src/HOL/Divides.thy
author nipkow
Sun Feb 15 22:58:02 2009 +0100 (2009-02-15)
changeset 29925 17d1e32ef867
parent 29667 53103fc8ffa3
child 29948 cdf12a1cb963
permissions -rw-r--r--
dvd and setprod 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 (unfold dvd_def, auto)
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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 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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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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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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class ring_div = semiring_div + comm_ring_1
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begin
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text {* Negation respects modular equivalence. *}
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lemma mod_minus_eq: "(- a) mod b = (- (a mod b)) mod b"
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proof -
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  have "(- a) mod b = (- (a div b * b + a mod b)) mod b"
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    by (simp only: mod_div_equality)
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  also have "\<dots> = (- (a mod b) + - (a div b) * b) mod b"
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    by (simp only: minus_add_distrib minus_mult_left add_ac)
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  also have "\<dots> = (- (a mod b)) mod b"
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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_minus_cong:
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  assumes "a mod b = a' mod b"
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  shows "(- a) mod b = (- a') mod b"
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proof -
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  have "(- (a mod b)) mod b = (- (a' mod b)) mod b"
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    unfolding assms ..
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  thus ?thesis
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    by (simp only: mod_minus_eq [symmetric])
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qed
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text {* Subtraction respects modular equivalence. *}
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lemma mod_diff_left_eq: "(a - b) mod c = (a mod c - b) mod c"
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  unfolding diff_minus
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  by (intro mod_add_cong mod_minus_cong) simp_all
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lemma mod_diff_right_eq: "(a - b) mod c = (a - b mod c) mod c"
huffman@29405
   319
  unfolding diff_minus
huffman@29405
   320
  by (intro mod_add_cong mod_minus_cong) simp_all
huffman@29405
   321
huffman@29405
   322
lemma mod_diff_eq: "(a - b) mod c = (a mod c - b mod c) mod c"
huffman@29405
   323
  unfolding diff_minus
huffman@29405
   324
  by (intro mod_add_cong mod_minus_cong) simp_all
huffman@29405
   325
huffman@29405
   326
lemma mod_diff_cong:
huffman@29405
   327
  assumes "a mod c = a' mod c"
huffman@29405
   328
  assumes "b mod c = b' mod c"
huffman@29405
   329
  shows "(a - b) mod c = (a' - b') mod c"
huffman@29405
   330
  unfolding diff_minus using assms
huffman@29405
   331
  by (intro mod_add_cong mod_minus_cong)
huffman@29405
   332
huffman@29405
   333
end
huffman@29405
   334
haftmann@25942
   335
haftmann@26100
   336
subsection {* Division on @{typ nat} *}
haftmann@26100
   337
haftmann@26100
   338
text {*
haftmann@26100
   339
  We define @{const div} and @{const mod} on @{typ nat} by means
haftmann@26100
   340
  of a characteristic relation with two input arguments
haftmann@26100
   341
  @{term "m\<Colon>nat"}, @{term "n\<Colon>nat"} and two output arguments
haftmann@26100
   342
  @{term "q\<Colon>nat"}(uotient) and @{term "r\<Colon>nat"}(emainder).
haftmann@26100
   343
*}
haftmann@26100
   344
haftmann@26100
   345
definition divmod_rel :: "nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> bool" where
haftmann@26100
   346
  "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
   347
haftmann@26100
   348
text {* @{const divmod_rel} is total: *}
haftmann@26100
   349
haftmann@26100
   350
lemma divmod_rel_ex:
haftmann@26100
   351
  obtains q r where "divmod_rel m n q r"
haftmann@26100
   352
proof (cases "n = 0")
haftmann@26100
   353
  case True with that show thesis
haftmann@26100
   354
    by (auto simp add: divmod_rel_def)
haftmann@26100
   355
next
haftmann@26100
   356
  case False
haftmann@26100
   357
  have "\<exists>q r. m = q * n + r \<and> r < n"
haftmann@26100
   358
  proof (induct m)
haftmann@26100
   359
    case 0 with `n \<noteq> 0`
haftmann@26100
   360
    have "(0\<Colon>nat) = 0 * n + 0 \<and> 0 < n" by simp
haftmann@26100
   361
    then show ?case by blast
haftmann@26100
   362
  next
haftmann@26100
   363
    case (Suc m) then obtain q' r'
haftmann@26100
   364
      where m: "m = q' * n + r'" and n: "r' < n" by auto
haftmann@26100
   365
    then show ?case proof (cases "Suc r' < n")
haftmann@26100
   366
      case True
haftmann@26100
   367
      from m n have "Suc m = q' * n + Suc r'" by simp
haftmann@26100
   368
      with True show ?thesis by blast
haftmann@26100
   369
    next
haftmann@26100
   370
      case False then have "n \<le> Suc r'" by auto
haftmann@26100
   371
      moreover from n have "Suc r' \<le> n" by auto
haftmann@26100
   372
      ultimately have "n = Suc r'" by auto
