src/HOL/Examples/Ackermann.thy

A further round of proof consolidation

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

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

theory Ackermann imports "HOL-Library.Multiset_Order" "HOL-Library.Product_Lexorder" 

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)"

subsection \<open>Example of proving termination by reasoning about the domain\<close>

text\<open>The stack-based version 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 "ackloop_dom (ack m n # l) \<Longrightarrow> ackloop_dom (n # m # l)"
proof (induction m arbitrary: n l)
  case 0
  then show ?case
    by auto
next
  case (Suc m)
  show ?case
    using Suc.prems
    by (induction n arbitrary: l) (simp_all add: Suc)
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

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)

subsection \<open>Example of proving termination using a multiset ordering\<close>

text \<open>This termination proof uses the argument from
Nachum Dershowitz and Zohar Manna. Proving termination with multiset orderings.
Communications of the ACM 22 (8) 1979, 465–476.\<close>

text\<open>Setting up the termination proof. Note that Dershowitz had @{term z} as a global variable.
The top two stack elements are treated differently from the rest.\<close>

fun ack_mset :: "nat list \<Rightarrow> (nat\<times>nat) multiset" where
  "ack_mset [] = {#}"
| "ack_mset [x] = {#}"
| "ack_mset (z#y#l) = mset ((y,z) # map (\<lambda>x. (Suc x, 0)) l)"

lemma case1: "ack_mset (Suc n # l) < add_mset (0,n) {# (Suc x, 0). x \<in># mset l #}"
proof (cases l)
  case (Cons m list)
  have "{#(m, Suc n)#} < {#(Suc m, 0)#}"
    by auto
  also have "\<dots> \<le> {#(Suc m, 0), (0,n)#}"
    by auto
  finally show ?thesis  
    by (simp add: Cons)
qed auto

text\<open>The stack-based version again. We need a fresh copy because 
  we've already proved the termination of @{term ackloop}.\<close>

function 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>In each recursive call, the function @{term ack_mset} decreases according to the multiset
ordering.\<close>
termination
  by (relation "inv_image {(x,y). x<y} ack_mset") (auto simp: wf case1)

text \<open>Another shortcut compared with before: equivalence follows directly from this lemma.\<close>
lemma Ackloop_ack: "Ackloop (n # m # l) = Ackloop (ack m n # l)"
  by (induction m n arbitrary: l rule: ack.induct) auto

theorem "ack m n = Ackloop [n,m]"
  by (simp add: Ackloop_ack)

end