src/HOL/Library/Discrete.thy
author haftmann
Fri Dec 11 11:31:57 2015 +0100 (2015-12-11)
changeset 61831 c43f87119d80
parent 61115 3a4400985780
child 61975 b4b11391c676
permissions -rw-r--r--
modernized
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(* Author: Florian Haftmann, TU Muenchen *)
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section \<open>Common discrete functions\<close>
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theory Discrete
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imports Main
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begin
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subsection \<open>Discrete logarithm\<close>
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context
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begin
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qualified fun log :: "nat \<Rightarrow> nat"
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  where [simp del]: "log n = (if n < 2 then 0 else Suc (log (n div 2)))"
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lemma log_induct [consumes 1, case_names one double]:
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  fixes n :: nat
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  assumes "n > 0"
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  assumes one: "P 1"
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  assumes double: "\<And>n. n \<ge> 2 \<Longrightarrow> P (n div 2) \<Longrightarrow> P n"
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  shows "P n"
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using `n > 0` proof (induct n rule: log.induct)
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  fix n
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  assume "\<not> n < 2 \<Longrightarrow>
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          0 < n div 2 \<Longrightarrow> P (n div 2)"
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  then have *: "n \<ge> 2 \<Longrightarrow> P (n div 2)" by simp
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  assume "n > 0"
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  show "P n"
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  proof (cases "n = 1")
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    case True with one show ?thesis by simp
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  next
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    case False with `n > 0` have "n \<ge> 2" by auto
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    moreover with * have "P (n div 2)" .
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    ultimately show ?thesis by (rule double)
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  qed
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qed
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lemma log_zero [simp]: "log 0 = 0"
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  by (simp add: log.simps)
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lemma log_one [simp]: "log 1 = 0"
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  by (simp add: log.simps)
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lemma log_Suc_zero [simp]: "log (Suc 0) = 0"
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  using log_one by simp
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lemma log_rec: "n \<ge> 2 \<Longrightarrow> log n = Suc (log (n div 2))"
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  by (simp add: log.simps)
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lemma log_twice [simp]: "n \<noteq> 0 \<Longrightarrow> log (2 * n) = Suc (log n)"
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  by (simp add: log_rec)
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lemma log_half [simp]: "log (n div 2) = log n - 1"
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proof (cases "n < 2")
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  case True
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  then have "n = 0 \<or> n = 1" by arith
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  then show ?thesis by (auto simp del: One_nat_def)
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next
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  case False
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  then show ?thesis by (simp add: log_rec)
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qed
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lemma log_exp [simp]: "log (2 ^ n) = n"
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  by (induct n) simp_all
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lemma log_mono: "mono log"
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proof
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  fix m n :: nat
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  assume "m \<le> n"
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  then show "log m \<le> log n"
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  proof (induct m arbitrary: n rule: log.induct)
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    case (1 m)
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    then have mn2: "m div 2 \<le> n div 2" by arith
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    show "log m \<le> log n"
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    proof (cases "m \<ge> 2")
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      case False
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      then have "m = 0 \<or> m = 1" by arith
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      then show ?thesis by (auto simp del: One_nat_def)
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    next
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      case True then have "\<not> m < 2" by simp
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      with mn2 have "n \<ge> 2" by arith
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      from True have m2_0: "m div 2 \<noteq> 0" by arith
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      with mn2 have n2_0: "n div 2 \<noteq> 0" by arith
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      from `\<not> m < 2` "1.hyps" mn2 have "log (m div 2) \<le> log (n div 2)" by blast
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      with m2_0 n2_0 have "log (2 * (m div 2)) \<le> log (2 * (n div 2))" by simp
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      with m2_0 n2_0 \<open>m \<ge> 2\<close> \<open>n \<ge> 2\<close> show ?thesis by (simp only: log_rec [of m] log_rec [of n]) simp
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    qed
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  qed
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qed
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lemma log_exp2_le:
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  assumes "n > 0"
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  shows "2 ^ log n \<le> n"
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using assms proof (induct n rule: log_induct)
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  show "2 ^ log 1 \<le> (1 :: nat)" by simp
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next
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  fix n :: nat
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  assume "n \<ge> 2"
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  with log_mono have "log n \<ge> Suc 0"
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    by (simp add: log.simps)
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  assume "2 ^ log (n div 2) \<le> n div 2"
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  with `n \<ge> 2` have "2 ^ (log n - Suc 0) \<le> n div 2" by simp
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  then have "2 ^ (log n - Suc 0) * 2 ^ 1 \<le> n div 2 * 2" by simp
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  with `log n \<ge> Suc 0` have "2 ^ log n \<le> n div 2 * 2"
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    unfolding power_add [symmetric] by simp
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  also have "n div 2 * 2 \<le> n" by (cases "even n") simp_all
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  finally show "2 ^ log n \<le> n" .
