diff --git a/src/HOL/Euclidean_Division.thy b/src/HOL/Euclidean_Division.thy
--- a/src/HOL/Euclidean_Division.thy
+++ b/src/HOL/Euclidean_Division.thy
@@ -1,3376 +1,3364 @@
 (*  Title:      HOL/Euclidean_Division.thy
     Author:     Manuel Eberl, TU Muenchen
     Author:     Florian Haftmann, TU Muenchen
 *)
 
 section \<open>Division in euclidean (semi)rings\<close>
 
 theory Euclidean_Division
   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 \<Rightarrow> nat"
   assumes size_0 [simp]: "euclidean_size 0 = 0"
   assumes mod_size_less:
     "b \<noteq> 0 \<Longrightarrow> euclidean_size (a mod b) < euclidean_size b"
   assumes size_mult_mono:
     "b \<noteq> 0 \<Longrightarrow> euclidean_size a \<le> euclidean_size (a * b)"
 begin
 
 lemma euclidean_size_eq_0_iff [simp]:
   "euclidean_size b = 0 \<longleftrightarrow> 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 \<noteq> 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 \<longleftrightarrow> b \<noteq> 0"
   using euclidean_size_eq_0_iff [symmetric, of b] by safe simp
 
 lemma size_mult_mono': "b \<noteq> 0 \<Longrightarrow> euclidean_size a \<le> euclidean_size (b * a)"
   by (subst mult.commute) (rule size_mult_mono)
 
 lemma dvd_euclidean_size_eq_imp_dvd:
   assumes "a \<noteq> 0" and "euclidean_size a = euclidean_size b"
     and "b dvd a"
   shows "a dvd b"
 proof (rule ccontr)
   assume "\<not> a dvd b"
   hence "b mod a \<noteq> 0" using mod_0_imp_dvd [of b a] by blast
   then have "b mod a \<noteq> 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 \<noteq> 0" by auto
   thus "euclidean_size (a * b) \<ge> euclidean_size b" by (rule size_mult_mono')
   from assms have "is_unit (1 div a)" by simp
   hence "1 div a \<noteq> 0" by (intro notI) simp_all
   hence "euclidean_size (a * b) \<le> 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) \<le> euclidean_size b" .
 qed
 
 lemma euclidean_size_unit:
   "is_unit a \<Longrightarrow> euclidean_size a = euclidean_size 1"
   using euclidean_size_times_unit [of a 1] by simp
 
 lemma unit_iff_euclidean_size:
   "is_unit a \<longleftrightarrow> euclidean_size a = euclidean_size 1 \<and> a \<noteq> 0"
 proof safe
   assume A: "a \<noteq> 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 \<noteq> 0" "b \<noteq> 0" "\<not> is_unit a"
   shows   "euclidean_size b < euclidean_size (a * b)"
 proof (rule ccontr)
   assume "\<not>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 \<noteq> 0"
   shows   "euclidean_size a \<le> euclidean_size b"
   using assms by (auto simp: size_mult_mono)
 
 lemma dvd_proper_imp_size_less:
   assumes "a dvd b" "\<not> b dvd a" "b \<noteq> 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 "\<not>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 \<longleftrightarrow> a div b = 0" (is "?P \<longleftrightarrow> ?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 \<longleftrightarrow> coprime a b" if "b \<noteq> 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) \<longleftrightarrow> coprime a b" if "a \<noteq> 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 \<longleftrightarrow> a dvd b - c"
 proof -
   have "a dvd c - b \<longleftrightarrow> 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 \<noteq> 0 \<Longrightarrow> (a + c * b) div b = c + a div b"
   and div_mult_mult1 [simp]: "c \<noteq> 0 \<Longrightarrow> (c * a) div (c * b) = a div b"
 begin
 
 lemma div_mult_self2 [simp]:
   assumes "b \<noteq> 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 \<noteq> 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 \<noteq> 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
     "\<dots> = (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 \<noteq> 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 \<noteq> 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 \<noteq> 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 "\<dots> = a div b"
     by (simp only: mod_div_mult_eq)
   also have "\<dots> = 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 "\<dots> = 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 "\<dots> = (a mod (c * k) + a div (c * k) * k * c) mod c"
     by (simp only: mod_mult_self1)
   also have "\<dots> = (a div (c * k) * (c * k) + a mod (c * k)) mod c"
     by (simp only: ac_simps)
   also have "\<dots> = a mod c"
     by (simp only: div_mult_mod_eq)
   finally show ?thesis .
 qed
 
 lemma div_mult_mult2 [simp]:
   "c \<noteq> 0 \<Longrightarrow> (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 \<Longrightarrow> k dvd n \<Longrightarrow> 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 \<Longrightarrow> (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 \<Longrightarrow> (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 "\<dots> = (a mod c + b + a div c * c) mod c"
     by (simp only: ac_simps)
   also have "\<dots> = (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]:
   "(\<Sum>i\<in>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 "(\<Sum>i\<in>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 "\<dots> = (f i + (\<Sum>i\<in>A. f i mod a) mod a) mod a"
     by (simp add: mod_simps)
   also have "\<dots> = (f i + (\<Sum>i\<in>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 "\<dots> = (a mod c * b + a div c * b * c) mod c"
     by (simp only: algebra_simps)
   also have "\<dots> = (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]:
   "(\<Prod>i\<in>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 "(\<Prod>i\<in>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 "\<dots> = (f i * ((\<Prod>i\<in>A. f i mod a) mod a)) mod a"
     by (simp add: mod_simps)
   also have "\<dots> = (f i * ((\<Prod>i\<in>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 "\<dots> = (- (a mod b) + - (a div b) * b) mod b"
     by (simp add: ac_simps)
   also have "\<dots> = (- (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 \<longleftrightarrow> c dvd a - b" (is "?P \<longleftrightarrow> ?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>Euclidean 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 \<Longrightarrow> x dvd a \<Longrightarrow> x dvd b \<Longrightarrow> 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: "\<exists>k. k * n \<le> l" for l :: nat
     by (rule exI [of _ 0]) simp
   have fin: "finite {k. k * n \<le> l}" if "n > 0" for l
   proof -
     from that have "{k. k * n \<le> l} \<subseteq> {k. k \<le> 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 \<le> m \<and> n dvd l}"
   proof (cases "n = 0")
     case True
     moreover have "{l. l = 0 \<and> l \<le> m} = {0::nat}"
       by auto
     ultimately show ?thesis
       by simp
   next
     case False
     with ex [of m] fin have "n * Max {k. k * n \<le> m} = Max (times n ` {k. k * n \<le> m})"
       by (auto simp add: nat_mult_max_right intro: hom_Max_commute)
     also have "times n ` {k. k * n \<le> m} = {l. l \<le> m \<and> 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 \<le> 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 \<noteq> 0"
   show "euclidean_size (m mod n) < euclidean_size n"
   proof -
     have "m < Suc (m div n) * n"
     proof (rule ccontr)
       assume "\<not> m < Suc (m div n) * n"
       then have "Suc (m div n) * n \<le> 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 \<le> 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 \<le> q" if "k * n \<le> q * n + r" for k
     proof (rule ccontr)
       assume "\<not> k \<le> 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 \<le> 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 \<le> 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) \<le> 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 \<le> n"
   using mod_less_divisor [of "Suc n" m] by arith
 
