Quelle Euclidean_Rings.thy Sprache: Isabelle

(*  Title:      HOL/Euclidean_Rgins.thy
    Author:     Manuel Eberl, TU Muenchen
    Author:     Florian Haftmann, TU Muenchen
*)

section \<open>Division in euclidean (semi)rings\<close>

theory Euclidean_Rings
  imports Int Lattices_Big
begin

subsection \<open>Euclidean (semi)rings with explicit division and remainder\<close>

class euclidean_semiring = semidom_modulo +
  fixes euclidean_size :: "'a \ nat"
  assumes size_0 [simp]: "euclidean_size 0 = 0"
  assumes mod_size_less:
    "b \ 0 \ euclidean_size (a mod b) < euclidean_size b"
  assumes size_mult_mono:
    "b \ 0 \ euclidean_size a \ euclidean_size (a * b)"
begin

lemma euclidean_size_eq_0_iff [simp]:
  "euclidean_size b = 0 \ b = 0"
proof
  assume "b = 0"
  then show "euclidean_size b = 0"
    by simp
next
  assume "euclidean_size b = 0"
  show "b = 0"
  proof (rule ccontr)
    assume "b \ 0"
    with mod_size_less have "euclidean_size (b mod b) < euclidean_size b" .
    with \<open>euclidean_size b = 0\<close> show False
      by simp
  qed
qed

lemma euclidean_size_greater_0_iff [simp]:
  "euclidean_size b > 0 \ b \ 0"
  using euclidean_size_eq_0_iff [symmetric, of b] by safe simp

lemma size_mult_mono': "b \ 0 \ euclidean_size a \ euclidean_size (b * a)"
  by (subst mult.commute) (rule size_mult_mono)

lemma dvd_euclidean_size_eq_imp_dvd:
  assumes "a \ 0" and "euclidean_size a = euclidean_size b"
    and "b dvd a"
  shows "a dvd b"
proof (rule ccontr)
  assume "\ a dvd b"
  hence "b mod a \ 0" using mod_0_imp_dvd [of b a] by blast
  then have "b mod a \ 0" by (simp add: mod_eq_0_iff_dvd)
  from \<open>b dvd a\<close> have "b dvd b mod a" by (simp add: dvd_mod_iff)
  then obtain c where "b mod a = b * c" unfolding dvd_def by blast
    with \<open>b mod a \<noteq> 0\<close> have "c \<noteq> 0" by auto
  with \<open>b mod a = b * c\<close> have "euclidean_size (b mod a) \<ge> euclidean_size b"
    using size_mult_mono by force
  moreover from \<open>\<not> a dvd b\<close> and \<open>a \<noteq> 0\<close>
  have "euclidean_size (b mod a) < euclidean_size a"
    using mod_size_less by blast
  ultimately show False using \<open>euclidean_size a = euclidean_size b\<close>
    by simp
qed

lemma euclidean_size_times_unit:
  assumes "is_unit a"
  shows   "euclidean_size (a * b) = euclidean_size b"
proof (rule antisym)
  from assms have [simp]: "a \ 0" by auto
  thus "euclidean_size (a * b) \ euclidean_size b" by (rule size_mult_mono')
  from assms have "is_unit (1 div a)" by simp
  hence "1 div a \ 0" by (intro notI) simp_all
  hence "euclidean_size (a * b) \ euclidean_size ((1 div a) * (a * b))"
    by (rule size_mult_mono')
  also from assms have "(1 div a) * (a * b) = b"
    by (simp add: algebra_simps unit_div_mult_swap)
  finally show "euclidean_size (a * b) \ euclidean_size b" .
qed

lemma euclidean_size_unit:
  "is_unit a \ euclidean_size a = euclidean_size 1"
  using euclidean_size_times_unit [of a 1] by simp

lemma unit_iff_euclidean_size:
  "is_unit a \ euclidean_size a = euclidean_size 1 \ a \ 0"
proof safe
  assume A: "a \ 0" and B: "euclidean_size a = euclidean_size 1"
  show "is_unit a"
    by (rule dvd_euclidean_size_eq_imp_dvd [OF A B]) simp_all
qed (auto intro: euclidean_size_unit)

lemma euclidean_size_times_nonunit:
  assumes "a \ 0" "b \ 0" "\ is_unit a"
  shows   "euclidean_size b < euclidean_size (a * b)"
proof (rule ccontr)
  assume "\euclidean_size b < euclidean_size (a * b)"
  with size_mult_mono'[OF assms(1), of b]
    have eq: "euclidean_size (a * b) = euclidean_size b" by simp
  have "a * b dvd b"
    by (rule dvd_euclidean_size_eq_imp_dvd [OF _ eq])
       (use assms in simp_all)
  hence "a * b dvd 1 * b" by simp
  with \<open>b \<noteq> 0\<close> have "is_unit a" by (subst (asm) dvd_times_right_cancel_iff)
  with assms(3) show False by contradiction
qed

lemma dvd_imp_size_le:
  assumes "a dvd b" "b \ 0"
  shows   "euclidean_size a \ euclidean_size b"
  using assms by (auto simp: size_mult_mono)

lemma dvd_proper_imp_size_less:
  assumes "a dvd b" "\ b dvd a" "b \ 0"
  shows   "euclidean_size a < euclidean_size b"
proof -
  from assms(1) obtain c where "b = a * c" by (erule dvdE)
  hence z: "b = c * a" by (simp add: mult.commute)
  from z assms have "\is_unit c" by (auto simp: mult.commute mult_unit_dvd_iff)
  with z assms show ?thesis
    by (auto intro!: euclidean_size_times_nonunit)
qed

lemma unit_imp_mod_eq_0:
  "a mod b = 0" if "is_unit b"
  using that by (simp add: mod_eq_0_iff_dvd unit_imp_dvd)

lemma mod_eq_self_iff_div_eq_0:
  "a mod b = a \ a div b = 0" (is "?P \ ?Q")
proof
  assume ?P
  with div_mult_mod_eq [of a b] show ?Q
    by auto
next
  assume ?Q
  with div_mult_mod_eq [of a b] show ?P
    by simp
qed

lemma coprime_mod_left_iff [simp]:
  "coprime (a mod b) b \ coprime a b" if "b \ 0"
  by (rule iffI; rule coprimeI)
    (use that in \<open>auto dest!: dvd_mod_imp_dvd coprime_common_divisor simp add: dvd_mod_iff\<close>)

lemma coprime_mod_right_iff [simp]:
  "coprime a (b mod a) \ coprime a b" if "a \ 0"
  using that coprime_mod_left_iff [of a b] by (simp add: ac_simps)

end

class euclidean_ring = idom_modulo + euclidean_semiring
begin

lemma dvd_diff_commute [ac_simps]:
  "a dvd c - b \ a dvd b - c"
proof -
  have "a dvd c - b \ a dvd (c - b) * - 1"
    by (subst dvd_mult_unit_iff) simp_all
  then show ?thesis
    by simp
qed

end

subsection \<open>Euclidean (semi)rings with cancel rules\<close>

class euclidean_semiring_cancel = euclidean_semiring +
  assumes div_mult_self1 [simp]: "b \ 0 \ (a + c * b) div b = c + a div b"
  and div_mult_mult1 [simp]: "c \ 0 \ (c * a) div (c * b) = a div b"
begin

lemma div_mult_self2 [simp]:
  assumes "b \ 0"
  shows "(a + b * c) div b = c + a div b"
  using assms div_mult_self1 [of b a c] by (simp add: mult.commute)

lemma div_mult_self3 [simp]:
  assumes "b \ 0"
  shows "(c * b + a) div b = c + a div b"
  using assms by (simp add: add.commute)

lemma div_mult_self4 [simp]:
  assumes "b \ 0"
  shows "(b * c + a) div b = c + a div b"
  using assms by (simp add: add.commute)

lemma mod_mult_self1 [simp]: "(a + c * b) mod b = a mod b"
proof (cases "b = 0")
  case True then show ?thesis by simp
next
  case False
  have "a + c * b = (a + c * b) div b * b + (a + c * b) mod b"
    by (simp add: div_mult_mod_eq)
  also from False div_mult_self1 [of b a c] have
    "\ = (c + a div b) * b + (a + c * b) mod b"
      by (simp add: algebra_simps)
  finally have "a = a div b * b + (a + c * b) mod b"
    by (simp add: add.commute [of a] add.assoc distrib_right)
  then have "a div b * b + (a + c * b) mod b = a div b * b + a mod b"
    by (simp add: div_mult_mod_eq)
  then show ?thesis by simp
qed

lemma mod_mult_self2 [simp]:
  "(a + b * c) mod b = a mod b"
  by (simp add: mult.commute [of b])

lemma mod_mult_self3 [simp]:
  "(c * b + a) mod b = a mod b"
  by (simp add: add.commute)

lemma mod_mult_self4 [simp]:
  "(b * c + a) mod b = a mod b"
  by (simp add: add.commute)

lemma mod_mult_self1_is_0 [simp]:
  "b * a mod b = 0"
  using mod_mult_self2 [of 0 b a] by simp

lemma mod_mult_self2_is_0 [simp]:
  "a * b mod b = 0"
  using mod_mult_self1 [of 0 a b] by simp

lemma div_add_self1:
  assumes "b \ 0"
  shows "(b + a) div b = a div b + 1"
  using assms div_mult_self1 [of b a 1] by (simp add: add.commute)

lemma div_add_self2:
  assumes "b \ 0"
  shows "(a + b) div b = a div b + 1"
  using assms div_add_self1 [of b a] by (simp add: add.commute)

lemma mod_add_self1 [simp]:
  "(b + a) mod b = a mod b"
  using mod_mult_self1 [of a 1 b] by (simp add: add.commute)

lemma mod_add_self2 [simp]:
  "(a + b) mod b = a mod b"
  using mod_mult_self1 [of a 1 b] by simp

lemma mod_div_trivial [simp]:
  "a mod b div b = 0"
proof (cases "b = 0")
  assume "b = 0"
  thus ?thesis by simp
next
  assume "b \ 0"
  hence "a div b + a mod b div b = (a mod b + a div b * b) div b"
    by (rule div_mult_self1 [symmetric])
  also have "\ = a div b"
    by (simp only: mod_div_mult_eq)
  also have "\ = a div b + 0"
    by simp
  finally show ?thesis
    by (rule add_left_imp_eq)
qed

