基本的Isabelle/Isar风格(练习4.6)

Question

我有兴趣使用Isabelle/Isar来编写既可读又经机器检查的校样，我希望改进我的风格并精简我的校样。

prog-证明有以下练习：

练习4.6.定义递归函数elems :: 'a list ⇒ 'a set并证明x ∈ elems xs ⟹ ∃ ys zs. xs = ys @ x # zs ∧ x ∉ elems ys。

模仿类似于我用钢笔和纸写的东西，我的解决方案是

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

但看起来挺长的。特别是，我认为这里引用被排除的中间规则是很麻烦的，我觉得应该有一些方便的示意图变量，比如?goal，它可以引用当前的目标或什么的。

我如何在不牺牲清晰性的情况下缩短这个证据？

Answer 1

你的具体问题并没有得到真正的回答，但我想指出，一个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)

Answer 2

这里还有一个比你自己的证据更短的证据：

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的一些特性来压缩证据：

1. 大括号中的块{...}允许您执行假设推理。
2. 可以使用note显式地命名事实。
3. moreover关键字启动一个计算，该计算在建立事实时隐含地“携带”事实。使用ultimately关键字的计算“到达一个头”。这种样式可以显著减少在验证过程中需要引入的显式命名事实的数量。
4. qed auto通过将auto应用于所有剩余的子目标来完成验证。一个注释指出，剩下的子目标是归纳的基本情况，这是微不足道的。