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