src/HOL/Divides.thy
author nipkow
Wed Apr 01 16:03:00 2009 +0200 (2009-04-01)
changeset 30837 3d4832d9f7e4
parent 30729 461ee3e49ad3
child 30840 98809b3f5e3c
permissions -rw-r--r--
added strong_setprod_cong[cong] (in analogy with setsum)
added some lemmas
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(*  Title:      HOL/Divides.thy
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    ID:         $Id$
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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 + 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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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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end
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lemma div_mult_div_if_dvd: "(y::'a::{semiring_div,no_zero_divisors}) dvd x \<Longrightarrow> 
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  z dvd w \<Longrightarrow> (x div y) * (w div z) = (x * w) div (y * z)"
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unfolding dvd_def
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  apply clarify
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  apply (case_tac "y = 0")
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  apply simp
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  apply (case_tac "z = 0")
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  apply simp
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  apply (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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   315
nipkow@30476
   316
lemma div_power: "(y::'a::{semiring_div,no_zero_divisors,recpower}) dvd x \<Longrightarrow>
nipkow@30476
   317
    (x div y)^n = x^n div y^n"
nipkow@30476
   318
apply (induct n)
nipkow@30476
   319
 apply simp
nipkow@30476
   320
apply(simp add: div_mult_div_if_dvd dvd_power_same)
nipkow@30476
   321
done
nipkow@30476
   322
huffman@29405
   323
class ring_div = semiring_div + comm_ring_1
huffman@29405
   324
begin
huffman@29405
   325
huffman@29405
   326
text {* Negation respects modular equivalence. *}
huffman@29405
   327
huffman@29405
   328
lemma mod_minus_eq: "(- a) mod b = (- (a mod b)) mod b"
huffman@29405
   329
proof -
huffman@29405
   330
  have "(- a) mod b = (- (a div b * b + a mod b)) mod b"
huffman@29405
   331
    by (simp only: mod_div_equality)
huffman@29405
   332
  also have "\<dots> = (- (a mod b) + - (a div b) * b) mod b"
huffman@29405
   333
    by (simp only: minus_add_distrib minus_mult_left add_ac)
huffman@29405
   334
  also have "\<dots> = (- (a mod b)) mod b"
huffman@29405
   335
    by (rule mod_mult_self1)
huffman@29405
   336
  finally show ?thesis .
huffman@29405
   337
qed
huffman@29405
   338
huffman@29405
   339
lemma mod_minus_cong:
huffman@29405
   340
  assumes "a mod b = a' mod b"
huffman@29405
   341
  shows "(- a) mod b = (- a') mod b"
huffman@29405
   342
proof -
huffman@29405
   343
  have "(- (a mod b)) mod b = (- (a' mod b)) mod b"
huffman@29405
   344
    unfolding assms ..
huffman@29405
   345
  thus ?thesis
huffman@29405
   346
    by (simp only: mod_minus_eq [symmetric])
huffman@29405
   347
qed
huffman@29405
   348
huffman@29405
   349
text {* Subtraction respects modular equivalence. *}
huffman@29405
   350
huffman@29405
   351
lemma mod_diff_left_eq: "(a - b) mod c = (a mod c - b) mod c"
huffman@29405
   352
  unfolding diff_minus
huffman@29405
   353
  by (intro mod_add_cong mod_minus_cong) simp_all
huffman@29405
   354
huffman@29405
   355
lemma mod_diff_right_eq: "(a - b) mod c = (a - b mod c) mod c"
huffman@29405
   356
  unfolding diff_minus
huffman@29405
   357
  by (intro mod_add_cong mod_minus_cong) simp_all
huffman@29405
   358
huffman@29405
   359
lemma mod_diff_eq: "(a - b) mod c = (a mod c - b mod c) mod c"
huffman@29405
   360
  unfolding diff_minus
huffman@29405
   361
  by (intro mod_add_cong mod_minus_cong) simp_all
huffman@29405
   362
huffman@29405
   363
lemma mod_diff_cong:
huffman@29405
   364
  assumes "a mod c = a' mod c"
huffman@29405
   365
  assumes "b mod c = b' mod c"
huffman@29405
   366
  shows "(a - b) mod c = (a' - b') mod c"
huffman@29405
   367
  unfolding diff_minus using assms
huffman@29405
   368
  by (intro mod_add_cong mod_minus_cong)
huffman@29405
   369
nipkow@30180
   370
lemma dvd_neg_div: "y dvd x \<Longrightarrow> -x div y = - (x div y)"
nipkow@30180
   371
apply (case_tac "y = 0") apply simp
nipkow@30180
   372
apply (auto simp add: dvd_def)
nipkow@30180
   373
apply (subgoal_tac "-(y * k) = y * - k")
nipkow@30180
   374
 apply (erule ssubst)
nipkow@30180
   375
 apply (erule div_mult_self1_is_id)
nipkow@30180
   376
apply simp
nipkow@30180
   377
done
nipkow@30180
   378
nipkow@30180
   379
lemma dvd_div_neg: "y dvd x \<Longrightarrow> x div -y = - (x div y)"
nipkow@30180
   380
apply (case_tac "y = 0") apply simp
nipkow@30180
   381
apply (auto simp add: dvd_def)
nipkow@30180
   382
apply (subgoal_tac "y * k = -y * -k")
nipkow@30180
   383
 apply (erule ssubst)
nipkow@30180
   384
 apply (rule div_mult_self1_is_id)
nipkow@30180
   385
 apply simp
nipkow@30180
   386
apply simp
nipkow@30180
   387
done
nipkow@30180
   388
huffman@29405
   389
end
huffman@29405
   390
haftmann@25942
   391
haftmann@26100
   392
subsection {* Division on @{typ nat} *}
haftmann@26100
   393
haftmann@26100
   394
text {*
haftmann@26100
   395
  We define @{const div} and @{const mod} on @{typ nat} by means
haftmann@26100
   396
  of a characteristic relation with two input arguments
haftmann@26100
   397
  @{term "m\<Colon>nat"}, @{term "n\<Colon>nat"} and two output arguments
haftmann@26100
   398
  @{term "q\<Colon>nat"}(uotient) and @{term "r\<Colon>nat"}(emainder).
