexp.
下面是计算这些表达式的函数:
Fixpoint aeval (a : aexp) : nat :=
match a with
| ANum n ⇒ n
| APlus a1 a2 ⇒ (aeval a1) + (aeval a2)
| AMinus a1 a2 ⇒ (aeval a1) - (aeval a2)
| AMult a1 a2 ⇒ (aeval a1) × (aeval a2)
end.
本练习旨在创建一个优化评估的函数。例如:
APlus a (ANum 0) --> a
这是我的优化功能:
Fixpoint optimizer_a (a:aexp) :aexp :=
match a with
| ANum n => ANum n
| APlus (ANum 0) e2 => optimizer_a e2
| APlus e1 (ANum 0) => optimizer_a e1
| APlus e1 e2 => APlus (optimizer_a e1) (optimizer_a e2)
| AMinus e1 (ANum 0) => optimizer_a e1
| AMinus e1 e2 => AMinus (optimizer_a e1) (optimizer_a e2)
| AMult (ANum 1) e2 => optimizer_a e2
| AMult e1 (ANum 1) => optimizer_a e1
| AMult e1 e2 => AMult (optimizer_a e1) (optimizer_a e2)
end.
现在,我要证明优化函数是正确的:
Theorem optimizer_a_sound : forall a, aeval (optimizer_a a) = aeval a.
这个证明很难。所以我试着用一些引理来分解证明
这里有一个引理:
Lemma optimizer_a_plus_sound : forall a b, aeval (optimizer_a (APlus a b)) = aeval (APlus (optimizer_a a) (optimizer_a b)).
我有证据,但很无聊。我在a上做一个归纳,然后,对于每种情况,我都会解构b,并解构exp来处理b为0的情况
我需要这样做是因为
n+0 = n
不会自动减少,我们需要一个加0的定理
现在,我想知道,我怎样才能用Coq建立一个更好的证明,以避免在证明过程中出现一些无聊的重复
这是我对这个引理的证明:
我想我应该使用“提示重写加上”但我不知道如何使用
顺便说一下,我还想知道一些技巧来展示初始定理(我的优化函数的奥秘)。 < P>当你在数据结构上使用归纳法,需要考虑案例时,这里有一种方法可以用来考虑变量<代码> x <代码>的情况,摆脱不可能的情况,解决一些琐碎的问题
destruct X; try congruence; auto.
在你的例子中,我们可以用它来证明关于优化\u a
函数的有用的重写引理
Lemma opt1: forall a b, a = ANum 0 -> optimizer_a (APlus a b) = optimizer_a b.
intros.
destruct a; try congruence; auto;
destruct n; try congruence; auto.
Qed.
Lemma opt2: forall a b, b = ANum 0 -> optimizer_a (APlus a b) = optimizer_a a.
intros;
destruct a; try congruence; auto;
destruct b; try congruence; auto;
destruct n; try congruence; auto;
destruct n0; try congruence; auto.
Qed.
Lemma opt3:
forall a b,
a <> ANum 0 -> b <> ANum 0 ->
optimizer_a (APlus a b) = APlus (optimizer_a a) (optimizer_a b).
Proof.
intros.
destruct a; try congruence; auto;
destruct b; try congruence; auto;
destruct n; try congruence; auto;
destruct n0; try congruence; auto.
Qed.
我们可以解构ANum0_dec a
并在上下文中获取a=ANum 0
,或者在上下文中获取ANum 0
Require Import Arith. (* for "auto with arith" to handle n = n + 0 *)
Lemma optimizer_a_plus_sound :
forall a b,
aeval (optimizer_a (APlus a b)) = aeval (APlus (optimizer_a a) (optimizer_a b)).
Proof.
intros a b.
destruct (ANum0_dec a); destruct (ANum0_dec b).
- rewrite opt1; subst; auto.
- rewrite opt1; subst; auto.
- rewrite opt2; subst; simpl; auto with arith.
- rewrite opt3; subst; auto.
Qed.
如果你使用上面的技巧,你可以定义你自己的战术,所以你不必输入太多。因为证明很短,你可以不用引理。(我为自毁同余自动调用了战术
dca
)
较短的证明不是可读的,但它本质上是:考虑变量的情况。
Lemma ANum0_dec: forall a, {a = ANum 0} + { a <> ANum 0}.
destruct a; try destruct n; try (right; discriminate); left; auto.
Qed.
Require Import Arith.
Ltac dca v := destruct v; try congruence; auto.
Lemma optimizer_a_plus_sound :
forall a b,
aeval (optimizer_a (APlus a b)) = aeval (APlus (optimizer_a a) (optimizer_a b)).
Proof.
intros a b;
destruct (ANum0_dec a), (ANum0_dec b).
- dca a; dca n.
- dca a; dca n0.
- dca b; dca n0; dca a; simpl; auto with arith; dca n0.
- dca a; dca b; dca n1; dca n2.
Qed.
你也可以用这种简洁的形式做完整的证明。在
|APlus e1 e2=>APlus(optimizer_a e1)(optimizer_a e2)
中,如果optimizer_a e1
返回ANum 0怎么办?
Lemma ANum0_dec: forall a, {a = ANum 0} + { a <> ANum 0}.
destruct a; try destruct n; try (right; discriminate); left; auto.
Qed.
Require Import Arith.
Ltac dca v := destruct v; try congruence; auto.
Lemma optimizer_a_plus_sound :
forall a b,
aeval (optimizer_a (APlus a b)) = aeval (APlus (optimizer_a a) (optimizer_a b)).
Proof.
intros a b;
destruct (ANum0_dec a), (ANum0_dec b).
- dca a; dca n.
- dca a; dca n0.
- dca b; dca n0; dca a; simpl; auto with arith; dca n0.
- dca a; dca b; dca n1; dca n2.
Qed.
Theorem optimizer_a_sound : forall a, aeval (optimizer_a a) = aeval a.
induction a.
* auto.
* rewrite optimizer_a_plus_sound; simpl; auto.
* (* ... and so on for Minus and Mult *)