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qed
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subsection \<open>Discrete square root\<close>
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qualified definition sqrt :: "nat \<Rightarrow> nat"
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  where "sqrt n = Max {m. m\<^sup>2 \<le> n}"
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lemma sqrt_aux:
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  fixes n :: nat
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  shows "finite {m. m\<^sup>2 \<le> n}" and "{m. m\<^sup>2 \<le> n} \<noteq> {}"
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proof -
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  { fix m
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    assume "m\<^sup>2 \<le> n"
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    then have "m \<le> n"
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      by (cases m) (simp_all add: power2_eq_square)
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  } note ** = this
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  then have "{m. m\<^sup>2 \<le> n} \<subseteq> {m. m \<le> n}" by auto
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  then show "finite {m. m\<^sup>2 \<le> n}" by (rule finite_subset) rule
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  have "0\<^sup>2 \<le> n" by simp
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  then show *: "{m. m\<^sup>2 \<le> n} \<noteq> {}" by blast
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qed
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lemma [code]: "sqrt n = Max (Set.filter (\<lambda>m. m\<^sup>2 \<le> n) {0..n})"
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proof -
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  from power2_nat_le_imp_le [of _ n] have "{m. m \<le> n \<and> m\<^sup>2 \<le> n} = {m. m\<^sup>2 \<le> n}" by auto
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  then show ?thesis by (simp add: sqrt_def Set.filter_def)
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qed
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lemma sqrt_inverse_power2 [simp]: "sqrt (n\<^sup>2) = n"
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proof -
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  have "{m. m \<le> n} \<noteq> {}" by auto
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  then have "Max {m. m \<le> n} \<le> n" by auto
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  then show ?thesis
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    by (auto simp add: sqrt_def power2_nat_le_eq_le intro: antisym)
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qed
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lemma sqrt_zero [simp]: "sqrt 0 = 0"
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  using sqrt_inverse_power2 [of 0] by simp
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lemma sqrt_one [simp]: "sqrt 1 = 1"
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  using sqrt_inverse_power2 [of 1] by simp
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lemma mono_sqrt: "mono sqrt"
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proof
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  fix m n :: nat
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  have *: "0 * 0 \<le> m" by simp
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  assume "m \<le> n"
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  then show "sqrt m \<le> sqrt n"
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    by (auto intro!: Max_mono \<open>0 * 0 \<le> m\<close> finite_less_ub simp add: power2_eq_square sqrt_def)
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qed
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lemma sqrt_greater_zero_iff [simp]: "sqrt n > 0 \<longleftrightarrow> n > 0"
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proof -
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  have *: "0 < Max {m. m\<^sup>2 \<le> n} \<longleftrightarrow> (\<exists>a\<in>{m. m\<^sup>2 \<le> n}. 0 < a)"
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    by (rule Max_gr_iff) (fact sqrt_aux)+
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  show ?thesis
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  proof
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    assume "0 < sqrt n"
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    then have "0 < Max {m. m\<^sup>2 \<le> n}" by (simp add: sqrt_def)
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    with * show "0 < n" by (auto dest: power2_nat_le_imp_le)
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  next
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    assume "0 < n"
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    then have "1\<^sup>2 \<le> n \<and> 0 < (1::nat)" by simp
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    then have "\<exists>q. q\<^sup>2 \<le> n \<and> 0 < q" ..
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    with * have "0 < Max {m. m\<^sup>2 \<le> n}" by blast
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    then show "0 < sqrt n" by  (simp add: sqrt_def)
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  qed
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qed
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lemma sqrt_power2_le [simp]: "(sqrt n)\<^sup>2 \<le> n" (* FIXME tune proof *)
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proof (cases "n > 0")
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  case False then show ?thesis by simp
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next
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  case True then have "sqrt n > 0" by simp
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  then have "mono (times (Max {m. m\<^sup>2 \<le> n}))" by (auto intro: mono_times_nat simp add: sqrt_def)
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  then have *: "Max {m. m\<^sup>2 \<le> n} * Max {m. m\<^sup>2 \<le> n} = Max (times (Max {m. m\<^sup>2 \<le> n}) ` {m. m\<^sup>2 \<le> n})"
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    using sqrt_aux [of n] by (rule mono_Max_commute)
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  have "Max (op * (Max {m. m * m \<le> n}) ` {m. m * m \<le> n}) \<le> n"
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    apply (subst Max_le_iff)
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    apply (metis (mono_tags) finite_imageI finite_less_ub le_square)
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    apply simp
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    apply (metis le0 mult_0_right)
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    apply auto
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    proof -
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      fix q
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      assume "q * q \<le> n"
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      show "Max {m. m * m \<le> n} * q \<le> n"
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      proof (cases "q > 0")
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        case False then show ?thesis by simp
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      next
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        case True then have "mono (times q)" by (rule mono_times_nat)
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        then have "q * Max {m. m * m \<le> n} = Max (times q ` {m. m * m \<le> n})"
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          using sqrt_aux [of n] by (auto simp add: power2_eq_square intro: mono_Max_commute)
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        then have "Max {m. m * m \<le> n} * q = Max (times q ` {m. m * m \<le> n})" by (simp add: ac_simps)
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        then show ?thesis
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          apply simp
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          apply (subst Max_le_iff)
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          apply auto
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          apply (metis (mono_tags) finite_imageI finite_less_ub le_square)
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          apply (metis \<open>q * q \<le> n\<close>)
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          apply (metis \<open>q * q \<le> n\<close> le_cases mult_le_mono1 mult_le_mono2 order_trans)
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          done
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      qed
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    qed
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  with * show ?thesis by (simp add: sqrt_def power2_eq_square)
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qed
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lemma sqrt_le: "sqrt n \<le> n"
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  using sqrt_aux [of n] by (auto simp add: sqrt_def intro: power2_nat_le_imp_le)
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end
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end