lemma mod_mod_trivial [simp]:
  "a mod b mod b = a mod b"
proof -
  have "a mod b mod b = (a mod b + a div b * b) mod b"
    by (simp only: mod_mult_self1)
  also have "\ = a mod b"
    by (simp only: mod_div_mult_eq)
  finally show ?thesis .
qed

lemma mod_mod_cancel:
  assumes "c dvd b"
  shows "a mod b mod c = a mod c"
proof -
  from \<open>c dvd b\<close> obtain k where "b = c * k"
    by (rule dvdE)
  have "a mod b mod c = a mod (c * k) mod c"
    by (simp only: \<open>b = c * k\<close>)
  also have "\ = (a mod (c * k) + a div (c * k) * k * c) mod c"
    by (simp only: mod_mult_self1)
  also have "\ = (a div (c * k) * (c * k) + a mod (c * k)) mod c"
    by (simp only: ac_simps)
  also have "\ = a mod c"
    by (simp only: div_mult_mod_eq)
  finally show ?thesis .
qed

lemma div_mult_mult2 [simp]:
  "c \ 0 \ (a * c) div (b * c) = a div b"
  by (drule div_mult_mult1) (simp add: mult.commute)

lemma div_mult_mult1_if [simp]:
  "(c * a) div (c * b) = (if c = 0 then 0 else a div b)"
  by simp_all

lemma mod_mult_mult1:
  "(c * a) mod (c * b) = c * (a mod b)"
proof (cases "c = 0")
  case True then show ?thesis by simp
next
  case False
  from div_mult_mod_eq
  have "((c * a) div (c * b)) * (c * b) + (c * a) mod (c * b) = c * a" .
  with False have "c * ((a div b) * b + a mod b) + (c * a) mod (c * b)
    = c * a + c * (a mod b)" by (simp add: algebra_simps)
  with div_mult_mod_eq show ?thesis by simp
qed

lemma mod_mult_mult2:
  "(a * c) mod (b * c) = (a mod b) * c"
  using mod_mult_mult1 [of c a b] by (simp add: mult.commute)

lemma mult_mod_left: "(a mod b) * c = (a * c) mod (b * c)"
  by (fact mod_mult_mult2 [symmetric])

lemma mult_mod_right: "c * (a mod b) = (c * a) mod (c * b)"
  by (fact mod_mult_mult1 [symmetric])

lemma dvd_mod: "k dvd m \ k dvd n \ k dvd (m mod n)"
  unfolding dvd_def by (auto simp add: mod_mult_mult1)

lemma div_plus_div_distrib_dvd_left:
  "c dvd a \ (a + b) div c = a div c + b div c"
  by (cases "c = 0") auto

lemma div_plus_div_distrib_dvd_right:
  "c dvd b \ (a + b) div c = a div c + b div c"
  using div_plus_div_distrib_dvd_left [of c b a]
  by (simp add: ac_simps)

lemma sum_div_partition:
  \<open>(\<Sum>a\<in>A. f a) div b = (\<Sum>a\<in>A \<inter> {a. b dvd f a}. f a div b) + (\<Sum>a\<in>A \<inter> {a. \<not> b dvd f a}. f a) div b\<close>
    if \<open>finite A\<close>
proof -
  have \<open>A = A \<inter> {a. b dvd f a} \<union> A \<inter> {a. \<not> b dvd f a}\<close>
    by auto
  then have \<open>(\<Sum>a\<in>A. f a) = (\<Sum>a\<in>A \<inter> {a. b dvd f a} \<union> A \<inter> {a. \<not> b dvd f a}. f a)\<close>
    by simp
  also have \<open>\<dots> = (\<Sum>a\<in>A \<inter> {a. b dvd f a}. f a) + (\<Sum>a\<in>A \<inter> {a. \<not> b dvd f a}. f a)\<close>
    using \<open>finite A\<close> by (auto intro: sum.union_inter_neutral)
  finally have *: \<open>sum f A = sum f (A \<inter> {a. b dvd f a}) + sum f (A \<inter> {a. \<not> b dvd f a})\<close> .
  define B where B: \<open>B = A \<inter> {a. b dvd f a}\<close>
  with \<open>finite A\<close> have \<open>finite B\<close> and \<open>a \<in> B \<Longrightarrow> b dvd f a\<close> for a
    by simp_all
  then have \<open>(\<Sum>a\<in>B. f a) div b = (\<Sum>a\<in>B. f a div b)\<close> and \<open>b dvd (\<Sum>a\<in>B. f a)\<close>
    by induction (simp_all add: div_plus_div_distrib_dvd_left)
  then show ?thesis using *
    by (simp add: B div_plus_div_distrib_dvd_left)
qed

named_theorems mod_simps

text \<open>Addition respects modular equivalence.\<close>

lemma mod_add_left_eq [mod_simps]:
  "(a mod c + b) mod c = (a + b) mod c"
proof -
  have "(a + b) mod c = (a div c * c + a mod c + b) mod c"
    by (simp only: div_mult_mod_eq)
  also have "\ = (a mod c + b + a div c * c) mod c"
    by (simp only: ac_simps)
  also have "\ = (a mod c + b) mod c"
    by (rule mod_mult_self1)
  finally show ?thesis
    by (rule sym)
qed

lemma mod_add_right_eq [mod_simps]:
  "(a + b mod c) mod c = (a + b) mod c"
  using mod_add_left_eq [of b c a] by (simp add: ac_simps)

lemma mod_add_eq:
  "(a mod c + b mod c) mod c = (a + b) mod c"
  by (simp add: mod_add_left_eq mod_add_right_eq)

lemma mod_sum_eq [mod_simps]:
  "(\i\A. f i mod a) mod a = sum f A mod a"
proof (induct A rule: infinite_finite_induct)
  case (insert i A)
  then have "(\i\insert i A. f i mod a) mod a
    = (f i mod a + (\<Sum>i\<in>A. f i mod a)) mod a"
    by simp
  also have "\ = (f i + (\i\A. f i mod a) mod a) mod a"
    by (simp add: mod_simps)
  also have "\ = (f i + (\i\A. f i) mod a) mod a"
    by (simp add: insert.hyps)
  finally show ?case
    by (simp add: insert.hyps mod_simps)
qed simp_all

lemma mod_add_cong:
  assumes "a mod c = a' mod c"
  assumes "b mod c = b' mod c"
  shows "(a + b) mod c = (a' + b') mod c"
proof -
  have "(a mod c + b mod c) mod c = (a' mod c + b' mod c) mod c"
    unfolding assms ..
  then show ?thesis
    by (simp add: mod_add_eq)
qed

text \<open>Multiplication respects modular equivalence.\<close>

lemma mod_mult_left_eq [mod_simps]:
  "((a mod c) * b) mod c = (a * b) mod c"
proof -
  have "(a * b) mod c = ((a div c * c + a mod c) * b) mod c"
    by (simp only: div_mult_mod_eq)
  also have "\ = (a mod c * b + a div c * b * c) mod c"
    by (simp only: algebra_simps)
  also have "\ = (a mod c * b) mod c"
    by (rule mod_mult_self1)
  finally show ?thesis
    by (rule sym)
qed

lemma mod_mult_right_eq [mod_simps]:
  "(a * (b mod c)) mod c = (a * b) mod c"
  using mod_mult_left_eq [of b c a] by (simp add: ac_simps)

lemma mod_mult_eq:
  "((a mod c) * (b mod c)) mod c = (a * b) mod c"
  by (simp add: mod_mult_left_eq mod_mult_right_eq)

lemma mod_prod_eq [mod_simps]:
  "(\i\A. f i mod a) mod a = prod f A mod a"
proof (induct A rule: infinite_finite_induct)
  case (insert i A)
  then have "(\i\insert i A. f i mod a) mod a
    = (f i mod a * (\<Prod>i\<in>A. f i mod a)) mod a"
    by simp
  also have "\ = (f i * ((\i\A. f i mod a) mod a)) mod a"
    by (simp add: mod_simps)
  also have "\ = (f i * ((\i\A. f i) mod a)) mod a"
    by (simp add: insert.hyps)
  finally show ?case
    by (simp add: insert.hyps mod_simps)
qed simp_all

lemma mod_mult_cong:
  assumes "a mod c = a' mod c"
  assumes "b mod c = b' mod c"
  shows "(a * b) mod c = (a' * b') mod c"
proof -
  have "(a mod c * (b mod c)) mod c = (a' mod c * (b' mod c)) mod c"
    unfolding assms ..
  then show ?thesis
    by (simp add: mod_mult_eq)
qed

text \<open>Exponentiation respects modular equivalence.\<close>

lemma power_mod [mod_simps]:
  "((a mod b) ^ n) mod b = (a ^ n) mod b"
proof (induct n)
  case 0
  then show ?case by simp
next
  case (Suc n)
  have "(a mod b) ^ Suc n mod b = (a mod b) * ((a mod b) ^ n mod b) mod b"
    by (simp add: mod_mult_right_eq)
  with Suc show ?case
    by (simp add: mod_mult_left_eq mod_mult_right_eq)
qed

lemma power_diff_power_eq:
  \<open>a ^ m div a ^ n = (if n \<le> m then a ^ (m - n) else 1 div a ^ (n - m))\<close>
    if \<open>a \<noteq> 0\<close>
proof (cases \<open>n \<le> m\<close>)
  case True
  with that power_diff [symmetric, of a n m] show ?thesis by simp
next
  case False
  then obtain q where n: \<open>n = m + Suc q\<close>
    by (auto simp add: not_le dest: less_imp_Suc_add)
  then have \<open>a ^ m div a ^ n = (a ^ m * 1) div (a ^ m * a ^ Suc q)\<close>
    by (simp add: power_add ac_simps)
  moreover from that have \<open>a ^ m \<noteq> 0\<close>
    by simp
  ultimately have \<open>a ^ m div a ^ n = 1 div a ^ Suc q\<close>
    by (subst (asm) div_mult_mult1) simp
  with False n show ?thesis
    by simp
qed

end

class euclidean_ring_cancel = euclidean_ring + euclidean_semiring_cancel
begin

subclass idom_divide ..