haftmann@26100
   399
*}
haftmann@26100
   400
haftmann@26100
   401
definition divmod_rel :: "nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> bool" where
haftmann@26100
   402
  "divmod_rel m n q r \<longleftrightarrow> m = q * n + r \<and> (if n > 0 then 0 \<le> r \<and> r < n else q = 0)"
haftmann@26100
   403
haftmann@26100
   404
text {* @{const divmod_rel} is total: *}
haftmann@26100
   405
haftmann@26100
   406
lemma divmod_rel_ex:
haftmann@26100
   407
  obtains q r where "divmod_rel m n q r"
haftmann@26100
   408
proof (cases "n = 0")
haftmann@26100
   409
  case True with that show thesis
haftmann@26100
   410
    by (auto simp add: divmod_rel_def)
haftmann@26100
   411
next
haftmann@26100
   412
  case False
haftmann@26100
   413
  have "\<exists>q r. m = q * n + r \<and> r < n"
haftmann@26100
   414
  proof (induct m)
haftmann@26100
   415
    case 0 with `n \<noteq> 0`
haftmann@26100
   416
    have "(0\<Colon>nat) = 0 * n + 0 \<and> 0 < n" by simp
haftmann@26100
   417
    then show ?case by blast
haftmann@26100
   418
  next
haftmann@26100
   419
    case (Suc m) then obtain q' r'
haftmann@26100
   420
      where m: "m = q' * n + r'" and n: "r' < n" by auto
haftmann@26100
   421
    then show ?case proof (cases "Suc r' < n")
haftmann@26100
   422
      case True
haftmann@26100
   423
      from m n have "Suc m = q' * n + Suc r'" by simp
haftmann@26100
   424
      with True show ?thesis by blast
haftmann@26100
   425
    next
haftmann@26100
   426
      case False then have "n \<le> Suc r'" by auto
haftmann@26100
   427
      moreover from n have "Suc r' \<le> n" by auto
haftmann@26100
   428
      ultimately have "n = Suc r'" by auto
haftmann@26100
   429
      with m have "Suc m = Suc q' * n + 0" by simp
haftmann@26100
   430
      with `n \<noteq> 0` show ?thesis by blast
haftmann@26100
   431
    qed
haftmann@26100
   432
  qed
haftmann@26100
   433
  with that show thesis
haftmann@26100
   434
    using `n \<noteq> 0` by (auto simp add: divmod_rel_def)
haftmann@26100
   435
qed
haftmann@26100
   436
haftmann@26100
   437
text {* @{const divmod_rel} is injective: *}
haftmann@26100
   438
haftmann@26100
   439
lemma divmod_rel_unique_div:
haftmann@26100
   440
  assumes "divmod_rel m n q r"
haftmann@26100
   441
    and "divmod_rel m n q' r'"
haftmann@26100
   442
  shows "q = q'"
haftmann@26100
   443
proof (cases "n = 0")
haftmann@26100
   444
  case True with assms show ?thesis
haftmann@26100
   445
    by (simp add: divmod_rel_def)
haftmann@26100
   446
next
haftmann@26100
   447
  case False
haftmann@26100
   448
  have aux: "\<And>q r q' r'. q' * n + r' = q * n + r \<Longrightarrow> r < n \<Longrightarrow> q' \<le> (q\<Colon>nat)"
haftmann@26100
   449
  apply (rule leI)
haftmann@26100
   450
  apply (subst less_iff_Suc_add)
haftmann@26100
   451
  apply (auto simp add: add_mult_distrib)
haftmann@26100
   452
  done
haftmann@26100
   453
  from `n \<noteq> 0` assms show ?thesis
haftmann@26100
   454
    by (auto simp add: divmod_rel_def
haftmann@26100
   455
      intro: order_antisym dest: aux sym)
haftmann@26100
   456
qed
haftmann@26100
   457
haftmann@26100
   458
lemma divmod_rel_unique_mod:
haftmann@26100
   459
  assumes "divmod_rel m n q r"
haftmann@26100
   460
    and "divmod_rel m n q' r'"
haftmann@26100
   461
  shows "r = r'"
haftmann@26100
   462
proof -
haftmann@26100
   463
  from assms have "q = q'" by (rule divmod_rel_unique_div)
haftmann@26100
   464
  with assms show ?thesis by (simp add: divmod_rel_def)
haftmann@26100
   465
qed
haftmann@26100
   466
haftmann@26100
   467
text {*
haftmann@26100
   468
  We instantiate divisibility on the natural numbers by
haftmann@26100
   469
  means of @{const divmod_rel}:
haftmann@26100
   470
*}
haftmann@25942
   471
haftmann@25942
   472
instantiation nat :: semiring_div
haftmann@25571
   473
begin
haftmann@25571
   474
haftmann@26100
   475
definition divmod :: "nat \<Rightarrow> nat \<Rightarrow> nat \<times> nat" where
haftmann@28562
   476
  [code del]: "divmod m n = (THE (q, r). divmod_rel m n q r)"
haftmann@26100
   477
haftmann@26100
   478
definition div_nat where
haftmann@26100
   479
  "m div n = fst (divmod m n)"
haftmann@26100
   480
haftmann@26100
   481
definition mod_nat where
haftmann@26100
   482
  "m mod n = snd (divmod m n)"
haftmann@25571
   483
haftmann@26100
   484
lemma divmod_div_mod:
haftmann@26100
   485
  "divmod m n = (m div n, m mod n)"
haftmann@26100
   486
  unfolding div_nat_def mod_nat_def by simp
haftmann@26100
   487
haftmann@26100
   488
lemma divmod_eq:
haftmann@26100
   489
  assumes "divmod_rel m n q r" 
haftmann@26100
   490
  shows "divmod m n = (q, r)"
haftmann@26100
   491
  using assms by (auto simp add: divmod_def
haftmann@26100
   492
    dest: divmod_rel_unique_div divmod_rel_unique_mod)
haftmann@25942
   493
haftmann@26100
   494
lemma div_eq:
haftmann@26100
   495
  assumes "divmod_rel m n q r" 
haftmann@26100
   496
  shows "m div n = q"
haftmann@26100
   497
  using assms by (auto dest: divmod_eq simp add: div_nat_def)
haftmann@26100
   498
haftmann@26100
   499
lemma mod_eq:
haftmann@26100
   500
  assumes "divmod_rel m n q r" 
haftmann@26100
   501
  shows "m mod n = r"
haftmann@26100
   502
  using assms by (auto dest: divmod_eq simp add: mod_nat_def)
haftmann@25571
   503
haftmann@26100
   504
lemma divmod_rel: "divmod_rel m n (m div n) (m mod n)"
haftmann@26100
   505
proof -
haftmann@26100
   506
  from divmod_rel_ex
haftmann@26100
   507
    obtain q r where rel: "divmod_rel m n q r" .
haftmann@26100
   508
  moreover with div_eq mod_eq have "m div n = q" and "m mod n = r"
haftmann@26100
   509
    by simp_all
haftmann@26100
   510
  ultimately show ?thesis by simp
haftmann@26100
   511
qed
paulson@14267
   512
haftmann@26100
   513
lemma divmod_zero:
haftmann@26100
   514
  "divmod m 0 = (0, m)"
haftmann@26100
   515
proof -
haftmann@26100
   516
  from divmod_rel [of m 0] show ?thesis
haftmann@26100
   517
    unfolding divmod_div_mod divmod_rel_def by simp
haftmann@26100
   518
qed
haftmann@25942
   519
haftmann@26100
   520
lemma divmod_base:
haftmann@26100
   521
  assumes "m < n"
haftmann@26100
   522
  shows "divmod m n = (0, m)"
haftmann@26100
   523
proof -
haftmann@26100
   524
  from divmod_rel [of m n] show ?thesis
haftmann@26100
   525
    unfolding divmod_div_mod divmod_rel_def
haftmann@26100
   526
    using assms by (cases "m div n = 0")
haftmann@26100
   527
      (auto simp add: gr0_conv_Suc [of "m div n"])
haftmann@26100
   528
qed
haftmann@25942
   529
haftmann@26100
   530
lemma divmod_step:
haftmann@26100
   531
  assumes "0 < n" and "n \<le> m"
haftmann@26100
   532
  shows "divmod m n = (Suc ((m - n) div n), (m - n) mod n)"
haftmann@26100
   533
proof -
haftmann@26100
   534
  from divmod_rel have divmod_m_n: "divmod_rel m n (m div n) (m mod n)" .
