# HG changeset patch
# User nipkow
# Date 1218180375 -7200
# Node ID 1bf827e3258d3daed9891a6b8ad7706bba5af618
# Parent  5def98c2646b2e9233c90281d1769853d4986e88
added lemmas

diff -r 5def98c2646b -r 1bf827e3258d src/HOL/Nat.thy
--- a/src/HOL/Nat.thy	Thu Aug 07 23:56:45 2008 +0200
+++ b/src/HOL/Nat.thy	Fri Aug 08 09:26:15 2008 +0200
@@ -1309,9 +1309,24 @@
 using `n < n'`
 by (induct n n' rule: less_Suc_induct[consumes 1]) (auto intro: mono)
 
+lemma lift_Suc_mono_less_iff:
+  "(!!n. f n < f(Suc n)) \<Longrightarrow> f(n) < f(m) \<longleftrightarrow> n<m"
+by(blast intro: less_asym' lift_Suc_mono_less[of f]
+         dest: linorder_not_less[THEN iffD1] le_eq_less_or_eq[THEN iffD1])
+
 end
 
 
+lemma mono_nat_linear_lb:
+  "(!!m n::nat. m<n \<Longrightarrow> f m < f n) \<Longrightarrow> f(m)+k \<le> f(m+k)"
+apply(induct_tac k)
+ apply simp
+apply(erule_tac x="m+n" in meta_allE)
+apply(erule_tac x="m+n+1" in meta_allE)
+apply simp
+done
+
+
 text{*Subtraction laws, mostly by Clemens Ballarin*}
 
 lemma diff_less_mono: "[| a < (b::nat); c \<le> a |] ==> a-c < b-c"
diff -r 5def98c2646b -r 1bf827e3258d src/HOL/ex/Puzzle.thy
--- a/src/HOL/ex/Puzzle.thy	Thu Aug 07 23:56:45 2008 +0200
+++ b/src/HOL/ex/Puzzle.thy	Fri Aug 08 09:26:15 2008 +0200
@@ -1,47 +1,139 @@
 (*  Title:      HOL/ex/Puzzle.thy
     ID:         $Id$
     Author:     Tobias Nipkow
-    Copyright   1993 TU Muenchen
-
-A question from "Bundeswettbewerb Mathematik"
-
-Proof due to Herbert Ehler
 *)
 
-header {* A question from ``Bundeswettbewerb Mathematik'' *}
+header {* Fun with Functions *}
 
 theory Puzzle imports Main begin
 
-consts f :: "nat => nat"
+text{* From \emph{Solving Mathematical Problems} by Terence Tao. He quotes
+Greitzer's \emph{Interational Mathematical Olympiads 1959--1977} as the
+original source. Was first brought to our attention by Herbert Ehler who
+provided a similar proof. *}
+
 
-specification (f)
-  f_ax [intro!]: "f(f(n)) < f(Suc(n))"
-    by (rule exI [of _ id], simp)
+theorem identity1: fixes f :: "nat \<Rightarrow> nat"
+assumes fff: "!!n. f(f(n)) < f(Suc(n))"
+shows "f(n) = n"
+proof -
+  { fix m n have key: "n \<le> m \<Longrightarrow> n \<le> f(m)"
+    proof(induct n arbitrary: m)
+      case 0 show ?case by simp
+    next
+      case (Suc n)
+      hence "m \<noteq> 0" by simp
+      then obtain k where [simp]: "m = Suc k" by (metis not0_implies_Suc)
+      have "n \<le> f(k)" using Suc by simp
+      hence "n \<le> f(f(k))" using Suc by simp
+      also have "\<dots> < f(m)" using fff by simp
+      finally show ?case by simp
+    qed }
+  hence "!!n. n \<le> f(n)" by simp
+  hence "!!n. f(n) < f(Suc n)" by(metis fff order_le_less_trans)
+  hence "f(n) < n+1" by (metis fff lift_Suc_mono_less_iff[of f] Suc_plus1)
+  with `n \<le> f(n)` show "f n = n" by arith
+qed
 
 
-lemma lemma0 [rule_format]: "\<forall>n. k=f(n) --> n <= f(n)"
-apply (induct_tac "k" rule: nat_less_induct)
-apply (rule allI)
-apply (rename_tac "i")
-apply (case_tac "i")
- apply simp
-apply (blast intro!: Suc_leI intro: le_less_trans)
-done
-
-lemma lemma1: "n <= f(n)"
-by (blast intro: lemma0)
+text{* From \emph{Solving Mathematical Problems} by Terence Tao. He quotes
+the \emph{Australian Mathematics Competition} 1984 as the original source.
+Possible extension:
+Should also hold if the range of @{text f} is the reals!
+*}
 