lemma div_minus_minus [simp]: "(- a) div (- b) = a div b"
  using div_mult_mult1 [of "- 1" a b] by simp

lemma mod_minus_minus [simp]: "(- a) mod (- b) = - (a mod b)"
  using mod_mult_mult1 [of "- 1" a b] by simp

lemma div_minus_right: "a div (- b) = (- a) div b"
  using div_minus_minus [of "- a" b] by simp

lemma mod_minus_right: "a mod (- b) = - ((- a) mod b)"
  using mod_minus_minus [of "- a" b] by simp

lemma div_minus1_right [simp]: "a div (- 1) = - a"
  using div_minus_right [of a 1] by simp

lemma mod_minus1_right [simp]: "a mod (- 1) = 0"
  using mod_minus_right [of a 1] by simp

text \<open>Negation respects modular equivalence.\<close>

lemma mod_minus_eq [mod_simps]:
  "(- (a mod b)) mod b = (- a) mod b"
proof -
  have "(- a) mod b = (- (a div b * b + a mod b)) mod b"
    by (simp only: div_mult_mod_eq)
  also have "\ = (- (a mod b) + - (a div b) * b) mod b"
    by (simp add: ac_simps)
  also have "\ = (- (a mod b)) mod b"
    by (rule mod_mult_self1)
  finally show ?thesis
    by (rule sym)
qed

lemma mod_minus_cong:
  assumes "a mod b = a' mod b"
  shows "(- a) mod b = (- a') mod b"
proof -
  have "(- (a mod b)) mod b = (- (a' mod b)) mod b"
    unfolding assms ..
  then show ?thesis
    by (simp add: mod_minus_eq)
qed

text \<open>Subtraction respects modular equivalence.\<close>

lemma mod_diff_left_eq [mod_simps]:
  "(a mod c - b) mod c = (a - b) mod c"
  using mod_add_cong [of a c "a mod c" "- b" "- b"]
  by simp

lemma mod_diff_right_eq [mod_simps]:
  "(a - b mod c) mod c = (a - b) mod c"
  using mod_add_cong [of a c a "- b" "- (b mod c)"] mod_minus_cong [of "b mod c" c b]
  by simp

lemma mod_diff_eq:
  "(a mod c - b mod c) mod c = (a - b) mod c"
  using mod_add_cong [of a c "a mod c" "- b" "- (b mod c)"] mod_minus_cong [of "b mod c" c b]
  by simp

lemma mod_diff_cong:
  assumes "a mod c = a' mod c"
  assumes "b mod c = b' mod c"
  shows "(a - b) mod c = (a' - b') mod c"
  using assms mod_add_cong [of a c a' "- b" "- b'"] mod_minus_cong [of b c "b'"]
  by simp

lemma minus_mod_self2 [simp]:
  "(a - b) mod b = a mod b"
  using mod_diff_right_eq [of a b b]
  by (simp add: mod_diff_right_eq)

lemma minus_mod_self1 [simp]:
  "(b - a) mod b = - a mod b"
  using mod_add_self2 [of "- a" b] by simp

lemma mod_eq_dvd_iff:
  "a mod c = b mod c \ c dvd a - b" (is "?P \ ?Q")
proof
  assume ?P
  then have "(a mod c - b mod c) mod c = 0"
    by simp
  then show ?Q
    by (simp add: dvd_eq_mod_eq_0 mod_simps)
next
  assume ?Q
  then obtain d where d: "a - b = c * d" ..
  then have "a = c * d + b"
    by (simp add: algebra_simps)
  then show ?P by simp
qed

lemma mod_eqE:
  assumes "a mod c = b mod c"
  obtains d where "b = a + c * d"
proof -
  from assms have "c dvd a - b"
    by (simp add: mod_eq_dvd_iff)
  then obtain d where "a - b = c * d" ..
  then have "b = a + c * - d"
    by (simp add: algebra_simps)
  with that show thesis .
qed

lemma invertible_coprime:
  "coprime a c" if "a * b mod c = 1"
  by (rule coprimeI) (use that dvd_mod_iff [of _ c "a * b"] in auto)

end

subsection \<open>Uniquely determined division\<close>

class unique_euclidean_semiring = euclidean_semiring +
  assumes euclidean_size_mult: \<open>euclidean_size (a * b) = euclidean_size a * euclidean_size b\<close>
  fixes division_segment :: \<open>'a \<Rightarrow> 'a\<close>
  assumes is_unit_division_segment [simp]: \<open>is_unit (division_segment a)\<close>
    and division_segment_mult:
    \<open>a \<noteq> 0 \<Longrightarrow> b \<noteq> 0 \<Longrightarrow> division_segment (a * b) = division_segment a * division_segment b\<close>
    and division_segment_mod:
    \<open>b \<noteq> 0 \<Longrightarrow> \<not> b dvd a \<Longrightarrow> division_segment (a mod b) = division_segment b\<close>
  assumes div_bounded:
    \<open>b \<noteq> 0 \<Longrightarrow> division_segment r = division_segment b
    \<Longrightarrow> euclidean_size r < euclidean_size b
    \<Longrightarrow> (q * b + r) div b = q\<close>
begin

lemma division_segment_not_0 [simp]:
  \<open>division_segment a \<noteq> 0\<close>
  using is_unit_division_segment [of a] is_unitE [of \<open>division_segment a\<close>] by blast

lemma euclidean_relationI [case_names by0 divides euclidean_relation]:
  \<open>(a div b, a mod b) = (q, r)\<close>
    if by0: \<open>b = 0 \<Longrightarrow> q = 0 \<and> r = a\<close>
    and divides: \<open>b \<noteq> 0 \<Longrightarrow> b dvd a \<Longrightarrow> r = 0 \<and> a = q * b\<close>
    and euclidean_relation: \<open>b \<noteq> 0 \<Longrightarrow> \<not> b dvd a \<Longrightarrow> division_segment r = division_segment b
      \<and> euclidean_size r < euclidean_size b \<and> a = q * b + r\<close>
proof (cases \<open>b = 0\<close>)
  case True
  with by0 show ?thesis
    by simp
next
  case False
  show ?thesis
  proof (cases \<open>b dvd a\<close>)
    case True
    with \<open>b \<noteq> 0\<close> divides
    show ?thesis
      by simp
  next
    case False
    with \<open>b \<noteq> 0\<close> euclidean_relation
    have \<open>division_segment r = division_segment b\<close>
      \<open>euclidean_size r < euclidean_size b\<close> \<open>a = q * b + r\<close>
      by simp_all
    from \<open>b \<noteq> 0\<close> \<open>division_segment r = division_segment b\<close>
      \<open>euclidean_size r < euclidean_size b\<close>
    have \<open>(q * b + r) div b = q\<close>
      by (rule div_bounded)
    with \<open>a = q * b + r\<close>
    have \<open>q = a div b\<close>
      by simp
    from \<open>a = q * b + r\<close>
    have \<open>a div b * b + a mod b = q * b + r\<close>
      by (simp add: div_mult_mod_eq)
    with \<open>q = a div b\<close>
    have \<open>q * b + a mod b = q * b + r\<close>
      by simp
    then have \<open>r = a mod b\<close>
      by simp
    with \<open>q = a div b\<close>
    show ?thesis
      by simp
  qed
qed

subclass euclidean_semiring_cancel
proof
  fix a b c
  assume \<open>b \<noteq> 0\<close>
  have \<open>((a + c * b) div b, (a + c * b) mod b) = (c + a div b, a mod b)\<close>
  proof (induction rule: euclidean_relationI)
    case by0
    with \<open>b \<noteq> 0\<close>
    show ?case
      by simp
  next
    case divides
    then show ?case
      by (simp add: algebra_simps dvd_add_left_iff)
  next
    case euclidean_relation
    then have \<open>\<not> b dvd a\<close>
      by (simp add: dvd_add_left_iff)
    have \<open>a mod b + (b * c + b * (a div b)) = b * c + ((a div b) * b + a mod b)\<close>
      by (simp add: ac_simps)
    with \<open>b \<noteq> 0\<close> have *: \<open>a mod b + (b * c + b * (a div b)) = b * c + a\<close>
      by (simp add: div_mult_mod_eq)
    from \<open>\<not> b dvd a\<close> euclidean_relation show ?case
      by (simp_all add: algebra_simps division_segment_mod mod_size_less *)
  qed
  then show \<open>(a + c * b) div b = c + a div b\<close>
    by simp
next
  fix a b c
  assume \<open>c \<noteq> 0\<close>
  have \<open>((c * a) div (c * b), (c * a) mod (c * b)) = (a div b, c * (a mod b))\<close>
  proof (induction rule: euclidean_relationI)
    case by0
    with \<open>c \<noteq> 0\<close> show ?case
      by simp
  next
    case divides
    then show ?case
      by (auto simp add: algebra_simps)
  next
    case euclidean_relation
    then have \<open>b \<noteq> 0\<close> \<open>a mod b \<noteq> 0\<close>
      by (simp_all add: mod_eq_0_iff_dvd)
    have \<open>c * (a mod b) + b * (c * (a div b)) = c * ((a div b) * b + a mod b)\<close>
      by (simp add: algebra_simps)
    with \<open>b \<noteq> 0\<close> have *: \<open>c * (a mod b) + b * (c * (a div b)) = c * a\<close>
      by (simp add: div_mult_mod_eq)
    from \<open>b \<noteq> 0\<close> \<open>c \<noteq> 0\<close> have \<open>euclidean_size c * euclidean_size (a mod b)
      < euclidean_size c * euclidean_size b\<close>
      using mod_size_less [of b a] by simp
    with euclidean_relation \<open>b \<noteq> 0\<close> \<open>a mod b \<noteq> 0\<close> show ?case
      by (simp add: algebra_simps division_segment_mult division_segment_mod euclidean_size_mult *)
  qed
  then show \<open>(c * a) div (c * b) = a div b\<close>
    by simp
qed

lemma div_eq_0_iff:
  \<open>a div b = 0 \<longleftrightarrow> euclidean_size a < euclidean_size b \<or> b = 0\<close> (is "_ \<longleftrightarrow> ?P")
  if \<open>division_segment a = division_segment b\<close>
proof (cases \<open>a = 0 \<or> b = 0\<close>)
  case True
  then show ?thesis by auto
next
  case False
  then have \<open>a \<noteq> 0\<close> \<open>b \<noteq> 0\<close>
    by simp_all
  have \<open>a div b = 0 \<longleftrightarrow> euclidean_size a < euclidean_size b\<close>
  proof
    assume \<open>a div b = 0\<close>
    then have \<open>a mod b = a\<close>
      using div_mult_mod_eq [of a b] by simp
    with \<open>b \<noteq> 0\<close> mod_size_less [of b a]
    show \<open>euclidean_size a < euclidean_size b\<close>
      by simp
  next
    assume \<open>euclidean_size a < euclidean_size b\<close>
    have \<open>(a div b, a mod b) = (0, a)\<close>
    proof (induction rule: euclidean_relationI)
      case by0
      show ?case
        by simp
    next
      case divides
      with \<open>euclidean_size a < euclidean_size b\<close> show ?case
        using dvd_imp_size_le [of b a] \<open>a \<noteq> 0\<close> by simp
    next
      case euclidean_relation
      with \<open>euclidean_size a < euclidean_size b\<close> that
      show ?case
        by simp
    qed
    then show \<open>a div b = 0\<close>
      by simp
  qed
  with \<open>b \<noteq> 0\<close> show ?thesis
    by simp
qed

lemma div_mult1_eq:
  \<open>(a * b) div c = a * (b div c) + a * (b mod c) div c\<close>
proof -
  have *: \<open>(a * b) mod c + (a * (c * (b div c)) + c * (a * (b mod c) div c)) = a * b\<close> (is \<open>?A + (?B + ?C) = _\<close>)
  proof -
    have \<open>?A = a * (b mod c) mod c\<close>
      by (simp add: mod_mult_right_eq)
    then have \<open>?C + ?A = a * (b mod c)\<close>
      by (simp add: mult_div_mod_eq)
    then have \<open>?B + (?C + ?A) = a * (c * (b div c) + (b mod c))\<close>
      by (simp add: algebra_simps)
    also have \<open>\<dots> = a * b\<close>
      by (simp add: mult_div_mod_eq)
    finally show ?thesis
      by (simp add: algebra_simps)
  qed
  have \<open>((a * b) div c, (a * b) mod c) = (a * (b div c) + a * (b mod c) div c, (a * b) mod c)\<close>
  proof (induction rule: euclidean_relationI)
    case by0
    then show ?case by simp
  next
    case divides
    with * show ?case
      by (simp add: algebra_simps)
  next
    case euclidean_relation
    with * show ?case
      by (simp add: division_segment_mod mod_size_less algebra_simps)
  qed
  then show ?thesis
    by simp
qed

lemma div_add1_eq:
  \<open>(a + b) div c = a div c + b div c + (a mod c + b mod c) div c\<close>
proof -
  have *: \<open>(a + b) mod c + (c * (a div c) + (c * (b div c) + c * ((a mod c + b mod c) div c))) = a + b\<close>
    (is \<open>?A + (?B + (?C + ?D)) = _\<close>)
  proof -
    have \<open>?A + (?B + (?C + ?D)) = ?A + ?D + (?B + ?C)\<close>
      by (simp add: ac_simps)
    also have \<open>?A + ?D = (a mod c + b mod c) mod c + ?D\<close>
      by (simp add: mod_add_eq)
    also have \<open>\<dots> = a mod c + b mod c\<close>
      by (simp add: mod_mult_div_eq)
    finally have \<open>?A + (?B + (?C + ?D)) = (a mod c + ?B) + (b mod c + ?C)\<close>
      by (simp add: ac_simps)
    then show ?thesis
      by (simp add: mod_mult_div_eq)
  qed
  have \<open>((a + b) div c, (a + b) mod c) = (a div c + b div c + (a mod c + b mod c) div c, (a + b) mod c)\<close>
  proof (induction rule: euclidean_relationI)
    case by0
    then show ?case
      by simp
  next
    case divides
    with * show ?case
      by (simp add: algebra_simps)
  next
    case euclidean_relation
    with * show ?case
      by (simp add: division_segment_mod mod_size_less algebra_simps)
  qed
  then show ?thesis
    by simp
qed

end

class unique_euclidean_ring = euclidean_ring + unique_euclidean_semiring
begin

subclass euclidean_ring_cancel ..

end

subsection \<open>Division on \<^typ>\<open>nat\<close>\<close>

instantiation nat :: normalization_semidom
begin

definition normalize_nat :: \<open>nat \<Rightarrow> nat\<close>
  where [simp]: \<open>normalize = (id :: nat \<Rightarrow> nat)\<close>

definition unit_factor_nat :: \<open>nat \<Rightarrow> nat\<close>
  where \<open>unit_factor n = of_bool (n > 0)\<close> for n :: nat

lemma unit_factor_simps [simp]:
  \<open>unit_factor 0 = (0::nat)\<close>
  \<open>unit_factor (Suc n) = 1\<close>
  by (simp_all add: unit_factor_nat_def)

definition divide_nat :: \<open>nat \<Rightarrow> nat \<Rightarrow> nat\<close>
  where \<open>m div n = (if n = 0 then 0 else Max {k. k * n \<le> m})\<close> for m n :: nat

instance
  by standard (auto simp add: divide_nat_def ac_simps unit_factor_nat_def intro: Max_eqI)

end

lemma coprime_Suc_0_left [simp]:
  "coprime (Suc 0) n"
  using coprime_1_left [of n] by simp

lemma coprime_Suc_0_right [simp]:
  "coprime n (Suc 0)"
  using coprime_1_right [of n] by simp

lemma coprime_common_divisor_nat: "coprime a b \ x dvd a \ x dvd b \ x = 1"
  for a b :: nat
  by (drule coprime_common_divisor [of _ _ x]) simp_all

instantiation nat :: unique_euclidean_semiring
begin

definition euclidean_size_nat :: \<open>nat \<Rightarrow> nat\<close>
  where [simp]: \<open>euclidean_size_nat = id\<close>

definition division_segment_nat :: \<open>nat \<Rightarrow> nat\<close>
  where [simp]: \<open>division_segment n = 1\<close> for n :: nat

definition modulo_nat :: \<open>nat \<Rightarrow> nat \<Rightarrow> nat\<close>
  where \<open>m mod n = m - (m div n * n)\<close> for m n :: nat

instance proof
  fix m n :: nat
  have ex: "\k. k * n \ l" for l :: nat
    by (rule exI [of _ 0]) simp
  have fin: "finite {k. k * n \ l}" if "n > 0" for l
  proof -
    from that have "{k. k * n \ l} \ {k. k \ l}"
      by (cases n) auto
    then show ?thesis
      by (rule finite_subset) simp
  qed
  have mult_div_unfold: "n * (m div n) = Max {l. l \ m \ n dvd l}"
  proof (cases "n = 0")
    case True
    moreover have "{l. l = 0 \ l \ m} = {0::nat}"
      by auto
    ultimately show ?thesis
      by simp
  next
    case False
    with ex [of m] fin have "n * Max {k. k * n \ m} = Max (times n ` {k. k * n \ m})"
      by (auto simp add: nat_mult_max_right intro: hom_Max_commute)
    also have "times n ` {k. k * n \ m} = {l. l \ m \ n dvd l}"
      by (auto simp add: ac_simps elim!: dvdE)
    finally show ?thesis
      using False by (simp add: divide_nat_def ac_simps)
  qed
  have less_eq: "m div n * n \ m"
    by (auto simp add: mult_div_unfold ac_simps intro: Max.boundedI)
  then show "m div n * n + m mod n = m"
    by (simp add: modulo_nat_def)
  assume "n \ 0"
  show "euclidean_size (m mod n) < euclidean_size n"
  proof -
    have "m < Suc (m div n) * n"
    proof (rule ccontr)
      assume "\ m < Suc (m div n) * n"
      then have "Suc (m div n) * n \ m"
        by (simp add: not_less)
      moreover from \<open>n \<noteq> 0\<close> have "Max {k. k * n \<le> m} < Suc (m div n)"
        by (simp add: divide_nat_def)
      with \<open>n \<noteq> 0\<close> ex fin have "\<And>k. k * n \<le> m \<Longrightarrow> k < Suc (m div n)"
        by auto
      ultimately have "Suc (m div n) < Suc (m div n)"
        by blast
      then show False
        by simp
    qed
    with \<open>n \<noteq> 0\<close> show ?thesis
      by (simp add: modulo_nat_def)
  qed
  show "euclidean_size m \ euclidean_size (m * n)"
    using \<open>n \<noteq> 0\<close> by (cases n) simp_all
  fix q r :: nat
  show "(q * n + r) div n = q" if "euclidean_size r < euclidean_size n"
  proof -
    from that have "r < n"
      by simp
    have "k \ q" if "k * n \ q * n + r" for k
    proof (rule ccontr)
      assume "\ k \ q"
      then have "q < k"
        by simp
      then obtain l where "k = Suc (q + l)"
        by (auto simp add: less_iff_Suc_add)
      with \<open>r < n\<close> that show False
        by (simp add: algebra_simps)
    qed
    with \<open>n \<noteq> 0\<close> ex fin show ?thesis
      by (auto simp add: divide_nat_def Max_eq_iff)
  qed
qed simp_all

end

lemma euclidean_relation_natI [case_names by0 divides euclidean_relation]:
  \<open>(m div n, m mod n) = (q, r)\<close>
    if by0: \<open>n = 0 \<Longrightarrow> q = 0 \<and> r = m\<close>
    and divides: \<open>n > 0 \<Longrightarrow> n dvd m \<Longrightarrow> r = 0 \<and> m = q * n\<close>
    and euclidean_relation: \<open>n > 0 \<Longrightarrow> \<not> n dvd m \<Longrightarrow> r < n \<and> m = q * n + r\<close> for m n q r :: nat
  by (rule euclidean_relationI) (use that in simp_all)