haftmann@26100
   535
  with assms have m_div_n: "m div n \<ge> 1"
haftmann@26100
   536
    by (cases "m div n") (auto simp add: divmod_rel_def)
huffman@30079
   537
  from assms divmod_m_n have "divmod_rel (m - n) n (m div n - Suc 0) (m mod n)"
haftmann@26100
   538
    by (cases "m div n") (auto simp add: divmod_rel_def)
huffman@30079
   539
  with divmod_eq have "divmod (m - n) n = (m div n - Suc 0, m mod n)" by simp
haftmann@26100
   540
  moreover from divmod_div_mod have "divmod (m - n) n = ((m - n) div n, (m - n) mod n)" .
haftmann@26100
   541
  ultimately have "m div n = Suc ((m - n) div n)"
haftmann@26100
   542
    and "m mod n = (m - n) mod n" using m_div_n by simp_all
haftmann@26100
   543
  then show ?thesis using divmod_div_mod by simp
haftmann@26100
   544
qed
haftmann@25942
   545
wenzelm@26300
   546
text {* The ''recursion'' equations for @{const div} and @{const mod} *}
haftmann@26100
   547
haftmann@26100
   548
lemma div_less [simp]:
haftmann@26100
   549
  fixes m n :: nat
haftmann@26100
   550
  assumes "m < n"
haftmann@26100
   551
  shows "m div n = 0"
haftmann@26100
   552
  using assms divmod_base divmod_div_mod by simp
haftmann@25942
   553
haftmann@26100
   554
lemma le_div_geq:
haftmann@26100
   555
  fixes m n :: nat
haftmann@26100
   556
  assumes "0 < n" and "n \<le> m"
haftmann@26100
   557
  shows "m div n = Suc ((m - n) div n)"
haftmann@26100
   558
  using assms divmod_step divmod_div_mod by simp
paulson@14267
   559
haftmann@26100
   560
lemma mod_less [simp]:
haftmann@26100
   561
  fixes m n :: nat
haftmann@26100
   562
  assumes "m < n"
haftmann@26100
   563
  shows "m mod n = m"
haftmann@26100
   564
  using assms divmod_base divmod_div_mod by simp
haftmann@26100
   565
haftmann@26100
   566
lemma le_mod_geq:
haftmann@26100
   567
  fixes m n :: nat
haftmann@26100
   568
  assumes "n \<le> m"
haftmann@26100
   569
  shows "m mod n = (m - n) mod n"
haftmann@26100
   570
  using assms divmod_step divmod_div_mod by (cases "n = 0") simp_all
paulson@14267
   571
haftmann@25942
   572
instance proof
haftmann@26100
   573
  fix m n :: nat show "m div n * n + m mod n = m"
haftmann@26100
   574
    using divmod_rel [of m n] by (simp add: divmod_rel_def)
haftmann@25942
   575
next
haftmann@26100
   576
  fix n :: nat show "n div 0 = 0"
haftmann@26100
   577
    using divmod_zero divmod_div_mod [of n 0] by simp
haftmann@25942
   578
next
haftmann@27651
   579
  fix n :: nat show "0 div n = 0"
haftmann@27651
   580
    using divmod_rel [of 0 n] by (cases n) (simp_all add: divmod_rel_def)
haftmann@27651
   581
next
haftmann@27651
   582
  fix m n q :: nat assume "n \<noteq> 0" then show "(q + m * n) div n = m + q div n"
haftmann@25942
   583
    by (induct m) (simp_all add: le_div_geq)
haftmann@25942
   584
qed
haftmann@26100
   585
haftmann@25942
   586
end
paulson@14267
   587
haftmann@26100
   588
text {* Simproc for cancelling @{const div} and @{const mod} *}
haftmann@25942
   589
haftmann@27651
   590
(*lemmas mod_div_equality_nat = semiring_div_class.times_div_mod_plus_zero_one.mod_div_equality [of "m\<Colon>nat" n, standard]
haftmann@27651
   591
lemmas mod_div_equality2_nat = mod_div_equality2 [of "n\<Colon>nat" m, standard*)
haftmann@25942
   592
haftmann@25942
   593
ML {*
haftmann@25942
   594
structure CancelDivModData =
haftmann@25942
   595
struct
haftmann@25942
   596
haftmann@26100
   597
val div_name = @{const_name div};
haftmann@26100
   598
val mod_name = @{const_name mod};
haftmann@25942
   599
val mk_binop = HOLogic.mk_binop;
haftmann@30496
   600
val mk_sum = Nat_Arith.mk_sum;
haftmann@30496
   601
val dest_sum = Nat_Arith.dest_sum;
haftmann@25942
   602
haftmann@25942
   603
(*logic*)
paulson@14267
   604
haftmann@25942
   605
val div_mod_eqs = map mk_meta_eq [@{thm div_mod_equality}, @{thm div_mod_equality2}]
haftmann@25942
   606
haftmann@25942
   607
val trans = trans
haftmann@25942
   608
haftmann@25942
   609
val prove_eq_sums =
haftmann@25942
   610
  let val simps = @{thm add_0} :: @{thm add_0_right} :: @{thms add_ac}
haftmann@30496
   611
  in Arith_Data.prove_conv2 all_tac (Arith_Data.simp_all_tac simps) end;
haftmann@25942
   612
haftmann@25942
   613
end;
haftmann@25942
   614
haftmann@25942
   615
structure CancelDivMod = CancelDivModFun(CancelDivModData);
haftmann@25942
   616
wenzelm@28262
   617
val cancel_div_mod_proc = Simplifier.simproc (the_context ())
haftmann@26100
   618
  "cancel_div_mod" ["(m::nat) + n"] (K CancelDivMod.proc);
haftmann@25942
   619
haftmann@25942
   620
Addsimprocs[cancel_div_mod_proc];
haftmann@25942
   621
*}
haftmann@25942
   622
haftmann@26100
   623
text {* code generator setup *}
haftmann@26100
   624
haftmann@26100
   625
lemma divmod_if [code]: "divmod m n = (if n = 0 \<or> m < n then (0, m) else
haftmann@26100
   626
  let (q, r) = divmod (m - n) n in (Suc q, r))"
nipkow@29667
   627
by (simp add: divmod_zero divmod_base divmod_step)
haftmann@26100
   628
    (simp add: divmod_div_mod)
haftmann@26100
   629
haftmann@26100
   630
code_modulename SML
haftmann@26100
   631
  Divides Nat
haftmann@26100
   632
haftmann@26100
   633
code_modulename OCaml
haftmann@26100
   634
  Divides Nat
haftmann@26100
   635
haftmann@26100
   636
code_modulename Haskell
haftmann@26100
   637
  Divides Nat
haftmann@26100
   638
haftmann@26100
   639