 lemma mod_less_eq_dividend [simp]:
   "m mod n \<le> 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 \<le> 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) \<longleftrightarrow> n = 0 \<and> P 0 \<or> (\<exists>q. (n * q \<le> m \<and> m < n * Suc q) \<and> P q)"
 proof (cases "n = 0")
   case True
   then show ?thesis
     by simp
 next
   case False
   then have "n * q \<le> m \<and> m < n * Suc q \<longleftrightarrow> 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 \<le> 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 \<le> 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 \<or> 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 \<longleftrightarrow> m < n \<or> n = 0" for m n :: nat
   by (simp add: div_eq_0_iff)
 
 lemma div_greater_zero_iff:
   "m div n > 0 \<longleftrightarrow> n \<le> m \<and> 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 \<longleftrightarrow> \<not> 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 \<longleftrightarrow> 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 \<or> 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 \<noteq> 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 \<le> n div k" if "m \<le> 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 \<le> k div m" if "0 < m" and "m \<le> n" for m n k :: nat
 using that proof (induct k arbitrary: m rule: less_induct)
   case (less k)
   show ?case
   proof (cases "n \<le> k")
     case False
     then show ?thesis
       by simp
   next
     case True
     have "(k - n) div n \<le> (k - m) div n"
       using less.prems
       by (blast intro: div_le_mono diff_le_mono2)
     also have "\<dots> \<le> (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 \<le> 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 \<longleftrightarrow> 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 \<noteq> 0"
       by (rule ccontr) (use that P in auto)
     moreover assume "n \<noteq> 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 \<le> 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>  (is "_ = ?rhs")
+  \<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 0 div n = 0\<close>
-    by (simp add: div_eq_0_iff)
-  have \<open>(m + 1) div n = ?rhs\<close>
+  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 True
-    then obtain q where \<open>Suc m = n * q\<close> ..
-    then have m: \<open>m = n * q - 1\<close>
-      by simp
-    have \<open>q > 0\<close> by (rule ccontr)
-      (use \<open>Suc m = n * q\<close> in simp)
-    from m have \<open>m mod n = (n * q - 1) mod n\<close>
-      by simp
-    also have \<open>\<dots> = (n * q - 1 + n) mod n\<close>
-      by simp
-    also have \<open>n * q - 1 + n = n * q + (n - 1)\<close>
-      using \<open>n > 1\<close> \<open>q > 0\<close> by (simp add: algebra_simps)
-    finally have \<open>m mod n = (n - 1) mod n\<close>
-      by simp
-    with \<open>n > 1\<close> have \<open>m mod n = n - 1\<close>
-      by simp
-    with True \<open>n > 1\<close> show ?thesis
-      by (subst div_add1_eq) auto
-  next
     case False
-    have \<open>Suc (m mod n) \<noteq> n\<close>
+    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 - 1\<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 have \<open>Suc (m mod n) < n\<close>
+    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
-    with False \<open>n > 1\<close> show ?thesis
-      by (subst div_add1_eq) (auto simp add: div_eq_0_iff mod_greater_zero_iff_not_dvd)
   qed
-  then show ?thesis
-    by simp
+  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>  (is "_ = ?rhs")
-proof (cases "n = 0")
+  \<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
-  have "Suc m mod n = Suc (m mod n) mod n"
+  moreover have \<open>Suc m mod n = Suc (m mod n) mod n\<close>
     by (simp add: mod_simps)
-  also have "\<dots> = ?rhs"
-    using False by (auto intro!: mod_nat_eqI intro: neq_le_trans simp add: Suc_le_eq)
-  finally show ?thesis .
+  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 \<noteq> (1::nat)"
   by (rule Suc_times_mod_eq) (use that in simp)
 
 lemma Suc_div_le_mono [simp]:
   "m div n \<le> 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: "\<And>n. n < p \<Longrightarrow> P n \<Longrightarrow> P (Suc n mod p)"
 using \<open>m < p\<close> proof (induct m)
   case 0
   show ?case
   proof (rule ccontr)
     assume "\<not> 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 "\<not> 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 \<ge> 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 \<ge> 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 \<longleftrightarrow> (\<exists>q1 q2. x + n * q1 = y + n * q2)"  (is "?lhs = ?rhs")
 proof
   assume H: "x mod n = y mod n"
   { assume xy: "x \<le> 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 "\<exists>q1 q2. x + n * q1 = y + n * q2"
       by blast
   }
   moreover
   { assume xy: "y \<le> 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 "\<exists>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>Elementary euclidean division on \<^typ>\<open>int\<close>\<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 \<noteq> 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
 
 
 subsubsection \<open>Algebraic foundations\<close>
 
 lemma coprime_int_iff [simp]:
   "coprime (int m) (int n) \<longleftrightarrow> coprime m n" (is "?P \<longleftrightarrow> ?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 \<bar>k\<bar> dvd m" "nat \<bar>k\<bar> 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 \<bar>k\<bar> l \<longleftrightarrow> 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 \<bar>l\<bar> \<longleftrightarrow> 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 \<bar>k\<bar>) n \<longleftrightarrow> coprime k (int n)"
 proof -
   define m where "m = nat \<bar>k\<bar>"
   then have "\<bar>k\<bar> = int m"
     by simp
   moreover have "coprime k (int n) \<longleftrightarrow> coprime \<bar>k\<bar> (int n)"
     by simp
   ultimately show ?thesis
     by simp
 qed
 
