# HG changeset patch
# User wenzelm
# Date 1002201931 -7200
# Node ID 79e5536af6c48129e53e6d56f4664a4755aebef0
# Parent  8e75b78f33f3ae353d70493023e539f20dbc59ba
Theory of the natural numbers: Peano's axioms, primitive recursion.
(Modernized version of Larry Paulson's theory "Nat".)

diff -r 8e75b78f33f3 -r 79e5536af6c4 src/FOL/ex/Natural_Numbers.thy
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/src/FOL/ex/Natural_Numbers.thy	Thu Oct 04 15:25:31 2001 +0200
@@ -0,0 +1,69 @@
+(*  Title:      FOL/ex/Natural_Numbers.thy
+    ID:         $Id$
+    Author:     Markus Wenzel, TU Munich
+
+Theory of the natural numbers: Peano's axioms, primitive recursion.
+(Modernized version of Larry Paulson's theory "Nat".)
+*)
+
+theory Natural_Numbers = FOL:
+
+typedecl nat
+arities nat :: "term"
+
+consts
+  Zero :: nat    ("0")
+  Suc :: "nat => nat"
+  rec :: "[nat, 'a, [nat, 'a] => 'a] => 'a"
+
+axioms
+  induct [induct type: nat]:
+    "P(0) \<Longrightarrow> (!!x. P(x) ==> P(Suc(x))) ==> P(n)"
+  Suc_inject: "Suc(m) = Suc(n) ==> m = n"
+  Suc_neq_0: "Suc(m) = 0 ==> R"
+  rec_0: "rec(0, a, f) = a"
+  rec_Suc: "rec(Suc(m), a, f) = f(m, rec(m, a, f))"
+
+lemma Suc_n_not_n: "Suc(k) \<noteq> k"
+proof (induct k)
+  show "Suc(0) \<noteq> 0"
+  proof
+    assume "Suc(0) = 0"
+    thus False by (rule Suc_neq_0)
+  qed
+  fix n assume hyp: "Suc(n) \<noteq> n"
+  show "Suc(Suc(n)) \<noteq> Suc(n)"
+  proof
+    assume "Suc(Suc(n)) = Suc(n)"
+    hence "Suc(n) = n" by (rule Suc_inject)
+    with hyp show False by contradiction
+  qed
+qed
+
+
+constdefs
+  add :: "[nat, nat] => nat"    (infixl "+" 60)
+  "m + n == rec(m, n, \<lambda>x y. Suc(y))"
+
+lemma add_0 [simp]: "0 + n = n"
+  by (unfold add_def) (rule rec_0)
+
+lemma add_Suc [simp]: "Suc(m) + n = Suc(m + n)"
+  by (unfold add_def) (rule rec_Suc)
+
+lemma add_assoc: "(k + m) + n = k + (m + n)"
+  by (induct k) simp_all
+
+lemma add_0_right: "m + 0 = m"
+  by (induct m) simp_all
+
+lemma add_Suc_right: "m + Suc(n) = Suc(m + n)"
+  by (induct m) simp_all
+
+lemma "(!!n. f(Suc(n)) = Suc(f(n))) ==> f(i + j) = i + f(j)"
+proof -
+  assume a: "!!n. f(Suc(n)) = Suc(f(n))"
+  show ?thesis by (induct i) (simp, simp add: a)  (* FIXME tune *)
+qed
+
+end