tin*_*lyx 5 coq induction coq-tactic
我读到一种类型的归纳原理只是一个关于命题的定理P.所以我构建了一个List基于右(或反向)列表构造函数的归纳原理.
Definition rcons {X:Type} (l:list X) (x:X) : list X :=
l ++ x::nil.
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归纳原理本身是:
Definition true_for_nil {X:Type}(P:list X -> Prop) : Prop :=
P nil.
Definition true_for_list {X:Type} (P:list X -> Prop) : Prop :=
forall xs, P xs.
Definition preserved_by_rcons {X:Type} (P: list X -> Prop): Prop :=
forall xs' x, P xs' -> P (rcons xs' x).
Theorem list_ind_rcons:
forall {X:Type} (P:list X -> Prop),
true_for_nil P ->
preserved_by_rcons P ->
true_for_list P.
Proof. Admitted.
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但是现在,我在使用该定理时遇到了麻烦.我不是如何调用它来实现与induction战术相同的.
例如,我尝试过:
Theorem rev_app_dist: forall {X} (l1 l2:list X), rev (l1 ++ l2) = rev l2 ++ rev l1.
Proof. intros X l1 l2.
induction l2 using list_ind_rcons.
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但在最后一行,我得到了:
Error: Cannot recognize an induction scheme.
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定义和应用自定义归纳原理的正确步骤是list_ind_rcons什么?
谢谢
你所做的大部分是正确的。问题是 Coq 在识别你写的是归纳原理时遇到了一些麻烦,因为中间定义。例如,这工作得很好:
Theorem list_ind_rcons:
forall {X:Type} (P:list X -> Prop),
P nil ->
(forall x l, P l -> P (rcons l x)) ->
forall l, P l.
Proof. Admitted.
Theorem rev_app_dist: forall {X} (l1 l2:list X), rev (l1 ++ l2) = rev l2 ++ rev l1.
Proof. intros X l1 l2.
induction l2 using @list_ind_rcons.
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我不知道 Coq 无法自动展开中间定义是否应该被视为错误,但至少有一个解决方法。
如果想保留中间定义,则可以使用该Section机制,如下所示:
Require Import Coq.Lists.List. Import ListNotations.
Definition rcons {X:Type} (l:list X) (x:X) : list X :=
l ++ [x].
Section custom_induction_principle.
Variable X : Type.
Variable P : list X -> Prop.
Hypothesis true_for_nil : P nil.
Hypothesis true_for_list : forall xs, P xs.
Hypothesis preserved_by_rcons : forall xs' x, P xs' -> P (rcons xs' x).
Fixpoint list_ind_rcons (xs : list X) : P xs. Admitted.
End custom_induction_principle.
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Coq 替换定义并list_ind_rcons具有所需的类型并induction ... using ...工作:
Theorem rev_app_dist: forall {X} (l1 l2:list X),
rev (l1 ++ l2) = rev l2 ++ rev l1.
Proof. intros X l1 l2.
induction l2 using list_ind_rcons.
Abort.
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顺便说一下,这个归纳原理存在于标准库(List模块)中:
Coq < Check rev_ind.
rev_ind
: forall (A : Type) (P : list A -> Prop),
P [] ->
(forall (x : A) (l : list A), P l -> P (l ++ [x])) ->
forall l : list A, P l
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