 lemma coprime_nat_abs_right_iff [simp]:
   "coprime n (nat \<bar>k\<bar>) \<longleftrightarrow> 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 \<Longrightarrow> x dvd a \<Longrightarrow> x dvd b \<Longrightarrow> \<bar>x\<bar> = 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:
   "\<bar>k\<bar> div \<bar>l\<bar> = int (nat \<bar>k\<bar> div nat \<bar>l\<bar>)"
   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:
   "\<bar>k\<bar> mod \<bar>l\<bar> = int (nat \<bar>k\<bar> mod nat \<bar>l\<bar>)"
   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 \<noteq> 0"
   shows "(sgn v * \<bar>k\<bar>) div (sgn v * \<bar>l\<bar>) = \<bar>k\<bar> div \<bar>l\<bar>"
   using assms by (simp add: sgn_mult abs_mult sgn_0_0
     divide_int_def [of "sgn v * \<bar>k\<bar>" "sgn v * \<bar>l\<bar>"] 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 = \<bar>k\<bar> div \<bar>l\<bar>"
   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) * (\<bar>k\<bar> div \<bar>l\<bar>)"
   using assms by (auto simp add: div_abs_eq ac_simps)
 
 lemma div_noneq_sgn_abs:
   fixes k l :: int
   assumes "l \<noteq> 0"
   assumes "sgn k \<noteq> sgn l"
   shows "k div l = - (\<bar>k\<bar> div \<bar>l\<bar>) - of_bool (\<not> 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 \<Rightarrow> nat"
   where [simp]: "euclidean_size_int = (nat \<circ> abs :: int \<Rightarrow> nat)"
 
 definition division_segment_int :: "int \<Rightarrow> int"
   where "division_segment_int k = (if k \<ge> 0 then 1 else - 1)"
 
 lemma division_segment_eq_sgn:
   "division_segment k = sgn k" if "k \<noteq> 0" for k :: int
   using that by (simp add: division_segment_int_def)
 
 lemma abs_division_segment [simp]:
   "\<bar>division_segment k\<bar> = 1" for k :: int
   by (simp add: division_segment_int_def)
 
 lemma abs_mod_less:
   "\<bar>k mod l\<bar> < \<bar>l\<bar>" if "l \<noteq> 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 \<noteq> 0" "\<not> 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 \<noteq> 0" and "\<not> 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 \<noteq> 0" and "n > 0"
     by (simp_all add: sgn_0_0)
   assume "division_segment r = division_segment l"
   moreover have "r = sgn r * \<bar>r\<bar>"
     by (simp add: sgn_mult_abs)
   moreover define u where "u = nat \<bar>r\<bar>"
   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 \<or> 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 \<noteq> 0" and "m > 0" and "s \<noteq> 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 \<longleftrightarrow> 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]:
   \<open>(k div l, k mod l) = (q, r)\<close>
     if by0': \<open>l = 0 \<Longrightarrow> q = 0 \<and> r = k\<close>
     and divides': \<open>l \<noteq> 0 \<Longrightarrow> l dvd k \<Longrightarrow> r = 0 \<and> k = q * l\<close>
     and euclidean_relation': \<open>l \<noteq> 0 \<Longrightarrow> \<not> l dvd k \<Longrightarrow> sgn r = sgn l
       \<and> \<bar>r\<bar> < \<bar>l\<bar> \<and> k = q * l + r\<close> for k l :: int
 proof (induction rule: euclidean_relationI)
   case by0
   then show ?case
     by (rule by0')
 next
   case divides
   then show ?case
     by (rule divides')
 next
   case euclidean_relation
   with euclidean_relation' have \<open>sgn r = sgn l\<close> \<open>\<bar>r\<bar> < \<bar>l\<bar>\<close> \<open>k = q * l + r\<close>
     by simp_all
   from \<open>sgn r = sgn l\<close> \<open>l \<noteq> 0\<close> have \<open>division_segment r = division_segment l\<close>
     by (simp add: division_segment_int_def sgn_if split: if_splits)
   with \<open>\<bar>r\<bar> < \<bar>l\<bar>\<close> \<open>k = q * l + r\<close>
   show ?case
     by simp
 qed
 
 
 subsection \<open>Special case: euclidean rings containing the natural numbers\<close>
 
 class unique_euclidean_semiring_with_nat = semidom + semiring_char_0 + unique_euclidean_semiring +
   assumes of_nat_div: "of_nat (m div n) = of_nat m div of_nat n"
     and division_segment_of_nat [simp]: "division_segment (of_nat n) = 1"
     and division_segment_euclidean_size [simp]: "division_segment a * of_nat (euclidean_size a) = a"
 begin
 
 lemma division_segment_eq_iff:
   "a = b" if "division_segment a = division_segment b"
     and "euclidean_size a = euclidean_size b"
   using that division_segment_euclidean_size [of a] by simp
 
 lemma euclidean_size_of_nat [simp]:
   "euclidean_size (of_nat n) = n"
 proof -
   have "division_segment (of_nat n) * of_nat (euclidean_size (of_nat n)) = of_nat n"
     by (fact division_segment_euclidean_size)
   then show ?thesis by simp
 qed
 
 lemma of_nat_euclidean_size:
   "of_nat (euclidean_size a) = a div division_segment a"
 proof -
   have "of_nat (euclidean_size a) = division_segment a * of_nat (euclidean_size a) div division_segment a"
     by (subst nonzero_mult_div_cancel_left) simp_all
   also have "\<dots> = a div division_segment a"
     by simp
   finally show ?thesis .
 qed
 
 lemma division_segment_1 [simp]:
   "division_segment 1 = 1"
   using division_segment_of_nat [of 1] by simp
 
 lemma division_segment_numeral [simp]:
   "division_segment (numeral k) = 1"
   using division_segment_of_nat [of "numeral k"] by simp
 
 lemma euclidean_size_1 [simp]:
   "euclidean_size 1 = 1"
   using euclidean_size_of_nat [of 1] by simp
 
 lemma euclidean_size_numeral [simp]:
   "euclidean_size (numeral k) = numeral k"
   using euclidean_size_of_nat [of "numeral k"] by simp
 
 lemma of_nat_dvd_iff:
   "of_nat m dvd of_nat n \<longleftrightarrow> m dvd n" (is "?P \<longleftrightarrow> ?Q")
 proof (cases "m = 0")
   case True
   then show ?thesis
     by simp
 next
   case False
   show ?thesis
   proof
     assume ?Q
     then show ?P
       by auto
   next
     assume ?P
     with False have "of_nat n = of_nat n div of_nat m * of_nat m"
       by simp
     then have "of_nat n = of_nat (n div m * m)"
       by (simp add: of_nat_div)
     then have "n = n div m * m"
       by (simp only: of_nat_eq_iff)
     then have "n = m * (n div m)"
       by (simp add: ac_simps)
     then show ?Q ..
   qed
 qed
 
 lemma of_nat_mod:
   "of_nat (m mod n) = of_nat m mod of_nat n"
 proof -
   have "of_nat m div of_nat n * of_nat n + of_nat m mod of_nat n = of_nat m"
     by (simp add: div_mult_mod_eq)
   also have "of_nat m = of_nat (m div n * n + m mod n)"
     by simp
   finally show ?thesis
     by (simp only: of_nat_div of_nat_mult of_nat_add) simp
 qed
 
 lemma one_div_two_eq_zero [simp]:
   "1 div 2 = 0"
 proof -
   from of_nat_div [symmetric] have "of_nat 1 div of_nat 2 = of_nat 0"
     by (simp only:) simp
   then show ?thesis
     by simp
 qed
 
 lemma one_mod_two_eq_one [simp]:
   "1 mod 2 = 1"
 proof -
   from of_nat_mod [symmetric] have "of_nat 1 mod of_nat 2 = of_nat 1"
     by (simp only:) simp
   then show ?thesis
     by simp
 qed
 
 lemma one_mod_2_pow_eq [simp]:
   "1 mod (2 ^ n) = of_bool (n > 0)"
 proof -
   have "1 mod (2 ^ n) = of_nat (1 mod (2 ^ n))"
     using of_nat_mod [of 1 "2 ^ n"] by simp
   also have "\<dots> = of_bool (n > 0)"
     by simp
   finally show ?thesis .
 qed
 
 lemma one_div_2_pow_eq [simp]:
   "1 div (2 ^ n) = of_bool (n = 0)"
   using div_mult_mod_eq [of 1 "2 ^ n"] by auto
 
 lemma div_mult2_eq':
   \<open>a div (of_nat m * of_nat n) = a div of_nat m div of_nat n\<close>
 proof (cases \<open>m = 0 \<or> n = 0\<close>)
   case True
   then show ?thesis
     by auto
 next
   case False
   then have \<open>m > 0\<close> \<open>n > 0\<close>
     by simp_all
   show ?thesis
   proof (cases \<open>of_nat m * of_nat n dvd a\<close>)
     case True
     then obtain b where \<open>a = (of_nat m * of_nat n) * b\<close> ..
     then have \<open>a = of_nat m * (of_nat n * b)\<close>
       by (simp add: ac_simps)
     then show ?thesis
       by simp
   next
     case False
     define q where \<open>q = a div (of_nat m * of_nat n)\<close>
     define r where \<open>r = a mod (of_nat m * of_nat n)\<close>
     from \<open>m > 0\<close> \<open>n > 0\<close> \<open>\<not> of_nat m * of_nat n dvd a\<close> r_def have "division_segment r = 1"
       using division_segment_of_nat [of "m * n"] by (simp add: division_segment_mod)
     with division_segment_euclidean_size [of r]
     have "of_nat (euclidean_size r) = r"
       by simp
     have "a mod (of_nat m * of_nat n) div (of_nat m * of_nat n) = 0"
       by simp
     with \<open>m > 0\<close> \<open>n > 0\<close> r_def have "r div (of_nat m * of_nat n) = 0"
       by simp
     with \<open>of_nat (euclidean_size r) = r\<close>
     have "of_nat (euclidean_size r) div (of_nat m * of_nat n) = 0"
       by simp
     then have "of_nat (euclidean_size r div (m * n)) = 0"
       by (simp add: of_nat_div)
     then have "of_nat (euclidean_size r div m div n) = 0"
       by (simp add: div_mult2_eq)
     with \<open>of_nat (euclidean_size r) = r\<close> have "r div of_nat m div of_nat n = 0"
       by (simp add: of_nat_div)
     with \<open>m > 0\<close> \<open>n > 0\<close> q_def
     have "q = (r div of_nat m + q * of_nat n * of_nat m div of_nat m) div of_nat n"
       by simp
     moreover have \<open>a = q * (of_nat m * of_nat n) + r\<close>
       by (simp add: q_def r_def div_mult_mod_eq)
     ultimately show \<open>a div (of_nat m * of_nat n) = a div of_nat m div of_nat n\<close>
       using q_def [symmetric] div_plus_div_distrib_dvd_right [of \<open>of_nat m\<close> \<open>q * (of_nat m * of_nat n)\<close> r]
       by (simp add: ac_simps)
   qed
 qed
 
 lemma mod_mult2_eq':
   "a mod (of_nat m * of_nat n) = of_nat m * (a div of_nat m mod of_nat n) + a mod of_nat m"
 proof -
   have "a div (of_nat m * of_nat n) * (of_nat m * of_nat n) + a mod (of_nat m * of_nat n) = a div of_nat m div of_nat n * of_nat n * of_nat m + (a div of_nat m mod of_nat n * of_nat m + a mod of_nat m)"
     by (simp add: combine_common_factor div_mult_mod_eq)
   moreover have "a div of_nat m div of_nat n * of_nat n * of_nat m = of_nat n * of_nat m * (a div of_nat m div of_nat n)"
     by (simp add: ac_simps)
   ultimately show ?thesis
     by (simp add: div_mult2_eq' mult_commute)
 qed
 
 lemma div_mult2_numeral_eq:
   "a div numeral k div numeral l = a div numeral (k * l)" (is "?A = ?B")
 proof -
   have "?A = a div of_nat (numeral k) div of_nat (numeral l)"
     by simp
   also have "\<dots> = a div (of_nat (numeral k) * of_nat (numeral l))"
     by (fact div_mult2_eq' [symmetric])
   also have "\<dots> = ?B"
     by simp
   finally show ?thesis .
 qed
 
 lemma numeral_Bit0_div_2:
   "numeral (num.Bit0 n) div 2 = numeral n"
 proof -
   have "numeral (num.Bit0 n) = numeral n + numeral n"
     by (simp only: numeral.simps)
   also have "\<dots> = numeral n * 2"
     by (simp add: mult_2_right)
   finally have "numeral (num.Bit0 n) div 2 = numeral n * 2 div 2"
     by simp
   also have "\<dots> = numeral n"
     by (rule nonzero_mult_div_cancel_right) simp
   finally show ?thesis .
 qed
 
 lemma numeral_Bit1_div_2:
   "numeral (num.Bit1 n) div 2 = numeral n"
 proof -
   have "numeral (num.Bit1 n) = numeral n + numeral n + 1"
     by (simp only: numeral.simps)
   also have "\<dots> = numeral n * 2 + 1"
     by (simp add: mult_2_right)
   finally have "numeral (num.Bit1 n) div 2 = (numeral n * 2 + 1) div 2"
     by simp
   also have "\<dots> = numeral n * 2 div 2 + 1 div 2"
     using dvd_triv_right by (rule div_plus_div_distrib_dvd_left)
   also have "\<dots> = numeral n * 2 div 2"
     by simp
   also have "\<dots> = numeral n"
     by (rule nonzero_mult_div_cancel_right) simp
   finally show ?thesis .
 qed
 
 lemma exp_mod_exp:
   \<open>2 ^ m mod 2 ^ n = of_bool (m < n) * 2 ^ m\<close>
 proof -
   have \<open>(2::nat) ^ m mod 2 ^ n = of_bool (m < n) * 2 ^ m\<close> (is \<open>?lhs = ?rhs\<close>)
     by (auto simp add: not_less monoid_mult_class.power_add dest!: le_Suc_ex)
   then have \<open>of_nat ?lhs = of_nat ?rhs\<close>
     by simp
   then show ?thesis
     by (simp add: of_nat_mod)
 qed
 
 lemma mask_mod_exp:
   \<open>(2 ^ n - 1) mod 2 ^ m = 2 ^ min m n - 1\<close>
 proof -
   have \<open>(2 ^ n - 1) mod 2 ^ m = 2 ^ min m n - (1::nat)\<close> (is \<open>?lhs = ?rhs\<close>)
   proof (cases \<open>n \<le> m\<close>)
     case True
     then show ?thesis
       by (simp add: Suc_le_lessD)
   next
     case False
     then have \<open>m < n\<close>
       by simp
     then obtain q where n: \<open>n = Suc q + m\<close>
       by (auto dest: less_imp_Suc_add)
     then have \<open>min m n = m\<close>
       by simp
     moreover have \<open>(2::nat) ^ m \<le> 2 * 2 ^ q * 2 ^ m\<close>
       using mult_le_mono1 [of 1 \<open>2 * 2 ^ q\<close> \<open>2 ^ m\<close>] by simp
     with n have \<open>2 ^ n - 1 = (2 ^ Suc q - 1) * 2 ^ m + (2 ^ m - (1::nat))\<close>
       by (simp add: monoid_mult_class.power_add algebra_simps)
     ultimately show ?thesis
       by (simp only: euclidean_semiring_cancel_class.mod_mult_self3) simp
   qed
   then have \<open>of_nat ?lhs = of_nat ?rhs\<close>
     by simp
   then show ?thesis
     by (simp add: of_nat_mod of_nat_diff)
 qed
 
 lemma of_bool_half_eq_0 [simp]:
   \<open>of_bool b div 2 = 0\<close>
   by simp
 
 end
 
 class unique_euclidean_ring_with_nat = ring + unique_euclidean_semiring_with_nat
 
 instance nat :: unique_euclidean_semiring_with_nat
   by standard (simp_all add: dvd_eq_mod_eq_0)
 
 instance int :: unique_euclidean_ring_with_nat
   by standard (auto simp add: divide_int_def division_segment_int_def elim: contrapos_np)
 
 
 subsection \<open>More on euclidean division on \<^typ>\<open>int\<close>\<close>
 
 subsubsection \<open>Trivial reduction steps\<close>
 
 lemma div_pos_pos_trivial [simp]:
   "k div l = 0" if "k \<ge> 0" and "k < l" for k l :: int
   using that by (simp add: unique_euclidean_semiring_class.div_eq_0_iff division_segment_int_def)
 
 lemma mod_pos_pos_trivial [simp]:
   "k mod l = k" if "k \<ge> 0" and "k < l" for k l :: int
   using that by (simp add: mod_eq_self_iff_div_eq_0)
 
 lemma div_neg_neg_trivial [simp]:
   "k div l = 0" if "k \<le> 0" and "l < k" for k l :: int
   using that by (cases "k = 0") (simp, simp add: unique_euclidean_semiring_class.div_eq_0_iff division_segment_int_def)
 
 lemma mod_neg_neg_trivial [simp]:
   "k mod l = k" if "k \<le> 0" and "l < k" for k l :: int
   using that by (simp add: mod_eq_self_iff_div_eq_0)
 
 lemma
   div_pos_neg_trivial: \<open>k div l = - 1\<close>  (is ?Q)
   and mod_pos_neg_trivial: \<open>k mod l = k + l\<close>  (is ?R)
     if \<open>0 < k\<close> and \<open>k + l \<le> 0\<close> for k l :: int
 proof -
   from that have \<open>l < 0\<close>
     by simp
   have \<open>(k div l, k mod l) = (- 1, k + l)\<close>
   proof (induction rule: euclidean_relation_intI)
     case by0
     with \<open>l < 0\<close> show ?case
       by simp
   next
     case divides
     from \<open>l dvd k\<close> obtain j where \<open>k = l * j\<close> ..
     with \<open>l < 0\<close> \<open>0 < k\<close> have \<open>j < 0\<close>
       by (simp add: zero_less_mult_iff)
     moreover from \<open>k + l \<le> 0\<close> \<open>k = l * j\<close> have \<open>l * (j + 1) \<le> 0\<close>
       by (simp add: algebra_simps)
     with \<open>l < 0\<close> have \<open>j + 1 \<ge> 0\<close>
       by (simp add: mult_le_0_iff)
     with \<open>j < 0\<close> have \<open>j = - 1\<close>
       by simp
     with \<open>k = l * j\<close> show ?case
       by simp
   next
     case euclidean_relation
     with \<open>k + l \<le> 0\<close> have \<open>k + l < 0\<close>
       by (auto simp add: less_le add_eq_0_iff)
     with \<open>0 < k\<close> show ?case
       by simp
   qed
   then show ?Q and ?R
     by simp_all
 qed
 
 text \<open>There is neither \<open>div_neg_pos_trivial\<close> nor \<open>mod_neg_pos_trivial\<close>
   because \<^term>\<open>0 div l = 0\<close> would supersede it.\<close>
 
