【问题标题】:Induction order for relation between three lists三个列表之间关系的归纳顺序
【发布时间】:2020-12-19 03:48:33
【问题描述】:

我正在研究字符串语法理论,但我完全被一个特定的定理所阻碍。我尝试过的每个归纳排序最终都陷入了荒谬和无用的归纳假设,我不确定我错过了什么。

我现在已经多次重读此"Varying the Induction Hypothesis" 部分,试图了解我做错了什么,但似乎我正在遵循他们关于一般性的建议。

简单地说,我很困惑。非常感谢您的指导!

这是我的定义和困难定理的概述。我稍后会给出我的完整脚本。

(* I plan to make these definitions more complex in the future *)
Definition TokenDefinition := String.string.
Definition Token := String.string.
Definition TokenMatches (def: TokenDefinition) (token: Token): Prop := def = token.

Definition TokenPath := list TokenDefinition.
Definition TokenStream := list Token.

Inductive PathMatchesStream: TokenPath -> TokenStream -> Prop :=
    | PathMatchesStream_base: forall def token,
        TokenMatches def token
        -> PathMatchesStream [def] [token]
    | PathMatchesStream_append: forall def token path stream,
        TokenMatches def token
        -> PathMatchesStream path stream
        -> PathMatchesStream (def :: path) (token :: stream)
.

(* the problematic theorem *)
Theorem PathMatchesStream_same_if_match_same:
    forall a b stream,
        PathMatchesStream a stream
        -> PathMatchesStream b stream
        -> a = b.
Proof.
    (* my full script has many failed attempts *)
Qed.

这是我的完整脚本,如果你有来自Certified Programming with Dependent TypesCpdt,它应该可以运行(我非常喜欢他的自动化风格)。

Set Implicit Arguments. Set Asymmetric Patterns.
Require Import List.
Import ListNotations.
Open Scope list_scope.
Require String.
Require Import Cpdt.CpdtTactics.
Require Import PeanoNat Lt.

Definition TokenDefinition := String.string.
Definition Token := String.string.

Definition TokenMatches (def: TokenDefinition) (token: Token): Prop := def = token.
Hint Unfold TokenMatches: core.

Ltac simpl_TokenMatches :=
    unfold TokenMatches in *; subst.

Theorem TokenDefinition_match_same_then_same:
    forall a b token, TokenMatches a token -> TokenMatches b token -> a = b.
Proof. crush. Qed.


Definition TokenPath := list TokenDefinition.
Definition TokenStream := list Token.

Inductive PathMatchesStream: TokenPath -> TokenStream -> Prop :=
    | PathMatchesStream_base: forall def token,
        TokenMatches def token
        -> PathMatchesStream [def] [token]
    | PathMatchesStream_append: forall def token path stream,
        TokenMatches def token
        -> PathMatchesStream path stream
        -> PathMatchesStream (def :: path) (token :: stream)
.
Hint Constructors PathMatchesStream: core.

Ltac invert_PathMatchesStream :=
    crush; repeat match goal with
        | [ H : TokenMatches _ _ |- _ ] =>
            simpl_TokenMatches; crush
        | [ H : PathMatchesStream _ _ |- _ ] =>
            solve [inversion H; clear H; crush]
    end.

Theorem PathMatchesStream_path_not_empty:
    forall stream, ~(PathMatchesStream [] stream).
Proof. invert_PathMatchesStream. Qed.
Hint Resolve PathMatchesStream_path_not_empty: core.

Theorem PathMatchesStream_stream_not_empty:
    forall path, ~(PathMatchesStream path []).
Proof. invert_PathMatchesStream. Qed.
Hint Resolve PathMatchesStream_stream_not_empty: core.

Theorem PathMatchesStream_length_non_zero:
    forall path stream, PathMatchesStream path stream -> 0 < (length path) /\ 0 < (length stream).
Proof. invert_PathMatchesStream. Qed.
Hint Resolve PathMatchesStream_length_non_zero: core.