haftmann@26100
   373
      with m have "Suc m = Suc q' * n + 0" by simp
haftmann@26100
   374
      with `n \<noteq> 0` show ?thesis by blast
haftmann@26100
   375
    qed
haftmann@26100
   376
  qed
haftmann@26100
   377
  with that show thesis
haftmann@26100
   378
    using `n \<noteq> 0` by (auto simp add: divmod_rel_def)
haftmann@26100
   379
qed
haftmann@26100
   380
haftmann@26100
   381
text {* @{const divmod_rel} is injective: *}
haftmann@26100
   382
haftmann@26100
   383
lemma divmod_rel_unique_div:
haftmann@26100
   384
  assumes "divmod_rel m n q r"
haftmann@26100
   385
    and "divmod_rel m n q' r'"
haftmann@26100
   386
  shows "q = q'"
haftmann@26100
   387
proof (cases "n = 0")
haftmann@26100
   388
  case True with assms show ?thesis
haftmann@26100
   389
    by (simp add: divmod_rel_def)
haftmann@26100
   390
next
haftmann@26100
   391
  case False
haftmann@26100
   392
  have aux: "\<And>q r q' r'. q' * n + r' = q * n + r \<Longrightarrow> r < n \<Longrightarrow> q' \<le> (q\<Colon>nat)"
haftmann@26100
   393
  apply (rule leI)
haftmann@26100
   394
  apply (subst less_iff_Suc_add)
haftmann@26100
   395
  apply (auto simp add: add_mult_distrib)
haftmann@26100
   396
  done
haftmann@26100
   397
  from `n \<noteq> 0` assms show ?thesis
haftmann@26100
   398
    by (auto simp add: divmod_rel_def
haftmann@26100
   399
      intro: order_antisym dest: aux sym)
haftmann@26100
   400
qed
haftmann@26100
   401
haftmann@26100
   402
lemma divmod_rel_unique_mod:
haftmann@26100
   403
  assumes "divmod_rel m n q r"
haftmann@26100
   404
    and "divmod_rel m n q' r'"
haftmann@26100
   405
  shows "r = r'"
haftmann@26100
   406
proof -
haftmann@26100
   407
  from assms have "q = q'" by (rule divmod_rel_unique_div)
haftmann@26100
   408
  with assms show ?thesis by (simp add: divmod_rel_def)
haftmann@26100
   409
qed
haftmann@26100
   410
haftmann@26100
   411
text {*
haftmann@26100
   412
  We instantiate divisibility on the natural numbers by
haftmann@26100
   413
  means of @{const divmod_rel}:
haftmann@26100
   414
*}
haftmann@25942
   415
haftmann@25942
   416
instantiation nat :: semiring_div
haftmann@25571
   417
begin
haftmann@25571
   418
haftmann@26100
   419
definition divmod :: "nat \<Rightarrow> nat \<Rightarrow> nat \<times> nat" where
haftmann@28562
   420
  [code del]: "divmod m n = (THE (q, r). divmod_rel m n q r)"
haftmann@26100
   421
haftmann@26100
   422
definition div_nat where
haftmann@26100
   423
  "m div n = fst (divmod m n)"
haftmann@26100
   424
haftmann@26100
   425
definition mod_nat where
haftmann@26100
   426
  "m mod n = snd (divmod m n)"
haftmann@25571
   427
haftmann@26100
   428
lemma divmod_div_mod:
haftmann@26100
   429
  "divmod m n = (m div n, m mod n)"
haftmann@26100
   430
  unfolding div_nat_def mod_nat_def by simp
haftmann@26100
   431
haftmann@26100
   432
lemma divmod_eq:
haftmann@26100
   433
  assumes "divmod_rel m n q r" 
haftmann@26100
   434
  shows "divmod m n = (q, r)"
haftmann@26100
   435
  using assms by (auto simp add: divmod_def
haftmann@26100
   436
    dest: divmod_rel_unique_div divmod_rel_unique_mod)
haftmann@25942
   437
haftmann@26100
   438
lemma div_eq:
haftmann@26100
   439
  assumes "divmod_rel m n q r" 
haftmann@26100
   440
  shows "m div n = q"
haftmann@26100
   441
  using assms by (auto dest: divmod_eq simp add: div_nat_def)
haftmann@26100
   442
haftmann@26100
   443
lemma mod_eq:
haftmann@26100
   444
  assumes "divmod_rel m n q r" 
haftmann@26100
   445
  shows "m mod n = r"
haftmann@26100
   446
  using assms by (auto dest: divmod_eq simp add: mod_nat_def)
haftmann@25571
   447
haftmann@26100
   448
lemma divmod_rel: "divmod_rel m n (m div n) (m mod n)"
haftmann@26100
   449
proof -
haftmann@26100
   450
  from divmod_rel_ex
haftmann@26100
   451
    obtain q r where rel: "divmod_rel m n q r" .
haftmann@26100
   452
  moreover with div_eq mod_eq have "m div n = q" and "m mod n = r"
haftmann@26100
   453
    by simp_all
haftmann@26100
   454
  ultimately show ?thesis by simp
haftmann@26100
   455
qed
paulson@14267
   456
haftmann@26100
   457
lemma divmod_zero:
haftmann@26100
   458
  "divmod m 0 = (0, m)"
haftmann@26100
   459
proof -
haftmann@26100
   460
  from divmod_rel [of m 0] show ?thesis
haftmann@26100
   461
    unfolding divmod_div_mod divmod_rel_def by simp
haftmann@26100
   462
qed
haftmann@25942
   463
haftmann@26100
   464
lemma divmod_base:
haftmann@26100
   465
  assumes "m < n"
haftmann@26100
   466
  shows "divmod m n = (0, m)"
haftmann@26100
   467
proof -
haftmann@26100
   468
  from divmod_rel [of m n] show ?thesis
haftmann@26100
   469
    unfolding divmod_div_mod divmod_rel_def
haftmann@26100
   470
    using assms by (cases "m div n = 0")
haftmann@26100
   471
      (auto simp add: gr0_conv_Suc [of "m div n"])
haftmann@26100
   472
qed
haftmann@25942
   473
haftmann@26100
   474
lemma divmod_step:
haftmann@26100
   475
  assumes "0 < n" and "n \<le> m"
haftmann@26100
   476
  shows "divmod m n = (Suc ((m - n) div n), (m - n) mod n)"
haftmann@26100
   477
proof -
haftmann@26100
   478
  from divmod_rel have divmod_m_n: "divmod_rel m n (m div n) (m mod n)" .
haftmann@26100
   479
  with assms have m_div_n: "m div n \<ge> 1"
haftmann@26100
   480
    by (cases "m div n") (auto simp add: divmod_rel_def)
haftmann@26100
   481
  from assms divmod_m_n have "divmod_rel (m - n) n (m div n - 1) (m mod n)"
haftmann@26100
   482
    by (cases "m div n") (auto simp add: divmod_rel_def)
haftmann@26100
   483
  with divmod_eq have "divmod (m - n) n = (m div n - 1, m mod n)" by simp
haftmann@26100
   484
  moreover from divmod_div_mod have "divmod (m - n) n = ((m - n) div n, (m - n) mod n)" .
