section \<open>Peano's axioms for Natural Numbers\<close>
theory Peano_Axioms
imports Main
begin
locale peano =
fixes zero :: 'n
fixes succ :: "'n \<Rightarrow> 'n"
assumes succ_neq_zero [simp]: "succ m \<noteq> zero"
assumes succ_inject [simp]: "succ m = succ n \<longleftrightarrow> m = n"
assumes induct [case_names zero succ, induct type: 'n]:
"P zero \<Longrightarrow> (\<And>n. P n \<Longrightarrow> P (succ n)) \<Longrightarrow> P n"
begin
lemma zero_neq_succ [simp]: "zero \<noteq> succ m"
by (rule succ_neq_zero [symmetric])
text \<open>\<^medskip> Primitive recursion as a (functional) relation -- polymorphic!\<close>
inductive Rec :: "'a \<Rightarrow> ('n \<Rightarrow> 'a \<Rightarrow> 'a) \<Rightarrow> 'n \<Rightarrow> 'a \<Rightarrow> bool"
for e :: 'a and r :: "'n \<Rightarrow> 'a \<Rightarrow> 'a"
where
Rec_zero: "Rec e r zero e"
| Rec_succ: "Rec e r m n \<Longrightarrow> Rec e r (succ m) (r m n)"
lemma Rec_functional: "\<exists>!y::'a. Rec e r x y" for x :: 'n
proof -
let ?R = "Rec e r"
show ?thesis
proof (induct x)
case zero
show "\<exists>!y. ?R zero y"
proof
show "?R zero e" ..
show "y = e" if "?R zero y" for y
using that by cases simp_all
qed
next
case (succ m)
from \<open>\<exists>!y. ?R m y\<close>
obtain y where y: "?R m y" and yy': "\<And>y'. ?R m y' \<Longrightarrow> y = y'"
by blast
show "\<exists>!z. ?R (succ m) z"
proof
from y show "?R (succ m) (r m y)" ..
next
fix z
assume "?R (succ m) z"
then obtain u where "z = r m u" and "?R m u"
by cases simp_all
with yy' show "z = r m y"
by (simp only:)
qed
qed
qed
text \<open>\<^medskip> The recursion operator -- polymorphic!\<close>
definition rec :: "'a \<Rightarrow> ('n \<Rightarrow> 'a \<Rightarrow> 'a) \<Rightarrow> 'n \<Rightarrow> 'a"
where "rec e r x = (THE y. Rec e r x y)"
lemma rec_eval:
assumes Rec: "Rec e r x y"
shows "rec e r x = y"
unfolding rec_def
using Rec_functional and Rec by (rule the1_equality)
lemma rec_zero [simp]: "rec e r zero = e"
proof (rule rec_eval)
show "Rec e r zero e" ..
qed
lemma rec_succ [simp]: "rec e r (succ m) = r m (rec e r m)"
proof (rule rec_eval)
let ?R = "Rec e r"
have "?R m (rec e r m)"
unfolding rec_def using Rec_functional by (rule theI')
then show "?R (succ m) (r m (rec e r m))" ..
qed
text \<open>\<^medskip> Example: addition (monomorphic)\<close>
definition add :: "'n \<Rightarrow> 'n \<Rightarrow> 'n"
where "add m n = rec n (\<lambda>_ k. succ k) m"
lemma add_zero [simp]: "add zero n = n"
and add_succ [simp]: "add (succ m) n = succ (add m n)"
unfolding add_def by simp_all
lemma add_assoc: "add (add k m) n = add k (add m n)"
by (induct k) simp_all
lemma add_zero_right: "add m zero = m"
by (induct m) simp_all
lemma add_succ_right: "add m (succ n) = succ (add m n)"
by (induct m) simp_all
lemma "add (succ (succ (succ zero))) (succ (succ zero)) =
succ (succ (succ (succ (succ zero))))"
by simp
text \<open>\<^medskip> Example: replication (polymorphic)\<close>
definition repl :: "'n \<Rightarrow> 'a \<Rightarrow> 'a list"
where "repl n x = rec [] (\<lambda>_ xs. x # xs) n"
lemma repl_zero [simp]: "repl zero x = []"
and repl_succ [simp]: "repl (succ n) x = x # repl n x"
unfolding repl_def by simp_all
lemma "repl (succ (succ (succ zero))) True = [True, True, True]"
by simp
end
text \<open>\<^medskip> Just see that our abstract specification makes sense \dots\<close>
interpretation peano 0 Suc
proof
fix m n
show "Suc m \<noteq> 0" by simp
show "Suc m = Suc n \<longleftrightarrow> m = n" by simp
show "P n"
if zero: "P 0"
and succ: "\<And>n. P n \<Longrightarrow> P (Suc n)"
for P
proof (induct n)
case 0
show ?case by (rule zero)
next
case Suc
then show ?case by (rule succ)
qed
qed
end