haftmann@26100
   640
subsubsection {* Quotient *}
haftmann@26100
   641
haftmann@26100
   642
lemma div_geq: "0 < n \<Longrightarrow>  \<not> m < n \<Longrightarrow> m div n = Suc ((m - n) div n)"
nipkow@29667
   643
by (simp add: le_div_geq linorder_not_less)
haftmann@26100
   644
haftmann@26100
   645
lemma div_if: "0 < n \<Longrightarrow> m div n = (if m < n then 0 else Suc ((m - n) div n))"
nipkow@29667
   646
by (simp add: div_geq)
haftmann@26100
   647
haftmann@26100
   648
lemma div_mult_self_is_m [simp]: "0<n ==> (m*n) div n = (m::nat)"
nipkow@29667
   649
by simp
haftmann@26100
   650
haftmann@26100
   651
lemma div_mult_self1_is_m [simp]: "0<n ==> (n*m) div n = (m::nat)"
nipkow@29667
   652
by simp
haftmann@26100
   653
haftmann@25942
   654
haftmann@25942
   655
subsubsection {* Remainder *}
haftmann@25942
   656
haftmann@26100
   657
lemma mod_less_divisor [simp]:
haftmann@26100
   658
  fixes m n :: nat
haftmann@26100
   659
  assumes "n > 0"
haftmann@26100
   660
  shows "m mod n < (n::nat)"
haftmann@26100
   661
  using assms divmod_rel unfolding divmod_rel_def by auto
paulson@14267
   662
haftmann@26100
   663
lemma mod_less_eq_dividend [simp]:
haftmann@26100
   664
  fixes m n :: nat
haftmann@26100
   665
  shows "m mod n \<le> m"
haftmann@26100
   666
proof (rule add_leD2)
haftmann@26100
   667
  from mod_div_equality have "m div n * n + m mod n = m" .
haftmann@26100
   668
  then show "m div n * n + m mod n \<le> m" by auto
haftmann@26100
   669
qed
haftmann@26100
   670
haftmann@26100
   671
lemma mod_geq: "\<not> m < (n\<Colon>nat) \<Longrightarrow> m mod n = (m - n) mod n"
nipkow@29667
   672
by (simp add: le_mod_geq linorder_not_less)
paulson@14267
   673
haftmann@26100
   674
lemma mod_if: "m mod (n\<Colon>nat) = (if m < n then m else (m - n) mod n)"
nipkow@29667
   675
by (simp add: le_mod_geq)
haftmann@26100
   676
paulson@14267
   677
lemma mod_1 [simp]: "m mod Suc 0 = 0"
nipkow@29667
   678
by (induct m) (simp_all add: mod_geq)
paulson@14267
   679
haftmann@26100
   680
lemma mod_mult_distrib: "(m mod n) * (k\<Colon>nat) = (m * k) mod (n * k)"
wenzelm@22718
   681
  apply (cases "n = 0", simp)
wenzelm@22718
   682
  apply (cases "k = 0", simp)
wenzelm@22718
   683
  apply (induct m rule: nat_less_induct)
wenzelm@22718
   684
  apply (subst mod_if, simp)
wenzelm@22718
   685
  apply (simp add: mod_geq diff_mult_distrib)
wenzelm@22718
   686
  done
paulson@14267
   687
paulson@14267
   688
lemma mod_mult_distrib2: "(k::nat) * (m mod n) = (k*m) mod (k*n)"
nipkow@29667
   689
by (simp add: mult_commute [of k] mod_mult_distrib)
paulson@14267
   690
paulson@14267
   691
(* a simple rearrangement of mod_div_equality: *)
paulson@14267
   692
lemma mult_div_cancel: "(n::nat) * (m div n) = m - (m mod n)"
nipkow@29667
   693
by (cut_tac a = m and b = n in mod_div_equality2, arith)
paulson@14267
   694
nipkow@15439
   695
lemma mod_le_divisor[simp]: "0 < n \<Longrightarrow> m mod n \<le> (n::nat)"
wenzelm@22718
   696
  apply (drule mod_less_divisor [where m = m])
wenzelm@22718
   697
  apply simp
wenzelm@22718
   698
  done
paulson@14267
   699
haftmann@26100
   700
subsubsection {* Quotient and Remainder *}
paulson@14267
   701
haftmann@26100
   702
lemma divmod_rel_mult1_eq:
haftmann@26100
   703
  "[| divmod_rel b c q r; c > 0 |]
haftmann@26100
   704
   ==> divmod_rel (a*b) c (a*q + a*r div c) (a*r mod c)"
nipkow@29667
   705
by (auto simp add: split_ifs divmod_rel_def algebra_simps)
paulson@14267
   706
paulson@14267
   707
lemma div_mult1_eq: "(a*b) div c = a*(b div c) + a*(b mod c) div (c::nat)"
nipkow@25134
   708
apply (cases "c = 0", simp)
haftmann@26100
   709
apply (blast intro: divmod_rel [THEN divmod_rel_mult1_eq, THEN div_eq])
nipkow@25134
   710
done
paulson@14267
   711
haftmann@26100
   712
lemma divmod_rel_add1_eq:
haftmann@26100
   713
  "[| divmod_rel a c aq ar; divmod_rel b c bq br;  c > 0 |]
haftmann@26100
   714
   ==> divmod_rel (a + b) c (aq + bq + (ar+br) div c) ((ar + br) mod c)"
nipkow@29667
   715
by (auto simp add: split_ifs divmod_rel_def algebra_simps)
paulson@14267
   716
paulson@14267
   717
(*NOT suitable for rewriting: the RHS has an instance of the LHS*)
paulson@14267
   718
lemma div_add1_eq:
nipkow@25134
   719
  "(a+b) div (c::nat) = a div c + b div c + ((a mod c + b mod c) div c)"
nipkow@25134
   720
apply (cases "c = 0", simp)
haftmann@26100
   721
apply (blast intro: divmod_rel_add1_eq [THEN div_eq] divmod_rel)
nipkow@25134
   722
done
paulson@14267
   723
paulson@14267
   724
lemma mod_lemma: "[| (0::nat) < c; r < b |] ==> b * (q mod c) + r < b * c"
wenzelm@22718
   725
  apply (cut_tac m = q and n = c in mod_less_divisor)
wenzelm@22718
   726
  apply (drule_tac [2] m = "q mod c" in less_imp_Suc_add, auto)
wenzelm@22718
   727
  apply (erule_tac P = "%x. ?lhs < ?rhs x" in ssubst)
wenzelm@22718
   728
  apply (simp add: add_mult_distrib2)
wenzelm@22718
   729
  done
paulson@10559
   730
haftmann@26100
   731
lemma divmod_rel_mult2_eq: "[| divmod_rel a b q r;  0 < b;  0 < c |]
haftmann@26100
   732
      ==> divmod_rel a (b*c) (q div c) (b*(q mod c) + r)"
nipkow@29667
   733
by (auto simp add: mult_ac divmod_rel_def add_mult_distrib2 [symmetric] mod_lemma)
paulson@14267
   734
paulson@14267
   735
lemma div_mult2_eq: "a div (b*c) = (a div b) div (c::nat)"
wenzelm@22718
   736
