我试图证明Coq中的"Practical Coinduction"中的第一个示例。第一个例子是证明无限流整数上的词典排序是可传递的。在Coq中证明Co-Inductive属性(词汇顺序是传递性的)
我一直没能制定的证明,以绕过Guardedness condition
这里是我发展至今。首先是无限流的通常定义。然后定义称为lex
的词典顺序。最后,传递性定理的失败证明。
Require Import Omega.
Section stream.
Variable A:Set.
CoInductive Stream : Set :=
| Cons : A -> Stream -> Stream.
Definition head (s : Stream) :=
match s with Cons a s' => a end.
Definition tail (s : Stream) :=
match s with Cons a s' => s' end.
Lemma cons_ht: forall s, Cons (head s) (tail s) = s.
intros. destruct s. reflexivity. Qed.
End stream.
Implicit Arguments Cons [A].
Implicit Arguments head [A].
Implicit Arguments tail [A].
Implicit Arguments cons_ht [A].
CoInductive lex s1 s2 : Prop :=
is_le : head s1 <= head s2 ->
(head s1 = head s2 -> lex (tail s1) (tail s2)) ->
lex s1 s2.
Lemma lex_helper: forall s1 s2,
head s1 = head s2 ->
lex (Cons (head s1) (tail s1)) (Cons (head s2) (tail s2)) ->
lex (tail s1) (tail s2).
Proof. intros; inversion H0; auto. Qed.
这里是我想证明的引理。我首先准备目标,以便我可以应用构造函数,希望最终能够使用cofix
中的“假设”。
Lemma lex_lemma : forall s1 s2 s3, lex s1 s2 -> lex s2 s3 -> lex s1 s3.
intros s1 s2 s3 lex12 lex23.
cofix.
rewrite <- (cons_ht s1).
rewrite <- (cons_ht s3).
assert (head s1 <= head s3) by (inversion lex12; inversion lex23; omega).
apply is_le; auto.
simpl; intros. inversion lex12; inversion lex23.
assert (head s2 = head s1) by omega.
rewrite <- H0, H5 in *.
assert (lex (tail s1) (tail s2)) by (auto).
assert (lex (tail s2) (tail s3)) by (auto).
apply lex_helper.
auto.
repeat rewrite cons_ht.
Guarded.
我该怎么做?感谢任何提示!
- 编辑
感谢亚瑟(一如既往!)有益和有启发答案,我也能完成的证明。我给我的版本以供参考。
Lemma lex_lemma : forall s1 s2 s3, lex s1 s2 -> lex s2 s3 -> lex s1 s3.
cofix.
intros s1 s2 s3 lex12 lex23.
inversion lex12; inversion lex23.
rewrite <- (cons_ht s1).
rewrite <- (cons_ht s3).
constructor; simpl.
inversion lex12; inversion lex23; omega.
intros; eapply lex_lemma; [apply H0 | apply H2]; omega.
Qed.
我用cons_ht
引理 “扩大” 的s1
和s3
值。这里的lex
(与head
和tail
)的定义更接近Practical Coinduction中的逐字制定。 Arthur使用更优雅的技术,让Coq自动扩展值 - 更好!