lemma div_nat_eqI:
  \<open>m div n = q\<close> if \<open>n * q \<le> m\<close> and \<open>m < n * Suc q\<close> for m n q :: nat
proof -
  have \<open>(m div n, m mod n) = (q, m - n * q)\<close>
  proof (induction rule: euclidean_relation_natI)
    case by0
    with that show ?case
      by simp
  next
    case divides
    from \<open>n dvd m\<close> obtain s where \<open>m = n * s\<close> ..
    with \<open>n > 0\<close> that have \<open>s < Suc q\<close>
      by (simp only: mult_less_cancel1)
    with \<open>m = n * s\<close> \<open>n > 0\<close> that have \<open>q = s\<close>
      by simp
    with \<open>m = n * s\<close> show ?case
      by (simp add: ac_simps)
  next
    case euclidean_relation
    with that show ?case
      by (simp add: ac_simps)
  qed
  then show ?thesis
    by simp
qed

lemma mod_nat_eqI:
  \<open>m mod n = r\<close> if \<open>r < n\<close> and \<open>r \<le> m\<close> and \<open>n dvd m - r\<close> for m n r :: nat
proof -
  have \<open>(m div n, m mod n) = ((m - r) div n, r)\<close>
  proof (induction rule: euclidean_relation_natI)
    case by0
    with that show ?case
      by simp
  next
    case divides
    from that dvd_minus_add [of r \<open>m\<close> 1 n]
    have \<open>n dvd m + (n - r)\<close>
      by simp
    with divides have \<open>n dvd n - r\<close>
      by (simp add: dvd_add_right_iff)
    then have \<open>n \<le> n - r\<close>
      by (rule dvd_imp_le) (use \<open>r < n\<close> in simp)
    with \<open>n > 0\<close> have \<open>r = 0\<close>
      by simp
    with \<open>n > 0\<close> that show ?case
      by simp
  next
    case euclidean_relation
    with that show ?case
      by (simp add: ac_simps)
  qed
  then show ?thesis
    by simp
qed

text \<open>Tool support\<close>

ML \<open>
structure Cancel_Div_Mod_Nat = Cancel_Div_Mod
(
  val div_name = \<^const_name>\<open>divide\<close>;
  val mod_name = \<^const_name>\<open>modulo\<close>;
  val mk_binop = HOLogic.mk_binop;
  val dest_plus = HOLogic.dest_bin \<^const_name>\<open>Groups.plus\<close> HOLogic.natT;
  val mk_sum = Arith_Data.mk_sum;
  fun dest_sum tm =
    if HOLogic.is_zero tm then []
    else
      (case try HOLogic.dest_Suc tm of
        SOME t => HOLogic.Suc_zero :: dest_sum t
      | NONE =>
          (case try dest_plus tm of
            SOME (t, u) => dest_sum t @ dest_sum u
          | NONE => [tm]));

  val div_mod_eqs = map mk_meta_eq @{thms cancel_div_mod_rules};

  val prove_eq_sums = Arith_Data.prove_conv2 all_tac
    (Arith_Data.simp_all_tac @{thms add_0_left add_0_right ac_simps})
)
\<close>

simproc_setup cancel_div_mod_nat ("(m::nat) + n") =
  \<open>K Cancel_Div_Mod_Nat.proc\<close>

lemma div_mult_self_is_m [simp]:
  "m * n div n = m" if "n > 0" for m n :: nat
  using that by simp

lemma div_mult_self1_is_m [simp]:
  "n * m div n = m" if "n > 0" for m n :: nat
  using that by simp

lemma mod_less_divisor [simp]:
  "m mod n < n" if "n > 0" for m n :: nat
  using mod_size_less [of n m] that by simp

lemma mod_le_divisor [simp]:
  "m mod n \ n" if "n > 0" for m n :: nat
  using that by (auto simp add: le_less)

lemma div_times_less_eq_dividend [simp]:
  "m div n * n \ m" for m n :: nat
  by (simp add: minus_mod_eq_div_mult [symmetric])

lemma times_div_less_eq_dividend [simp]:
  "n * (m div n) \ m" for m n :: nat
  using div_times_less_eq_dividend [of m n]
  by (simp add: ac_simps)

lemma dividend_less_div_times:
  "m < n + (m div n) * n" if "0 < n" for m n :: nat
proof -
  from that have "m mod n < n"
    by simp
  then show ?thesis
    by (simp add: minus_mod_eq_div_mult [symmetric])
qed

lemma dividend_less_times_div:
  "m < n + n * (m div n)" if "0 < n" for m n :: nat
  using dividend_less_div_times [of n m] that
  by (simp add: ac_simps)

lemma mod_Suc_le_divisor [simp]:
  "m mod Suc n \ n"
  using mod_less_divisor [of "Suc n" m] by arith

lemma mod_less_eq_dividend [simp]:
  "m mod n \ m" for m n :: nat
proof (rule add_leD2)
  from div_mult_mod_eq have "m div n * n + m mod n = m" .
  then show "m div n * n + m mod n \ m" by auto
qed

lemma
  div_less [simp]: "m div n = 0"
  and mod_less [simp]: "m mod n = m"
  if "m < n" for m n :: nat
  using that by (auto intro: div_nat_eqI mod_nat_eqI)

lemma split_div:
  \<open>P (m div n) \<longleftrightarrow>
    (n = 0 \<longrightarrow> P 0) \<and>
    (n \<noteq> 0 \<longrightarrow> (\<forall>i j. j < n \<and> m = n * i + j \<longrightarrow> P i))\<close> (is ?div)
  and split_mod:
  \<open>Q (m mod n) \<longleftrightarrow>
    (n = 0 \<longrightarrow> Q m) \<and>
    (n \<noteq> 0 \<longrightarrow> (\<forall>i j. j < n \<and> m = n * i + j \<longrightarrow> Q j))\<close> (is ?mod)
  for m n :: nat
proof -
  have *: \<open>R (m div n) (m mod n) \<longleftrightarrow>
    (n = 0 \<longrightarrow> R 0 m) \<and>
    (n \<noteq> 0 \<longrightarrow> (\<forall>i j. j < n \<and> m = n * i + j \<longrightarrow> R i j))\<close> for R
    by (cases \<open>n = 0\<close>) auto
  from * [of \<open>\<lambda>q _. P q\<close>] show ?div .
  from * [of \<open>\<lambda>_ r. Q r\<close>] show ?mod .
qed

declare split_div [of _ _ \<open>numeral n\<close>, linarith_split] for n
declare split_mod [of _ _ \<open>numeral n\<close>, linarith_split] for n

lemma split_div':
  "P (m div n) \ n = 0 \ P 0 \ (\q. (n * q \ m \ m < n * Suc q) \ P q)"
proof (cases "n = 0")
  case True
  then show ?thesis
    by simp
next
  case False
  then have "n * q \ m \ m < n * Suc q \ m div n = q" for q
    by (auto intro: div_nat_eqI dividend_less_times_div)
  then show ?thesis
    by auto
qed

lemma le_div_geq:
  "m div n = Suc ((m - n) div n)" if "0 < n" and "n \ m" for m n :: nat
proof -
  from \<open>n \<le> m\<close> obtain q where "m = n + q"
    by (auto simp add: le_iff_add)
  with \<open>0 < n\<close> show ?thesis
    by (simp add: div_add_self1)
qed

lemma le_mod_geq:
  "m mod n = (m - n) mod n" if "n \ m" for m n :: nat
proof -
  from \<open>n \<le> m\<close> obtain q where "m = n + q"
    by (auto simp add: le_iff_add)
  then show ?thesis
    by simp
qed

lemma div_if:
  "m div n = (if m < n \ n = 0 then 0 else Suc ((m - n) div n))"
  by (simp add: le_div_geq)

lemma mod_if:
  "m mod n = (if m < n then m else (m - n) mod n)" for m n :: nat
  by (simp add: le_mod_geq)

lemma div_eq_0_iff:
  "m div n = 0 \ m < n \ n = 0" for m n :: nat
  by (simp add: div_eq_0_iff)

lemma div_greater_zero_iff:
  "m div n > 0 \ n \ m \ n > 0" for m n :: nat
  using div_eq_0_iff [of m n] by auto

lemma mod_greater_zero_iff_not_dvd:
  "m mod n > 0 \ \ n dvd m" for m n :: nat
  by (simp add: dvd_eq_mod_eq_0)

lemma div_by_Suc_0 [simp]:
  "m div Suc 0 = m"
  using div_by_1 [of m] by simp

lemma mod_by_Suc_0 [simp]:
  "m mod Suc 0 = 0"
  using mod_by_1 [of m] by simp

lemma div2_Suc_Suc [simp]:
  "Suc (Suc m) div 2 = Suc (m div 2)"
  by (simp add: numeral_2_eq_2 le_div_geq)

lemma Suc_n_div_2_gt_zero [simp]:
  "0 < Suc n div 2" if "n > 0" for n :: nat
  using that by (cases n) simp_all

lemma div_2_gt_zero [simp]:
  "0 < n div 2" if "Suc 0 < n" for n :: nat
  using that Suc_n_div_2_gt_zero [of "n - 1"] by simp

lemma mod2_Suc_Suc [simp]:
  "Suc (Suc m) mod 2 = m mod 2"
  by (simp add: numeral_2_eq_2 le_mod_geq)

lemma add_self_div_2 [simp]:
  "(m + m) div 2 = m" for m :: nat
  by (simp add: mult_2 [symmetric])

lemma add_self_mod_2 [simp]:
  "(m + m) mod 2 = 0" for m :: nat
  by (simp add: mult_2 [symmetric])

lemma mod2_gr_0 [simp]:
  "0 < m mod 2 \ m mod 2 = 1" for m :: nat
proof -
  have "m mod 2 < 2"
    by (rule mod_less_divisor) simp
  then have "m mod 2 = 0 \ m mod 2 = 1"
    by arith
  then show ?thesis
    by auto
qed

lemma mod_Suc_eq [mod_simps]:
  "Suc (m mod n) mod n = Suc m mod n"
proof -
  have "(m mod n + 1) mod n = (m + 1) mod n"
    by (simp only: mod_simps)
  then show ?thesis
    by simp
qed