haftmann@26100
   485
  ultimately have "m div n = Suc ((m - n) div n)"
haftmann@26100
   486
    and "m mod n = (m - n) mod n" using m_div_n by simp_all
haftmann@26100
   487
  then show ?thesis using divmod_div_mod by simp
haftmann@26100
   488
qed
haftmann@25942
   489
wenzelm@26300
   490
text {* The ''recursion'' equations for @{const div} and @{const mod} *}
haftmann@26100
   491
haftmann@26100
   492
lemma div_less [simp]:
haftmann@26100
   493
  fixes m n :: nat
haftmann@26100
   494
  assumes "m < n"
haftmann@26100
   495
  shows "m div n = 0"
haftmann@26100
   496
  using assms divmod_base divmod_div_mod by simp
haftmann@25942
   497
haftmann@26100
   498
lemma le_div_geq:
haftmann@26100
   499
  fixes m n :: nat
haftmann@26100
   500
  assumes "0 < n" and "n \<le> m"
haftmann@26100
   501
  shows "m div n = Suc ((m - n) div n)"
haftmann@26100
   502
  using assms divmod_step divmod_div_mod by simp
paulson@14267
   503
haftmann@26100
   504
lemma mod_less [simp]:
haftmann@26100
   505
  fixes m n :: nat
haftmann@26100
   506
  assumes "m < n"
haftmann@26100
   507
  shows "m mod n = m"
haftmann@26100
   508
  using assms divmod_base divmod_div_mod by simp
haftmann@26100
   509
haftmann@26100
   510
lemma le_mod_geq:
haftmann@26100
   511
  fixes m n :: nat
haftmann@26100
   512
  assumes "n \<le> m"
haftmann@26100
   513
  shows "m mod n = (m - n) mod n"
haftmann@26100
   514
  using assms divmod_step divmod_div_mod by (cases "n = 0") simp_all
paulson@14267
   515
haftmann@25942
   516
instance proof
haftmann@26100
   517
  fix m n :: nat show "m div n * n + m mod n = m"
haftmann@26100
   518
    using divmod_rel [of m n] by (simp add: divmod_rel_def)
haftmann@25942
   519
next
haftmann@26100
   520
  fix n :: nat show "n div 0 = 0"
haftmann@26100
   521
    using divmod_zero divmod_div_mod [of n 0] by simp
haftmann@25942
   522
next
haftmann@27651
   523
  fix n :: nat show "0 div n = 0"
haftmann@27651
   524
    using divmod_rel [of 0 n] by (cases n) (simp_all add: divmod_rel_def)
haftmann@27651
   525
next
haftmann@27651
   526
  fix m n q :: nat assume "n \<noteq> 0" then show "(q + m * n) div n = m + q div n"
haftmann@25942
   527
    by (induct m) (simp_all add: le_div_geq)
haftmann@25942
   528
qed
haftmann@26100
   529
haftmann@25942
   530
end
paulson@14267
   531
haftmann@26100
   532
text {* Simproc for cancelling @{const div} and @{const mod} *}
haftmann@25942
   533
haftmann@27651
   534
(*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
   535
lemmas mod_div_equality2_nat = mod_div_equality2 [of "n\<Colon>nat" m, standard*)
haftmann@25942
   536
haftmann@25942
   537
ML {*
haftmann@25942
   538
structure CancelDivModData =
haftmann@25942
   539
struct
haftmann@25942
   540
haftmann@26100
   541
val div_name = @{const_name div};
haftmann@26100
   542
val mod_name = @{const_name mod};
haftmann@25942
   543
val mk_binop = HOLogic.mk_binop;
haftmann@26100
   544
val mk_sum = ArithData.mk_sum;
haftmann@26100
   545
val dest_sum = ArithData.dest_sum;
haftmann@25942
   546
haftmann@25942
   547
(*logic*)
paulson@14267
   548
haftmann@25942
   549
val div_mod_eqs = map mk_meta_eq [@{thm div_mod_equality}, @{thm div_mod_equality2}]
haftmann@25942
   550
haftmann@25942
   551
val trans = trans
haftmann@25942
   552
haftmann@25942
   553
val prove_eq_sums =
haftmann@25942
   554
  let val simps = @{thm add_0} :: @{thm add_0_right} :: @{thms add_ac}
haftmann@26100
   555
  in ArithData.prove_conv all_tac (ArithData.simp_all_tac simps) end;
haftmann@25942
   556
haftmann@25942
   557
end;
haftmann@25942
   558
haftmann@25942
   559
structure CancelDivMod = CancelDivModFun(CancelDivModData);
haftmann@25942
   560
wenzelm@28262
   561
val cancel_div_mod_proc = Simplifier.simproc (the_context ())
haftmann@26100
   562
  "cancel_div_mod" ["(m::nat) + n"] (K CancelDivMod.proc);
haftmann@25942
   563
haftmann@25942
   564
Addsimprocs[cancel_div_mod_proc];
haftmann@25942
   565
*}
haftmann@25942
   566
haftmann@26100
   567
text {* code generator setup *}
haftmann@26100
   568
haftmann@26100
   569
lemma divmod_if [code]: "divmod m n = (if n = 0 \<or> m < n then (0, m) else
haftmann@26100
   570
  let (q, r) = divmod (m - n) n in (Suc q, r))"
nipkow@29667
   571
by (simp add: divmod_zero divmod_base divmod_step)
haftmann@26100
   572
    (simp add: divmod_div_mod)
haftmann@26100
   573
haftmann@26100
   574
code_modulename SML
haftmann@26100
   575
  Divides Nat
haftmann@26100
   576
haftmann@26100
   577
code_modulename OCaml
haftmann@26100
   578
  Divides Nat
haftmann@26100
   579
haftmann@26100
   580
code_modulename Haskell
haftmann@26100
   581
  Divides Nat
haftmann@26100
   582
haftmann@26100
   583
haftmann@26100
   584
subsubsection {* Quotient *}
haftmann@26100
   585
haftmann@26100
   586
lemma div_geq: "0 < n \<Longrightarrow>  \<not> m < n \<Longrightarrow> m div n = Suc ((m - n) div n)"
nipkow@29667
   587
by (simp add: le_div_geq linorder_not_less)
haftmann@26100
   588
haftmann@26100
   589
lemma div_if: "0 < n \<Longrightarrow> m div n = (if m < n then 0 else Suc ((m - n) div n))"
nipkow@29667
   590
by (simp add: div_geq)
haftmann@26100
   591
haftmann@26100
   592
lemma div_mult_self_is_m [simp]: "0<n ==> (m*n) div n = (m::nat)"
nipkow@29667
   593
by simp
haftmann@26100
   594
haftmann@26100
   595
lemma div_mult_self1_is_m [simp]: "0<n ==> (n*m) div n = (m::nat)"
nipkow@29667
   596
by simp
haftmann@26100
   597
haftmann@25942
   598
haftmann@25942
   599
subsubsection {* Remainder *}
haftmann@25942
   600
haftmann@26100
   601
lemma mod_less_divisor [simp]:
haftmann@26100
   602
  fixes m n :: nat
haftmann@26100
   603
  assumes "n > 0"
haftmann@26100
   604
  shows "m mod n < (n::nat)"
haftmann@26100
   605
  using assms divmod_rel unfolding divmod_rel_def by auto
paulson@14267
   606
haftmann@26100
   607
lemma mod_less_eq_dividend [simp]:
haftmann@26100
   608
  fixes m n :: nat
haftmann@26100
   609
  shows "m mod n \<le> m"
haftmann@26100
   610
proof (rule add_leD2)
haftmann@26100
   611
  from mod_div_equality have "m div n * n + m mod n = m" .
haftmann@26100
   612
  then show "m div n * n + m mod n \<le> m" by auto
haftmann@26100
   613
qed
haftmann@26100
   614
haftmann@26100
   615
lemma mod_geq: "\<not> m < (n\<Colon>nat) \<Longrightarrow> m mod n = (m - n) mod n"
nipkow@29667
   616
by (simp add: le_mod_geq linorder_not_less)
paulson@14267
   617
haftmann@26100
   618
lemma mod_if: "m mod (n\<Colon>nat) = (if m < n then m else (m - n) mod n)"
nipkow@29667
   619
by (simp add: le_mod_geq)
haftmann@26100
   620
paulson@14267
   621
lemma mod_1 [simp]: "m mod Suc 0 = 0"
nipkow@29667
   622
by (induct m) (simp_all add: mod_geq)
paulson@14267
   623
haftmann@26100
   624
lemma mod_mult_distrib: "(m mod n) * (k\<Colon>nat) = (m * k) mod (n * k)"
wenzelm@22718
   625
  apply (cases "n = 0", simp)
wenzelm@22718
   626
  apply (cases "k = 0", simp)
wenzelm@22718
   627
  apply (induct m rule: nat_less_induct)
wenzelm@22718
   628
  apply (subst mod_if, simp)
wenzelm@22718
   629
  apply (simp add: mod_geq diff_mult_distrib)
wenzelm@22718
   630
  done
paulson@14267
   631
paulson@14267
   632
lemma mod_mult_distrib2: "(k::nat) * (m mod n) = (k*m) mod (k*n)"
nipkow@29667
   633
by (simp add: mult_commute [of k] mod_mult_distrib)
paulson@14267
   634
paulson@14267
   635
(* a simple rearrangement of mod_div_equality: *)
paulson@14267
   636
lemma mult_div_cancel: "(n::nat) * (m div n) = m - (m mod n)"
nipkow@29667
   637
by (cut_tac a = m and b = n in mod_div_equality2, arith)
paulson@14267
   638
nipkow@15439
   639
lemma mod_le_divisor[simp]: "0 < n \<Longrightarrow> m mod n \<le> (n::nat)"
wenzelm@22718
   640
  apply (drule mod_less_divisor [where m = m])
wenzelm@22718
   641
  apply simp
wenzelm@22718
   642
  done
paulson@14267
   643
haftmann@26100
   644
subsubsection {* Quotient and Remainder *}
paulson@14267
   645
haftmann@26100
   646
lemma divmod_rel_mult1_eq:
haftmann@26100
   647
  "[| divmod_rel b c q r; c > 0 |]
haftmann@26100
   648
   ==> divmod_rel (a*b) c (a*q + a*r div c) (a*r mod c)"
nipkow@29667
   649
by (auto simp add: split_ifs divmod_rel_def algebra_simps)
paulson@14267
   650
paulson@14267
   651
lemma div_mult1_eq: "(a*b) div c = a*(b div c) + a*(b mod c) div (c::nat)"
nipkow@25134
   652
apply (cases "c = 0", simp)
haftmann@26100
   653
apply (blast intro: divmod_rel [THEN divmod_rel_mult1_eq, THEN div_eq])
nipkow@25134
   654
done
paulson@14267
   655
paulson@14267
   656
lemma mod_mult1_eq: "(a*b) mod c = a*(b mod c) mod (c::nat)"
nipkow@29667
   657
by (rule mod_mult_right_eq)
paulson@14267
   658
paulson@14267
   659
lemma mod_mult1_eq': "(a*b) mod (c::nat) = ((a mod c) * b) mod c"
nipkow@29667
   660
by (rule mod_mult_left_eq)
paulson@14267
   661
nipkow@25162
   662
lemma mod_mult_distrib_mod:
nipkow@25162
   663
  "(a*b) mod (c::nat) = ((a mod c) * (b mod c)) mod c"
nipkow@29667
   664
by (rule mod_mult_eq)
paulson@14267
   665
haftmann@26100
   666
lemma divmod_rel_add1_eq:
haftmann@26100
   667
  "[| divmod_rel a c aq ar; divmod_rel b c bq br;  c > 0 |]
haftmann@26100
   668
   ==> divmod_rel (a + b) c (aq + bq + (ar+br) div c) ((ar + br) mod c)"
nipkow@29667
   669
by (auto simp add: split_ifs divmod_rel_def algebra_simps)
paulson@14267
   670
paulson@14267
   671
(*NOT suitable for rewriting: the RHS has an instance of the LHS*)
paulson@14267
   672
lemma div_add1_eq:
nipkow@25134
   673
  "(a+b) div (c::nat) = a div c + b div c + ((a mod c + b mod c) div c)"
nipkow@25134
   674
apply (cases "c = 0", simp)
haftmann@26100
   675
apply (blast intro: divmod_rel_add1_eq [THEN div_eq] divmod_rel)
nipkow@25134
   676
done
paulson@14267
   677
paulson@14267
   678
lemma mod_add1_eq: "(a+b) mod (c::nat) = (a mod c + b mod c) mod c"
nipkow@29667
   679
by (rule mod_add_eq)
paulson@14267
   680
paulson@14267
   681
lemma mod_lemma: "[| (0::nat) < c; r < b |] ==> b * (q mod c) + r < b * c"
wenzelm@22718
   682
  apply (cut_tac m = q and n = c in mod_less_divisor)
wenzelm@22718
   683
  apply (drule_tac [2] m = "q mod c" in less_imp_Suc_add, auto)
wenzelm@22718
   684
  apply (erule_tac P = "%x. ?lhs < ?rhs x" in ssubst)
wenzelm@22718
   685
  apply (simp add: add_mult_distrib2)
wenzelm@22718
   686
  done
paulson@10559
   687
haftmann@26100
   688
lemma divmod_rel_mult2_eq: "[| divmod_rel a b q r;  0 < b;  0 < c |]
haftmann@26100
   689
      ==> divmod_rel a (b*c) (q div c) (b*(q mod c) + r)"
nipkow@29667
   690
by (auto simp add: mult_ac divmod_rel_def add_mult_distrib2 [symmetric] mod_lemma)
paulson@14267
   691
paulson@14267
   692
lemma div_mult2_eq: "a div (b*c) = (a div b) div (c::nat)"
wenzelm@22718
   693
  apply (cases "b = 0", simp)
wenzelm@22718
   694
  apply (cases "c = 0", simp)
haftmann@26100
   695
  apply (force simp add: divmod_rel [THEN divmod_rel_mult2_eq, THEN div_eq])
wenzelm@22718
   696
  done
paulson@14267
   697
paulson@14267
   698
lemma mod_mult2_eq: "a mod (b*c) = b*(a div b mod c) + a mod (b::nat)"
wenzelm@22718
   699
  apply (cases "b = 0", simp)
wenzelm@22718
   700
  apply (cases "c = 0", simp)
haftmann@26100
   701
  apply (auto simp add: mult_commute divmod_rel [THEN divmod_rel_mult2_eq, THEN mod_eq])
wenzelm@22718
   702
  done
paulson@14267
   703
paulson@14267
   704
haftmann@25942
   705
subsubsection{*Cancellation of Common Factors in Division*}
paulson@14267
   706
paulson@14267
   707
lemma div_mult_mult_lemma:
wenzelm@22718
   708
    "[| (0::nat) < b;  0 < c |] ==> (c*a) div (c*b) = a div b"
nipkow@29667
   709
by (auto simp add: div_mult2_eq)
paulson@14267
   710
paulson@14267
   711
lemma div_mult_mult1 [simp]: "(0::nat) < c ==> (c*a) div (c*b) = a div b"
wenzelm@22718
   712
  apply (cases "b = 0")
wenzelm@22718
   713
  apply (auto simp add: linorder_neq_iff [of b] div_mult_mult_lemma)
wenzelm@22718
   714
  done
paulson@14267
   715
paulson@14267
   716
lemma div_mult_mult2 [simp]: "(0::nat) < c ==> (a*c) div (b*c) = a div b"
wenzelm@22718
   717
  apply (drule div_mult_mult1)
wenzelm@22718
   718
  apply (auto simp add: mult_commute)
wenzelm@22718
   719
  done
paulson@14267
   720
paulson@14267
   721
haftmann@25942
   722
subsubsection{*Further Facts about Quotient and Remainder*}
paulson@14267
   723
paulson@14267
   724
lemma div_1 [simp]: "m div Suc 0 = m"
nipkow@29667
   725
by (induct m) (simp_all add: div_geq)
paulson@14267
   726
paulson@14267
   727
paulson@14267
   728
(* Monotonicity of div in first argument *)
paulson@14267
   729
lemma div_le_mono [rule_format (no_asm)]:
wenzelm@22718
   730
    "\<forall>m::nat. m \<le> n --> (m div k) \<le> (n div k)"
paulson@14267
   731
apply (case_tac "k=0", simp)
paulson@15251
   732
apply (induct "n" rule: nat_less_induct, clarify)
paulson@14267
   733
apply (case_tac "n<k")
paulson@14267
   734
(* 1  case n<k *)
paulson@14267
   735
apply simp
paulson@14267
   736
(* 2  case n >= k *)
paulson@14267
   737
apply (case_tac "m<k")
paulson@14267
   738
(* 2.1  case m<k *)
paulson@14267
   739
apply simp
paulson@14267
   740
(* 2.2  case m>=k *)
nipkow@15439
   741
apply (simp add: div_geq diff_le_mono)
paulson@14267
   742
done
paulson@14267
   743
paulson@14267
   744
(* Antimonotonicity of div in second argument *)
paulson@14267
   745
lemma div_le_mono2: "!!m::nat. [| 0<m; m\<le>n |] ==> (k div n) \<le> (k div m)"
paulson@14267
   746
apply (subgoal_tac "0<n")
wenzelm@22718
   747
 prefer 2 apply simp
paulson@15251
   748
apply (induct_tac k rule: nat_less_induct)
paulson@14267
   749
apply (rename_tac "k")
paulson@14267
   750
apply (case_tac "k<n", simp)
paulson@14267
   751
apply (subgoal_tac "~ (k<m) ")
wenzelm@22718
   752
 prefer 2 apply simp
paulson@14267
   753
apply (simp add: div_geq)
paulson@15251
   754
apply (subgoal_tac "(k-n) div n \<le> (k-m) div n")
paulson@14267
   755
 prefer 2
paulson@14267
   756
 apply (blast intro: div_le_mono diff_le_mono2)
paulson@14267
   757
apply (rule le_trans, simp)
nipkow@15439
   758
apply (simp)
paulson@14267
   759
done
paulson@14267
   760
paulson@14267
   761
lemma div_le_dividend [simp]: "m div n \<le> (m::nat)"
paulson@14267
   762
apply (case_tac "n=0", simp)
paulson@14267
   763
apply (subgoal_tac "m div n \<le> m div 1", simp)
paulson@14267
   764
apply (rule div_le_mono2)
paulson@14267
   765
apply (simp_all (no_asm_simp))
paulson@14267
   766
done
paulson@14267
   767
wenzelm@22718
   768
(* Similar for "less than" *)
paulson@17085
   769
lemma div_less_dividend [rule_format]:
paulson@14267
   770
     "!!n::nat. 1<n ==> 0 < m --> m div n < m"
paulson@15251
   771
apply (induct_tac m rule: nat_less_induct)
paulson@14267
   772
apply (rename_tac "m")
paulson@14267
   773
apply (case_tac "m<n", simp)
paulson@14267
   774
apply (subgoal_tac "0<n")
wenzelm@22718
   775
 prefer 2 apply simp
paulson@14267
   776
apply (simp add: div_geq)
paulson@14267
   777
apply (case_tac "n<m")
paulson@15251
   778
 apply (subgoal_tac "(m-n) div n < (m-n) ")
paulson@14267
   779
  apply (rule impI less_trans_Suc)+
paulson@14267
   780
apply assumption
nipkow@15439
   781
  apply (simp_all)
paulson@14267
   782
done
paulson@14267
   783
paulson@17085
   784
declare div_less_dividend [simp]
paulson@17085
   785
paulson@14267
   786
text{*A fact for the mutilated chess board*}
paulson@14267
   787
lemma mod_Suc: "Suc(m) mod n = (if Suc(m mod n) = n then 0 else Suc(m mod n))"
paulson@14267
   788
apply (case_tac "n=0", simp)
paulson@15251
   789
apply (induct "m" rule: nat_less_induct)
paulson@14267
   790
apply (case_tac "Suc (na) <n")
paulson@14267
   791
(* case Suc(na) < n *)
paulson@14267
   792
apply (frule lessI [THEN less_trans], simp add: less_not_refl3)
paulson@14267
   793
(* case n \<le> Suc(na) *)
paulson@16796
   794
apply (simp add: linorder_not_less le_Suc_eq mod_geq)
nipkow@15439
   795
apply (auto simp add: Suc_diff_le le_mod_geq)
paulson@14267
   796
done
paulson@14267
   797
huffman@29403
   798
lemma nat_mod_div_trivial: "m mod n div n = (0 :: nat)"
nipkow@29667
   799
by simp
paulson@14437
   800
huffman@29403
   801
lemma nat_mod_mod_trivial: "m mod n mod n = (m mod n :: nat)"
nipkow@29667
   802
by simp
paulson@14437
   803
paulson@14267
   804
haftmann@27651
   805
subsubsection {* The Divides Relation *}
paulson@24286
   806
paulson@14267
   807
lemma dvd_1_left [iff]: "Suc 0 dvd k"
wenzelm@22718
   808
  unfolding dvd_def by simp
paulson@14267
   809
paulson@14267
   810
lemma dvd_1_iff_1 [simp]: "(m dvd Suc 0) = (m = Suc 0)"
nipkow@29667
   811
by (simp add: dvd_def)
paulson@14267
   812
paulson@14267
   813
lemma dvd_anti_sym: "[| m dvd n; n dvd m |] ==> m = (n::nat)"
wenzelm@22718
   814
  unfolding dvd_def
wenzelm@22718
   815
  by (force dest: mult_eq_self_implies_10 simp add: mult_assoc mult_eq_1_iff)
paulson@14267
   816
haftmann@23684
   817
text {* @{term "op dvd"} is a partial order *}
haftmann@23684
   818
haftmann@29509
