src/HOL/Divides.thy
author haftmann
Mon Apr 27 10:11:44 2009 +0200 (2009-04-27)
changeset 31001 7e6ffd8f51a9
parent 30934 ed5377c2b0a3
child 31009 41fd307cab30
permissions -rw-r--r--
cleaned up theory power further
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(*  Title:      HOL/Divides.thy
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    Author:     Lawrence C Paulson, Cambridge University Computer Laboratory
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    Copyright   1999  University of Cambridge
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*)
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header {* The division operators div and mod *}
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theory Divides
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imports Nat Power Product_Type
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uses "~~/src/Provers/Arith/cancel_div_mod.ML"
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begin
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subsection {* Syntactic division operations *}
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class div = dvd +
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  fixes div :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixl "div" 70)
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    and mod :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixl "mod" 70)
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subsection {* Abstract division in commutative semirings. *}
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class semiring_div = comm_semiring_1_cancel + no_zero_divisors + div +
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  assumes mod_div_equality: "a div b * b + a mod b = a"
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    and div_by_0 [simp]: "a div 0 = 0"
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    and div_0 [simp]: "0 div a = 0"
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    and div_mult_self1 [simp]: "b \<noteq> 0 \<Longrightarrow> (a + c * b) div b = c + a div b"
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    and div_mult_mult1 [simp]: "c \<noteq> 0 \<Longrightarrow> (c * a) div (c * b) = a div b"
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begin
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text {* @{const div} and @{const mod} *}
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lemma mod_div_equality2: "b * (a div b) + a mod b = a"
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  unfolding mult_commute [of b]
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  by (rule mod_div_equality)
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lemma mod_div_equality': "a mod b + a div b * b = a"
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  using mod_div_equality [of a b]
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  by (simp only: add_ac)
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lemma div_mod_equality: "((a div b) * b + a mod b) + c = a + c"
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  by (simp add: mod_div_equality)
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lemma div_mod_equality2: "(b * (a div b) + a mod b) + c = a + c"
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  by (simp add: mod_div_equality2)
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lemma mod_by_0 [simp]: "a mod 0 = a"
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  using mod_div_equality [of a zero] by simp
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lemma mod_0 [simp]: "0 mod a = 0"
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  using mod_div_equality [of zero a] div_0 by simp
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lemma div_mult_self2 [simp]:
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  assumes "b \<noteq> 0"
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  shows "(a + b * c) div b = c + a div b"
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  using assms div_mult_self1 [of b a c] by (simp add: mult_commute)
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lemma mod_mult_self1 [simp]: "(a + c * b) mod b = a mod b"
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proof (cases "b = 0")
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  case True then show ?thesis by simp
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next
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  case False
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  have "a + c * b = (a + c * b) div b * b + (a + c * b) mod b"
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    by (simp add: mod_div_equality)
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  also from False div_mult_self1 [of b a c] have
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    "\<dots> = (c + a div b) * b + (a + c * b) mod b"
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      by (simp add: algebra_simps)
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  finally have "a = a div b * b + (a + c * b) mod b"
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    by (simp add: add_commute [of a] add_assoc left_distrib)
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  then have "a div b * b + (a + c * b) mod b = a div b * b + a mod b"
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    by (simp add: mod_div_equality)
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  then show ?thesis by simp
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qed
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lemma mod_mult_self2 [simp]: "(a + b * c) mod b = a mod b"
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  by (simp add: mult_commute [of b])
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lemma div_mult_self1_is_id [simp]: "b \<noteq> 0 \<Longrightarrow> b * a div b = a"
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  using div_mult_self2 [of b 0 a] by simp
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lemma div_mult_self2_is_id [simp]: "b \<noteq> 0 \<Longrightarrow> a * b div b = a"
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  using div_mult_self1 [of b 0 a] by simp
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lemma mod_mult_self1_is_0 [simp]: "b * a mod b = 0"
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  using mod_mult_self2 [of 0 b a] by simp
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lemma mod_mult_self2_is_0 [simp]: "a * b mod b = 0"
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  using mod_mult_self1 [of 0 a b] by simp
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lemma div_by_1 [simp]: "a div 1 = a"
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  using div_mult_self2_is_id [of 1 a] zero_neq_one by simp
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lemma mod_by_1 [simp]: "a mod 1 = 0"
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proof -
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  from mod_div_equality [of a one] div_by_1 have "a + a mod 1 = a" by simp
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  then have "a + a mod 1 = a + 0" by simp
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  then show ?thesis by (rule add_left_imp_eq)
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qed
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lemma mod_self [simp]: "a mod a = 0"
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  using mod_mult_self2_is_0 [of 1] by simp
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lemma div_self [simp]: "a \<noteq> 0 \<Longrightarrow> a div a = 1"
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  using div_mult_self2_is_id [of _ 1] by simp
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lemma div_add_self1 [simp]:
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  assumes "b \<noteq> 0"
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  shows "(b + a) div b = a div b + 1"
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  using assms div_mult_self1 [of b a 1] by (simp add: add_commute)
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lemma div_add_self2 [simp]:
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  assumes "b \<noteq> 0"
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  shows "(a + b) div b = a div b + 1"
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  using assms div_add_self1 [of b a] by (simp add: add_commute)
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lemma mod_add_self1 [simp]:
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  "(b + a) mod b = a mod b"
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  using mod_mult_self1 [of a 1 b] by (simp add: add_commute)
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lemma mod_add_self2 [simp]:
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  "(a + b) mod b = a mod b"
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  using mod_mult_self1 [of a 1 b] by simp
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lemma mod_div_decomp:
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  fixes a b
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  obtains q r where "q = a div b" and "r = a mod b"
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    and "a = q * b + r"
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proof -
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  from mod_div_equality have "a = a div b * b + a mod b" by simp
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  moreover have "a div b = a div b" ..
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  moreover have "a mod b = a mod b" ..
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  note that ultimately show thesis by blast
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qed
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lemma dvd_eq_mod_eq_0 [code unfold]: "a dvd b \<longleftrightarrow> b mod a = 0"
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proof
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  assume "b mod a = 0"
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  with mod_div_equality [of b a] have "b div a * a = b" by simp
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  then have "b = a * (b div a)" unfolding mult_commute ..
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  then have "\<exists>c. b = a * c" ..
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  then show "a dvd b" unfolding dvd_def .
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next
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  assume "a dvd b"
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  then have "\<exists>c. b = a * c" unfolding dvd_def .
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  then obtain c where "b = a * c" ..
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  then have "b mod a = a * c mod a" by simp
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  then have "b mod a = c * a mod a" by (simp add: mult_commute)
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  then show "b mod a = 0" by simp
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qed
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lemma mod_div_trivial [simp]: "a mod b div b = 0"
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proof (cases "b = 0")
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  assume "b = 0"
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  thus ?thesis by simp
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next
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  assume "b \<noteq> 0"
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  hence "a div b + a mod b div b = (a mod b + a div b * b) div b"
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    by (rule div_mult_self1 [symmetric])
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  also have "\<dots> = a div b"
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    by (simp only: mod_div_equality')
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  also have "\<dots> = a div b + 0"
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    by simp
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  finally show ?thesis
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    by (rule add_left_imp_eq)
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qed
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lemma mod_mod_trivial [simp]: "a mod b mod b = a mod b"
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proof -
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  have "a mod b mod b = (a mod b + a div b * b) mod b"
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    by (simp only: mod_mult_self1)
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  also have "\<dots> = a mod b"
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    by (simp only: mod_div_equality')
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  finally show ?thesis .
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qed
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lemma dvd_imp_mod_0: "a dvd b \<Longrightarrow> b mod a = 0"
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by (rule dvd_eq_mod_eq_0[THEN iffD1])
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lemma dvd_div_mult_self: "a dvd b \<Longrightarrow> (b div a) * a = b"
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by (subst (2) mod_div_equality [of b a, symmetric]) (simp add:dvd_imp_mod_0)
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lemma dvd_div_mult: "a dvd b \<Longrightarrow> (b div a) * c = b * c div a"
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apply (cases "a = 0")
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 apply simp
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apply (auto simp: dvd_def mult_assoc)
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done
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lemma div_dvd_div[simp]:
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  "a dvd b \<Longrightarrow> a dvd c \<Longrightarrow> (b div a dvd c div a) = (b dvd c)"
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apply (cases "a = 0")
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 apply simp
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apply (unfold dvd_def)
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apply auto
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 apply(blast intro:mult_assoc[symmetric])
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apply(fastsimp simp add: mult_assoc)
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done
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lemma dvd_mod_imp_dvd: "[| k dvd m mod n;  k dvd n |] ==> k dvd m"
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  apply (subgoal_tac "k dvd (m div n) *n + m mod n")
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   apply (simp add: mod_div_equality)
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  apply (simp only: dvd_add dvd_mult)
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  done
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text {* Addition respects modular equivalence. *}
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lemma mod_add_left_eq: "(a + b) mod c = (a mod c + b) mod c"
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proof -
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  have "(a + b) mod c = (a div c * c + a mod c + b) mod c"
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    by (simp only: mod_div_equality)
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  also have "\<dots> = (a mod c + b + a div c * c) mod c"
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    by (simp only: add_ac)
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  also have "\<dots> = (a mod c + b) mod c"
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    by (rule mod_mult_self1)
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  finally show ?thesis .
