src/HOL/Examples/Ackermann.thy

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(*  Title:      HOL/Examples/Ackermann.thy
    Author:     Larry Paulson
*)

section \<open>A Tail-Recursive, Stack-Based Ackermann's Function\<close>

theory Ackermann imports Main

begin

text\<open>This theory investigates a stack-based implementation of Ackermann's function.
Let's recall the traditional definition,
as modified by R{\'o}zsa P\'eter and Raphael Robinson.\<close>

fun ack :: "[nat,nat] \<Rightarrow> nat" where
  "ack 0 n             = Suc n"
| "ack (Suc m) 0       = ack m 1"
| "ack (Suc m) (Suc n) = ack m (ack (Suc m) n)"

text\<open>Here is the stack-based version, which uses lists.\<close>

function (domintros) ackloop :: "nat list \<Rightarrow> nat" where
  "ackloop (n # 0 # l)         = ackloop (Suc n # l)"
| "ackloop (0 # Suc m # l)     = ackloop (1 # m # l)"
| "ackloop (Suc n # Suc m # l) = ackloop (n # Suc m # m # l)"
| "ackloop [m] = m"
| "ackloop [] =  0"
  by pat_completeness auto

text\<open>
The key task is to prove termination. In the first recursive call, the head of the list gets bigger
while the list gets shorter, suggesting that the length of the list should be the primary
termination criterion. But in the third recursive call, the list gets longer. The idea of trying
a multiset-based termination argument is frustrated by the second recursive call when m = 0:
the list elements are simply permuted.

Fortunately, the function definition package allows us to define a function and only later identify its domain of termination.
Instead, it makes all the recursion equations conditional on satisfying
the function's domain predicate. Here we shall eventually be able
to show that the predicate is always satisfied.\<close>

text\<open>@{thm [display] ackloop.domintros[no_vars]}\<close>
declare ackloop.domintros [simp]

text \<open>Termination is trivial if the length of the list is less then two.
The following lemma is the key to proving termination for longer lists.\<close>
lemma "\<And>n l. ackloop_dom (ack m n # l) \<Longrightarrow> ackloop_dom (n # m # l)"
proof (induction m)
  case 0
  then show ?case
    by auto
next
  case (Suc m)
  note IH = Suc
  have "\<And>l. ackloop_dom (ack (Suc m) n # l) \<Longrightarrow> ackloop_dom (n # Suc m # l)"
  proof (induction n)
    case 0
    then show ?case
      by (simp add: IH)
  next
    case (Suc n)
    then show ?case
      by (auto simp: IH)
  qed
  then show ?case
    using Suc.prems by blast
qed

text \<open>The proof above (which actually is unused) can be expressed concisely as follows.\<close>
lemma ackloop_dom_longer:
  "ackloop_dom (ack m n # l) \<Longrightarrow> ackloop_dom (n # m # l)"
  by (induction m n arbitrary: l rule: ack.induct) auto

lemma "ackloop_dom (ack m n # l) \<Longrightarrow> ackloop_dom (n # m # l)"
  by (induction m n arbitrary: l rule: ack.induct) auto

text\<open>This function codifies what @{term ackloop} is designed to do.
Proving the two functions equivalent also shows that @{term ackloop} can be used
to compute Ackermann's function.\<close>
fun acklist :: "nat list \<Rightarrow> nat" where
  "acklist (n#m#l) = acklist (ack m n # l)"
| "acklist [m] = m"
| "acklist [] =  0"

text\<open>The induction rule for @{term acklist} is @{thm [display] acklist.induct[no_vars]}.\<close>

lemma ackloop_dom: "ackloop_dom l"
  by (induction l rule: acklist.induct) (auto simp: ackloop_dom_longer)

termination ackloop
  by (simp add: ackloop_dom)

text\<open>This result is trivial even by inspection of the function definitions
(which faithfully follow the definition of Ackermann's function).
All that we needed was termination.\<close>
lemma ackloop_acklist: "ackloop l = acklist l"
  by (induction l rule: ackloop.induct) auto

theorem ack: "ack m n = ackloop [n,m]"
  by (simp add: ackloop_acklist)

end