  apply (cases "b = 0", simp)
wenzelm@22718
   737
  apply (cases "c = 0", simp)
haftmann@26100
   738
  apply (force simp add: divmod_rel [THEN divmod_rel_mult2_eq, THEN div_eq])
wenzelm@22718
   739
  done
paulson@14267
   740
paulson@14267
   741
lemma mod_mult2_eq: "a mod (b*c) = b*(a div b mod c) + a mod (b::nat)"
wenzelm@22718
   742
  apply (cases "b = 0", simp)
wenzelm@22718
   743
  apply (cases "c = 0", simp)
haftmann@26100
   744
  apply (auto simp add: mult_commute divmod_rel [THEN divmod_rel_mult2_eq, THEN mod_eq])
wenzelm@22718
   745
  done
paulson@14267
   746
paulson@14267
   747
haftmann@25942
   748
subsubsection{*Cancellation of Common Factors in Division*}
paulson@14267
   749
paulson@14267
   750
lemma div_mult_mult_lemma:
wenzelm@22718
   751
    "[| (0::nat) < b;  0 < c |] ==> (c*a) div (c*b) = a div b"
nipkow@29667
   752
by (auto simp add: div_mult2_eq)
paulson@14267
   753
paulson@14267
   754
lemma div_mult_mult1 [simp]: "(0::nat) < c ==> (c*a) div (c*b) = a div b"
wenzelm@22718
   755
  apply (cases "b = 0")
wenzelm@22718
   756
  apply (auto simp add: linorder_neq_iff [of b] div_mult_mult_lemma)
wenzelm@22718
   757
  done
paulson@14267
   758
paulson@14267
   759
lemma div_mult_mult2 [simp]: "(0::nat) < c ==> (a*c) div (b*c) = a div b"
wenzelm@22718
   760
  apply (drule div_mult_mult1)
wenzelm@22718
   761
  apply (auto simp add: mult_commute)
wenzelm@22718
   762
  done
paulson@14267
   763
paulson@14267
   764
haftmann@25942
   765
subsubsection{*Further Facts about Quotient and Remainder*}
paulson@14267
   766
paulson@14267
   767
lemma div_1 [simp]: "m div Suc 0 = m"
nipkow@29667
   768
by (induct m) (simp_all add: div_geq)
paulson@14267
   769
paulson@14267
   770
paulson@14267
   771
(* Monotonicity of div in first argument *)
nipkow@30837
   772
lemma div_le_mono [rule_format]:
wenzelm@22718
   773
    "\<forall>m::nat. m \<le> n --> (m div k) \<le> (n div k)"
paulson@14267
   774
apply (case_tac "k=0", simp)
paulson@15251
   775
apply (induct "n" rule: nat_less_induct, clarify)
paulson@14267
   776
apply (case_tac "n<k")
paulson@14267
   777
(* 1  case n<k *)
paulson@14267
   778
apply simp
paulson@14267
   779
(* 2  case n >= k *)
paulson@14267
   780
apply (case_tac "m<k")
paulson@14267
   781
(* 2.1  case m<k *)
paulson@14267
   782
apply simp
paulson@14267
   783
(* 2.2  case m>=k *)
nipkow@15439
   784
apply (simp add: div_geq diff_le_mono)
paulson@14267
   785
done
paulson@14267
   786
paulson@14267
   787
(* Antimonotonicity of div in second argument *)
paulson@14267
   788
lemma div_le_mono2: "!!m::nat. [| 0<m; m\<le>n |] ==> (k div n) \<le> (k div m)"
paulson@14267
   789
apply (subgoal_tac "0<n")
wenzelm@22718
   790
 prefer 2 apply simp
paulson@15251
   791
apply (induct_tac k rule: nat_less_induct)
paulson@14267
   792
apply (rename_tac "k")
paulson@14267
   793
apply (case_tac "k<n", simp)
paulson@14267
   794
apply (subgoal_tac "~ (k<m) ")
wenzelm@22718
   795
 prefer 2 apply simp
paulson@14267
   796
apply (simp add: div_geq)
paulson@15251
   797
apply (subgoal_tac "(k-n) div n \<le> (k-m) div n")
paulson@14267
   798
 prefer 2
paulson@14267
   799
 apply (blast intro: div_le_mono diff_le_mono2)
paulson@14267
   800
apply (rule le_trans, simp)
nipkow@15439
   801
apply (simp)
paulson@14267
   802
done
paulson@14267
   803
paulson@14267
   804
lemma div_le_dividend [simp]: "m div n \<le> (m::nat)"
paulson@14267
   805
apply (case_tac "n=0", simp)
paulson@14267
   806
apply (subgoal_tac "m div n \<le> m div 1", simp)
paulson@14267
   807
apply (rule div_le_mono2)
paulson@14267
   808
apply (simp_all (no_asm_simp))
paulson@14267
   809
done
paulson@14267
   810
wenzelm@22718
   811
(* Similar for "less than" *)
paulson@17085
   812
lemma div_less_dividend [rule_format]:
paulson@14267
   813
     "!!n::nat. 1<n ==> 0 < m --> m div n < m"
paulson@15251
   814
apply (induct_tac m rule: nat_less_induct)
paulson@14267
   815
apply (rename_tac "m")
paulson@14267
   816
apply (case_tac "m<n", simp)
paulson@14267
   817
apply (subgoal_tac "0<n")
wenzelm@22718
   818
 prefer 2 apply simp
paulson@14267
   819
apply (simp add: div_geq)
paulson@14267
   820
apply (case_tac "n<m")
paulson@15251
   821
 apply (subgoal_tac "(m-n) div n < (m-n) ")
paulson@14267
   822
  apply (rule impI less_trans_Suc)+
paulson@14267
   823
apply assumption
nipkow@15439
   824
  apply (simp_all)
paulson@14267
   825
done
paulson@14267
   826
nipkow@30837
   827
lemma nat_div_eq_0 [simp]: "(n::nat) > 0 ==> ((m div n) = 0) = (m < n)"
nipkow@30837
   828
by(auto, subst mod_div_equality [of m n, symmetric], auto)
nipkow@30837
   829
paulson@17085
   830
declare div_less_dividend [simp]
paulson@17085
   831
paulson@14267
   832
text{*A fact for the mutilated chess board*}
paulson@14267
   833
lemma mod_Suc: "Suc(m) mod n = (if Suc(m mod n) = n then 0 else Suc(m mod n))"
paulson@14267
   834
apply (case_tac "n=0", simp)
paulson@15251
   835
apply (induct "m" rule: nat_less_induct)
paulson@14267
   836
apply (case_tac "Suc (na) <n")
paulson@14267
   837
(* case Suc(na) < n *)
paulson@14267
   838
apply (frule lessI [THEN less_trans], simp add: less_not_refl3)
paulson@14267
   839
(* case n \<le> Suc(na) *)
paulson@16796
   840
apply (simp add: linorder_not_less le_Suc_eq mod_geq)
nipkow@15439
   841
apply (auto simp add: Suc_diff_le le_mod_geq)