 
 subsubsection \<open>More uniqueness rules\<close>
 
 lemma
   fixes a b q r :: int
   assumes \<open>a = b * q + r\<close> \<open>0 \<le> r\<close> \<open>r < b\<close>
   shows int_div_pos_eq:
       \<open>a div b = q\<close> (is ?Q)
     and int_mod_pos_eq:
       \<open>a mod b = r\<close> (is ?R)
 proof -
   have \<open>(a div b, a mod b) = (q, r)\<close>
     by (induction rule: euclidean_relation_intI)
       (use assms in \<open>auto simp add: ac_simps dvd_add_left_iff sgn_1_pos le_less dest: zdvd_imp_le\<close>)
   then show ?Q and ?R
     by simp_all
 qed
 
 lemma int_div_neg_eq:
   \<open>a div b = q\<close> if \<open>a = b * q + r\<close> \<open>r \<le> 0\<close> \<open>b < r\<close> for a b q r :: int
   using that int_div_pos_eq [of a \<open>- b\<close> \<open>- q\<close> \<open>- r\<close>] by simp_all
 
 lemma int_mod_neg_eq:
   \<open>a mod b = r\<close> if \<open>a = b * q + r\<close> \<open>r \<le> 0\<close> \<open>b < r\<close> for a b q r :: int
   using that int_div_neg_eq [of a b q r] by simp
 
 
 subsubsection \<open>Laws for unary minus\<close>
 
 lemma zmod_zminus1_not_zero:
   fixes k l :: int
   shows "- k mod l \<noteq> 0 \<Longrightarrow> k mod l \<noteq> 0"
   by (simp add: mod_eq_0_iff_dvd)
 
 lemma zmod_zminus2_not_zero:
   fixes k l :: int
   shows "k mod - l \<noteq> 0 \<Longrightarrow> k mod l \<noteq> 0"
   by (simp add: mod_eq_0_iff_dvd)
 
 lemma zdiv_zminus1_eq_if:
   \<open>(- a) div b = (if a mod b = 0 then - (a div b) else - (a div b) - 1)\<close>
   if \<open>b \<noteq> 0\<close> for a b :: int
   using that sgn_not_eq_imp [of b \<open>- a\<close>]
   by (cases \<open>a = 0\<close>) (auto simp add: div_eq_div_abs [of \<open>- a\<close> b] div_eq_div_abs [of a b] sgn_eq_0_iff)
 
 lemma zdiv_zminus2_eq_if:
   \<open>a div (- b) = (if a mod b = 0 then - (a div b) else - (a div b) - 1)\<close>
   if \<open>b \<noteq> 0\<close> for a b :: int
   using that by (auto simp add: zdiv_zminus1_eq_if div_minus_right)
 
 lemma zmod_zminus1_eq_if:
   \<open>(- a) mod b = (if a mod b = 0 then 0 else b - (a mod b))\<close>
   for a b :: int
   by (cases \<open>b = 0\<close>)
     (auto simp flip: minus_div_mult_eq_mod simp add: zdiv_zminus1_eq_if algebra_simps)
 
 lemma zmod_zminus2_eq_if:
   \<open>a mod (- b) = (if a mod b = 0 then 0 else (a mod b) - b)\<close>
   for a b :: int
   by (auto simp add: zmod_zminus1_eq_if mod_minus_right)
 
 
 subsubsection \<open>Borders\<close>
 
 lemma pos_mod_bound [simp]:
   "k mod l < l" if "l > 0" for k l :: int
 proof -
   obtain m and s where "k = sgn s * int m"
     by (rule int_sgnE)
   moreover from that obtain n where "l = sgn 1 * int n"
     by (cases l) simp_all
   moreover from this that have "n > 0"
     by simp
   ultimately show ?thesis
     by (simp only: modulo_int_unfold)
       (auto simp add: mod_greater_zero_iff_not_dvd sgn_1_pos)
 qed
 
 lemma neg_mod_bound [simp]:
   "l < k mod l" if "l < 0" for k l :: int
 proof -
   obtain m and s where "k = sgn s * int m"
     by (rule int_sgnE)
   moreover from that obtain q where "l = sgn (- 1) * int (Suc q)"
     by (cases l) simp_all
   moreover define n where "n = Suc q"
   then have "Suc q = n"
     by simp
   ultimately show ?thesis
     by (simp only: modulo_int_unfold)
       (auto simp add: mod_greater_zero_iff_not_dvd sgn_1_neg)
 qed
 
 lemma pos_mod_sign [simp]:
   "0 \<le> k mod l" if "l > 0" for k l :: int
 proof -
   obtain m and s where "k = sgn s * int m"
     by (rule int_sgnE)
   moreover from that obtain n where "l = sgn 1 * int n"
     by (cases l) auto
   moreover from this that have "n > 0"
     by simp
   ultimately show ?thesis
     by (simp only: modulo_int_unfold) (auto simp add: sgn_1_pos)
 qed
 
 lemma neg_mod_sign [simp]:
   "k mod l \<le> 0" if "l < 0" for k l :: int
 proof -
   obtain m and s where "k = sgn s * int m"
     by (rule int_sgnE)
   moreover from that obtain q where "l = sgn (- 1) * int (Suc q)"
     by (cases l) simp_all
   moreover define n where "n = Suc q"
   then have "Suc q = n"
     by simp
   moreover have \<open>int (m mod n) \<le> int n\<close>
     using \<open>Suc q = n\<close> by simp
   then have \<open>sgn s * int (m mod n) \<le> int n\<close>
     by (cases s \<open>0::int\<close> rule: linorder_cases) simp_all
   ultimately show ?thesis
     by (simp only: modulo_int_unfold) auto
 qed
 
 
 subsubsection \<open>Splitting Rules for div and mod\<close>
 
 lemma split_zdiv:
   \<open>P (n div k) \<longleftrightarrow>
     (k = 0 \<longrightarrow> P 0) \<and>
     (0 < k \<longrightarrow> (\<forall>i j. 0 \<le> j \<and> j < k \<and> n = k * i + j \<longrightarrow> P i)) \<and>
     (k < 0 \<longrightarrow> (\<forall>i j. k < j \<and> j \<le> 0 \<and> n = k * i + j \<longrightarrow> P i))\<close> (is ?div)
   and split_zmod:
   \<open>Q (n mod k) \<longleftrightarrow>
     (k = 0 \<longrightarrow> Q n) \<and>
     (0 < k \<longrightarrow> (\<forall>i j. 0 \<le> j \<and> j < k \<and> n = k * i + j \<longrightarrow> Q j)) \<and>
     (k < 0 \<longrightarrow> (\<forall>i j. k < j \<and> j \<le> 0 \<and> n = k * i + j \<longrightarrow> Q j))\<close> (is ?mod)
   for n k :: int
 proof -
   have *: \<open>R (n div k) (n mod k) \<longleftrightarrow>
     (k = 0 \<longrightarrow> R 0 n) \<and>
     (0 < k \<longrightarrow> (\<forall>i j. 0 \<le> j \<and> j < k \<and> n = k * i + j \<longrightarrow> R i j)) \<and>
     (k < 0 \<longrightarrow> (\<forall>i j. k < j \<and> j \<le> 0 \<and> n = k * i + j \<longrightarrow> R i j))\<close> for R
     by (cases \<open>k = 0\<close>)
       (auto simp add: linorder_class.neq_iff)
   from * [of \<open>\<lambda>q _. P q\<close>] show ?div .
   from * [of \<open>\<lambda>_ r. Q r\<close>] show ?mod .
 qed
 
