Question

我试图自动决定ASCII字符是否为空白的决策程序。这是我现在拥有的。

Require Import Ascii String.

Scheme Equality for ascii.

Definition IsWhitespace (c : ascii) := (c = "009"%char) \/ (c = "032"%char).

Definition isWhitespace (c : ascii) : {IsWhitespace c} + {not (IsWhitespace c)}.
Proof.
  unfold IsWhitespace.
  pose proof (ascii_eq_dec c "009"%char) as [H1|H1];
  pose proof (ascii_eq_dec c "032"%char) as [H2|H2];
  auto.
  right. intros [H3|H3]; auto.
Admitted.

什么是使证明更简洁的好方法？

Answer 1

通常，使证明更加自动化需要编写比您开始时更多的代码，以便您可以处理更多案例。采用这种方法，我改编了fiat-crypto的一些样板：

<div class="my-container">

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 <blockquote class="twitter-tweet" data-lang="en">
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 <blockquote class="twitter-tweet" data-lang="en">
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</div>

使用此样板，证明变为

Require Import Coq.Strings.Ascii Coq.Strings.String.

Class Decidable (P : Prop) := dec : {P} + {~P}.
Arguments dec _ {_}.
Notation DecidableRel R := (forall x y, Decidable (R x y)).

Global Instance dec_or {A B} {HA : Decidable A} {HB : Decidable B} : Decidable (A \/ B).
Proof. hnf in *; tauto. Defined.
Global Instance dec_eq_ascii : DecidableRel (@eq ascii) := ascii_dec.

与证据一样短。（请注意，Definition IsWhitespace (c : ascii) := (c = "009"%char) \/ (c = "032"%char). Definition isWhitespace (c : ascii) : Decidable (IsWhitespace c) := _.与:= _相同，. Proof. exact _. Defined.本身与. Proof. typeclasses eauto. Defined.相同。）

请注意，这与ejgallego的证明非常相似，因为tauto与intuition fail相同。

另请注意the original boilerplate in fiat-crypto比使用hnf in *; tauto更长，但也更强大，并处理几十种不同类型的可判定命题。

Answer 2

证明几乎是最简洁的！至多你可以做的是调用更强大的策略，例如intuition：

Definition isWhitespace (c : ascii) : {IsWhitespace c} + {not (IsWhitespace c)}.
Proof.
now unfold IsWhitespace;
    case (ascii_eq_dec c "009"%char);
    case (ascii_eq_dec c " "%char); intuition.

Answer 3

遵循杰森的回答精神，我们当然可以使用一些处理可判定平等的图书馆来达到你的结果：

这会将ascii声明为具有可判定等式的类型：

From Coq Require Import Ascii String ssreflect ssrfun ssrbool.
From mathcomp Require Import eqtype ssrnat.

Lemma ascii_NK : cancel N_of_ascii ascii_of_N.
Proof. exact: ascii_N_embedding. Qed.

Definition ascii_eqMixin := CanEqMixin ascii_NK.
Canonical ascii_eqType := EqType _ ascii_eqMixin.

在这种风格中，通常你说你的属性是可判定的命题，所以没有什么可以证明的：

Definition IsWhitespaceb (c : ascii) := [|| c == "009"%char | c == " "%char].

但是如果你愿意，你当然可以恢复非计算性的：

Definition IsWhitespace (c : ascii) := (c = "009"%char) \/ (c = "032"%char).

Lemma whitespaceP c : reflect (IsWhitespace c) (IsWhitespaceb c).
Proof. exact: pred2P. Qed.

当然可以使用更多自动化。

战术自动化：简单的决策程序

3 个答案: