Adapted to changes in induct method.
header {* An old chestnut *}
theory Puzzle
imports Main
begin
text_raw {*
\footnote{A question from ``Bundeswettbewerb Mathematik''. Original
pen-and-paper proof due to Herbert Ehler; Isabelle tactic script by
Tobias Nipkow.}
*}
text {*
\textbf{Problem.} Given some function $f\colon \Nat \to \Nat$ such
that $f \ap (f \ap n) < f \ap (\idt{Suc} \ap n)$ for all $n$.
Demonstrate that $f$ is the identity.
*}
theorem
assumes f_ax: "\<And>n. f (f n) < f (Suc n)"
shows "f n = n"
proof (rule order_antisym)
{
fix n show "n \<le> f n"
proof (induct "f n" arbitrary: n rule: less_induct)
case less
show "n \<le> f n"
proof (cases n)
case (Suc m)
from f_ax have "f (f m) < f n" by (simp only: Suc)
with less have "f m \<le> f (f m)" .
also from f_ax have "\<dots> < f n" by (simp only: Suc)
finally have "f m < f n" .
with less have "m \<le> f m" .
also note `\<dots> < f n`
finally have "m < f n" .
then have "n \<le> f n" by (simp only: Suc)
then show ?thesis .
next
case 0
then show ?thesis by simp
qed
qed
} note ge = this
{
fix m n :: nat
assume "m \<le> n"
then have "f m \<le> f n"
proof (induct n)
case 0
then have "m = 0" by simp
then show ?case by simp
next
case (Suc n)
from Suc.prems show "f m \<le> f (Suc n)"
proof (rule le_SucE)
assume "m \<le> n"
with Suc.hyps have "f m \<le> f n" .
also from ge f_ax have "\<dots> < f (Suc n)"
by (rule le_less_trans)
finally show ?thesis by simp
next
assume "m = Suc n"
then show ?thesis by simp
qed
qed
} note mono = this
show "f n \<le> n"
proof -
have "\<not> n < f n"
proof
assume "n < f n"
then have "Suc n \<le> f n" by simp
then have "f (Suc n) \<le> f (f n)" by (rule mono)
also have "\<dots> < f (Suc n)" by (rule f_ax)
finally have "\<dots> < \<dots>" . then show False ..
qed
then show ?thesis by simp
qed
qed
end