基本Isabelle/Isar风格（练习4.6）

我对使用Isabelle/Isar编写既可供人阅读又可由机器检查的校样感兴趣，我希望改进我的风格并简化我的校样

prog prove有以下练习：

练习4.6.定义递归函数

elems:：'列表⇒ '一套

和证明x∈ 元素xs⟹ ∃ ys zs。xs=ys@x#zs∧ x∉ elems ys

我的解决方案是模仿我用钢笔和纸写的东西

fun elems :: "'a list ⇒ 'a set" where
"elems [] = {}" |
"elems (x # xs) = {x} ∪ elems xs"

fun takeUntil :: "('a ⇒ bool) ⇒ 'a list ⇒ 'a list" where
"takeUntil f [] = []" |
"takeUntil f (x # xs) = (case (f x) of False ⇒ x # takeUntil f xs | True ⇒ [])"

theorem "x ∈ elems xs ⟹ ∃ ys zs. xs = ys @ x # zs ∧ x ∉ elems ys"
proof -
  assume 1: "x ∈ elems xs"
  let ?ys = "takeUntil (λ z. z = x) xs"
  let ?zs = "drop (length ?ys + 1) xs"
  have "xs = ?ys @ x # ?zs ∧ x ∉ elems ?ys"
  proof
    have 2: "x ∉ elems ?ys"
    proof (induction xs)
      case Nil
      thus ?case by simp
    next
      case (Cons a xs)
      thus ?case
      proof -
        {
          assume "a = x"
          hence "takeUntil (λz. z = x) (a # xs) = []" by simp
          hence A: ?thesis by simp
        }
        note eq = this
        {
          assume "a ≠ x"
          hence "takeUntil (λz. z = x) (a # xs) = a # takeUntil (λz. z = x) xs" by simp
          hence ?thesis using Cons.IH by auto
        }
        note noteq = this
        have "a = x ∨ a ≠ x" by simp
        thus ?thesis using eq noteq by blast
      qed
    qed

    from 1 have "xs = ?ys @ x # ?zs"
    proof (induction xs)
      case Nil
      hence False by simp
      thus ?case by simp
    next
      case (Cons a xs)
      {
        assume 1: "a = x"
        hence 2: "takeUntil (λz. z = x) (a # xs) = []" by simp
        hence "length (takeUntil (λz. z = x) (a # xs)) + 1 = 1" by simp
        hence 3: "drop (length (takeUntil (λz. z = x) (a # xs)) + 1) (a # xs) = xs" by simp
        from 1 2 3 have ?case by simp
      }
      note eq = this
      {
        assume 1: "a ≠ x"
        with Cons.prems have "x ∈ elems xs" by simp
        with Cons.IH
         have IH: "xs = takeUntil (λz. z = x) xs @ x # drop (length (takeUntil (λz. z = x) xs) + 1) xs" by simp
        from 1 have 2: "takeUntil (λz. z = x) (a # xs) = a # takeUntil (λz. z = x) (xs)" by simp
        from 1 have "drop (length (takeUntil (λz. z = x) (a # xs)) + 1) (a # xs) = drop (length (takeUntil (λz. z = x) xs) + 1) xs" by simp
        hence ?case using IH 2 by simp
      }
      note noteq = this
      have "a = x ∨ a ≠ x" by simp
      thus ?case using eq noteq by blast
    qed
    with 2 have 3: ?thesis by blast
    thus "xs = takeUntil (λz. z = x) xs @ x # drop (length (takeUntil (λz. z = x) xs) + 1) xs" by simp
    from 3 show "x ∉ elems (takeUntil (λz. z = x) xs)" by simp
  qed
  thus ?thesis by blast
qed

但它似乎相当长。特别是，我认为在这里调用排除中间法则是很麻烦的，我觉得应该有一些方便的图解变量，比如

？goal

，它可以引用当前的目标或其他东西

---

如何在不牺牲清晰性的情况下缩短这个证明？

并不是对你的具体问题的回答，但我想指出，更简洁的证明仍然是可以理解的

lemma "x ∈ elems xs ⟹ ∃ ys zs. xs = ys @ x # zs ∧ x ∉ elems ys"
proof (induction)
  case (Cons l ls)
  thus ?case
  proof (cases "x ≠ l")
    case True
    hence "∃ys zs. ls = ys @ x # zs ∧ x ∉ elems ys" using Cons by simp
    thus ?thesis using ‹x ≠ l› Cons_eq_appendI by fastforce
  qed (fastforce)
qed (simp)

---

这是另一个比你自己的短的证据：

fun elems :: ‹'a list ⇒ 'a set› where
  ‹elems [] = {}› |
  ‹elems (x#xs) = {x} ∪ elems xs›

lemma elems_prefix_suffix:
  assumes ‹x ∈ elems xs›
  shows ‹∃pre suf. xs = pre @ [x] @ suf ∧ x ∉ elems pre›
using assms proof(induction xs)
  fix y ys
  assume *: ‹x ∈ elems (y#ys)›
    and IH: ‹x ∈ elems ys ⟹ ∃pre suf. ys = pre @ [x] @ suf ∧ x ∉ elems pre›
  {
    assume ‹x = y›
    from this have ‹∃pre suf. y#ys = pre @ [x] @ suf ∧ x ∉ elems pre›
      using * by fastforce
  }
  note L = this
  {
    assume ‹x ≠ y› and ‹x ∈ elems ys›
    moreover from this obtain pre and suf where ‹ys = pre @ [x] @ suf› and ‹x ∉ elems pre›
      using IH by auto
    moreover have ‹y#ys = y#pre @ [x] @ suf› and ‹x ∉ elems (y#pre)›
      by(simp add: calculation)+
    ultimately have ‹∃pre suf. y#ys = pre @ [x] @ suf ∧ x ∉ elems pre›
      by(metis append_Cons)
  }
  from this and L show ‹∃pre suf. y#ys = pre @ [x] @ suf ∧ x ∉ elems pre›
    using * by auto
qed auto ― ‹Base case trivial›

我使用了Isar的一些功能来压缩证据：

大括号中的块允许您执行假设推理

事实可以使用

注释

明确命名

此外，

关键字启动一个计算，该计算隐式地“携带”已建立的事实。计算“到达头部”，并带有
最终
关键字。这种风格可以显著减少在证明过程中需要引入的显式命名事实的数量

qed auto

通过对所有剩余的子目标应用

auto

来完成证明。一条评论指出，剩下的子目标是归纳的基本情况，这是微不足道的