paulson@14267
   842
done
paulson@14267
   843
paulson@14267
   844
haftmann@27651
   845
subsubsection {* The Divides Relation *}
paulson@24286
   846
paulson@14267
   847
lemma dvd_1_left [iff]: "Suc 0 dvd k"
wenzelm@22718
   848
  unfolding dvd_def by simp
paulson@14267
   849
paulson@14267
   850
lemma dvd_1_iff_1 [simp]: "(m dvd Suc 0) = (m = Suc 0)"
nipkow@29667
   851
by (simp add: dvd_def)
paulson@14267
   852
huffman@30079
   853
lemma nat_dvd_1_iff_1 [simp]: "m dvd (1::nat) \<longleftrightarrow> m = 1"
huffman@30079
   854
by (simp add: dvd_def)
huffman@30079
   855
paulson@14267
   856
lemma dvd_anti_sym: "[| m dvd n; n dvd m |] ==> m = (n::nat)"
wenzelm@22718
   857
  unfolding dvd_def
wenzelm@22718
   858
  by (force dest: mult_eq_self_implies_10 simp add: mult_assoc mult_eq_1_iff)
paulson@14267
   859
haftmann@23684
   860
text {* @{term "op dvd"} is a partial order *}
haftmann@23684
   861
wenzelm@30729
   862
interpretation dvd: order "op dvd" "\<lambda>n m \<Colon> nat. n dvd m \<and> \<not> m dvd n"
haftmann@28823
   863
  proof qed (auto intro: dvd_refl dvd_trans dvd_anti_sym)
paulson@14267
   864
nipkow@30042
   865
lemma nat_dvd_diff[simp]: "[| k dvd m; k dvd n |] ==> k dvd (m-n :: nat)"
nipkow@30042
   866
unfolding dvd_def
nipkow@30042
   867
by (blast intro: diff_mult_distrib2 [symmetric])
paulson@14267
   868
paulson@14267
   869
lemma dvd_diffD: "[| k dvd m-n; k dvd n; n\<le>m |] ==> k dvd (m::nat)"
wenzelm@22718
   870
  apply (erule linorder_not_less [THEN iffD2, THEN add_diff_inverse, THEN subst])
wenzelm@22718
   871
  apply (blast intro: dvd_add)
wenzelm@22718
   872
  done
paulson@14267
   873
paulson@14267
   874
lemma dvd_diffD1: "[| k dvd m-n; k dvd m; n\<le>m |] ==> k dvd (n::nat)"
nipkow@30042
   875
by (drule_tac m = m in nat_dvd_diff, auto)
paulson@14267
   876
paulson@14267
   877
lemma dvd_reduce: "(k dvd n + k) = (k dvd (n::nat))"
wenzelm@22718
   878
  apply (rule iffI)
wenzelm@22718
   879
   apply (erule_tac [2] dvd_add)
wenzelm@22718
   880
   apply (rule_tac [2] dvd_refl)
wenzelm@22718
   881
  apply (subgoal_tac "n = (n+k) -k")
wenzelm@22718
   882
   prefer 2 apply simp
wenzelm@22718
   883
  apply (erule ssubst)
nipkow@30042
   884
  apply (erule nat_dvd_diff)
wenzelm@22718
   885
  apply (rule dvd_refl)
wenzelm@22718
   886
  done
paulson@14267
   887
paulson@14267
   888
lemma dvd_mod: "!!n::nat. [| f dvd m; f dvd n |] ==> f dvd m mod n"
wenzelm@22718
   889
  unfolding dvd_def
wenzelm@22718
   890
  apply (case_tac "n = 0", auto)
wenzelm@22718
   891
  apply (blast intro: mod_mult_distrib2 [symmetric])
wenzelm@22718
   892
  done
paulson@14267
   893
paulson@14267
   894
lemma dvd_mod_iff: "k dvd n ==> ((k::nat) dvd m mod n) = (k dvd m)"
nipkow@29667
   895
by (blast intro: dvd_mod_imp_dvd dvd_mod)
paulson@14267
   896
paulson@14267
   897
lemma dvd_mult_cancel: "!!k::nat. [| k*m dvd k*n; 0<k |] ==> m dvd n"
wenzelm@22718
   898
  unfolding dvd_def
wenzelm@22718
   899
  apply (erule exE)
wenzelm@22718
   900
  apply (simp add: mult_ac)
wenzelm@22718
   901
  done
paulson@14267
   902
paulson@14267
   903
lemma dvd_mult_cancel1: "0<m ==> (m*n dvd m) = (n = (1::nat))"
wenzelm@22718
   904
  apply auto
wenzelm@22718
   905
   apply (subgoal_tac "m*n dvd m*1")
wenzelm@22718
   906
   apply (drule dvd_mult_cancel, auto)
wenzelm@22718
   907
  done
paulson@14267
   908
paulson@14267
   909
lemma dvd_mult_cancel2: "0<m ==> (n*m dvd m) = (n = (1::nat))"
wenzelm@22718
   910
  apply (subst mult_commute)
wenzelm@22718
   911
  apply (erule dvd_mult_cancel1)
wenzelm@22718
   912
  done
paulson@14267
   913
paulson@14267
   914
lemma dvd_imp_le: "[| k dvd n; 0 < n |] ==> k \<le> (n::nat)"
nipkow@30837
   915
by (auto elim!: dvdE) (auto simp add: gr0_conv_Suc)
nipkow@30837
   916
nipkow@30837
   917
lemma nat_dvd_not_less: "(0::nat) < m \<Longrightarrow> m < n \<Longrightarrow> \<not> n dvd m"
nipkow@30837
   918
by (auto elim!: dvdE) (auto simp add: gr0_conv_Suc)
paulson@14267
   919
paulson@14267
   920
lemma dvd_mult_div_cancel: "n dvd m ==> n * (m div n) = (m::nat)"
wenzelm@22718
   921
  apply (subgoal_tac "m mod n = 0")
wenzelm@22718
   922
   apply (simp add: mult_div_cancel)
wenzelm@22718
   923
  apply (simp only: dvd_eq_mod_eq_0)
wenzelm@22718
   924
  done
paulson@14267
   925
nipkow@25162
   926
lemma nat_zero_less_power_iff [simp]: "(x^n > 0) = (x > (0::nat) | n=0)"
wenzelm@22718
   927
  by (induct n) auto
haftmann@21408
   928
haftmann@21408
   929
lemma power_dvd_imp_le: "[|i^m dvd i^n;  (1::nat) < i|] ==> m \<le> n"
wenzelm@22718
   930
  apply (rule power_le_imp_le_exp, assumption)
wenzelm@22718
   931
  apply (erule dvd_imp_le, simp)
wenzelm@22718
   932
  done
haftmann@21408
   933
paulson@14267
   934
lemma mod_eq_0_iff: "(m mod d = 0) = (\<exists>q::nat. m = d*q)"
nipkow@29667
   935
by (auto simp add: dvd_eq_mod_eq_0 [symmetric] dvd_def)
paulson@17084
   936
wenzelm@22718
   937
lemmas mod_eq_0D [dest!] = mod_eq_0_iff [THEN iffD1]
paulson@14267
   938
paulson@14267
   939
(*Loses information, namely we also have r<d provided d is nonzero*)
paulson@14267
   940
lemma mod_eqD: "(m mod d = r) ==> \<exists>q::nat. m = r + q*d"
haftmann@27651
   941
  apply (cut_tac a = m in mod_div_equality)
wenzelm@22718
   942
  apply (simp only: add_ac)
wenzelm@22718
   943