 text \<open>Enable (lin)arith to deal with \<^const>\<open>divide\<close> and \<^const>\<open>modulo\<close>
   when these are applied to some constant that is of the form
   \<^term>\<open>numeral k\<close>:\<close>
 declare split_zdiv [of _ _ \<open>numeral n\<close>, linarith_split] for n
 declare split_zdiv [of _ _ \<open>- numeral n\<close>, linarith_split] for n
 declare split_zmod [of _ _ \<open>numeral n\<close>, linarith_split] for n
 declare split_zmod [of _ _ \<open>- numeral n\<close>, linarith_split] for n
 
 lemma zdiv_eq_0_iff:
   "i div k = 0 \<longleftrightarrow> k = 0 \<or> 0 \<le> i \<and> i < k \<or> i \<le> 0 \<and> k < i" (is "?L = ?R")
   for i k :: int
 proof
   assume ?L
   moreover have "?L \<longrightarrow> ?R"
     by (rule split_zdiv [THEN iffD2]) simp
   ultimately show ?R
     by blast
 next
   assume ?R then show ?L
     by auto
 qed
 
 lemma zmod_trivial_iff:
   fixes i k :: int
   shows "i mod k = i \<longleftrightarrow> k = 0 \<or> 0 \<le> i \<and> i < k \<or> i \<le> 0 \<and> k < i"
 proof -
   have "i mod k = i \<longleftrightarrow> i div k = 0"
     using div_mult_mod_eq [of i k] by safe auto
   with zdiv_eq_0_iff
   show ?thesis
     by simp
 qed
 
 
 subsubsection \<open>Algebraic rewrites\<close>
 
 lemma zdiv_zmult2_eq:
   \<open>a div (b * c) = (a div b) div c\<close> if \<open>c \<ge> 0\<close> for a b c :: int
 proof (cases \<open>b \<ge> 0\<close>)
   case True
   with that show ?thesis
     using div_mult2_eq' [of a \<open>nat b\<close> \<open>nat c\<close>] by simp
 next
   case False
   with that show ?thesis
     using div_mult2_eq' [of \<open>- a\<close> \<open>nat (- b)\<close> \<open>nat c\<close>] by simp
 qed
 
 lemma zdiv_zmult2_eq':
   \<open>k div (l * j) = ((sgn j * k) div l) div \<bar>j\<bar>\<close> for k l j :: int
 proof -
   have \<open>k div (l * j) = (sgn j * k) div (sgn j * (l * j))\<close>
     by (simp add: sgn_0_0)
   also have \<open>sgn j * (l * j) = l * \<bar>j\<bar>\<close>
     by (simp add: mult.left_commute [of _ l] abs_sgn) (simp add: ac_simps)
   also have \<open>(sgn j * k) div (l * \<bar>j\<bar>) = ((sgn j * k) div l) div \<bar>j\<bar>\<close>
     by (simp add: zdiv_zmult2_eq)
   finally show ?thesis .
 qed
 
 lemma zmod_zmult2_eq:
   \<open>a mod (b * c) = b * (a div b mod c) + a mod b\<close> if \<open>c \<ge> 0\<close> for a b c :: int
 proof (cases \<open>b \<ge> 0\<close>)
   case True
   with that show ?thesis
     using mod_mult2_eq' [of a \<open>nat b\<close> \<open>nat c\<close>] by simp
 next
   case False
   with that show ?thesis
     using mod_mult2_eq' [of \<open>- a\<close> \<open>nat (- b)\<close> \<open>nat c\<close>] by simp
 qed
 
 lemma half_nonnegative_int_iff [simp]:
   \<open>k div 2 \<ge> 0 \<longleftrightarrow> k \<ge> 0\<close> for k :: int
   by auto
 
 lemma half_negative_int_iff [simp]:
   \<open>k div 2 < 0 \<longleftrightarrow> k < 0\<close> for k :: int
   by auto
 
 
 subsubsection \<open>Distributive laws for conversions.\<close>
 
 lemma zdiv_int:
   "int (a div b) = int a div int b"
   by (fact of_nat_div)
 
 lemma zmod_int:
   "int (a mod b) = int a mod int b"
   by (fact of_nat_mod)
 
 lemma nat_div_distrib:
   \<open>nat (x div y) = nat x div nat y\<close> if \<open>0 \<le> x\<close>
   using that by (simp add: divide_int_def sgn_if)
 
 lemma nat_div_distrib':
   \<open>nat (x div y) = nat x div nat y\<close> if \<open>0 \<le> y\<close>
   using that by (simp add: divide_int_def sgn_if)
 
 lemma nat_mod_distrib: \<comment> \<open>Fails if y<0: the LHS collapses to (nat z) but the RHS doesn't\<close>
   \<open>nat (x mod y) = nat x mod nat y\<close> if \<open>0 \<le> x\<close> \<open>0 \<le> y\<close>
   using that by (simp add: modulo_int_def sgn_if)
 
 
 subsubsection \<open>Monotonicity in the First Argument (Dividend)\<close>
 
 lemma zdiv_mono1:
   \<open>a div b \<le> a' div b\<close>
     if \<open>a \<le> a'\<close> \<open>0 < b\<close>
     for a b b' :: int
 proof -
   from \<open>a \<le> a'\<close> have \<open>b * (a div b) + a mod b \<le> b * (a' div b) + a' mod b\<close>
     by simp
   then have \<open>b * (a div b) \<le> (a' mod b - a mod b) + b * (a' div b)\<close>
     by (simp add: algebra_simps)
   moreover have \<open>a' mod b < b + a mod b\<close>
     by (rule less_le_trans [of _ b]) (use \<open>0 < b\<close> in simp_all)
   ultimately have \<open>b * (a div b) < b * (1 + a' div b)\<close>
     by (simp add: distrib_left)
   with \<open>0 < b\<close> have \<open>a div b < 1 + a' div b\<close>
     by (simp add: mult_less_cancel_left)
   then show ?thesis
     by simp
 qed
 