   819
interpretation dvd!: order "op dvd" "\<lambda>n m \<Colon> nat. n dvd m \<and> \<not> m dvd n"
haftmann@28823
   820
  proof qed (auto intro: dvd_refl dvd_trans dvd_anti_sym)
paulson@14267
   821
paulson@14267
   822
lemma dvd_diff: "[| k dvd m; k dvd n |] ==> k dvd (m-n :: nat)"
wenzelm@22718
   823
  unfolding dvd_def
wenzelm@22718
   824
  by (blast intro: diff_mult_distrib2 [symmetric])
paulson@14267
   825
paulson@14267
   826
lemma dvd_diffD: "[| k dvd m-n; k dvd n; n\<le>m |] ==> k dvd (m::nat)"
wenzelm@22718
   827
  apply (erule linorder_not_less [THEN iffD2, THEN add_diff_inverse, THEN subst])
wenzelm@22718
   828
  apply (blast intro: dvd_add)
wenzelm@22718
   829
  done
paulson@14267
   830
paulson@14267
   831
lemma dvd_diffD1: "[| k dvd m-n; k dvd m; n\<le>m |] ==> k dvd (n::nat)"
nipkow@29667
   832
by (drule_tac m = m in dvd_diff, auto)
paulson@14267
   833
paulson@14267
   834
lemma dvd_reduce: "(k dvd n + k) = (k dvd (n::nat))"
wenzelm@22718
   835
  apply (rule iffI)
wenzelm@22718
   836
   apply (erule_tac [2] dvd_add)
wenzelm@22718
   837
   apply (rule_tac [2] dvd_refl)
wenzelm@22718
   838
  apply (subgoal_tac "n = (n+k) -k")
wenzelm@22718
   839
   prefer 2 apply simp
wenzelm@22718
   840
  apply (erule ssubst)
wenzelm@22718
   841
  apply (erule dvd_diff)
wenzelm@22718
   842
  apply (rule dvd_refl)
wenzelm@22718
   843
  done
paulson@14267
   844
paulson@14267
   845
lemma dvd_mod: "!!n::nat. [| f dvd m; f dvd n |] ==> f dvd m mod n"
wenzelm@22718
   846
  unfolding dvd_def
wenzelm@22718
   847
  apply (case_tac "n = 0", auto)
wenzelm@22718
   848
  apply (blast intro: mod_mult_distrib2 [symmetric])
wenzelm@22718
   849
  done
paulson@14267
   850
paulson@14267
   851
lemma dvd_mod_imp_dvd: "[| (k::nat) dvd m mod n;  k dvd n |] ==> k dvd m"
wenzelm@22718
   852
  apply (subgoal_tac "k dvd (m div n) *n + m mod n")
wenzelm@22718
   853
   apply (simp add: mod_div_equality)
wenzelm@22718
   854
  apply (simp only: dvd_add dvd_mult)
wenzelm@22718
   855
  done
paulson@14267
   856
paulson@14267
   857
lemma dvd_mod_iff: "k dvd n ==> ((k::nat) dvd m mod n) = (k dvd m)"
nipkow@29667
   858
by (blast intro: dvd_mod_imp_dvd dvd_mod)
paulson@14267
   859
paulson@14267
   860
lemma dvd_mult_cancel: "!!k::nat. [| k*m dvd k*n; 0<k |] ==> m dvd n"
wenzelm@22718
   861
  unfolding dvd_def
wenzelm@22718
   862
  apply (erule exE)
wenzelm@22718
   863
  apply (simp add: mult_ac)
wenzelm@22718
   864
  done
paulson@14267
   865
paulson@14267
   866
lemma dvd_mult_cancel1: "0<m ==> (m*n dvd m) = (n = (1::nat))"
wenzelm@22718
   867
  apply auto
wenzelm@22718
   868
   apply (subgoal_tac "m*n dvd m*1")
wenzelm@22718
   869
   apply (drule dvd_mult_cancel, auto)
wenzelm@22718
   870
  done
paulson@14267
   871
paulson@14267
   872
lemma dvd_mult_cancel2: "0<m ==> (n*m dvd m) = (n = (1::nat))"
wenzelm@22718
   873
  apply (subst mult_commute)
wenzelm@22718
   874
  apply (erule dvd_mult_cancel1)
wenzelm@22718
   875
  done
paulson@14267
   876
paulson@14267
   877
lemma dvd_imp_le: "[| k dvd n; 0 < n |] ==> k \<le> (n::nat)"
wenzelm@22718
   878
  apply (unfold dvd_def, clarify)
wenzelm@22718
   879
  apply (simp_all (no_asm_use) add: zero_less_mult_iff)
wenzelm@22718
   880
  apply (erule conjE)
wenzelm@22718
   881
  apply (rule le_trans)
wenzelm@22718
   882
   apply (rule_tac [2] le_refl [THEN mult_le_mono])
wenzelm@22718
   883
   apply (erule_tac [2] Suc_leI, simp)
wenzelm@22718
   884
  done
paulson@14267
   885
paulson@14267
   886
lemma dvd_mult_div_cancel: "n dvd m ==> n * (m div n) = (m::nat)"
wenzelm@22718
   887
  apply (subgoal_tac "m mod n = 0")
wenzelm@22718
   888
   apply (simp add: mult_div_cancel)
wenzelm@22718
   889
  apply (simp only: dvd_eq_mod_eq_0)
wenzelm@22718
   890
  done
paulson@14267
   891
haftmann@21408
   892
lemma le_imp_power_dvd: "!!i::nat. m \<le> n ==> i^m dvd i^n"
wenzelm@22718
   893
  apply (unfold dvd_def)
wenzelm@22718
   894
  apply (erule linorder_not_less [THEN iffD2, THEN add_diff_inverse, THEN subst])
wenzelm@22718
   895
  apply (simp add: power_add)
wenzelm@22718
   896
  done
haftmann@21408
   897
nipkow@25162
   898
lemma nat_zero_less_power_iff [simp]: "(x^n > 0) = (x > (0::nat) | n=0)"
wenzelm@22718
   899
  by (induct n) auto
haftmann@21408
   900
haftmann@21408
   901
lemma power_le_dvd [rule_format]: "k^j dvd n --> i\<le>j --> k^i dvd (n::nat)"
wenzelm@22718
   902
  apply (induct j)
wenzelm@22718
   903
   apply (simp_all add: le_Suc_eq)
wenzelm@22718
   904
  apply (blast dest!: dvd_mult_right)
wenzelm@22718
   905
  done
haftmann@21408
   906
haftmann@21408
   907
lemma power_dvd_imp_le: "[|i^m dvd i^n;  (1::nat) < i|] ==> m \<le> n"
wenzelm@22718
   908
  apply (rule power_le_imp_le_exp, assumption)
wenzelm@22718
   909
  apply (erule dvd_imp_le, simp)
wenzelm@22718
   910
  done
haftmann@21408
   911
paulson@14267
   912
lemma mod_eq_0_iff: "(m mod d = 0) = (\<exists>q::nat. m = d*q)"
nipkow@29667
   913
by (auto simp add: dvd_eq_mod_eq_0 [symmetric] dvd_def)
paulson@17084
   914
wenzelm@22718
   915
lemmas mod_eq_0D [dest!] = mod_eq_0_iff [THEN iffD1]
paulson@14267
   916
paulson@14267
   917
(*Loses information, namely we also have r<d provided d is nonzero*)
paulson@14267
   918
lemma mod_eqD: "(m mod d = r) ==> \<exists>q::nat. m = r + q*d"
haftmann@27651
   919
  apply (cut_tac a = m in mod_div_equality)
wenzelm@22718
