是否可以向 Isabelle 中的函数域添加假设？

Question

我不确定 post 这样的后续问题是否可以，但无论如何我都会这样做。

所以几天前我post编辑了这个问题：

我认为答案很好，但在试图证明引理时

lemma "applied1 {#''a'',''a'',''d'',''c'',''a'',''a'',''d'',''c''#} {#''a'',''a'',''c''#} ''f'' = {#''f'',''f'',''d'',''d''#}"

我真的卡住了。我发现在展开 def 并应用一些简单的自动化后我不能简单地做到这一点。所以我回到我原来的函数并对它做了一些调整，这样如果输入导致无限循环，它 returns 什么都没有。我以为这次会成功，但伊莎贝尔仍然无法证明终止。我敢肯定，很明显 size x 不断减少 size y 并且不能为负数，因此它最终必须在 size x = 0 或 y 不再是时终止x 的子集。

fun applied2 :: "'a multiset ⇒ 'a multiset ⇒ 'a ⇒ 'a multiset option" where
"applied2 x y z = (if z ∈# y ∨ y = {#} then None else (if y ⊆# x then Some (plus {#z#} (the (applied2 (x - y) y z))) else Some x))"

是否可以说服 Isabelle 使用 function 而不是 fun 来终止此函数？还是我必须考虑其他限制条件？

如果我不应该 post 提出这样的问题，我真的很抱歉。我对 Isabelle 仍然缺乏经验，我希望我能坚持自己的目标，尽我所能地学习这门语言。提前致谢！

Answer 1

我相信查看 documentation 会给您正确的语法。

function applied2 :: "'a multiset ⇒ 'a multiset ⇒ 'a ⇒ 'a multiset option" where
"applied2 x y z = (if z ∈# y ∨ y = {#} then None else (if y ⊆# x then Some (plus {#z#} (the (applied2 (x - y) y z))) else Some x))"
  by pat_completeness auto
termination
  by (relation "measure (λ(x,y,z). size x)")
    (auto simp: mset_subset_eq_exists_conv nonempty_has_size)

如果问题是证明，sledgehammer 会为您找到它。

但是，我看不出您打算如何从 applied2 转到您真正想要的功能。真正的问题是确定性：你需要一个命令来查看子集。 Manuel 的解决方案是使用 Sup，但这确实不可执行。

Answer 2

如果非递归定义的唯一问题是如何将其应用于具体输入，我仍然认为我所说的可执行的替代定义是可行的方法。下面证明我给的两个非递归定义是等价的，以及对你上面给的例子的应用：

definition applied :: "'a multiset ⇒ 'a multiset ⇒ 'a ⇒ 'a multiset" where
  "applied ms xs y = (if xs = {#} then ms else
     (let n = Max {n. repeat_mset n xs ⊆# ms}
      in ms - repeat_mset n xs + replicate_mset n y))"

lemma count_le_size: "count M x ≤ size M"
  by (induction M) auto

lemma applied_code [code]:
  "applied ms xs y = (if xs = {#} then ms else
     (let n = (MIN x ∈set_mset xs. count ms x div count xs x)
      in ms - repeat_mset n xs + replicate_mset n y))"
  unfolding applied_def
proof (intro if_cong let_cong refl)
  assume ne: "xs ≠ {#}"
  have subset: "{n. repeat_mset n xs ⊆# ms} ⊆ {..size ms}"
  proof safe
    fix n assume n: "repeat_mset n xs ⊆# ms"
    from ne obtain x where x: "x ∈# xs"
      by auto
    have "n * 1 ≤ n * count xs x"
      using x by (intro mult_left_mono) auto
    also have "… = count (repeat_mset n xs) x"
      by simp
    also have "… ≤ count ms x"
      using n by (intro mset_subset_eq_count)
    also have "… ≤ size ms"
      by (rule count_le_size)
    finally show "n ≤ size ms" by simp
  qed
  hence finite: "finite {n. repeat_mset n xs ⊆# ms}"
    by (rule finite_subset) auto

  show "Max {n. repeat_mset n xs ⊆# ms} = (MIN x∈set_mset xs. count ms x div count xs x)"
  proof (intro antisym)
    show "Max {n. repeat_mset n xs ⊆# ms} ≤ (MIN x∈set_mset xs. count ms x div count xs x)"
    proof (rule Max.boundedI)
      show "{n. repeat_mset n xs ⊆# ms} ≠ {}"
        by (auto intro: exI[of _ 0])
    next
      fix n assume n: "n ∈ {n. repeat_mset n xs ⊆# ms}"
      show "n ≤ (MIN x∈set_mset xs. count ms x div count xs x)"
      proof (safe intro!: Min.boundedI)
        fix x assume x: "x ∈# xs"
        have "count (repeat_mset n xs) x ≤ count ms x"
          using n by (intro mset_subset_eq_count) auto
        also have "count (repeat_mset n xs) x = n * count xs x"
          by simp
        finally show "n ≤ count ms x div count xs x"
          by (metis count_eq_zero_iff div_le_mono nonzero_mult_div_cancel_right x)
      qed (use ne in auto)
    qed (fact finite)
  next
    define m where "m = (MIN x∈set_mset xs. count ms x div count xs x)"
    show "m ≤ Max {n. repeat_mset n xs ⊆# ms}"
    proof (rule Max.coboundedI[OF finite], safe)
      show "repeat_mset m xs ⊆# ms"
      proof (rule mset_subset_eqI)
        fix x
        show "count (repeat_mset m xs) x ≤ count ms x"
        proof (cases "x ∈# xs")
          case True
          have "count (repeat_mset m xs) x = m * count xs x"
            by simp
          also have "… ≤ (count ms x div count xs x) * count xs x"
            unfolding m_def using ‹x ∈# xs› by (intro mult_right_mono Min.coboundedI) auto
          also have "… ≤ count ms x"
            by simp
          finally show ?thesis .
        next
          case False
          hence "count xs x = 0"
            by (meson not_in_iff)
          thus ?thesis by simp
        qed
      qed
    qed
  qed
qed

lemma replicate_mset_unfold:
  assumes "n > 0"
  shows   "replicate_mset n x = {#x#} + replicate_mset (n - 1) x"
  using assms by (cases n) auto

lemma
  assumes "a ≠ c" "a ≠ f" "c ≠ f"
  shows   "applied {#a,a,c,a,a,c#} {#a,a,c#} f = mset [f, f]"
  using assms
  by (simp add: applied_code replicate_mset_unfold flip: One_nat_def)

value 命令对该示例不起作用，因为 a、c 等是自由变量。但是如果你例如为他们制作一个特别的数据类型，它有效：

datatype test = a | b | c | f

value "applied {#a,a,c,a,a,c#} {#a,a,c#} f"
(* "mset [f, f]" :: "test multiset" *)

是否可以向 Isabelle 中的函数域添加假设？

Is it possible to add assumptions to the domain of a function in Isabelle?

function

functional-programming

isabelle

multiset