diff --git a/src/HOL/Examples/Ackermann.thy b/src/HOL/Examples/Ackermann.thy --- a/src/HOL/Examples/Ackermann.thy +++ b/src/HOL/Examples/Ackermann.thy @@ -1,102 +1,92 @@ (* Title: HOL/Examples/Ackermann.thy Author: Larry Paulson *) section \A Tail-Recursive, Stack-Based Ackermann's Function\ theory Ackermann imports Main begin text\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.\ fun ack :: "[nat,nat] \ 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\Here is the stack-based version, which uses lists.\ function (domintros) ackloop :: "nat list \ 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\ 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.\ text\@{thm [display] ackloop.domintros[no_vars]}\ declare ackloop.domintros [simp] text \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.\ -lemma "\n l. ackloop_dom (ack m n # l) \ ackloop_dom (n # m # l)" -proof (induction m) +lemma "ackloop_dom (ack m n # l) \ ackloop_dom (n # m # l)" +proof (induction m arbitrary: n l) case 0 then show ?case by auto next case (Suc m) - note IH = Suc - have "\l. ackloop_dom (ack (Suc m) n # l) \ 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 + show ?case + using Suc.prems + by (induction n arbitrary: l) (simp_all add: Suc) qed text \The proof above (which actually is unused) can be expressed concisely as follows.\ lemma ackloop_dom_longer: "ackloop_dom (ack m n # l) \ ackloop_dom (n # m # l)" by (induction m n arbitrary: l rule: ack.induct) auto lemma "ackloop_dom (ack m n # l) \ ackloop_dom (n # m # l)" by (induction m n arbitrary: l rule: ack.induct) auto text\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.\ fun acklist :: "nat list \ nat" where "acklist (n#m#l) = acklist (ack m n # l)" | "acklist [m] = m" | "acklist [] = 0" text\The induction rule for @{term acklist} is @{thm [display] acklist.induct[no_vars]}.\ 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\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.\ 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