-lemma f_mono [rule_format (no_asm)]: "m <= n --> f(m) <= f(n)"
-apply (induct_tac "n")
- apply simp 
- apply (metis f_ax le_SucE le_trans lemma0 nat_le_linear nat_less_le)
-done
+lemma identity2: fixes f :: "nat \<Rightarrow> nat"
+assumes "f(k) = k" and "k \<ge> 2"
+and f_times: "\<And>m n. f(m*n) = f(m)*f(n)"
+and f_mono: "\<And>m n. m<n \<Longrightarrow> f m < f n"
+shows "f(n) = n"
+proof -
+  have 0: "f(0) = 0"
+    by (metis f_mono f_times mult_1_right mult_is_0 nat_less_le nat_mult_eq_cancel_disj not_less_eq)
+  have 1: "f(1) = 1"
+    by (metis f_mono f_times gr_implies_not0 mult_eq_self_implies_10 nat_mult_1_right zero_less_one)
+  have 2: "f 2 = 2"
+  proof -
+    have "2 + (k - 2) = k" using `k \<ge> 2` by arith
+    hence "f(2) \<le> 2"
+      using mono_nat_linear_lb[of f 2 "k - 2",OF f_mono] `f k = k`
+      by simp arith
+    thus "f 2 = 2" using 1 f_mono[of 1 2] by arith
+  qed
+  show ?thesis
+  proof(induct rule:less_induct)
+    case (less i)
+    show ?case
+    proof cases
+      assume "i\<le>1" thus ?case using 0 1 by (auto simp add:le_Suc_eq)
+    next
+      assume "~i\<le>1"
+      show ?case
+      proof cases
+	assume "i mod 2 = 0"
+	hence "EX k. i=2*k" by arith
+	then obtain k where "i = 2*k" ..
+	hence "0 < k" and "k<i" using `~i\<le>1` by arith+
+	hence "f(k) = k" using less(1) by blast
+	thus "f(i) = i" using `i = 2*k` by(simp add:f_times 2)
+      next
+	assume "i mod 2 \<noteq> 0"
+	hence "EX k. i=2*k+1" by arith
+	then obtain k where "i = 2*k+1" ..
+	hence "0<k" and "k+1<i" using `~i\<le>1` by arith+
+	have "2*k < f(2*k+1)"
+	proof -
+	  have "2*k = 2*f(k)" using less(1) `i=2*k+1` by simp
+	  also have "\<dots> = f(2*k)" by(simp add:f_times 2)
+	  also have "\<dots> < f(2*k+1)" using f_mono[of "2*k" "2*k+1"] by simp
+	  finally show ?thesis .
+	qed
+	moreover
+	have "f(2*k+1) < 2*(k+1)"
+	proof -
+	  have "f(2*k+1) < f(2*k+2)" using f_mono[of "2*k+1" "2*k+2"] by simp
+	  also have "\<dots> = f(2*(k+1))" by simp
+	  also have "\<dots> = 2*f(k+1)" by(simp only:f_times 2)
+	  also have "f(k+1) = k+1" using less(1) `i=2*k+1` `~i\<le>1` by simp
+	  finally show ?thesis .
+	qed
+	ultimately show "f(i) = i" using `i = 2*k+1` by arith
+      qed
+    qed
+  qed
+qed
 
-lemma f_id: "f(n) = n"
-apply (rule order_antisym)
-apply (rule_tac [2] lemma1) 
-apply (blast intro: leI dest: leD f_mono Suc_leI)
-done
+
+text{* The only total model of a naive recursion equation of factorial on
+integers is 0 for all negative arguments: *}
+
+theorem ifac_neg0: fixes ifac :: "int \<Rightarrow> int"
+assumes ifac_rec: "!!i. ifac i = (if i=0 then 1 else i*ifac(i - 1))"
+shows "i<0 \<Longrightarrow> ifac i = 0"
+proof(rule ccontr)
+  assume 0: "i<0" "ifac i \<noteq> 0"
+  { fix j assume "j \<le> i"
+    have "ifac j \<noteq> 0"
+      apply(rule int_le_induct[OF `j\<le>i`])
+       apply(rule `ifac i \<noteq> 0`)
+      apply (metis `i<0` ifac_rec linorder_not_le mult_eq_0_iff)
+      done
+  } note below0 = this
+  { fix j assume "j<i"
+    have "1 < -j" using `j<i` `i<0` by arith
+    have "ifac(j - 1) \<noteq> 0" using `j<i` by(simp add: below0)
+    then have "\<bar>ifac (j - 1)\<bar> < (-j) * \<bar>ifac (j - 1)\<bar>" using `j<i`
+      mult_le_less_imp_less[OF order_refl[of "abs(ifac(j - 1))"] `1 < -j`]
+      by(simp add:mult_commute)
+    hence "abs(ifac(j - 1)) < abs(ifac j)"
+      using `1 < -j` by(simp add: ifac_rec[of "j"] abs_mult)
+  } note not_wf = this
+  let ?f = "%j. nat(abs(ifac(i - int(j+1))))"
+  obtain k where "\<not> ?f (Suc k) < ?f k"
+    using wf_no_infinite_down_chainE[OF wf_less, of "?f"] by blast
+  moreover have "i - int (k + 1) - 1 = i - int (Suc k + 1)" by arith
+  ultimately show False using not_wf[of "i - int(k+1)"]
+    by (simp only:) arith
+qed
 
 end
-