lemma mod_Suc_Suc_eq [mod_simps]:
  "Suc (Suc (m mod n)) mod n = Suc (Suc m) mod n"
proof -
  have "(m mod n + 2) mod n = (m + 2) mod n"
    by (simp only: mod_simps)
  then show ?thesis
    by simp
qed

lemma
  Suc_mod_mult_self1 [simp]: "Suc (m + k * n) mod n = Suc m mod n"
  and Suc_mod_mult_self2 [simp]: "Suc (m + n * k) mod n = Suc m mod n"
  and Suc_mod_mult_self3 [simp]: "Suc (k * n + m) mod n = Suc m mod n"
  and Suc_mod_mult_self4 [simp]: "Suc (n * k + m) mod n = Suc m mod n"
  by (subst mod_Suc_eq [symmetric], simp add: mod_simps)+

lemma Suc_0_mod_eq [simp]:
  "Suc 0 mod n = of_bool (n \ Suc 0)"
  by (cases n) simp_all

lemma div_mult2_eq:
    \<open>m div (n * q) = (m div n) div q\<close> (is ?Q)
  and mod_mult2_eq:
    \<open>m mod (n * q) = n * (m div n mod q) + m mod n\<close> (is ?R)
  for m n q :: nat
proof -
  have \<open>(m div (n * q), m mod (n * q)) = ((m div n) div q, n * (m div n mod q) + m mod n)\<close>
  proof (induction rule: euclidean_relation_natI)
    case by0
    then show ?case
      by auto
  next
    case divides
    from \<open>n * q dvd m\<close> obtain t where \<open>m = n * q * t\<close> ..
    with \<open>n * q > 0\<close> show ?case
      by (simp add: algebra_simps)
  next
    case euclidean_relation
    then have \<open>n > 0\<close> \<open>q > 0\<close>
      by simp_all
    from \<open>n > 0\<close> have \<open>m mod n < n\<close>
      by (rule mod_less_divisor)
    from \<open>q > 0\<close> have \<open>m div n mod q < q\<close>
      by (rule mod_less_divisor)
    then obtain s where \<open>q = Suc (m div n mod q + s)\<close>
      by (blast dest: less_imp_Suc_add)
    moreover have \<open>m mod n + n * (m div n mod q) < n * Suc (m div n mod q + s)\<close>
      using \<open>m mod n < n\<close> by (simp add: add_mult_distrib2)
    ultimately have \<open>m mod n + n * (m div n mod q) < n * q\<close>
      by simp
    then show ?case
      by (simp add: algebra_simps flip: add_mult_distrib2)
  qed
  then show ?Q and ?R
    by simp_all
qed

lemma div_le_mono:
  "m div k \ n div k" if "m \ n" for m n k :: nat
proof -
  from that obtain q where "n = m + q"
    by (auto simp add: le_iff_add)
  then show ?thesis
    by (simp add: div_add1_eq [of m q k])
qed

text \<open>Antimonotonicity of \<^const>\<open>divide\<close> in second argument\<close>

lemma div_le_mono2:
  "k div n \ k div m" if "0 < m" and "m \ n" for m n k :: nat
using that proof (induct k arbitrary: m rule: less_induct)
  case (less k)
  show ?case
  proof (cases "n \ k")
    case False
    then show ?thesis
      by simp
  next
    case True
    have "(k - n) div n \ (k - m) div n"
      using less.prems
      by (blast intro: div_le_mono diff_le_mono2)
    also have "\ \ (k - m) div m"
      using \<open>n \<le> k\<close> less.prems less.hyps [of "k - m" m]
      by simp
    finally show ?thesis
      using \<open>n \<le> k\<close> less.prems
      by (simp add: le_div_geq)
  qed
qed

lemma div_le_dividend [simp]:
  "m div n \ m" for m n :: nat
  using div_le_mono2 [of 1 n m] by (cases "n = 0") simp_all

lemma div_less_dividend [simp]:
  "m div n < m" if "1 < n" and "0 < m" for m n :: nat
using that proof (induct m rule: less_induct)
  case (less m)
  show ?case
  proof (cases "n < m")
    case False
    with less show ?thesis
      by (cases "n = m") simp_all
  next
    case True
    then show ?thesis
      using less.hyps [of "m - n"] less.prems
      by (simp add: le_div_geq)
  qed
qed

lemma div_eq_dividend_iff:
  "m div n = m \ n = 1" if "m > 0" for m n :: nat
proof
  assume "n = 1"
  then show "m div n = m"
    by simp
next
  assume P: "m div n = m"
  show "n = 1"
  proof (rule ccontr)
    have "n \ 0"
      by (rule ccontr) (use that P in auto)
    moreover assume "n \ 1"
    ultimately have "n > 1"
      by simp
    with that have "m div n < m"
      by simp
    with P show False
      by simp
  qed
qed

lemma less_mult_imp_div_less:
  "m div n < i" if "m < i * n" for m n i :: nat
proof -
  from that have "i * n > 0"
    by (cases "i * n = 0") simp_all
  then have "i > 0" and "n > 0"
    by simp_all
  have "m div n * n \ m"
    by simp
  then have "m div n * n < i * n"
    using that by (rule le_less_trans)
  with \<open>n > 0\<close> show ?thesis
    by simp
qed

lemma div_less_iff_less_mult:
  \<open>m div q < n \<longleftrightarrow> m < n * q\<close> (is \<open>?P \<longleftrightarrow> ?Q\<close>)
  if \<open>q > 0\<close> for m n q :: nat
proof
  assume ?Q then show ?P
    by (rule less_mult_imp_div_less)
next
  assume ?P
  then obtain h where \<open>n = Suc (m div q + h)\<close>
    using less_natE by blast
  moreover have \<open>m < m + (Suc h * q - m mod q)\<close>
    using that by (simp add: trans_less_add1)
  ultimately show ?Q
    by (simp add: algebra_simps flip: minus_mod_eq_mult_div)
qed

lemma less_eq_div_iff_mult_less_eq:
  \<open>m \<le> n div q \<longleftrightarrow> m * q \<le> n\<close> if \<open>q > 0\<close> for m n q :: nat
  using div_less_iff_less_mult [of q n m] that by auto

lemma div_Suc:
  \<open>Suc m div n = (if Suc m mod n = 0 then Suc (m div n) else m div n)\<close>
proof (cases \<open>n = 0 \<or> n = 1\<close>)
  case True
  then show ?thesis by auto
next
  case False
  then have \<open>n > 1\<close>
    by simp
  then have \<open>Suc m div n = m div n + Suc (m mod n) div n\<close>
    using div_add1_eq [of m 1 n] by simp
  also have \<open>Suc (m mod n) div n = of_bool (n dvd Suc m)\<close>
  proof (cases \<open>n dvd Suc m\<close>)
    case False
    moreover have \<open>Suc (m mod n) \<noteq> n\<close>
    proof (rule ccontr)
      assume \<open>\<not> Suc (m mod n) \<noteq> n\<close>
      then have \<open>m mod n = n - Suc 0\<close>
        by simp
      with \<open>n > 1\<close> have \<open>(m + 1) mod n = 0\<close>
        by (subst mod_add_left_eq [symmetric]) simp
      then have \<open>n dvd Suc m\<close>
        by auto
      with False show False ..
    qed
    moreover have \<open>Suc (m mod n) \<le> n\<close>
      using \<open>n > 1\<close> by (simp add: Suc_le_eq)
    ultimately show ?thesis
      by (simp add: div_eq_0_iff)
  next
    case True
    then obtain q where q: \<open>Suc m = n * q\<close> ..
    moreover have \<open>q > 0\<close> by (rule ccontr)
      (use q in simp)
    ultimately have \<open>m mod n = n - Suc 0\<close>
      using \<open>n > 1\<close> mult_le_cancel1 [of n \<open>Suc 0\<close> q]
      by (auto intro: mod_nat_eqI)
    with True \<open>n > 1\<close> show ?thesis
      by simp
  qed
  finally show ?thesis
    by (simp add: mod_greater_zero_iff_not_dvd)
qed

lemma mod_Suc:
  \<open>Suc m mod n = (if Suc (m mod n) = n then 0 else Suc (m mod n))\<close>
proof (cases \<open>n = 0\<close>)
  case True
  then show ?thesis
    by simp
next
  case False
  moreover have \<open>Suc m mod n = Suc (m mod n) mod n\<close>
    by (simp add: mod_simps)
  ultimately show ?thesis
    by (auto intro!: mod_nat_eqI intro: neq_le_trans simp add: Suc_le_eq)
qed

lemma Suc_times_mod_eq:
  "Suc (m * n) mod m = 1" if "Suc 0 < m"
  using that by (simp add: mod_Suc)

lemma Suc_times_numeral_mod_eq [simp]:
  "Suc (numeral k * n) mod numeral k = 1" if "numeral k \ (1::nat)"
  by (rule Suc_times_mod_eq) (use that in simp)

lemma Suc_div_le_mono [simp]:
  "m div n \ Suc m div n"
  by (simp add: div_le_mono)

text \<open>These lemmas collapse some needless occurrences of Suc:
  at least three Sucs, since two and fewer are rewritten back to Suc again!
  We already have some rules to simplify operands smaller than 3.\<close>

lemma div_Suc_eq_div_add3 [simp]:
  "m div Suc (Suc (Suc n)) = m div (3 + n)"
  by (simp add: Suc3_eq_add_3)

lemma mod_Suc_eq_mod_add3 [simp]:
  "m mod Suc (Suc (Suc n)) = m mod (3 + n)"
  by (simp add: Suc3_eq_add_3)

lemma Suc_div_eq_add3_div:
  "Suc (Suc (Suc m)) div n = (3 + m) div n"
  by (simp add: Suc3_eq_add_3)

lemma Suc_mod_eq_add3_mod:
  "Suc (Suc (Suc m)) mod n = (3 + m) mod n"
  by (simp add: Suc3_eq_add_3)

lemmas Suc_div_eq_add3_div_numeral [simp] =
  Suc_div_eq_add3_div [of _ "numeral v"] for v

