(* Title: FOL/ex/Natural_Numbers.thy
Author: Markus Wenzel, TU Munich
*)
section \<open>Natural numbers\<close>
theory Natural_Numbers
imports FOL
begin
text \<open>
Theory of the natural numbers: Peano's axioms, primitive recursion.
(Modernized version of Larry Paulson's theory "Nat".) \medskip
\<close>
typedecl nat
instance 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) \<noteq> k"
proof (induct k)
show "Suc(0) \<noteq> 0"
proof
assume "Suc(0) = 0"
then show False by (rule Suc_neq_0)
qed
next
fix n assume hyp: "Suc(n) \<noteq> n"
show "Suc(Suc(n)) \<noteq> 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, \<lambda>x 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