  apply (blast intro: sym)
wenzelm@22718
   944
  done
paulson@14267
   945
nipkow@13152
   946
lemma split_div:
nipkow@13189
   947
 "P(n div k :: nat) =
nipkow@13189
   948
 ((k = 0 \<longrightarrow> P 0) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P i)))"
nipkow@13189
   949
 (is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")
nipkow@13189
   950
proof
nipkow@13189
   951
  assume P: ?P
nipkow@13189
   952
  show ?Q
nipkow@13189
   953
  proof (cases)
nipkow@13189
   954
    assume "k = 0"
haftmann@27651
   955
    with P show ?Q by simp
nipkow@13189
   956
  next
nipkow@13189
   957
    assume not0: "k \<noteq> 0"
nipkow@13189
   958
    thus ?Q
nipkow@13189
   959
    proof (simp, intro allI impI)
nipkow@13189
   960
      fix i j
nipkow@13189
   961
      assume n: "n = k*i + j" and j: "j < k"
nipkow@13189
   962
      show "P i"
nipkow@13189
   963
      proof (cases)
wenzelm@22718
   964
        assume "i = 0"
wenzelm@22718
   965
        with n j P show "P i" by simp
nipkow@13189
   966
      next
wenzelm@22718
   967
        assume "i \<noteq> 0"
wenzelm@22718
   968
        with not0 n j P show "P i" by(simp add:add_ac)
nipkow@13189
   969
      qed
nipkow@13189
   970
    qed
nipkow@13189
   971
  qed
nipkow@13189
   972
next
nipkow@13189
   973
  assume Q: ?Q
nipkow@13189
   974
  show ?P
nipkow@13189
   975
  proof (cases)
nipkow@13189
   976
    assume "k = 0"
haftmann@27651
   977
    with Q show ?P by simp
nipkow@13189
   978
  next
nipkow@13189
   979
    assume not0: "k \<noteq> 0"
nipkow@13189
   980
    with Q have R: ?R by simp
nipkow@13189
   981
    from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]
nipkow@13517
   982
    show ?P by simp
nipkow@13189
   983
  qed
nipkow@13189
   984
qed
nipkow@13189
   985
berghofe@13882
   986
lemma split_div_lemma:
haftmann@26100
   987
  assumes "0 < n"
haftmann@26100
   988
  shows "n * q \<le> m \<and> m < n * Suc q \<longleftrightarrow> q = ((m\<Colon>nat) div n)" (is "?lhs \<longleftrightarrow> ?rhs")
haftmann@26100
   989
proof
haftmann@26100
   990
  assume ?rhs
haftmann@26100
   991
  with mult_div_cancel have nq: "n * q = m - (m mod n)" by simp
haftmann@26100
   992
  then have A: "n * q \<le> m" by simp
haftmann@26100
   993
  have "n - (m mod n) > 0" using mod_less_divisor assms by auto
haftmann@26100
   994
  then have "m < m + (n - (m mod n))" by simp
haftmann@26100
   995
  then have "m < n + (m - (m mod n))" by simp
haftmann@26100
   996
  with nq have "m < n + n * q" by simp
haftmann@26100
   997
  then have B: "m < n * Suc q" by simp
haftmann@26100
   998
  from A B show ?lhs ..
haftmann@26100
   999
next
haftmann@26100
  1000
  assume P: ?lhs
haftmann@26100
  1001
  then have "divmod_rel m n q (m - n * q)"
haftmann@26100
  1002
    unfolding divmod_rel_def by (auto simp add: mult_ac)
haftmann@26100
  1003
  then show ?rhs using divmod_rel by (rule divmod_rel_unique_div)
haftmann@26100
  1004
qed
berghofe@13882
  1005
berghofe@13882
  1006
theorem split_div':
berghofe@13882
  1007
  "P ((m::nat) div n) = ((n = 0 \<and> P 0) \<or>
paulson@14267
  1008
   (\<exists>q. (n * q \<le> m \<and> m < n * (Suc q)) \<and> P q))"
berghofe@13882
  1009
  apply (case_tac "0 < n")
berghofe@13882
  1010
  apply (simp only: add: split_div_lemma)
haftmann@27651
  1011
  apply simp_all
berghofe@13882
  1012
  done
berghofe@13882
  1013
nipkow@13189
  1014
lemma split_mod:
nipkow@13189
  1015
 "P(n mod k :: nat) =
nipkow@13189
  1016
 ((k = 0 \<longrightarrow> P n) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P j)))"
nipkow@13189
  1017
 (is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")
nipkow@13189
  1018
proof
nipkow@13189
  1019
  assume P: ?P
nipkow@13189
  1020
  show ?Q
nipkow@13189
  1021
  proof (cases)
nipkow@13189
  1022
    assume "k = 0"
haftmann@27651
  1023
    with P show ?Q by simp
nipkow@13189
  1024
  next
nipkow@13189
  1025
    assume not0: "k \<noteq> 0"
nipkow@13189
  1026
    thus ?Q
nipkow@13189
  1027
    proof (simp, intro allI impI)
nipkow@13189
  1028
      fix i j
nipkow@13189
  1029
      assume "n = k*i + j" "j < k"
nipkow@13189
  1030
      thus "P j" using not0 P by(simp add:add_ac mult_ac)
nipkow@13189
  1031
    qed
nipkow@13189
  1032
  qed
nipkow@13189
  1033
next
nipkow@13189
  1034
  assume Q: ?Q
nipkow@13189
  1035
  show ?P
nipkow@13189
  1036
  proof (cases)
nipkow@13189
  1037
    assume "k = 0"
haftmann@27651
  1038
    with Q show ?P by simp
nipkow@13189
  1039
  next
nipkow@13189
  1040
    assume not0: "k \<noteq> 0"
nipkow@13189
  1041
    with Q have R: ?R by simp
nipkow@13189
  1042
    from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]
nipkow@13517
  1043
    show ?P by simp
nipkow@13189
  1044
  qed
nipkow@13189
  1045
qed
nipkow@13189
  1046
berghofe@13882
  1047
theorem mod_div_equality': "(m::nat) mod n = m - (m div n) * n"
berghofe@13882
  1048
  apply (rule_tac P="%x. m mod n = x - (m div n) * n" in
berghofe@13882
  1049
    subst [OF mod_div_equality [of _ n]])
berghofe@13882
  1050
  apply arith
berghofe@13882
  1051
  done
berghofe@13882
  1052
haftmann@22800
  1053
lemma div_mod_equality':
haftmann@22800
  1054
  fixes m n :: nat
haftmann@22800
  1055
  shows "m div n * n = m - m mod n"
haftmann@22800
  1056
proof -
haftmann@22800
  1057
  have "m mod n \<le> m mod n" ..