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qed
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lemma mod_add_right_eq: "(a + b) mod c = (a + b mod c) mod c"
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proof -
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  have "(a + b) mod c = (a + (b div c * c + b mod c)) mod c"
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    by (simp only: mod_div_equality)
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  also have "\<dots> = (a + b mod c + b div c * c) mod c"
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    by (simp only: add_ac)
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  also have "\<dots> = (a + b mod c) mod c"
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    by (rule mod_mult_self1)
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  finally show ?thesis .
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qed
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lemma mod_add_eq: "(a + b) mod c = (a mod c + b mod c) mod c"
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by (rule trans [OF mod_add_left_eq mod_add_right_eq])
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lemma mod_add_cong:
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  assumes "a mod c = a' mod c"
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  assumes "b mod c = b' mod c"
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  shows "(a + b) mod c = (a' + b') mod c"
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proof -
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  have "(a mod c + b mod c) mod c = (a' mod c + b' mod c) mod c"
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    unfolding assms ..
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  thus ?thesis
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    by (simp only: mod_add_eq [symmetric])
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qed
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lemma div_add [simp]: "z dvd x \<Longrightarrow> z dvd y
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  \<Longrightarrow> (x + y) div z = x div z + y div z"
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by (cases "z = 0", simp, unfold dvd_def, auto simp add: algebra_simps)
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text {* Multiplication respects modular equivalence. *}
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lemma mod_mult_left_eq: "(a * b) mod c = ((a mod c) * b) mod c"
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proof -
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  have "(a * b) mod c = ((a div c * c + a mod c) * b) mod c"
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    by (simp only: mod_div_equality)
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  also have "\<dots> = (a mod c * b + a div c * b * c) mod c"
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    by (simp only: algebra_simps)
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  also have "\<dots> = (a mod c * b) mod c"
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    by (rule mod_mult_self1)
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  finally show ?thesis .
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qed
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lemma mod_mult_right_eq: "(a * b) mod c = (a * (b mod c)) mod c"
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proof -
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  have "(a * b) mod c = (a * (b div c * c + b mod c)) mod c"
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    by (simp only: mod_div_equality)
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  also have "\<dots> = (a * (b mod c) + a * (b div c) * c) mod c"
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    by (simp only: algebra_simps)
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  also have "\<dots> = (a * (b mod c)) mod c"
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    by (rule mod_mult_self1)
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  finally show ?thesis .
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qed
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lemma mod_mult_eq: "(a * b) mod c = ((a mod c) * (b mod c)) mod c"
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by (rule trans [OF mod_mult_left_eq mod_mult_right_eq])
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lemma mod_mult_cong:
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  assumes "a mod c = a' mod c"
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  assumes "b mod c = b' mod c"
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  shows "(a * b) mod c = (a' * b') mod c"
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proof -
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  have "(a mod c * (b mod c)) mod c = (a' mod c * (b' mod c)) mod c"
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    unfolding assms ..
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  thus ?thesis
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    by (simp only: mod_mult_eq [symmetric])
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qed
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lemma mod_mod_cancel:
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  assumes "c dvd b"
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  shows "a mod b mod c = a mod c"
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proof -
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  from `c dvd b` obtain k where "b = c * k"
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    by (rule dvdE)
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  have "a mod b mod c = a mod (c * k) mod c"
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    by (simp only: `b = c * k`)
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  also have "\<dots> = (a mod (c * k) + a div (c * k) * k * c) mod c"
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    by (simp only: mod_mult_self1)
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  also have "\<dots> = (a div (c * k) * (c * k) + a mod (c * k)) mod c"
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    by (simp only: add_ac mult_ac)
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  also have "\<dots> = a mod c"
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    by (simp only: mod_div_equality)
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  finally show ?thesis .
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qed
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lemma div_mult_div_if_dvd:
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  "y dvd x \<Longrightarrow> z dvd w \<Longrightarrow> (x div y) * (w div z) = (x * w) div (y * z)"
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  apply (cases "y = 0", simp)
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  apply (cases "z = 0", simp)
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  apply (auto elim!: dvdE simp add: algebra_simps)
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  apply (subst mult_assoc [symmetric])
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  apply (simp add: no_zero_divisors)
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  done
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lemma div_mult_mult2 [simp]:
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  "c \<noteq> 0 \<Longrightarrow> (a * c) div (b * c) = a div b"
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   311
  by (drule div_mult_mult1) (simp add: mult_commute)
haftmann@30930
   312
haftmann@30930
   313
lemma div_mult_mult1_if [simp]:
haftmann@30930
   314
  "(c * a) div (c * b) = (if c = 0 then 0 else a div b)"
haftmann@30930
   315
  by simp_all
nipkow@30476
   316
haftmann@30930
   317
lemma mod_mult_mult1:
haftmann@30930
   318
  "(c * a) mod (c * b) = c * (a mod b)"
haftmann@30930
   319
proof (cases "c = 0")
haftmann@30930
   320
  case True then show ?thesis by simp
haftmann@30930
   321
next
haftmann@30930
   322
  case False
haftmann@30930
   323
  from mod_div_equality
haftmann@30930
   324
  have "((c * a) div (c * b)) * (c * b) + (c * a) mod (c * b) = c * a" .
haftmann@30930
   325
  with False have "c * ((a div b) * b + a mod b) + (c * a) mod (c * b)
haftmann@30930
   326
    = c * a + c * (a mod b)" by (simp add: algebra_simps)
haftmann@30930
   327
  with mod_div_equality show ?thesis by simp 
haftmann@30930
   328
qed
haftmann@30930
   329
  
haftmann@30930
   330
lemma mod_mult_mult2:
haftmann@30930
   331
  "(a * c) mod (b * c) = (a mod b) * c"
haftmann@30930
   332
  using mod_mult_mult1 [of c a b] by (simp add: mult_commute)
haftmann@30930
   333
haftmann@30930
   334
end
nipkow@30476
   335
nipkow@30476
   336
lemma div_power: "(y::'a::{semiring_div,no_zero_divisors,recpower}) dvd x \<Longrightarrow>
nipkow@30476
   337
    (x div y)^n = x^n div y^n"
nipkow@30476
   338
apply (induct n)
nipkow@30476
   339
 apply simp
nipkow@30476
   340
apply(simp add: div_mult_div_if_dvd dvd_power_same)
nipkow@30476
   341
done
nipkow@30476
   342
huffman@29405
   343
class ring_div = semiring_div + comm_ring_1
huffman@29405
   344
begin
huffman@29405
   345
huffman@29405
   346
text {* Negation respects modular equivalence. *}
huffman@29405
   347
huffman@29405
   348
lemma mod_minus_eq: "(- a) mod b = (- (a mod b)) mod b"
huffman@29405
   349
proof -
huffman@29405
   350
  have "(- a) mod b = (- (a div b * b + a mod b)) mod b"
huffman@29405
   351
    by (simp only: mod_div_equality)
huffman@29405
   352
  also have "\<dots> = (- (a mod b) + - (a div b) * b) mod b"
huffman@29405
   353
    by (simp only: minus_add_distrib minus_mult_left add_ac)
huffman@29405
   354
  also have "\<dots> = (- (a mod b)) mod b"
huffman@29405
   355
    by (rule mod_mult_self1)
huffman@29405
   356
  finally show ?thesis .
huffman@29405
   357
qed
huffman@29405
   358
huffman@29405
   359
lemma mod_minus_cong:
huffman@29405
   360
  assumes "a mod b = a' mod b"
huffman@29405
   361
  shows "(- a) mod b = (- a') mod b"
huffman@29405
   362
proof -
huffman@29405
   363
  have "(- (a mod b)) mod b = (- (a' mod b)) mod b"
huffman@29405
   364
    unfolding assms ..