   920
  apply (simp only: add_ac)
wenzelm@22718
   921
  apply (blast intro: sym)
wenzelm@22718
   922
  done
paulson@14267
   923
nipkow@13152
   924
lemma split_div:
nipkow@13189
   925
 "P(n div k :: nat) =
nipkow@13189
   926
 ((k = 0 \<longrightarrow> P 0) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P i)))"
nipkow@13189
   927
 (is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")
nipkow@13189
   928
proof
nipkow@13189
   929
  assume P: ?P
nipkow@13189
   930
  show ?Q
nipkow@13189
   931
  proof (cases)
nipkow@13189
   932
    assume "k = 0"
haftmann@27651
   933
    with P show ?Q by simp
nipkow@13189
   934
  next
nipkow@13189
   935
    assume not0: "k \<noteq> 0"
nipkow@13189
   936
    thus ?Q
nipkow@13189
   937
    proof (simp, intro allI impI)
nipkow@13189
   938
      fix i j
nipkow@13189
   939
      assume n: "n = k*i + j" and j: "j < k"
nipkow@13189
   940
      show "P i"
nipkow@13189
   941
      proof (cases)
wenzelm@22718
   942
        assume "i = 0"
wenzelm@22718
   943
        with n j P show "P i" by simp
nipkow@13189
   944
      next
wenzelm@22718
   945
        assume "i \<noteq> 0"
wenzelm@22718
   946
        with not0 n j P show "P i" by(simp add:add_ac)
nipkow@13189
   947
      qed
nipkow@13189
   948
    qed
nipkow@13189
   949
  qed
nipkow@13189
   950
next
nipkow@13189
   951
  assume Q: ?Q
nipkow@13189
   952
  show ?P
nipkow@13189
   953
  proof (cases)
nipkow@13189
   954
    assume "k = 0"
haftmann@27651
   955
    with Q show ?P by simp
nipkow@13189
   956
  next
nipkow@13189
   957
    assume not0: "k \<noteq> 0"
nipkow@13189
   958
    with Q have R: ?R by simp
nipkow@13189
   959
    from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]
nipkow@13517
   960
    show ?P by simp
nipkow@13189
   961
  qed
nipkow@13189
   962
qed
nipkow@13189
   963
berghofe@13882
   964
lemma split_div_lemma:
haftmann@26100
   965
  assumes "0 < n"
haftmann@26100
   966
  shows "n * q \<le> m \<and> m < n * Suc q \<longleftrightarrow> q = ((m\<Colon>nat) div n)" (is "?lhs \<longleftrightarrow> ?rhs")
haftmann@26100
   967
proof
haftmann@26100
   968
  assume ?rhs
haftmann@26100
   969
  with mult_div_cancel have nq: "n * q = m - (m mod n)" by simp
haftmann@26100
   970
  then have A: "n * q \<le> m" by simp
haftmann@26100
   971
  have "n - (m mod n) > 0" using mod_less_divisor assms by auto
haftmann@26100
   972
  then have "m < m + (n - (m mod n))" by simp
haftmann@26100
   973
  then have "m < n + (m - (m mod n))" by simp
haftmann@26100
   974
  with nq have "m < n + n * q" by simp
haftmann@26100
   975
  then have B: "m < n * Suc q" by simp
haftmann@26100
   976
  from A B show ?lhs ..
haftmann@26100
   977
next
haftmann@26100
   978
  assume P: ?lhs
haftmann@26100
   979
  then have "divmod_rel m n q (m - n * q)"
haftmann@26100
   980
    unfolding divmod_rel_def by (auto simp add: mult_ac)
haftmann@26100
   981
  then show ?rhs using divmod_rel by (rule divmod_rel_unique_div)
haftmann@26100
   982
qed
berghofe@13882
   983
berghofe@13882
   984
theorem split_div':
berghofe@13882
   985
  "P ((m::nat) div n) = ((n = 0 \<and> P 0) \<or>
paulson@14267
   986
   (\<exists>q. (n * q \<le> m \<and> m < n * (Suc q)) \<and> P q))"
berghofe@13882
   987
  apply (case_tac "0 < n")
berghofe@13882
   988
  apply (simp only: add: split_div_lemma)
haftmann@27651
   989
  apply simp_all
berghofe@13882
   990
  done
berghofe@13882
   991
nipkow@13189
   992
lemma split_mod:
nipkow@13189
   993
 "P(n mod k :: nat) =
nipkow@13189
   994
 ((k = 0 \<longrightarrow> P n) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P j)))"
nipkow@13189
   995
 (is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")
nipkow@13189
   996
proof
nipkow@13189
   997
  assume P: ?P
nipkow@13189
   998
  show ?Q
nipkow@13189
   999
  proof (cases)
nipkow@13189
  1000
    assume "k = 0"
haftmann@27651
  1001
    with P show ?Q by simp
nipkow@13189
  1002
  next
nipkow@13189
  1003
    assume not0: "k \<noteq> 0"
nipkow@13189
  1004
    thus ?Q
nipkow@13189
  1005
    proof (simp, intro allI impI)
nipkow@13189
  1006
      fix i j
nipkow@13189
  1007
      assume "n = k*i + j" "j < k"
nipkow@13189
  1008
      thus "P j" using not0 P by(simp add:add_ac mult_ac)
nipkow@13189
  1009
    qed
nipkow@13189
  1010
  qed
nipkow@13189
  1011
next
nipkow@13189
  1012
  assume Q: ?Q
nipkow@13189
  1013
  show ?P
nipkow@13189
  1014
  proof (cases)
nipkow@13189
  1015
    assume "k = 0"
haftmann@27651
  1016
    with Q show ?P by simp
nipkow@13189
  1017
  next
nipkow@13189
  1018
    assume not0: "k \<noteq> 0"
nipkow@13189
  1019
    with Q have R: ?R by simp
nipkow@13189
  1020
    from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]
nipkow@13517
  1021
    show ?P by simp
nipkow@13189
  1022
  qed
nipkow@13189
  1023
qed
nipkow@13189
  1024
berghofe@13882
  1025
theorem mod_div_equality': "(m::nat) mod n = m - (m div n) * n"
berghofe@13882
  1026
  apply (rule_tac P="%x. m mod n = x - (m div n) * n" in
berghofe@13882
  1027
    subst [OF mod_div_equality [of _ n]])
berghofe@13882
  1028
  apply arith
berghofe@13882
  1029
  done
berghofe@13882
  1030
haftmann@22800
  1031
lemma div_mod_equality':
haftmann@22800
  1032
  fixes m n :: nat
haftmann@22800
  1033
  shows "m div n * n = m - m mod n"
haftmann@22800
  1034
proof -
haftmann@22800
  1035
  have "m mod n \<le> m mod n" ..