lemmas Suc_mod_eq_add3_mod_numeral [simp] =
  Suc_mod_eq_add3_mod [of _ "numeral v"] for v

lemma (in field_char_0) of_nat_div:
  "of_nat (m div n) = ((of_nat m - of_nat (m mod n)) / of_nat n)"
proof -
  have "of_nat (m div n) = ((of_nat (m div n * n + m mod n) - of_nat (m mod n)) / of_nat n :: 'a)"
    unfolding of_nat_add by (cases "n = 0") simp_all
  then show ?thesis
    by simp
qed

text \<open>An ``induction'' law for modulus arithmetic.\<close>

lemma mod_induct [consumes 3, case_names step]:
  "P m" if "P n" and "n < p" and "m < p"
    and step: "\n. n < p \ P n \ P (Suc n mod p)"
using \<open>m < p\<close> proof (induct m)
  case 0
  show ?case
  proof (rule ccontr)
    assume "\ P 0"
    from \<open>n < p\<close> have "0 < p"
      by simp
    from \<open>n < p\<close> obtain m where "0 < m" and "p = n + m"
      by (blast dest: less_imp_add_positive)
    with \<open>P n\<close> have "P (p - m)"
      by simp
    moreover have "\ P (p - m)"
    using \<open>0 < m\<close> proof (induct m)
      case 0
      then show ?case
        by simp
    next
      case (Suc m)
      show ?case
      proof
        assume P: "P (p - Suc m)"
        with \<open>\<not> P 0\<close> have "Suc m < p"
          by (auto intro: ccontr)
        then have "Suc (p - Suc m) = p - m"
          by arith
        moreover from \<open>0 < p\<close> have "p - Suc m < p"
          by arith
        with P step have "P ((Suc (p - Suc m)) mod p)"
          by blast
        ultimately show False
          using \<open>\<not> P 0\<close> Suc.hyps by (cases "m = 0") simp_all
      qed
    qed
    ultimately show False
      by blast
  qed
next
  case (Suc m)
  then have "m < p" and mod: "Suc m mod p = Suc m"
    by simp_all
  from \<open>m < p\<close> have "P m"
    by (rule Suc.hyps)
  with \<open>m < p\<close> have "P (Suc m mod p)"
    by (rule step)
  with mod show ?case
    by simp
qed

lemma funpow_mod_eq: \<^marker>\<open>contributor \<open>Lars Noschinski\<close>\<close>
  \<open>(f ^^ (m mod n)) x = (f ^^ m) x\<close> if \<open>(f ^^ n) x = x\<close>
proof -
  have \<open>(f ^^ m) x = (f ^^ (m mod n + m div n * n)) x\<close>
    by simp
  also have \<open>\<dots> = (f ^^ (m mod n)) (((f ^^ n) ^^ (m div n)) x)\<close>
    by (simp only: funpow_add funpow_mult ac_simps) simp
  also have \<open>((f ^^ n) ^^ q) x = x\<close> for q
    by (induction q) (use \<open>(f ^^ n) x = x\<close> in simp_all)
  finally show ?thesis
    by simp
qed

lemma mod_eq_dvd_iff_nat:
  \<open>m mod q = n mod q \<longleftrightarrow> q dvd m - n\<close> (is \<open>?P \<longleftrightarrow> ?Q\<close>)
    if \<open>m \<ge> n\<close> for m n q :: nat
proof
  assume ?Q
  then obtain s where \<open>m - n = q * s\<close> ..
  with that have \<open>m = q * s + n\<close>
    by simp
  then show ?P
    by simp
next
  assume ?P
  have \<open>m - n = m div q * q + m mod q - (n div q * q + n mod q)\<close>
    by simp
  also have \<open>\<dots> = q * (m div q - n div q)\<close>
    by (simp only: algebra_simps \<open>?P\<close>)
  finally show ?Q ..
qed

lemma mod_eq_iff_dvd_symdiff_nat:
  \<open>m mod q = n mod q \<longleftrightarrow> q dvd nat \<bar>int m - int n\<bar>\<close>
  by (auto simp add: abs_if mod_eq_dvd_iff_nat nat_diff_distrib dest: sym intro: sym)

lemma mod_eq_nat1E:
  fixes m n q :: nat
  assumes "m mod q = n mod q" and "m \ n"
  obtains s where "m = n + q * s"
proof -
  from assms have "q dvd m - n"
    by (simp add: mod_eq_dvd_iff_nat)
  then obtain s where "m - n = q * s" ..
  with \<open>m \<ge> n\<close> have "m = n + q * s"
    by simp
  with that show thesis .
qed

lemma mod_eq_nat2E:
  fixes m n q :: nat
  assumes "m mod q = n mod q" and "n \ m"
  obtains s where "n = m + q * s"
  using assms mod_eq_nat1E [of n q m] by (auto simp add: ac_simps)

lemma nat_mod_eq_iff:
  "(x::nat) mod n = y mod n \ (\q1 q2. x + n * q1 = y + n * q2)" (is "?lhs = ?rhs")
proof
  assume H: "x mod n = y mod n"
  { assume xy: "x \ y"
    from H have th: "y mod n = x mod n" by simp
    from mod_eq_nat1E [OF th xy] obtain q where "y = x + n * q" .
    then have "x + n * q = y + n * 0"
      by simp
    then have "\q1 q2. x + n * q1 = y + n * q2"
      by blast
  }
  moreover
  { assume xy: "y \ x"
    from mod_eq_nat1E [OF H xy] obtain q where "x = y + n * q" .
    then have "x + n * 0 = y + n * q"
      by simp
    then have "\q1 q2. x + n * q1 = y + n * q2"
      by blast
  }
  ultimately show ?rhs using linear[of x y] by blast
next
  assume ?rhs then obtain q1 q2 where q12: "x + n * q1 = y + n * q2" by blast
  hence "(x + n * q1) mod n = (y + n * q2) mod n" by simp
  thus  ?lhs by simp
qed

subsection \<open>Division on \<^typ>\<open>int\<close>\<close>

text \<open>
  The following specification of integer division rounds towards minus infinity
  and is advocated by Donald Knuth. See \cite{leijen01} for an overview and
  terminology of different possibilities to specify integer division;
  there division rounding towards minus infinitiy is named ``F-division''.
\<close>

subsubsection \<open>Basic instantiation\<close>

instantiation int :: "{normalization_semidom, idom_modulo}"
begin

definition normalize_int :: \<open>int \<Rightarrow> int\<close>
  where [simp]: \<open>normalize = (abs :: int \<Rightarrow> int)\<close>

definition unit_factor_int :: \<open>int \<Rightarrow> int\<close>
  where [simp]: \<open>unit_factor = (sgn :: int \<Rightarrow> int)\<close>

definition divide_int :: \<open>int \<Rightarrow> int \<Rightarrow> int\<close>
  where \<open>k div l = (sgn k * sgn l * int (nat \<bar>k\<bar> div nat \<bar>l\<bar>)
    - of_bool (l \<noteq> 0 \<and> sgn k \<noteq> sgn l \<and> \<not> l dvd k))\<close>

lemma divide_int_unfold:
  \<open>(sgn k * int m) div (sgn l * int n) = (sgn k * sgn l * int (m div n)
    - of_bool ((k = 0 \<longleftrightarrow> m = 0) \<and> l \<noteq> 0 \<and> n \<noteq> 0 \<and> sgn k \<noteq> sgn l \<and> \<not> n dvd m))\<close>
  by (simp add: divide_int_def sgn_mult nat_mult_distrib abs_mult sgn_eq_0_iff ac_simps)

definition modulo_int :: \<open>int \<Rightarrow> int \<Rightarrow> int\<close>
  where \<open>k mod l = sgn k * int (nat \<bar>k\<bar> mod nat \<bar>l\<bar>) + l * of_bool (sgn k \<noteq> sgn l \<and> \<not> l dvd k)\<close>

lemma modulo_int_unfold:
  \<open>(sgn k * int m) mod (sgn l * int n) =
    sgn k * int (m mod (of_bool (l \<noteq> 0) * n)) + (sgn l * int n) * of_bool ((k = 0 \<longleftrightarrow> m = 0) \<and> sgn k \<noteq> sgn l \<and> \<not> n dvd m)\<close>
  by (auto simp add: modulo_int_def sgn_mult abs_mult)

instance proof
  fix k :: int show "k div 0 = 0"
  by (simp add: divide_int_def)
next
  fix k l :: int
  assume "l \ 0"
  obtain n m and s t where k: "k = sgn s * int n" and l: "l = sgn t * int m"
    by (blast intro: int_sgnE elim: that)
  then have "k * l = sgn (s * t) * int (n * m)"
    by (simp add: ac_simps sgn_mult)
  with k l \<open>l \<noteq> 0\<close> show "k * l div l = k"
    by (simp only: divide_int_unfold)
      (auto simp add: algebra_simps sgn_mult sgn_1_pos sgn_0_0)
next
  fix k l :: int
  obtain n m and s t where "k = sgn s * int n" and "l = sgn t * int m"
    by (blast intro: int_sgnE elim: that)
  then show "k div l * l + k mod l = k"
    by (simp add: divide_int_unfold modulo_int_unfold algebra_simps modulo_nat_def of_nat_diff)
qed (auto simp add: sgn_mult mult_sgn_abs abs_eq_iff')

end

lemma of_int_div: "b dvd a \ of_int (a div b) = (of_int a / of_int b :: 'a :: field_char_0)"
  by (elim dvdE) (auto simp: divide_simps mult_ac)

subsubsection \<open>Algebraic foundations\<close>

lemma coprime_int_iff [simp]:
  "coprime (int m) (int n) \ coprime m n" (is "?P \ ?Q")
proof
  assume ?P
  show ?Q
  proof (rule coprimeI)
    fix q
    assume "q dvd m" "q dvd n"
    then have "int q dvd int m" "int q dvd int n"
      by simp_all
    with \<open>?P\<close> have "is_unit (int q)"
      by (rule coprime_common_divisor)
    then show "is_unit q"
      by simp
  qed
next
  assume ?Q
  show ?P
  proof (rule coprimeI)
    fix k
    assume "k dvd int m" "k dvd int n"
    then have "nat \k\ dvd m" "nat \k\ dvd n"
      by simp_all
    with \<open>?Q\<close> have "is_unit (nat \<bar>k\<bar>)"
      by (rule coprime_common_divisor)
    then show "is_unit k"
      by simp
  qed
qed