huffman@29405
   365
  thus ?thesis
huffman@29405
   366
    by (simp only: mod_minus_eq [symmetric])
huffman@29405
   367
qed
huffman@29405
   368
huffman@29405
   369
text {* Subtraction respects modular equivalence. *}
huffman@29405
   370
huffman@29405
   371
lemma mod_diff_left_eq: "(a - b) mod c = (a mod c - b) mod c"
huffman@29405
   372
  unfolding diff_minus
huffman@29405
   373
  by (intro mod_add_cong mod_minus_cong) simp_all
huffman@29405
   374
huffman@29405
   375
lemma mod_diff_right_eq: "(a - b) mod c = (a - b mod c) mod c"
huffman@29405
   376
  unfolding diff_minus
huffman@29405
   377
  by (intro mod_add_cong mod_minus_cong) simp_all
huffman@29405
   378
huffman@29405
   379
lemma mod_diff_eq: "(a - b) mod c = (a mod c - b mod c) mod c"
huffman@29405
   380
  unfolding diff_minus
huffman@29405
   381
  by (intro mod_add_cong mod_minus_cong) simp_all
huffman@29405
   382
huffman@29405
   383
lemma mod_diff_cong:
huffman@29405
   384
  assumes "a mod c = a' mod c"
huffman@29405
   385
  assumes "b mod c = b' mod c"
huffman@29405
   386
  shows "(a - b) mod c = (a' - b') mod c"
huffman@29405
   387
  unfolding diff_minus using assms
huffman@29405
   388
  by (intro mod_add_cong mod_minus_cong)
huffman@29405
   389
nipkow@30180
   390
lemma dvd_neg_div: "y dvd x \<Longrightarrow> -x div y = - (x div y)"
nipkow@30180
   391
apply (case_tac "y = 0") apply simp
nipkow@30180
   392
apply (auto simp add: dvd_def)
nipkow@30180
   393
apply (subgoal_tac "-(y * k) = y * - k")
nipkow@30180
   394
 apply (erule ssubst)
nipkow@30180
   395
 apply (erule div_mult_self1_is_id)
nipkow@30180
   396
apply simp
nipkow@30180
   397
done
nipkow@30180
   398
nipkow@30180
   399
lemma dvd_div_neg: "y dvd x \<Longrightarrow> x div -y = - (x div y)"
nipkow@30180
   400
apply (case_tac "y = 0") apply simp
nipkow@30180
   401
apply (auto simp add: dvd_def)
nipkow@30180
   402
apply (subgoal_tac "y * k = -y * -k")
nipkow@30180
   403
 apply (erule ssubst)
nipkow@30180
   404
 apply (rule div_mult_self1_is_id)
nipkow@30180
   405
 apply simp
nipkow@30180
   406
apply simp
nipkow@30180
   407
done
nipkow@30180
   408
huffman@29405
   409
end
huffman@29405
   410
haftmann@25942
   411
haftmann@26100
   412
subsection {* Division on @{typ nat} *}
haftmann@26100
   413
haftmann@26100
   414
text {*
haftmann@26100
   415
  We define @{const div} and @{const mod} on @{typ nat} by means
haftmann@26100
   416
  of a characteristic relation with two input arguments
haftmann@26100
   417
  @{term "m\<Colon>nat"}, @{term "n\<Colon>nat"} and two output arguments
haftmann@26100
   418
  @{term "q\<Colon>nat"}(uotient) and @{term "r\<Colon>nat"}(emainder).
haftmann@26100
   419
*}
haftmann@26100
   420
haftmann@30923
   421
definition divmod_rel :: "nat \<Rightarrow> nat \<Rightarrow> nat \<times> nat \<Rightarrow> bool" where
haftmann@30923
   422
  "divmod_rel m n qr \<longleftrightarrow>
haftmann@30923
   423
    m = fst qr * n + snd qr \<and>
haftmann@30923
   424
      (if n = 0 then fst qr = 0 else if n > 0 then 0 \<le> snd qr \<and> snd qr < n else n < snd qr \<and> snd qr \<le> 0)"
haftmann@26100
   425
haftmann@26100
   426
text {* @{const divmod_rel} is total: *}
haftmann@26100
   427
haftmann@26100
   428
lemma divmod_rel_ex:
haftmann@30923
   429
  obtains q r where "divmod_rel m n (q, r)"
haftmann@26100
   430
proof (cases "n = 0")
haftmann@30923
   431
  case True  with that show thesis
haftmann@26100
   432
    by (auto simp add: divmod_rel_def)
haftmann@26100
   433
next
haftmann@26100
   434
  case False
haftmann@26100
   435
  have "\<exists>q r. m = q * n + r \<and> r < n"
haftmann@26100
   436
  proof (induct m)
haftmann@26100
   437
    case 0 with `n \<noteq> 0`
haftmann@26100
   438
    have "(0\<Colon>nat) = 0 * n + 0 \<and> 0 < n" by simp
haftmann@26100
   439
    then show ?case by blast
haftmann@26100
   440
  next
haftmann@26100
   441
    case (Suc m) then obtain q' r'
haftmann@26100
   442
      where m: "m = q' * n + r'" and n: "r' < n" by auto
haftmann@26100
   443
    then show ?case proof (cases "Suc r' < n")
haftmann@26100
   444
      case True
haftmann@26100
   445
      from m n have "Suc m = q' * n + Suc r'" by simp
haftmann@26100
   446
      with True show ?thesis by blast
haftmann@26100
   447
    next
haftmann@26100
   448
      case False then have "n \<le> Suc r'" by auto
haftmann@26100
   449
      moreover from n have "Suc r' \<le> n" by auto
haftmann@26100
   450
      ultimately have "n = Suc r'" by auto
haftmann@26100
   451
      with m have "Suc m = Suc q' * n + 0" by simp
haftmann@26100
   452
      with `n \<noteq> 0` show ?thesis by blast
haftmann@26100
   453
    qed
haftmann@26100
   454
  qed
haftmann@26100
   455
  with that show thesis
haftmann@26100
   456
    using `n \<noteq> 0` by (auto simp add: divmod_rel_def)
haftmann@26100
   457
qed
haftmann@26100
   458
haftmann@26100
   459
text {* @{const divmod_rel} is injective: *}
haftmann@26100
   460
haftmann@30923
   461
lemma divmod_rel_unique:
haftmann@30923
   462
  assumes "divmod_rel m n qr"
haftmann@30923
   463
    and "divmod_rel m n qr'"
haftmann@30923
   464
  shows "qr = qr'"
haftmann@26100
   465
proof (cases "n = 0")
haftmann@26100
   466
  case True with assms show ?thesis
haftmann@30923
   467
    by (cases qr, cases qr')
haftmann@30923
   468
      (simp add: divmod_rel_def)
haftmann@26100
   469
next
haftmann@26100
   470
  case False
haftmann@26100
   471
  have aux: "\<And>q r q' r'. q' * n + r' = q * n + r \<Longrightarrow> r < n \<Longrightarrow> q' \<le> (q\<Colon>nat)"
haftmann@26100
   472
  apply (rule leI)
haftmann@26100
   473
  apply (subst less_iff_Suc_add)
haftmann@26100
   474
  apply (auto simp add: add_mult_distrib)
haftmann@26100
   475
  done
haftmann@30923
   476
  from `n \<noteq> 0` assms have "fst qr = fst qr'"
haftmann@30923
   477
    by (auto simp add: divmod_rel_def intro: order_antisym dest: aux sym)
haftmann@30923
   478
  moreover from this assms have "snd qr = snd qr'"
haftmann@30923
   479
    by (simp add: divmod_rel_def)
haftmann@30923
   480
  ultimately show ?thesis by (cases qr, cases qr') simp
haftmann@26100
   481
qed
haftmann@26100
   482
haftmann@26100
   483
text {*
haftmann@26100
   484
  We instantiate divisibility on the natural numbers by
haftmann@26100
   485
  means of @{const divmod_rel}:
haftmann@26100
   486
*}
haftmann@25942
   487
haftmann@25942
   488
instantiation nat :: semiring_div
haftmann@25571
   489
begin
haftmann@25571
   490
haftmann@26100
   491
definition divmod :: "nat \<Rightarrow> nat \<Rightarrow> nat \<times> nat" where
haftmann@30923
   492
  [code del]: "divmod m n = (THE qr. divmod_rel m n qr)"
haftmann@30923
   493
haftmann@30923
   494
lemma divmod_rel_divmod:
haftmann@30923
   495
  "divmod_rel m n (divmod m n)"
haftmann@30923
   496
proof -
haftmann@30923
   497
  from divmod_rel_ex
haftmann@30923
   498
    obtain qr where rel: "divmod_rel m n qr" .
