(* Title: FOL/ex/Nat.thy ID: $Id: Nat.thy,v 1.7 2006/06/07 21:21:56 wenzelm Exp $ Author: Lawrence C Paulson, Cambridge University Computer Laboratory Copyright 1992 University of Cambridge *) header {* Theory of the natural numbers: Peano's axioms, primitive recursion *} theory Nat imports FOL begin typedecl nat arities nat :: "term" consts 0 :: nat ("0") Suc :: "nat => nat" rec :: "[nat, 'a, [nat,'a]=>'a] => 'a" add :: "[nat, nat] => nat" (infixl "+" 60) axioms induct: "[| P(0); !!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))" defs add_def: "m+n == rec(m, n, %x y. Suc(y))" subsection {* Proofs about the natural numbers *} lemma Suc_n_not_n: "Suc(k) ~= k" apply (rule_tac n = k in induct) apply (rule notI) apply (erule Suc_neq_0) apply (rule notI) apply (erule notE) apply (erule Suc_inject) done lemma "(k+m)+n = k+(m+n)" apply (rule induct) back back back back back back oops lemma add_0 [simp]: "0+n = n" apply (unfold add_def) apply (rule rec_0) done lemma add_Suc [simp]: "Suc(m)+n = Suc(m+n)" apply (unfold add_def) apply (rule rec_Suc) done lemma add_assoc: "(k+m)+n = k+(m+n)" apply (rule_tac n = k in induct) apply simp apply simp done lemma add_0_right: "m+0 = m" apply (rule_tac n = m in induct) apply simp apply simp done lemma add_Suc_right: "m+Suc(n) = Suc(m+n)" apply (rule_tac n = m in induct) apply simp_all done lemma assumes prem: "!!n. f(Suc(n)) = Suc(f(n))" shows "f(i+j) = i+f(j)" apply (rule_tac n = i in induct) apply simp apply (simp add: prem) done 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)