lemma coprime_abs_left_iff [simp]:
  "coprime \k\ l \ coprime k l" for k l :: int
  using coprime_normalize_left_iff [of k l] by simp

lemma coprime_abs_right_iff [simp]:
  "coprime k \l\ \ coprime k l" for k l :: int
  using coprime_abs_left_iff [of l k] by (simp add: ac_simps)

lemma coprime_nat_abs_left_iff [simp]:
  "coprime (nat \k\) n \ coprime k (int n)"
proof -
  define m where "m = nat \k\"
  then have "\k\ = int m"
    by simp
  moreover have "coprime k (int n) \ coprime \k\ (int n)"
    by simp
  ultimately show ?thesis
    by simp
qed

lemma coprime_nat_abs_right_iff [simp]:
  "coprime n (nat \k\) \ coprime (int n) k"
  using coprime_nat_abs_left_iff [of k n] by (simp add: ac_simps)

lemma coprime_common_divisor_int: "coprime a b \ x dvd a \ x dvd b \ \x\ = 1"
  for a b :: int
  by (drule coprime_common_divisor [of _ _ x]) simp_all

subsubsection \<open>Basic conversions\<close>

lemma div_abs_eq_div_nat:
  "\k\ div \l\ = int (nat \k\ div nat \l\)"
  by (auto simp add: divide_int_def)

lemma div_eq_div_abs:
  \<open>k div l = sgn k * sgn l * (\<bar>k\<bar> div \<bar>l\<bar>)
    - of_bool (l \<noteq> 0 \<and> sgn k \<noteq> sgn l \<and> \<not> l dvd k)\<close>
  for k l :: int
  by (simp add: divide_int_def [of k l] div_abs_eq_div_nat)

lemma div_abs_eq:
  \<open>\<bar>k\<bar> div \<bar>l\<bar> = sgn k * sgn l * (k div l + of_bool (sgn k \<noteq> sgn l \<and> \<not> l dvd k))\<close>
  for k l :: int
  by (simp add: div_eq_div_abs [of k l] ac_simps)

lemma mod_abs_eq_div_nat:
  "\k\ mod \l\ = int (nat \k\ mod nat \l\)"
  by (simp add: modulo_int_def)

lemma mod_eq_mod_abs:
  \<open>k mod l = sgn k * (\<bar>k\<bar> mod \<bar>l\<bar>) + l * of_bool (sgn k \<noteq> sgn l \<and> \<not> l dvd k)\<close>
  for k l :: int
  by (simp add: modulo_int_def [of k l] mod_abs_eq_div_nat)

lemma mod_abs_eq:
  \<open>\<bar>k\<bar> mod \<bar>l\<bar> = sgn k * (k mod l - l * of_bool (sgn k \<noteq> sgn l \<and> \<not> l dvd k))\<close>
  for k l :: int
  by (auto simp: mod_eq_mod_abs [of k l])

lemma div_sgn_abs_cancel:
  fixes k l v :: int
  assumes "v \ 0"
  shows "(sgn v * \k\) div (sgn v * \l\) = \k\ div \l\"
  using assms by (simp add: sgn_mult abs_mult sgn_0_0
    divide_int_def [of "sgn v * \k\" "sgn v * \l\"] flip: div_abs_eq_div_nat)

lemma div_eq_sgn_abs:
  fixes k l v :: int
  assumes "sgn k = sgn l"
  shows "k div l = \k\ div \l\"
  using assms by (auto simp add: div_abs_eq)

lemma div_dvd_sgn_abs:
  fixes k l :: int
  assumes "l dvd k"
  shows "k div l = (sgn k * sgn l) * (\k\ div \l\)"
  using assms by (auto simp add: div_abs_eq ac_simps)

lemma div_noneq_sgn_abs:
  fixes k l :: int
  assumes "l \ 0"
  assumes "sgn k \ sgn l"
  shows "k div l = - (\k\ div \l\) - of_bool (\ l dvd k)"
  using assms by (auto simp add: div_abs_eq ac_simps sgn_0_0 dest!: sgn_not_eq_imp)

subsubsection \<open>Euclidean division\<close>

instantiation int :: unique_euclidean_ring
begin

definition euclidean_size_int :: "int \ nat"
  where [simp]: "euclidean_size_int = (nat \ abs :: int \ nat)"

definition division_segment_int :: "int \ int"
  where "division_segment_int k = (if k \ 0 then 1 else - 1)"

lemma division_segment_eq_sgn:
  "division_segment k = sgn k" if "k \ 0" for k :: int
  using that by (simp add: division_segment_int_def)

lemma abs_division_segment [simp]:
  "\division_segment k\ = 1" for k :: int
  by (simp add: division_segment_int_def)

lemma abs_mod_less:
  "\k mod l\ < \l\" if "l \ 0" for k l :: int
proof -
  obtain n m and s t where "k = sgn s * int n" and "l = sgn t * int m"
    by (blast intro: int_sgnE elim: that)
  with that show ?thesis
    by (auto simp add: modulo_int_unfold abs_mult mod_greater_zero_iff_not_dvd
        simp flip: right_diff_distrib dest!: sgn_not_eq_imp)
      (simp add: sgn_0_0)
qed

lemma sgn_mod:
  "sgn (k mod l) = sgn l" if "l \ 0" "\ l dvd k" for k l :: int
proof -
  obtain n m and s t where "k = sgn s * int n" and "l = sgn t * int m"
    by (blast intro: int_sgnE elim: that)
  with that show ?thesis
    by (auto simp add: modulo_int_unfold sgn_mult mod_greater_zero_iff_not_dvd
      simp flip: right_diff_distrib dest!: sgn_not_eq_imp)
qed

instance proof
  fix k l :: int
  show "division_segment (k mod l) = division_segment l" if
    "l \ 0" and "\ l dvd k"
    using that by (simp add: division_segment_eq_sgn dvd_eq_mod_eq_0 sgn_mod)
next
  fix l q r :: int
  obtain n m and s t
     where l: "l = sgn s * int n" and q: "q = sgn t * int m"
    by (blast intro: int_sgnE elim: that)
  assume \<open>l \<noteq> 0\<close>
  with l have "s \ 0" and "n > 0"
    by (simp_all add: sgn_0_0)
  assume "division_segment r = division_segment l"
  moreover have "r = sgn r * \r\"
    by (simp add: sgn_mult_abs)
  moreover define u where "u = nat \r\"
  ultimately have "r = sgn l * int u"
    using division_segment_eq_sgn \<open>l \<noteq> 0\<close> by (cases "r = 0") simp_all
  with l \<open>n > 0\<close> have r: "r = sgn s * int u"
    by (simp add: sgn_mult)
  assume "euclidean_size r < euclidean_size l"
  with l r \<open>s \<noteq> 0\<close> have "u < n"
    by (simp add: abs_mult)
  show "(q * l + r) div l = q"
  proof (cases "q = 0 \ r = 0")
    case True
    then show ?thesis
    proof
      assume "q = 0"
      then show ?thesis
        using l r \<open>u < n\<close> by (simp add: divide_int_unfold)
    next
      assume "r = 0"
      from \<open>r = 0\<close> have *: "q * l + r = sgn (t * s) * int (n * m)"
        using q l by (simp add: ac_simps sgn_mult)
      from \<open>s \<noteq> 0\<close> \<open>n > 0\<close> show ?thesis
        by (simp only: *, simp only: * q l divide_int_unfold)
          (auto simp add: sgn_mult ac_simps)
    qed
  next
    case False
    with q r have "t \ 0" and "m > 0" and "s \ 0" and "u > 0"
      by (simp_all add: sgn_0_0)
    moreover from \<open>0 < m\<close> \<open>u < n\<close> have "u \<le> m * n"
      using mult_le_less_imp_less [of 1 m u n] by simp
    ultimately have *: "q * l + r = sgn (s * t)
      * int (if t < 0 then m * n - u else m * n + u)"
      using l q r
      by (simp add: sgn_mult algebra_simps of_nat_diff)
    have "(m * n - u) div n = m - 1" if "u > 0"
      using \<open>0 < m\<close> \<open>u < n\<close> that
      by (auto intro: div_nat_eqI simp add: algebra_simps)
    moreover have "n dvd m * n - u \ n dvd u"
      using \<open>u \<le> m * n\<close> dvd_diffD1 [of n "m * n" u]
      by auto
    ultimately show ?thesis
      using \<open>s \<noteq> 0\<close> \<open>m > 0\<close> \<open>u > 0\<close> \<open>u < n\<close> \<open>u \<le> m * n\<close>
      by (simp only: *, simp only: l q divide_int_unfold)
        (auto simp add: sgn_mult sgn_0_0 sgn_1_pos algebra_simps dest: dvd_imp_le)
  qed
qed (use mult_le_mono2 [of 1] in \<open>auto simp add: division_segment_int_def not_le zero_less_mult_iff mult_less_0_iff abs_mult sgn_mult abs_mod_less sgn_mod nat_mult_distrib\<close>)

end

lemma euclidean_relation_intI [case_names by0 divides euclidean_relation]:
--> --------------------

--> maximum size reached

--> --------------------

quality100%

¤ Die Informationen auf dieser Webseite wurden nach bestem Wissen sorgfältig zusammengestellt. Es wird jedoch weder Vollständigkeit, noch Richtigkeit, noch Qualität der bereit gestellten Informationen zugesichert.0.46Bemerkung: (vorverarbeitet) ¤

Wurzel

Suchen

Beweissystem der NASA

Beweissystem Isabelle

NIST Cobol Testsuite

Cephes Mathematical Library

Wiener Entwicklungsmethode

Haftungshinweis

Die Informationen auf dieser Webseite wurden nach bestem Wissen sorgfältig zusammengestellt. Es wird jedoch weder Vollständigkeit, noch Richtigkeit, noch Qualität der bereit gestellten Informationen zugesichert.

Bemerkung:

Die farbliche Syntaxdarstellung ist noch experimentell.