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首页 > 解决方案 > 如何使用 Isar 证明消除规则？

isabelle - 如何使用 Isar 证明消除规则？

问题描述

这是一个简单的理论：

datatype t1 = A | B | C
datatype t2 = D | E t1 | F | G

inductive R where
  "R A B"
| "R B C"

inductive_cases [elim]: "R x B" "R x A" "R x C"

inductive S where
  "S D (E _)"
| "R x y ⟹ S (E x) (E y)"

inductive_cases [elim]: "S x D" "S x (E y)"

elim我可以使用两个辅助引理来证明引理：

lemma tranclp_S_x_E:
  "S⇧+⇧+ x (E y) ⟹ x = D ∨ (∃z. x = E z)"
  by (induct rule: converse_tranclp_induct; auto)

(* Let's assume that it's proven *)
lemma reflect_tranclp_E:
  "S⇧+⇧+ (E x) (E y) ⟹ R⇧+⇧+ x y"
  sorry

lemma elim:
  "S⇧+⇧+ x (E y) ⟹
   (x = D ⟹ P) ⟹ (⋀z. x = E z ⟹ R⇧+⇧+ z y ⟹ P) ⟹ P"
  using reflect_tranclp_E tranclp_S_x_E by blast

我需要证明elim使用 Isar：

lemma elim:
  assumes "S⇧+⇧+ x (E y)"
    shows "(x = D ⟹ P) ⟹ (⋀z. x = E z ⟹ R⇧+⇧+ z y ⟹ P) ⟹ P"
proof -
  assume "S⇧+⇧+ x (E y)"
  then obtain z where "x = D ∨ x = E z"
    by (induct rule: converse_tranclp_induct; auto)
  also have "S⇧+⇧+ (E z) (E y) ⟹ R⇧+⇧+ z y"
    sorry
  finally show ?thesis

但我收到以下错误：

No matching trans rules for calculation:
    x = D ∨ x = E z
    S⇧+⇧+ (E z) (E y) ⟹ R⇧+⇧+ z y

Failed to refine any pending goal 
Local statement fails to refine any pending goal
Failed attempt to solve goal by exported rule:
  (S⇧+⇧+ x (E y)) ⟹ P

如何修复它们？

我想这个引理可以有一个更简单的证明。但我需要分两步证明：

  1. 显示可能的值x
  2. E反映传递闭包的显示

我还认为这个引理可以通过案例来证明x。但是我的真实数据类型有太多的案例。因此，这不是首选解决方案。

标签： isabelle

解决方案

此变体似乎有效：

lemma elim:
  assumes "S⇧+⇧+ x (E y)"
      and "x = D ⟹ P"
      and "⋀z. x = E z ⟹ R⇧+⇧+ z y ⟹ P"
    shows "P"
proof -
  have "S⇧+⇧+ x (E y)" by (simp add: assms(1))
  then obtain z where "x = D ∨ x = E z"
    by (induct rule: converse_tranclp_induct; auto)
  moreover
  have "S⇧+⇧+ (E z) (E y) ⟹ R⇧+⇧+ z y"
    sorry
  ultimately show ?thesis
    using assms by auto
qed
  • 假设应该与目标分开。
  • 作为第一个声明，我应该使用have而不是assume. 这不是一个新的假设，只是现有的假设。
  • 而不是finally我应该使用ultimately. 似乎后者的应用逻辑更简单。

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