haftmann@30923
   499
  then show ?thesis
haftmann@30923
   500
  by (auto simp add: divmod_def intro: theI elim: divmod_rel_unique)
haftmann@30923
   501
qed
haftmann@30923
   502
haftmann@30923
   503
lemma divmod_eq:
haftmann@30923
   504
  assumes "divmod_rel m n qr" 
haftmann@30923
   505
  shows "divmod m n = qr"
haftmann@30923
   506
  using assms by (auto intro: divmod_rel_unique divmod_rel_divmod)
haftmann@26100
   507
haftmann@26100
   508
definition div_nat where
haftmann@26100
   509
  "m div n = fst (divmod m n)"
haftmann@26100
   510
haftmann@26100
   511
definition mod_nat where
haftmann@26100
   512
  "m mod n = snd (divmod m n)"
haftmann@25571
   513
haftmann@26100
   514
lemma divmod_div_mod:
haftmann@26100
   515
  "divmod m n = (m div n, m mod n)"
haftmann@26100
   516
  unfolding div_nat_def mod_nat_def by simp
haftmann@26100
   517
haftmann@26100
   518
lemma div_eq:
haftmann@30923
   519
  assumes "divmod_rel m n (q, r)" 
haftmann@26100
   520
  shows "m div n = q"
haftmann@30923
   521
  using assms by (auto dest: divmod_eq simp add: divmod_div_mod)
haftmann@26100
   522
haftmann@26100
   523
lemma mod_eq:
haftmann@30923
   524
  assumes "divmod_rel m n (q, r)" 
haftmann@26100
   525
  shows "m mod n = r"
haftmann@30923
   526
  using assms by (auto dest: divmod_eq simp add: divmod_div_mod)
haftmann@25571
   527
haftmann@30923
   528
lemma divmod_rel: "divmod_rel m n (m div n, m mod n)"
haftmann@30923
   529
  by (simp add: div_nat_def mod_nat_def divmod_rel_divmod)
paulson@14267
   530
haftmann@26100
   531
lemma divmod_zero:
haftmann@26100
   532
  "divmod m 0 = (0, m)"
haftmann@26100
   533
proof -
haftmann@26100
   534
  from divmod_rel [of m 0] show ?thesis
haftmann@26100
   535
    unfolding divmod_div_mod divmod_rel_def by simp
haftmann@26100
   536
qed
haftmann@25942
   537
haftmann@26100
   538
lemma divmod_base:
haftmann@26100
   539
  assumes "m < n"
haftmann@26100
   540
  shows "divmod m n = (0, m)"
haftmann@26100
   541
proof -
haftmann@26100
   542
  from divmod_rel [of m n] show ?thesis
haftmann@26100
   543
    unfolding divmod_div_mod divmod_rel_def
haftmann@26100
   544
    using assms by (cases "m div n = 0")
haftmann@26100
   545
      (auto simp add: gr0_conv_Suc [of "m div n"])
haftmann@26100
   546
qed
haftmann@25942
   547
haftmann@26100
   548
lemma divmod_step:
haftmann@26100
   549
  assumes "0 < n" and "n \<le> m"
haftmann@26100
   550
  shows "divmod m n = (Suc ((m - n) div n), (m - n) mod n)"
haftmann@26100
   551
proof -
haftmann@30923
   552
  from divmod_rel have divmod_m_n: "divmod_rel m n (m div n, m mod n)" .
haftmann@26100
   553
  with assms have m_div_n: "m div n \<ge> 1"
haftmann@26100
   554
    by (cases "m div n") (auto simp add: divmod_rel_def)
haftmann@30923
   555
  from assms divmod_m_n have "divmod_rel (m - n) n (m div n - Suc 0, m mod n)"
haftmann@26100
   556
    by (cases "m div n") (auto simp add: divmod_rel_def)
huffman@30079
   557
  with divmod_eq have "divmod (m - n) n = (m div n - Suc 0, m mod n)" by simp
haftmann@26100
   558
  moreover from divmod_div_mod have "divmod (m - n) n = ((m - n) div n, (m - n) mod n)" .
haftmann@26100
   559
  ultimately have "m div n = Suc ((m - n) div n)"
haftmann@26100
   560
    and "m mod n = (m - n) mod n" using m_div_n by simp_all
haftmann@26100
   561
  then show ?thesis using divmod_div_mod by simp
haftmann@26100
   562
qed
haftmann@25942
   563
wenzelm@26300
   564
text {* The ''recursion'' equations for @{const div} and @{const mod} *}
haftmann@26100
   565
haftmann@26100
   566
lemma div_less [simp]:
haftmann@26100
   567
  fixes m n :: nat
haftmann@26100
   568
  assumes "m < n"
haftmann@26100
   569
  shows "m div n = 0"
haftmann@26100
   570
  using assms divmod_base divmod_div_mod by simp
haftmann@25942
   571
haftmann@26100
   572
lemma le_div_geq:
haftmann@26100
   573
  fixes m n :: nat
haftmann@26100
   574
  assumes "0 < n" and "n \<le> m"
haftmann@26100
   575
  shows "m div n = Suc ((m - n) div n)"
haftmann@26100
   576
  using assms divmod_step divmod_div_mod by simp
paulson@14267
   577
haftmann@26100
   578
lemma mod_less [simp]:
haftmann@26100
   579
  fixes m n :: nat
haftmann@26100
   580
  assumes "m < n"
haftmann@26100
   581
  shows "m mod n = m"
haftmann@26100
   582
  using assms divmod_base divmod_div_mod by simp
haftmann@26100
   583
haftmann@26100
   584
lemma le_mod_geq:
haftmann@26100
   585
  fixes m n :: nat
haftmann@26100
   586
  assumes "n \<le> m"
haftmann@26100
   587
  shows "m mod n = (m - n) mod n"
haftmann@26100
   588
  using assms divmod_step divmod_div_mod by (cases "n = 0") simp_all
paulson@14267
   589
haftmann@30930
   590
instance proof -
haftmann@30930
   591
  have [simp]: "\<And>n::nat. n div 0 = 0"
haftmann@30930
   592
    by (simp add: div_nat_def divmod_zero)
haftmann@30930
   593
  have [simp]: "\<And>n::nat. 0 div n = 0"
haftmann@30930
   594
  proof -
haftmann@30930
   595
    fix n :: nat
haftmann@30930
   596
    show "0 div n = 0"
haftmann@30930
   597
      by (cases "n = 0") simp_all
haftmann@30930
   598
  qed
haftmann@30930
   599
  show "OFCLASS(nat, semiring_div_class)" proof
haftmann@30930
   600
    fix m n :: nat
haftmann@30930
   601
    show "m div n * n + m mod n = m"
haftmann@30930
   602
      using divmod_rel [of m n] by (simp add: divmod_rel_def)
haftmann@30930
   603
  next
haftmann@30930
   604
    fix m n q :: nat
haftmann@30930
   605
    assume "n \<noteq> 0"
haftmann@30930
   606
    then show "(q + m * n) div n = m + q div n"
haftmann@30930
   607
      by (induct m) (simp_all add: le_div_geq)
haftmann@30930
   608
  next
haftmann@30930
   609
    fix m n q :: nat
haftmann@30930
   610
    assume "m \<noteq> 0"
haftmann@30930
   611
    then show "(m * n) div (m * q) = n div q"
haftmann@30930
   612
    proof (cases "n \<noteq> 0 \<and> q \<noteq> 0")
haftmann@30930
   613
      case False then show ?thesis by auto
haftmann@30930
   614
    next
haftmann@30930
   615
      case True with `m \<noteq> 0`
haftmann@30930
   616
        have "m > 0" and "n > 0" and "q > 0" by auto
haftmann@30930
   617
      then have "\<And>a b. divmod_rel n q (a, b) \<Longrightarrow> divmod_rel (m * n) (m * q) (a, m * b)"
haftmann@30930
   618
        by (auto simp add: divmod_rel_def) (simp_all add: algebra_simps)
haftmann@30930
   619
      moreover from divmod_rel have "divmod_rel n q (n div q, n mod q)" .
haftmann@30930
   620
      ultimately have "divmod_rel (m * n) (m * q) (n div q, m * (n mod q))" .