Theorem PathMatchesStream_same_if_match_same:
    forall a b stream,
        PathMatchesStream a stream
        -> PathMatchesStream b stream
        -> a = b.
Proof.

intros a; induction a as [| atok a IHa]; intros b; induction b as [| btok b IHb].

- invert_PathMatchesStream.
- invert_PathMatchesStream.
- invert_PathMatchesStream.

-
induction stream as [| tok stream IHstream].

+ invert_PathMatchesStream.


+
intros Ha Hb.

apply IHa.

(*intros a b stream Ha.
generalize dependent b.
induction Ha.
-
intros b Hb.
unfold TokenMatches in *.
induction Hb.
+

-




intros a.

induction a as [| atok a IHa].

- invert_PathMatchesStream.

-
intros b.
induction b as [| btok b IHb].
-- invert_PathMatchesStream.
--
intros stream.
induction stream as [| tok stream IHstream].
+ invert_PathMatchesStream.
+
intros Ha.
induction Ha; intros Hb; induction Hb.
++ apply IHb.
++ invert_PathMatchesStream.
++
inversion Ha; clear Ha; inversion Hb; clear Hb; invert_PathMatchesStream.
++
apply IHa.
++
*)

(*intros a b stream.
generalize dependent b.
generalize dependent a.
induction stream as [| tok stream IHstream].
- invert_PathMatchesStream.
-
intros a.
induction a as [| atok a IHa].
-- invert_PathMatchesStream.
--
intros b.
induction b as [| btok b IHb].
++ invert_PathMatchesStream.
++
intros Ha Hb.

induction Ha; induction Hb; simpl_TokenMatches.
crush.
invert_PathMatchesStream.*)

Qed.

【问题讨论】:

    标签: coq induction


    【解决方案1】:

    对于这个定理,你实际上不需要太花哨的东西。只是一个辅助引理:

    Require Import Coq.Lists.List.
    Require Import Coq.Strings.String.
    Import ListNotations.
    
    
    (* I plan to make these definitions more complex in the future *)
    Definition TokenDefinition := String.string.
    Definition Token := String.string.
    Definition TokenMatches (def: TokenDefinition) (token: Token): Prop := def = token.
    
    Definition TokenPath := list TokenDefinition.
    Definition TokenStream := list Token.
    
    Inductive PathMatchesStream: TokenPath -> TokenStream -> Prop :=
        | PathMatchesStream_base: forall def token,
            TokenMatches def token
            -> PathMatchesStream [def] [token]
        | PathMatchesStream_append: forall def token path stream,
            TokenMatches def token
            -> PathMatchesStream path stream
            -> PathMatchesStream (def :: path) (token :: stream)
    .
    
    Theorem PathMatchesStream_same_if_match_same_aux :
        forall a stream, PathMatchesStream a stream -> a = stream.
    Proof.
    intros a b H.
    now induction H; unfold TokenMatches in *; subst.
    Qed.
    
    (* the problematic theorem *)
    Theorem PathMatchesStream_same_if_match_same:
        forall a b stream,
            PathMatchesStream a stream
            -> PathMatchesStream b stream
            -> a = b.
    Proof.
    now intros
       ? ? ? ->%PathMatchesStream_same_if_match_same_aux ->%PathMatchesStream_same_if_match_same_aux.
    Qed.
    

    不过,我不知道这是否仍然足以满足您的想法……

    【讨论】:

    • 你真的在教我东西! ?-&gt;% 语法我都不熟悉,我得查一下!
    • 我也担心这对于我的改进来说还不够:/我可能需要将我的属性重述为某种折叠列表并证明一些关于它们的更一般的定理。我希望我很快就会出现另一个问题哈。不过感谢您的帮助!提醒一下,并非每个定理都需要归纳是非常有帮助的。
    • @blaineh 很高兴为您提供帮助!但请记住,这些运算符只是方便的语法。可以通过更明确的方式(例如将辅助引理应用于假设)来实现相同的效果。
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