haftmann@22800
  1058
  from div_mod_equality have 
haftmann@22800
  1059
    "m div n * n + m mod n - m mod n = m - m mod n" by simp
haftmann@22800
  1060
  with diff_add_assoc [OF `m mod n \<le> m mod n`, of "m div n * n"] have
haftmann@22800
  1061
    "m div n * n + (m mod n - m mod n) = m - m mod n"
haftmann@22800
  1062
    by simp
haftmann@22800
  1063
  then show ?thesis by simp
haftmann@22800
  1064
qed
haftmann@22800
  1065
haftmann@22800
  1066
haftmann@25942
  1067
subsubsection {*An ``induction'' law for modulus arithmetic.*}
paulson@14640
  1068
paulson@14640
  1069
lemma mod_induct_0:
paulson@14640
  1070
  assumes step: "\<forall>i<p. P i \<longrightarrow> P ((Suc i) mod p)"
paulson@14640
  1071
  and base: "P i" and i: "i<p"
paulson@14640
  1072
  shows "P 0"
paulson@14640
  1073
proof (rule ccontr)
paulson@14640
  1074
  assume contra: "\<not>(P 0)"
paulson@14640
  1075
  from i have p: "0<p" by simp
paulson@14640
  1076
  have "\<forall>k. 0<k \<longrightarrow> \<not> P (p-k)" (is "\<forall>k. ?A k")
paulson@14640
  1077
  proof
paulson@14640
  1078
    fix k
paulson@14640
  1079
    show "?A k"
paulson@14640
  1080
    proof (induct k)
paulson@14640
  1081
      show "?A 0" by simp  -- "by contradiction"
paulson@14640
  1082
    next
paulson@14640
  1083
      fix n
paulson@14640
  1084
      assume ih: "?A n"
paulson@14640
  1085
      show "?A (Suc n)"
paulson@14640
  1086
      proof (clarsimp)
wenzelm@22718
  1087
        assume y: "P (p - Suc n)"
wenzelm@22718
  1088
        have n: "Suc n < p"
wenzelm@22718
  1089
        proof (rule ccontr)
wenzelm@22718
  1090
          assume "\<not>(Suc n < p)"
wenzelm@22718
  1091
          hence "p - Suc n = 0"
wenzelm@22718
  1092
            by simp
wenzelm@22718
  1093
          with y contra show "False"
wenzelm@22718
  1094
            by simp
wenzelm@22718
  1095
        qed
wenzelm@22718
  1096
        hence n2: "Suc (p - Suc n) = p-n" by arith
wenzelm@22718
  1097
        from p have "p - Suc n < p" by arith
wenzelm@22718
  1098
        with y step have z: "P ((Suc (p - Suc n)) mod p)"
wenzelm@22718
  1099
          by blast
wenzelm@22718
  1100
        show "False"
wenzelm@22718
  1101
        proof (cases "n=0")
wenzelm@22718
  1102
          case True
wenzelm@22718
  1103
          with z n2 contra show ?thesis by simp
wenzelm@22718
  1104
        next
wenzelm@22718
  1105
          case False
wenzelm@22718
  1106
          with p have "p-n < p" by arith
wenzelm@22718
  1107
          with z n2 False ih show ?thesis by simp
wenzelm@22718
  1108
        qed
paulson@14640
  1109
      qed
paulson@14640
  1110
    qed
paulson@14640
  1111
  qed
paulson@14640
  1112
  moreover
paulson@14640
  1113
  from i obtain k where "0<k \<and> i+k=p"
paulson@14640
  1114
    by (blast dest: less_imp_add_positive)
paulson@14640
  1115
  hence "0<k \<and> i=p-k" by auto
paulson@14640
  1116
  moreover
paulson@14640
  1117
  note base
paulson@14640
  1118
  ultimately
paulson@14640
  1119
  show "False" by blast
paulson@14640
  1120
qed
paulson@14640
  1121
paulson@14640
  1122
lemma mod_induct:
paulson@14640
  1123
  assumes step: "\<forall>i<p. P i \<longrightarrow> P ((Suc i) mod p)"
paulson@14640
  1124
  and base: "P i" and i: "i<p" and j: "j<p"
paulson@14640
  1125
  shows "P j"
paulson@14640
  1126
proof -
paulson@14640
  1127
  have "\<forall>j<p. P j"
paulson@14640
  1128
  proof
paulson@14640
  1129
    fix j
paulson@14640
  1130
    show "j<p \<longrightarrow> P j" (is "?A j")
paulson@14640
  1131
    proof (induct j)
paulson@14640
  1132
      from step base i show "?A 0"
wenzelm@22718
  1133
        by (auto elim: mod_induct_0)
paulson@14640
  1134
    next
paulson@14640
  1135
      fix k
paulson@14640
  1136
      assume ih: "?A k"
paulson@14640
  1137
      show "?A (Suc k)"
paulson@14640
  1138
      proof
wenzelm@22718
  1139
        assume suc: "Suc k < p"
wenzelm@22718
  1140
        hence k: "k<p" by simp
wenzelm@22718
  1141
        with ih have "P k" ..
wenzelm@22718
  1142
        with step k have "P (Suc k mod p)"
wenzelm@22718
  1143
          by blast
wenzelm@22718
  1144
        moreover
wenzelm@22718
  1145
        from suc have "Suc k mod p = Suc k"
wenzelm@22718
  1146
          by simp
wenzelm@22718
  1147
        ultimately
wenzelm@22718
  1148
        show "P (Suc k)" by simp
paulson@14640
  1149
      qed
paulson@14640
  1150
    qed
paulson@14640
  1151
  qed
paulson@14640
  1152
  with j show ?thesis by blast
paulson@14640
  1153
qed
paulson@14640
  1154
paulson@3366
  1155
end