haftmann@30930
   621
      then show ?thesis by (simp add: div_eq)
haftmann@30930
   622
    qed
haftmann@30930
   623
  qed simp_all
haftmann@25942
   624
qed
haftmann@26100
   625
haftmann@25942
   626
end
paulson@14267
   627
haftmann@26100
   628
text {* Simproc for cancelling @{const div} and @{const mod} *}
haftmann@25942
   629
haftmann@30934
   630
ML {*
haftmann@30934
   631
local
haftmann@30934
   632
haftmann@30934
   633
structure CancelDivMod = CancelDivModFun(struct
haftmann@25942
   634
haftmann@30934
   635
  val div_name = @{const_name div};
haftmann@30934
   636
  val mod_name = @{const_name mod};
haftmann@30934
   637
  val mk_binop = HOLogic.mk_binop;
haftmann@30934
   638
  val mk_sum = Nat_Arith.mk_sum;
haftmann@30934
   639
  val dest_sum = Nat_Arith.dest_sum;
haftmann@25942
   640
haftmann@30934
   641
  val div_mod_eqs = map mk_meta_eq [@{thm div_mod_equality}, @{thm div_mod_equality2}];
paulson@14267
   642
haftmann@30934
   643
  val trans = trans;
haftmann@25942
   644
haftmann@30934
   645
  val prove_eq_sums = Arith_Data.prove_conv2 all_tac (Arith_Data.simp_all_tac
haftmann@30934
   646
    (@{thm monoid_add_class.add_0_left} :: @{thm monoid_add_class.add_0_right} :: @{thms add_ac}))
haftmann@25942
   647
haftmann@30934
   648
end)
haftmann@25942
   649
haftmann@30934
   650
in
haftmann@25942
   651
haftmann@30934
   652
val cancel_div_mod_nat_proc = Simplifier.simproc (the_context ())
haftmann@26100
   653
  "cancel_div_mod" ["(m::nat) + n"] (K CancelDivMod.proc);
haftmann@25942
   654
haftmann@30934
   655
val _ = Addsimprocs [cancel_div_mod_nat_proc];
haftmann@30934
   656
haftmann@30934
   657
end
haftmann@25942
   658
*}
haftmann@25942
   659
haftmann@26100
   660
text {* code generator setup *}
haftmann@26100
   661
haftmann@26100
   662
lemma divmod_if [code]: "divmod m n = (if n = 0 \<or> m < n then (0, m) else
haftmann@26100
   663
  let (q, r) = divmod (m - n) n in (Suc q, r))"
nipkow@29667
   664
by (simp add: divmod_zero divmod_base divmod_step)
haftmann@26100
   665
    (simp add: divmod_div_mod)
haftmann@26100
   666
haftmann@26100
   667
code_modulename SML
haftmann@26100
   668
  Divides Nat
haftmann@26100
   669
haftmann@26100
   670
code_modulename OCaml
haftmann@26100
   671
  Divides Nat
haftmann@26100
   672
haftmann@26100
   673
code_modulename Haskell
haftmann@26100
   674
  Divides Nat
haftmann@26100
   675
haftmann@26100
   676
haftmann@26100
   677
subsubsection {* Quotient *}
haftmann@26100
   678
haftmann@26100
   679
lemma div_geq: "0 < n \<Longrightarrow>  \<not> m < n \<Longrightarrow> m div n = Suc ((m - n) div n)"
nipkow@29667
   680
by (simp add: le_div_geq linorder_not_less)
haftmann@26100
   681
haftmann@26100
   682
lemma div_if: "0 < n \<Longrightarrow> m div n = (if m < n then 0 else Suc ((m - n) div n))"
nipkow@29667
   683
by (simp add: div_geq)
haftmann@26100
   684
haftmann@26100
   685
lemma div_mult_self_is_m [simp]: "0<n ==> (m*n) div n = (m::nat)"
nipkow@29667
   686
by simp
haftmann@26100
   687
haftmann@26100
   688
lemma div_mult_self1_is_m [simp]: "0<n ==> (n*m) div n = (m::nat)"
nipkow@29667
   689
by simp
haftmann@26100
   690
haftmann@25942
   691
haftmann@25942
   692
subsubsection {* Remainder *}
haftmann@25942
   693
haftmann@26100
   694
lemma mod_less_divisor [simp]:
haftmann@26100
   695
  fixes m n :: nat
haftmann@26100
   696
  assumes "n > 0"
haftmann@26100
   697
  shows "m mod n < (n::nat)"
haftmann@30923
   698
  using assms divmod_rel [of m n] unfolding divmod_rel_def by auto
paulson@14267
   699
haftmann@26100
   700
lemma mod_less_eq_dividend [simp]:
haftmann@26100
   701
  fixes m n :: nat
haftmann@26100
   702
  shows "m mod n \<le> m"
haftmann@26100
   703
proof (rule add_leD2)
haftmann@26100
   704
  from mod_div_equality have "m div n * n + m mod n = m" .
haftmann@26100
   705
  then show "m div n * n + m mod n \<le> m" by auto
haftmann@26100
   706
qed
haftmann@26100
   707
haftmann@26100
   708
lemma mod_geq: "\<not> m < (n\<Colon>nat) \<Longrightarrow> m mod n = (m - n) mod n"
nipkow@29667
   709
by (simp add: le_mod_geq linorder_not_less)
paulson@14267
   710
haftmann@26100
   711
lemma mod_if: "m mod (n\<Colon>nat) = (if m < n then m else (m - n) mod n)"
nipkow@29667
   712
by (simp add: le_mod_geq)
haftmann@26100
   713
paulson@14267
   714
lemma mod_1 [simp]: "m mod Suc 0 = 0"
nipkow@29667
   715
by (induct m) (simp_all add: mod_geq)
paulson@14267
   716
haftmann@26100
   717
lemma mod_mult_distrib: "(m mod n) * (k\<Colon>nat) = (m * k) mod (n * k)"
wenzelm@22718
   718
  apply (cases "n = 0", simp)
wenzelm@22718
   719
  apply (cases "k = 0", simp)
wenzelm@22718
   720
  apply (induct m rule: nat_less_induct)
wenzelm@22718
   721
  apply (subst mod_if, simp)
wenzelm@22718
   722
  apply (simp add: mod_geq diff_mult_distrib)
wenzelm@22718
   723
  done
paulson@14267
   724
paulson@14267
   725
lemma mod_mult_distrib2: "(k::nat) * (m mod n) = (k*m) mod (k*n)"
nipkow@29667
   726
by (simp add: mult_commute [of k] mod_mult_distrib)
paulson@14267
   727
paulson@14267
   728
(* a simple rearrangement of mod_div_equality: *)
paulson@14267
   729
lemma mult_div_cancel: "(n::nat) * (m div n) = m - (m mod n)"
nipkow@29667
   730
by (cut_tac a = m and b = n in mod_div_equality2, arith)
paulson@14267
   731
nipkow@15439
   732
lemma mod_le_divisor[simp]: "0 < n \<Longrightarrow> m mod n \<le> (n::nat)"
wenzelm@22718
   733
  apply (drule mod_less_divisor [where m = m])
wenzelm@22718
   734
  apply simp
wenzelm@22718
   735
  done
paulson@14267
   736
haftmann@26100
   737
subsubsection {* Quotient and Remainder *}
paulson@14267
   738
haftmann@26100
   739
lemma divmod_rel_mult1_eq:
haftmann@30923
   740
  "divmod_rel b c (q, r) \<Longrightarrow> c > 0
haftmann@30923
   741
   \<Longrightarrow> divmod_rel (a * b) c (a * q + a * r div c, a * r mod c)"
nipkow@29667
   742
by (auto simp add: split_ifs divmod_rel_def algebra_simps)
paulson@14267
   743
haftmann@30923
   744
lemma div_mult1_eq:
haftmann@30923
   745
  "(a * b) div c = a * (b div c) + a * (b mod c) div (c::nat)"
nipkow@25134
   746
apply (cases "c = 0", simp)
haftmann@26100
   747
apply (blast intro: divmod_rel [THEN divmod_rel_mult1_eq, THEN div_eq])
nipkow@25134
   748
done
paulson@14267
   749
haftmann@26100
   750
lemma divmod_rel_add1_eq:
haftmann@30923
   751
  "divmod_rel a c (aq, ar) \<Longrightarrow> divmod_rel b c (bq, br) \<Longrightarrow>  c > 0
haftmann@30923
   752
   \<Longrightarrow> divmod_rel (a + b) c (aq + bq + (ar + br) div c, (ar + br) mod c)"
nipkow@29667
   753
by (auto simp add: split_ifs divmod_rel_def algebra_simps)
