问题描述
Isabelle 是否可以定义一个终止递归函数 f
where
例如考虑理论Trie_Fun中定义的数据类型trie
:
datatype 'a trie = Nd bool "'a ⇒ 'a trie option"
以及我尝试使用一个简单的函数 height
来计算尝试的高度(具有有限多的出边):
theory Scratch
imports "HOL-Data_Structures.Trie_Fun"
begin
function height :: "'a trie ⇒ nat" where
"height (Nd _ edges) = (if dom edges = Set.empty ∨ ¬ finite (dom edges)
then 0
else 1 + Max (height ` ran edges))"
by pat_completeness auto
termination (* ??? *)
end
此处 lexicographic_order
不足以证明该函数是终止的,但到目前为止,我还无法在 trie
(用于终止)上制定任何本身不需要类似的度量递归。
在此我必须承认,我不确定我是否正确理解了 Isabelle/HOL 中的数据类型(即,上述定义的 trie
是否实际上总是有限高度)。
是否可以证明 height
终止?
解决方法
根据 Peter Zeller 的评论,我能够通过在定义中添加 height
然后使用事实 (domintros)
对 trie
进行归纳来证明 height.domintros
的终止1}},产生以下终止证明:
function (domintros) height :: "'a trie ⇒ nat" where
"height (Nd _ edges) = (if dom edges = Set.empty ∨ ¬ finite (dom edges)
then 0
else 1 + Max (height ` ran edges))"
by pat_completeness auto
termination apply auto
proof -
fix x :: "'a trie"
show "height_dom x"
proof (induction)
case (Nd b edges)
have "(⋀x. x ∈ ran edges ⟹ height_dom x)"
proof -
fix x assume "x ∈ ran edges"
then have "∃a. edges a = Some x"
unfolding ran_def by blast
then have "∃a. Some x = edges a"
by (metis (no_types))
then have "Some x ∈ range edges"
by blast
then show "height_dom x"
using Nd by auto
qed
then show ?case
using height.domintros by blast
qed
qed