author | wenzelm |
Tue, 12 Sep 2000 22:13:23 +0200 | |
changeset 9941 | fe05af7ec816 |
parent 9906 | 5c027cca6262 |
child 10007 | 64bf7da1994a |
permissions | -rw-r--r-- |
8020 | 1 |
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header {* An old chestnut *}; |
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theory Puzzle = Main:; |
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text_raw {* |
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\footnote{A question from ``Bundeswettbewerb Mathematik''. |
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Original pen-and-paper proof due to Herbert Ehler; Isabelle tactic |
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script by Tobias Nipkow.} |
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*}; |
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subsection {* Generalized mathematical induction *}; |
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text {* |
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The following derived rule admits induction over some expression |
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$f(x)$ wrt.\ the ${<}$ relation on natural numbers. |
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*}; |
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lemma gen_less_induct: |
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"(!!x. ALL y. f y < f x --> P y (f y) ==> P x (f x)) |
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==> P x (f x :: nat)" |
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(is "(!!x. ?H x ==> ?C x) ==> _"); |
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proof -; |
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assume asm: "!!x. ?H x ==> ?C x"; |
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{; |
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fix k; |
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have "ALL x. k = f x --> ?C x" (is "?Q k"); |
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proof (rule nat_less_induct); |
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fix k; assume hyp: "ALL m<k. ?Q m"; |
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show "?Q k"; |
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proof; |
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fix x; show "k = f x --> ?C x"; |
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proof; |
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assume "k = f x"; |
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with hyp; have "?H x"; by blast; |
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thus "?C x"; by (rule asm); |
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qed; |
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qed; |
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qed; |
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}; |
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thus "?C x"; by simp; |
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qed; |
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subsection {* The problem *}; |
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text {* |
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Given some function $f\colon \Nat \to \Nat$ such that $f \ap (f \ap |
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n) < f \ap (\idt{Suc} \ap n)$ for all $n$. Demonstrate that $f$ is |
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the identity. |
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*}; |
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consts f :: "nat => nat"; |
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axioms f_ax: "f (f n) < f (Suc n)"; |
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theorem "f n = n"; |
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proof (rule order_antisym); |
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txt {* |
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Note that the generalized form of $n \le f \ap n$ is required |
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later for monotonicity as well. |
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*}; |
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show ge: "!!n. n <= f n"; |
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proof -; |
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fix n; |
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show "?thesis n" (is "?P n (f n)"); |
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proof (rule gen_less_induct [of f ?P]); |
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fix n; assume hyp: "ALL m. f m < f n --> ?P m (f m)"; |
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show "?P n (f n)"; |
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proof (rule nat.exhaust); |
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assume "n = 0"; thus ?thesis; by simp; |
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next; |
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fix m; assume n_Suc: "n = Suc m"; |
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from f_ax; have "f (f m) < f (Suc m)"; .; |
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with hyp n_Suc; have "f m <= f (f m)"; by blast; |
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also; from f_ax; have "... < f (Suc m)"; .; |
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finally; have lt: "f m < f (Suc m)"; .; |
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with hyp n_Suc; have "m <= f m"; by blast; |
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also; note lt; |
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finally; have "m < f (Suc m)"; .; |
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thus "n <= f n"; by (simp only: n_Suc); |
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qed; |
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qed; |
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qed; |
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txt {* |
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In order to show the other direction, we first establish |
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monotonicity of $f$. |
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*}; |
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have mono: "!!m n. m <= n --> f m <= f n"; |
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proof -; |
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fix m n; |
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show "?thesis m n" (is "?P n"); |
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proof (induct n); |
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show "?P 0"; by simp; |
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fix n; assume hyp: "?P n"; |
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show "?P (Suc n)"; |
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proof; |
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assume "m <= Suc n"; |
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thus "f m <= f (Suc n)"; |
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proof (rule le_SucE); |
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assume "m <= n"; |
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with hyp; have "f m <= f n"; ..; |
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also; from ge f_ax; have "... < 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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thus ?thesis; by simp; |
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qed; |
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qed; |
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qed; |
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qed; |
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show "f n <= n"; |
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proof (rule leI); |
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show "~ n < f n"; |
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proof; |
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assume "n < f n"; |
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hence "Suc n <= f n"; by (rule Suc_leI); |
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9941
fe05af7ec816
renamed atts: rulify to rule_format, elimify to elim_format;
wenzelm
parents:
9906
diff
changeset
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hence "f (Suc n) <= f (f n)"; by (rule mono [rule_format]); |
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also; have "... < f (Suc n)"; by (rule f_ax); |
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finally; have "... < ..."; .; thus False; ..; |
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qed; |
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qed; |
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qed; |
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end; |