paulson@14267
   754
paulson@14267
   755
(*NOT suitable for rewriting: the RHS has an instance of the LHS*)
paulson@14267
   756
lemma div_add1_eq:
nipkow@25134
   757
  "(a+b) div (c::nat) = a div c + b div c + ((a mod c + b mod c) div c)"
nipkow@25134
   758
apply (cases "c = 0", simp)
haftmann@26100
   759
apply (blast intro: divmod_rel_add1_eq [THEN div_eq] divmod_rel)
nipkow@25134
   760
done
paulson@14267
   761
paulson@14267
   762
lemma mod_lemma: "[| (0::nat) < c; r < b |] ==> b * (q mod c) + r < b * c"
wenzelm@22718
   763
  apply (cut_tac m = q and n = c in mod_less_divisor)
wenzelm@22718
   764
  apply (drule_tac [2] m = "q mod c" in less_imp_Suc_add, auto)
wenzelm@22718
   765
  apply (erule_tac P = "%x. ?lhs < ?rhs x" in ssubst)
wenzelm@22718
   766
  apply (simp add: add_mult_distrib2)
wenzelm@22718
   767
  done
paulson@10559
   768
haftmann@30923
   769
lemma divmod_rel_mult2_eq:
haftmann@30923
   770
  "divmod_rel a b (q, r) \<Longrightarrow> 0 < b \<Longrightarrow> 0 < c
haftmann@30923
   771
   \<Longrightarrow> divmod_rel a (b * c) (q div c, b *(q mod c) + r)"
nipkow@29667
   772
by (auto simp add: mult_ac divmod_rel_def add_mult_distrib2 [symmetric] mod_lemma)
paulson@14267
   773
paulson@14267
   774
lemma div_mult2_eq: "a div (b*c) = (a div b) div (c::nat)"
wenzelm@22718
   775
  apply (cases "b = 0", simp)
wenzelm@22718
   776
  apply (cases "c = 0", simp)
haftmann@26100
   777
  apply (force simp add: divmod_rel [THEN divmod_rel_mult2_eq, THEN div_eq])
wenzelm@22718
   778
  done
paulson@14267
   779
paulson@14267
   780
lemma mod_mult2_eq: "a mod (b*c) = b*(a div b mod c) + a mod (b::nat)"
wenzelm@22718
   781
  apply (cases "b = 0", simp)
wenzelm@22718
   782
  apply (cases "c = 0", simp)
haftmann@26100
   783
  apply (auto simp add: mult_commute divmod_rel [THEN divmod_rel_mult2_eq, THEN mod_eq])
wenzelm@22718
   784
  done
paulson@14267
   785
paulson@14267
   786
haftmann@25942
   787
subsubsection{*Further Facts about Quotient and Remainder*}
paulson@14267
   788
paulson@14267
   789
lemma div_1 [simp]: "m div Suc 0 = m"
nipkow@29667
   790
by (induct m) (simp_all add: div_geq)
paulson@14267
   791
paulson@14267
   792
paulson@14267
   793
(* Monotonicity of div in first argument *)
haftmann@30923
   794
lemma div_le_mono [rule_format (no_asm)]:
wenzelm@22718
   795
    "\<forall>m::nat. m \<le> n --> (m div k) \<le> (n div k)"
paulson@14267
   796
apply (case_tac "k=0", simp)
paulson@15251
   797
apply (induct "n" rule: nat_less_induct, clarify)
paulson@14267
   798
apply (case_tac "n<k")
paulson@14267
   799
(* 1  case n<k *)
paulson@14267
   800
apply simp
paulson@14267
   801
(* 2  case n >= k *)
paulson@14267
   802
apply (case_tac "m<k")
paulson@14267
   803
(* 2.1  case m<k *)
paulson@14267
   804
apply simp
paulson@14267
   805
(* 2.2  case m>=k *)
nipkow@15439
   806
apply (simp add: div_geq diff_le_mono)
paulson@14267
   807
done
paulson@14267
   808
paulson@14267
   809
(* Antimonotonicity of div in second argument *)
paulson@14267
   810
lemma div_le_mono2: "!!m::nat. [| 0<m; m\<le>n |] ==> (k div n) \<le> (k div m)"
paulson@14267
   811
apply (subgoal_tac "0<n")
wenzelm@22718
   812
 prefer 2 apply simp
paulson@15251
   813
apply (induct_tac k rule: nat_less_induct)
paulson@14267
   814
apply (rename_tac "k")
paulson@14267
   815
apply (case_tac "k<n", simp)
paulson@14267
   816
apply (subgoal_tac "~ (k<m) ")
wenzelm@22718
   817
 prefer 2 apply simp
paulson@14267
   818
apply (simp add: div_geq)
paulson@15251
   819
apply (subgoal_tac "(k-n) div n \<le> (k-m) div n")
paulson@14267
   820
 prefer 2
paulson@14267
   821
 apply (blast intro: div_le_mono diff_le_mono2)
paulson@14267
   822
apply (rule le_trans, simp)
nipkow@15439
   823
apply (simp)
paulson@14267
   824
done
paulson@14267
   825
paulson@14267
   826
lemma div_le_dividend [simp]: "m div n \<le> (m::nat)"
paulson@14267
   827
apply (case_tac "n=0", simp)
paulson@14267
   828
apply (subgoal_tac "m div n \<le> m div 1", simp)
paulson@14267
   829
apply (rule div_le_mono2)
paulson@14267
   830
apply (simp_all (no_asm_simp))
paulson@14267
   831
done
paulson@14267
   832
wenzelm@22718
   833
(* Similar for "less than" *)
paulson@17085
   834
lemma div_less_dividend [rule_format]:
paulson@14267
   835
     "!!n::nat. 1<n ==> 0 < m --> m div n < m"
paulson@15251
   836
apply (induct_tac m rule: nat_less_induct)
paulson@14267
   837
apply (rename_tac "m")
paulson@14267
   838
apply (case_tac "m<n", simp)
paulson@14267
   839
apply (subgoal_tac "0<n")
wenzelm@22718
   840
 prefer 2 apply simp
paulson@14267
   841
apply (simp add: div_geq)
paulson@14267
   842
apply (case_tac "n<m")
paulson@15251
   843
 apply (subgoal_tac "(m-n) div n < (m-n) ")
paulson@14267
   844
  apply (rule impI less_trans_Suc)+
paulson@14267
   845
apply assumption
nipkow@15439
   846
  apply (simp_all)
paulson@14267
   847
done
paulson@14267
   848
paulson@17085
   849
declare div_less_dividend [simp]
paulson@17085
   850
paulson@14267
   851
text{*A fact for the mutilated chess board*}
paulson@14267
   852
lemma mod_Suc: "Suc(m) mod n = (if Suc(m mod n) = n then 0 else Suc(m mod n))"
paulson@14267
   853
apply (case_tac "n=0", simp)
paulson@15251
   854
apply (induct "m" rule: nat_less_induct)
paulson@14267
   855
apply (case_tac "Suc (na) <n")
paulson@14267
   856
(* case Suc(na) < n *)
paulson@14267
   857
apply (frule lessI [THEN less_trans], simp add: less_not_refl3)
paulson@14267
   858
(* case n \<le> Suc(na) *)
paulson@16796
   859
apply (simp add: linorder_not_less le_Suc_eq mod_geq)
nipkow@15439
   860
apply (auto simp add: Suc_diff_le le_mod_geq)
paulson@14267
   861
done
paulson@14267
   862
paulson@14267
   863
haftmann@27651
   864
subsubsection {* The Divides Relation *}
paulson@24286
   865
paulson@14267
   866
lemma dvd_1_left [iff]: "Suc 0 dvd k"
wenzelm@22718
   867
  unfolding dvd_def by simp
paulson@14267
   868
paulson@14267
   869
lemma dvd_1_iff_1 [simp]: "(m dvd Suc 0) = (m = Suc 0)"
nipkow@29667
   870
by (simp add: dvd_def)
paulson@14267
   871
huffman@30079
   872
lemma nat_dvd_1_iff_1 [simp]: "m dvd (1::nat) \<longleftrightarrow> m = 1"
huffman@30079
   873
by (simp add: dvd_def)
huffman@30079
   874
paulson@14267
   875
lemma dvd_anti_sym: "[| m dvd n; n dvd m |] ==> m = (n::nat)"
wenzelm@22718
   876
  unfolding dvd_def
wenzelm@22718
   877
  by (force dest: mult_eq_self_implies_10 simp add: mult_assoc mult_eq_1_iff)
paulson@14267
   878
haftmann@23684
   879
text {* @{term "op dvd"} is a partial order *}
haftmann@23684
   880
wenzelm@30729
   881
