(* Title: FOL/ex/Natural_Numbers.thy ID: $Id: Natural_Numbers.thy,v 1.10 2008/04/18 21:49:40 wenzelm Exp $ Author: Markus Wenzel, TU Munich *) header {* Natural numbers *} theory Natural_Numbers imports FOL begin text {* Theory of the natural numbers: Peano's axioms, primitive recursion. (Modernized version of Larry Paulson's theory "Nat".) \medskip *} typedecl nat arities nat :: "term" axiomatization Zero :: nat ("0") and Suc :: "nat => nat" and rec :: "[nat, 'a, [nat, 'a] => 'a] => 'a" where induct [case_names 0 Suc, induct type: nat]: "P(0) ==> (!!x. P(x) ==> P(Suc(x))) ==> P(n)" and Suc_inject: "Suc(m) = Suc(n) ==> m = n" and Suc_neq_0: "Suc(m) = 0 ==> R" and rec_0: "rec(0, a, f) = a" and rec_Suc: "rec(Suc(m), a, f) = f(m, rec(m, a, f))" lemma Suc_n_not_n: "Suc(k) ≠ k" proof (induct k) show "Suc(0) ≠ 0" proof assume "Suc(0) = 0" then show False by (rule Suc_neq_0) qed next fix n assume hyp: "Suc(n) ≠ n" show "Suc(Suc(n)) ≠ Suc(n)" proof assume "Suc(Suc(n)) = Suc(n)" then have "Suc(n) = n" by (rule Suc_inject) with hyp show False by contradiction qed qed definition add :: "[nat, nat] => nat" (infixl "+" 60) where "m + n = rec(m, n, λx y. Suc(y))" lemma add_0 [simp]: "0 + n = n" unfolding add_def by (rule rec_0) lemma add_Suc [simp]: "Suc(m) + n = Suc(m + n)" unfolding add_def by (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 assumes "!!n. f(Suc(n)) = Suc(f(n))" shows "f(i + j) = i + f(j)" using assms by (induct i) simp_all end
lemma Suc_n_not_n:
Suc(k) ≠ k
lemma add_0:
0 + n = n
lemma add_Suc:
Suc(m) + n = Suc(m + n)
lemma add_assoc:
k + m + n = k + (m + n)
lemma add_0_right:
m + 0 = m
lemma add_Suc_right:
m + Suc(n) = Suc(m + n)
lemma
(!!n. f(Suc(n)) = Suc(f(n))) ==> f(i + j) = i + f(j)