haftmann@22800
  1036
  from div_mod_equality have 
haftmann@22800
  1037
    "m div n * n + m mod n - m mod n = m - m mod n" by simp
haftmann@22800
  1038
  with diff_add_assoc [OF `m mod n \<le> m mod n`, of "m div n * n"] have
haftmann@22800
  1039
    "m div n * n + (m mod n - m mod n) = m - m mod n"
haftmann@22800
  1040
    by simp
haftmann@22800
  1041
  then show ?thesis by simp
haftmann@22800
  1042
qed
haftmann@22800
  1043
haftmann@22800
  1044
haftmann@25942
  1045
subsubsection {*An ``induction'' law for modulus arithmetic.*}
paulson@14640
  1046
paulson@14640
  1047
lemma mod_induct_0:
paulson@14640
  1048
  assumes step: "\<forall>i<p. P i \<longrightarrow> P ((Suc i) mod p)"
paulson@14640
  1049
  and base: "P i" and i: "i<p"
paulson@14640
  1050
  shows "P 0"
paulson@14640
  1051
proof (rule ccontr)
paulson@14640
  1052
  assume contra: "\<not>(P 0)"
paulson@14640
  1053
  from i have p: "0<p" by simp
paulson@14640
  1054
  have "\<forall>k. 0<k \<longrightarrow> \<not> P (p-k)" (is "\<forall>k. ?A k")
paulson@14640
  1055
  proof
paulson@14640
  1056
    fix k
paulson@14640
  1057
    show "?A k"
paulson@14640
  1058
    proof (induct k)
paulson@14640
  1059
      show "?A 0" by simp  -- "by contradiction"
paulson@14640
  1060
    next
paulson@14640
  1061
      fix n
paulson@14640
  1062
      assume ih: "?A n"
paulson@14640
  1063
      show "?A (Suc n)"
paulson@14640
  1064
      proof (clarsimp)
wenzelm@22718
  1065
        assume y: "P (p - Suc n)"
wenzelm@22718
  1066
        have n: "Suc n < p"
wenzelm@22718
  1067
        proof (rule ccontr)
wenzelm@22718
  1068
          assume "\<not>(Suc n < p)"
wenzelm@22718
  1069
          hence "p - Suc n = 0"
wenzelm@22718
  1070
            by simp
wenzelm@22718
  1071
          with y contra show "False"
wenzelm@22718
  1072
            by simp
wenzelm@22718
  1073
        qed
wenzelm@22718
  1074
        hence n2: "Suc (p - Suc n) = p-n" by arith
wenzelm@22718
  1075
        from p have "p - Suc n < p" by arith
wenzelm@22718
  1076
        with y step have z: "P ((Suc (p - Suc n)) mod p)"
wenzelm@22718
  1077
          by blast
wenzelm@22718
  1078
        show "False"
wenzelm@22718
  1079
        proof (cases "n=0")
wenzelm@22718
  1080
          case True
wenzelm@22718
  1081
          with z n2 contra show ?thesis by simp
wenzelm@22718
  1082
        next
wenzelm@22718
  1083
          case False
wenzelm@22718
  1084
          with p have "p-n < p" by arith
wenzelm@22718
  1085
          with z n2 False ih show ?thesis by simp
wenzelm@22718
  1086
        qed
paulson@14640
  1087
      qed
paulson@14640
  1088
    qed
paulson@14640
  1089
  qed
paulson@14640
  1090
  moreover
paulson@14640
  1091
  from i obtain k where "0<k \<and> i+k=p"
paulson@14640
  1092
    by (blast dest: less_imp_add_positive)
paulson@14640
  1093
  hence "0<k \<and> i=p-k" by auto
paulson@14640
  1094
  moreover
paulson@14640
  1095
  note base
paulson@14640
  1096
  ultimately
paulson@14640
  1097
  show "False" by blast
paulson@14640
  1098
qed
paulson@14640
  1099
paulson@14640
  1100
lemma mod_induct:
paulson@14640
  1101
  assumes step: "\<forall>i<p. P i \<longrightarrow> P ((Suc i) mod p)"
paulson@14640
  1102
  and base: "P i" and i: "i<p" and j: "j<p"
paulson@14640
  1103
  shows "P j"
paulson@14640
  1104
proof -
paulson@14640
  1105
  have "\<forall>j<p. P j"
paulson@14640
  1106
  proof
paulson@14640
  1107
    fix j
paulson@14640
  1108
    show "j<p \<longrightarrow> P j" (is "?A j")
paulson@14640
  1109
    proof (induct j)
paulson@14640
  1110
      from step base i show "?A 0"
wenzelm@22718
  1111
        by (auto elim: mod_induct_0)
paulson@14640
  1112
    next
paulson@14640
  1113
      fix k
paulson@14640
  1114
      assume ih: "?A k"
paulson@14640
  1115
      show "?A (Suc k)"
paulson@14640
  1116
      proof
wenzelm@22718
  1117
        assume suc: "Suc k < p"
wenzelm@22718
  1118
        hence k: "k<p" by simp
wenzelm@22718
  1119
        with ih have "P k" ..
wenzelm@22718
  1120
        with step k have "P (Suc k mod p)"
wenzelm@22718
  1121
          by blast
wenzelm@22718
  1122
        moreover
wenzelm@22718
  1123
        from suc have "Suc k mod p = Suc k"
wenzelm@22718
  1124
          by simp
wenzelm@22718
  1125
        ultimately
wenzelm@22718
  1126
        show "P (Suc k)" by simp
paulson@14640
  1127
      qed
paulson@14640
  1128
    qed
paulson@14640
  1129
  qed
paulson@14640
  1130
  with j show ?thesis by blast
paulson@14640
  1131
qed
paulson@14640
  1132
paulson@3366
  1133
end