interpretation dvd: order "op dvd" "\<lambda>n m \<Colon> nat. n dvd m \<and> \<not> m dvd n"
haftmann@28823
   882
  proof qed (auto intro: dvd_refl dvd_trans dvd_anti_sym)
paulson@14267
   883
nipkow@30042
   884
lemma nat_dvd_diff[simp]: "[| k dvd m; k dvd n |] ==> k dvd (m-n :: nat)"
nipkow@30042
   885
unfolding dvd_def
nipkow@30042
   886
by (blast intro: diff_mult_distrib2 [symmetric])
paulson@14267
   887
paulson@14267
   888
lemma dvd_diffD: "[| k dvd m-n; k dvd n; n\<le>m |] ==> k dvd (m::nat)"
wenzelm@22718
   889
  apply (erule linorder_not_less [THEN iffD2, THEN add_diff_inverse, THEN subst])
wenzelm@22718
   890
  apply (blast intro: dvd_add)
wenzelm@22718
   891
  done
paulson@14267
   892
paulson@14267
   893
lemma dvd_diffD1: "[| k dvd m-n; k dvd m; n\<le>m |] ==> k dvd (n::nat)"
nipkow@30042
   894
by (drule_tac m = m in nat_dvd_diff, auto)
paulson@14267
   895
paulson@14267
   896
lemma dvd_reduce: "(k dvd n + k) = (k dvd (n::nat))"
wenzelm@22718
   897
  apply (rule iffI)
wenzelm@22718
   898
   apply (erule_tac [2] dvd_add)
wenzelm@22718
   899
   apply (rule_tac [2] dvd_refl)
wenzelm@22718
   900
  apply (subgoal_tac "n = (n+k) -k")
wenzelm@22718
   901
   prefer 2 apply simp
wenzelm@22718
   902
  apply (erule ssubst)
nipkow@30042
   903
  apply (erule nat_dvd_diff)
wenzelm@22718
   904
  apply (rule dvd_refl)
wenzelm@22718
   905
  done
paulson@14267
   906
paulson@14267
   907
lemma dvd_mod: "!!n::nat. [| f dvd m; f dvd n |] ==> f dvd m mod n"
wenzelm@22718
   908
  unfolding dvd_def
wenzelm@22718
   909
  apply (case_tac "n = 0", auto)
wenzelm@22718
   910
  apply (blast intro: mod_mult_distrib2 [symmetric])
wenzelm@22718
   911
  done
paulson@14267
   912
paulson@14267
   913
lemma dvd_mod_iff: "k dvd n ==> ((k::nat) dvd m mod n) = (k dvd m)"
nipkow@29667
   914
by (blast intro: dvd_mod_imp_dvd dvd_mod)
paulson@14267
   915
paulson@14267
   916
lemma dvd_mult_cancel: "!!k::nat. [| k*m dvd k*n; 0<k |] ==> m dvd n"
wenzelm@22718
   917
  unfolding dvd_def
wenzelm@22718
   918
  apply (erule exE)
wenzelm@22718
   919
  apply (simp add: mult_ac)
wenzelm@22718
   920
  done
paulson@14267
   921
paulson@14267
   922
lemma dvd_mult_cancel1: "0<m ==> (m*n dvd m) = (n = (1::nat))"
wenzelm@22718
   923
  apply auto
wenzelm@22718
   924
   apply (subgoal_tac "m*n dvd m*1")
wenzelm@22718
   925
   apply (drule dvd_mult_cancel, auto)
wenzelm@22718
   926
  done
paulson@14267
   927
paulson@14267
   928
lemma dvd_mult_cancel2: "0<m ==> (n*m dvd m) = (n = (1::nat))"
wenzelm@22718
   929
  apply (subst mult_commute)
wenzelm@22718
   930
  apply (erule dvd_mult_cancel1)
wenzelm@22718
   931
  done
paulson@14267
   932
paulson@14267
   933
lemma dvd_imp_le: "[| k dvd n; 0 < n |] ==> k \<le> (n::nat)"
haftmann@30923
   934
  by (auto elim!: dvdE) (auto simp add: gr0_conv_Suc)
paulson@14267
   935
paulson@14267
   936
lemma dvd_mult_div_cancel: "n dvd m ==> n * (m div n) = (m::nat)"
haftmann@30923
   937
  by (simp add: dvd_eq_mod_eq_0 mult_div_cancel)
paulson@14267
   938
nipkow@25162
   939
lemma nat_zero_less_power_iff [simp]: "(x^n > 0) = (x > (0::nat) | n=0)"
wenzelm@22718
   940
  by (induct n) auto
haftmann@21408
   941
haftmann@21408
   942
lemma power_dvd_imp_le: "[|i^m dvd i^n;  (1::nat) < i|] ==> m \<le> n"
wenzelm@22718
   943
  apply (rule power_le_imp_le_exp, assumption)
wenzelm@22718
   944
  apply (erule dvd_imp_le, simp)
wenzelm@22718
   945
  done
haftmann@21408
   946
paulson@14267
   947
lemma mod_eq_0_iff: "(m mod d = 0) = (\<exists>q::nat. m = d*q)"
nipkow@29667
   948
by (auto simp add: dvd_eq_mod_eq_0 [symmetric] dvd_def)
paulson@17084
   949
wenzelm@22718
   950
lemmas mod_eq_0D [dest!] = mod_eq_0_iff [THEN iffD1]
paulson@14267
   951
paulson@14267
   952
(*Loses information, namely we also have r<d provided d is nonzero*)
paulson@14267
   953
lemma mod_eqD: "(m mod d = r) ==> \<exists>q::nat. m = r + q*d"
haftmann@27651
   954
  apply (cut_tac a = m in mod_div_equality)
wenzelm@22718
   955
  apply (simp only: add_ac)
wenzelm@22718
   956
  apply (blast intro: sym)
wenzelm@22718
   957
  done
paulson@14267
   958
nipkow@13152
   959
lemma split_div:
nipkow@13189
   960
 "P(n div k :: nat) =
nipkow@13189
   961
 ((k = 0 \<longrightarrow> P 0) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P i)))"
nipkow@13189
   962
 (is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")
nipkow@13189
   963
proof
nipkow@13189
   964
  assume P: ?P
nipkow@13189
   965
  show ?Q
nipkow@13189
   966
  proof (cases)
nipkow@13189
   967
    assume "k = 0"
haftmann@27651
   968
    with P show ?Q by simp
nipkow@13189
   969
  next
nipkow@13189
   970
    assume not0: "k \<noteq> 0"
nipkow@13189
   971
    thus ?Q
nipkow@13189
   972
    proof (simp, intro allI impI)
nipkow@13189
   973
      fix i j
nipkow@13189
   974
      assume n: "n = k*i + j" and j: "j < k"
nipkow@13189
   975
      show "P i"
nipkow@13189
   976
      proof (cases)
wenzelm@22718
   977
        assume "i = 0"
wenzelm@22718
   978
        with n j P show "P i" by simp
nipkow@13189
   979
      next
wenzelm@22718
   980
        assume "i \<noteq> 0"
wenzelm@22718
   981
        with not0 n j P show "P i" by(simp add:add_ac)
nipkow@13189
   982
      qed
nipkow@13189
   983
    qed
nipkow@13189
   984
  qed
nipkow@13189
   985
next
nipkow@13189
   986
  assume Q: ?Q
nipkow@13189
   987
  show ?P
nipkow@13189
   988
  proof (cases)
nipkow@13189
   989
    assume "k = 0"
haftmann@27651
   990
    with Q show ?P by simp
nipkow@13189
   991
  next
nipkow@13189
   992
    assume not0: "k \<noteq> 0"
nipkow@13189
   993
    with Q have R: ?R by simp
nipkow@13189
   994
    from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]
nipkow@13517
   995
    show ?P by simp
nipkow@13189
   996
  qed
nipkow@13189
   997
qed
nipkow@13189
   998
berghofe@13882
   999
lemma split_div_lemma:
haftmann@26100
  1000
  assumes "0 < n"
haftmann@26100
  1001
  shows "n * q \<le> m \<and> m < n * Suc q \<longleftrightarrow> q = ((m\<Colon>nat) div n)" (is "?lhs \<longleftrightarrow> ?rhs")
haftmann@26100
  1002
proof
haftmann@26100
  1003
  assume ?rhs
haftmann@26100
  1004
  with mult_div_cancel have nq: "n * q = m - (m mod n)" by simp
haftmann@26100
  1005
  then have A: "n * q \<le> m" by simp
haftmann@26100
  1006
  have "n - (m mod n) > 0" using mod_less_divisor assms by auto
haftmann@26100
  1007
  then have "m < m + (n - (m mod n))" by simp
haftmann@26100
  1008
  then have "m < n + (m - (m mod n))" by simp
haftmann@26100
  1009
  with nq have "m < n + n * q" by simp
haftmann@26100
  1010
  then have B: "m < n * Suc q" by simp
haftmann@26100
  1011
  from A B show ?lhs ..
haftmann@26100
  1012
next
haftmann@26100
  1013
  assume P: ?lhs
haftmann@30923
  1014
  then have "divmod_rel m n (q, m - n * q)"
haftmann@26100
  1015
    unfolding divmod_rel_def by (auto simp add: mult_ac)
haftmann@30923
  1016
  with divmod_rel_unique divmod_rel [of m n]
haftmann@30923
  1017
  have "(q, m - n * q) = (m div n, m mod n)" by auto
haftmann@30923
  1018
  then show ?rhs by simp
haftmann@26100
  1019
qed
berghofe@13882
  1020
berghofe@13882
  1021
theorem split_div':
berghofe@13882
  1022
  "P ((m::nat) div n) = ((n = 0 \<and> P 0) \<or>
paulson@14267
  1023
   (\<exists>q. (n * q \<le> m \<and> m < n * (Suc q)) \<and> P q))"
berghofe@13882
  1024
  apply (case_tac "0 < n")
berghofe@13882
  1025
  apply (simp only: add: split_div_lemma)
haftmann@27651
  1026
  apply simp_all
berghofe@13882
  1027
  done
berghofe@13882
  1028
nipkow@13189
  1029
lemma split_mod:
nipkow@13189
  1030
 "P(n mod k :: nat) =
nipkow@13189
  1031
 ((k = 0 \<longrightarrow> P n) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P j)))"
nipkow@13189
  1032
 (is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")
nipkow@13189
  1033
proof
nipkow@13189
  1034
  assume P: ?P
nipkow@13189
  1035
  show ?Q
nipkow@13189
  1036
  proof (cases)
nipkow@13189
  1037
    assume "k = 0"
haftmann@27651
  1038
    with P show ?Q by simp
nipkow@13189
  1039
  next
nipkow@13189
  1040
    assume not0: "k \<noteq> 0"
nipkow@13189
  1041
    thus ?Q
nipkow@13189
  1042
    proof (simp, intro allI impI)
nipkow@13189
  1043
      fix i j
nipkow@13189
  1044
      assume "n = k*i + j" "j < k"
nipkow@13189
  1045
      thus "P j" using not0 P by(simp add:add_ac mult_ac)
nipkow@13189
  1046
    qed
nipkow@13189
  1047
  qed
nipkow@13189
  1048
next
nipkow@13189
  1049
  assume Q: ?Q
nipkow@13189
  1050
  show ?P
nipkow@13189
  1051
  proof (cases)
nipkow@13189
  1052
    assume "k = 0"
haftmann@27651
  1053
    with Q show ?P by simp
nipkow@13189
  1054
  next
nipkow@13189
  1055
    assume not0: "k \<noteq> 0"
nipkow@13189
  1056
    with Q have R: ?R by simp
nipkow@13189
  1057
    from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]
nipkow@13517
  1058
    show ?P by simp
nipkow@13189
  1059
  qed
nipkow@13189
  1060
qed
nipkow@13189
  1061
berghofe@13882
  1062
theorem mod_div_equality': "(m::nat) mod n = m - (m div n) * n"
berghofe@13882
  1063
  apply (rule_tac P="%x. m mod n = x - (m div n) * n" in
berghofe@13882
  1064
    subst [OF mod_div_equality [of _ n]])
berghofe@13882
  1065
  apply arith
berghofe@13882
  1066
  done
berghofe@13882
  1067
haftmann@22800
  1068
lemma div_mod_equality':
haftmann@22800
  1069
  fixes m n :: nat
haftmann@22800
  1070
  shows "m div n * n = m - m mod n"
haftmann@22800
  1071
proof -
haftmann@22800
  1072
  have "m mod n \<le> m mod n" ..
haftmann@22800
  1073
  from div_mod_equality have 
haftmann@22800
  1074
    "m div n * n + m mod n - m mod n = m - m mod n" by simp
haftmann@22800
  1075
  with diff_add_assoc [OF `m mod n \<le> m mod n`, of "m div n * n"] have
haftmann@22800
  1076
    "m div n * n + (m mod n - m mod n) = m - m mod n"
haftmann@22800
  1077
    by simp
haftmann@22800
  1078
  then show ?thesis by simp
haftmann@22800
  1079
qed
haftmann@22800
  1080
haftmann@22800
  1081
haftmann@25942
  1082
subsubsection {*An ``induction'' law for modulus arithmetic.*}
paulson@14640
  1083
paulson@14640
  1084
lemma mod_induct_0:
paulson@14640
  1085
  assumes step: "\<forall>i<p. P i \<longrightarrow> P ((Suc i) mod p)"
paulson@14640
  1086
  and base: "P i" and i: "i<p"
paulson@14640
  1087
  shows "P 0"
paulson@14640
  1088
proof (rule ccontr)
paulson@14640
  1089
  assume contra: "\<not>(P 0)"
paulson@14640
  1090
  from i have p: "0<p" by simp
paulson@14640
  1091
  have "\<forall>k. 0<k \<longrightarrow> \<not> P (p-k)" (is "\<forall>k. ?A k")
paulson@14640
  1092
  proof
paulson@14640
  1093
    fix k
paulson@14640
  1094
    show "?A k"
paulson@14640
  1095
    proof (induct k)
paulson@14640
  1096
      show "?A 0" by simp  -- "by contradiction"
paulson@14640
  1097
    next
paulson@14640
  1098
      fix n
paulson@14640
  1099
      assume ih: "?A n"
paulson@14640
  1100
      show "?A (Suc n)"
paulson@14640
  1101
      proof (clarsimp)
wenzelm@22718
  1102
        assume y: "P (p - Suc n)"
wenzelm@22718
  1103
        have n: "Suc n < p"
wenzelm@22718
  1104
        proof (rule ccontr)
wenzelm@22718
  1105
          assume "\<not>(Suc n < p)"
wenzelm@22718
  1106
          hence "p - Suc n = 0"
wenzelm@22718
  1107
            by simp
wenzelm@22718
  1108
          with y contra show "False"
wenzelm@22718
  1109
            by simp
wenzelm@22718
  1110
        qed
wenzelm@22718
  1111
        hence n2: "Suc (p - Suc n) = p-n" by arith
wenzelm@22718
  1112
        from p have "p - Suc n < p" by arith
wenzelm@22718
  1113
        with y step have z: "P ((Suc (p - Suc n)) mod p)"
wenzelm@22718
  1114
          by blast
wenzelm@22718
  1115
        show "False"
wenzelm@22718
  1116
        proof (cases "n=0")
wenzelm@22718
  1117
          case True
wenzelm@22718
  1118
          with z n2 contra show ?thesis by simp
wenzelm@22718
  1119
        next
wenzelm@22718
  1120
          case False
wenzelm@22718
  1121
          with p have "p-n < p" by arith
wenzelm@22718
  1122
          with z n2 False ih show ?thesis by simp
wenzelm@22718
  1123
        qed
paulson@14640
  1124
      qed
paulson@14640
  1125
    qed
paulson@14640
  1126
  qed
paulson@14640
  1127
  moreover
paulson@14640
  1128
  from i obtain k where "0<k \<and> i+k=p"
paulson@14640
  1129
    by (blast dest: less_imp_add_positive)
paulson@14640
  1130
  hence "0<k \<and> i=p-k" by auto
paulson@14640
  1131
  moreover
paulson@14640
  1132
  note base
paulson@14640
  1133
  ultimately
paulson@14640
  1134
  show "False" by blast
paulson@14640
  1135
qed
paulson@14640
  1136
paulson@14640
  1137
lemma mod_induct:
paulson@14640
  1138
  assumes step: "\<forall>i<p. P i \<longrightarrow> P ((Suc i) mod p)"
paulson@14640
  1139
  and base: "P i" and i: "i<p" and j: "j<p"
paulson@14640
  1140
  shows "P j"
paulson@14640
  1141
proof -
paulson@14640
  1142
  have "\<forall>j<p. P j"
paulson@14640
  1143
  proof
paulson@14640
  1144
    fix j
paulson@14640
  1145
    show "j<p \<longrightarrow> P j" (is "?A j")
paulson@14640
  1146
    proof (induct j)
paulson@14640
  1147
      from step base i show "?A 0"
wenzelm@22718
  1148
        by (auto elim: mod_induct_0)
paulson@14640
  1149
    next
paulson@14640
  1150
      fix k
paulson@14640
  1151
      assume ih: "?A k"
paulson@14640
  1152
      show "?A (Suc k)"
paulson@14640
  1153
      proof
wenzelm@22718
  1154
        assume suc: "Suc k < p"
wenzelm@22718
  1155
        hence k: "k<p" by simp
wenzelm@22718
  1156
        with ih have "P k" ..
wenzelm@22718
  1157
        with step k have "P (Suc k mod p)"
wenzelm@22718
  1158
          by blast
wenzelm@22718
  1159
        moreover
wenzelm@22718
  1160
        from suc have "Suc k mod p = Suc k"
wenzelm@22718
  1161
          by simp
wenzelm@22718
  1162
        ultimately
wenzelm@22718
  1163
        show "P (Suc k)" by simp
paulson@14640
  1164
      qed
paulson@14640
  1165
    qed
paulson@14640
  1166
  qed
paulson@14640
  1167
  with j show ?thesis by blast
paulson@14640
  1168
qed
paulson@14640
  1169
haftmann@30923
  1170
lemma nat_dvd_not_less:
haftmann@30923
  1171
  fixes m n :: nat
haftmann@30923
  1172
  shows "0 < m \<Longrightarrow> m < n \<Longrightarrow> \<not> n dvd m"
haftmann@30923
  1173
by (auto elim!: dvdE) (auto simp add: gr0_conv_Suc)
haftmann@30923
  1174
paulson@3366
  1175
end