 lemma zdiv_mono1_neg:
   \<open>a' div b \<le> a div b\<close>
     if \<open>a \<le> a'\<close> \<open>b < 0\<close>
     for a a' b :: int
   using that zdiv_mono1 [of \<open>- a'\<close> \<open>- a\<close> \<open>- b\<close>] by simp
 
 
 subsubsection \<open>Monotonicity in the Second Argument (Divisor)\<close>
 
 lemma zdiv_mono2:
   \<open>a div b \<le> a div b'\<close> if \<open>0 \<le> a\<close> \<open>0 < b'\<close> \<open>b' \<le> b\<close> for a b b' :: int
 proof -
   define q q' r r' where **: \<open>q = a div b\<close> \<open>q' = a div b'\<close> \<open>r = a mod b\<close> \<open>r' = a mod b'\<close>
   then have *: \<open>b * q + r = b' * q' + r'\<close> \<open>0 \<le> b' * q' + r'\<close>
     \<open>r' < b'\<close> \<open>0 \<le> r\<close> \<open>0 < b'\<close> \<open>b' \<le> b\<close>
     using that by simp_all
   have \<open>0 < b' * (q' + 1)\<close>
     using * by (simp add: distrib_left)
   with * have \<open>0 \<le> q'\<close>
     by (simp add: zero_less_mult_iff)
   moreover have \<open>b * q = r' - r + b' * q'\<close>
     using * by linarith
   ultimately have \<open>b * q < b * (q' + 1)\<close>
     using mult_right_mono * unfolding distrib_left by fastforce
   with * have \<open>q \<le> q'\<close>
     by (simp add: mult_less_cancel_left_pos)
   with ** show ?thesis
     by simp
 qed
 
 lemma zdiv_mono2_neg:
   \<open>a div b' \<le> a div b\<close> if \<open>a < 0\<close> \<open>0 < b'\<close> \<open>b' \<le> b\<close> for a b b' :: int
 proof -
   define q q' r r' where **: \<open>q = a div b\<close> \<open>q' = a div b'\<close> \<open>r = a mod b\<close> \<open>r' = a mod b'\<close>
   then have *: \<open>b * q + r = b' * q' + r'\<close> \<open>b' * q' + r' < 0\<close>
     \<open>r < b\<close> \<open>0 \<le> r'\<close> \<open>0 < b'\<close> \<open>b' \<le> b\<close>
     using that by simp_all
   have \<open>b' * q' < 0\<close>
     using * by linarith
   with * have \<open>q' \<le> 0\<close>
     by (simp add: mult_less_0_iff)
   have \<open>b * q' \<le> b' * q'\<close>
     by (simp add: \<open>q' \<le> 0\<close> * mult_right_mono_neg)
   then have "b * q' < b * (q + 1)"
     using * by (simp add: distrib_left)
   then have \<open>q' \<le> q\<close>
     using * by (simp add: mult_less_cancel_left)
   then show ?thesis
     by (simp add: **)
 qed
 
 
 subsubsection \<open>Quotients of Signs\<close>
 
 lemma div_eq_minus1:
   \<open>0 < b \<Longrightarrow> - 1 div b = - 1\<close> for b :: int
   by (simp add: divide_int_def)
 
 lemma zmod_minus1:
   \<open>0 < b \<Longrightarrow> - 1 mod b = b - 1\<close> for b :: int
   by (auto simp add: modulo_int_def)
 
 lemma minus_mod_int_eq:
   \<open>- k mod l = l - 1 - (k - 1) mod l\<close> if \<open>l \<ge> 0\<close> for k l :: int
 proof (cases \<open>l = 0\<close>)
   case True
   then show ?thesis
     by simp
 next
   case False
   with that have \<open>l > 0\<close>
     by simp
   then show ?thesis
   proof (cases \<open>l dvd k\<close>)
     case True
     then obtain j where \<open>k = l * j\<close> ..
     moreover have \<open>(l * j mod l - 1) mod l = l - 1\<close>
       using \<open>l > 0\<close> by (simp add: zmod_minus1)
     then have \<open>(l * j - 1) mod l = l - 1\<close>
       by (simp only: mod_simps)
     ultimately show ?thesis
       by simp
   next
     case False
     moreover have 1: \<open>0 < k mod l\<close>
       using \<open>0 < l\<close> False le_less by fastforce
     moreover have 2: \<open>k mod l < 1 + l\<close>
       using \<open>0 < l\<close> pos_mod_bound[of l k] by linarith
     from 1 2 \<open>l > 0\<close> have \<open>(k mod l - 1) mod l = k mod l - 1\<close>
       by (simp add: zmod_trivial_iff)
     ultimately show ?thesis
       by (simp only: zmod_zminus1_eq_if)
          (simp add: mod_eq_0_iff_dvd algebra_simps mod_simps)
   qed
 qed
 
 lemma div_neg_pos_less0:
   \<open>a div b < 0\<close> if \<open>a < 0\<close> \<open>0 < b\<close> for a b :: int
 proof -
   have "a div b \<le> - 1 div b"
     using zdiv_mono1 that by auto
   also have "... \<le> -1"
     by (simp add: that(2) div_eq_minus1)
   finally show ?thesis
     by force
 qed
 
 lemma div_nonneg_neg_le0:
   \<open>a div b \<le> 0\<close> if \<open>0 \<le> a\<close> \<open>b < 0\<close> for a b :: int
   using that by (auto dest: zdiv_mono1_neg)
 
 lemma div_nonpos_pos_le0:
   \<open>a div b \<le> 0\<close> if \<open>a \<le> 0\<close> \<open>0 < b\<close> for a b :: int
   using that by (auto dest: zdiv_mono1)
 
 text\<open>Now for some equivalences of the form \<open>a div b >=< 0 \<longleftrightarrow> \<dots>\<close>
 conditional upon the sign of \<open>a\<close> or \<open>b\<close>. There are many more.
 They should all be simp rules unless that causes too much search.\<close>
 
 lemma pos_imp_zdiv_nonneg_iff:
   \<open>0 \<le> a div b \<longleftrightarrow> 0 \<le> a\<close>
   if \<open>0 < b\<close> for a b :: int
 proof
   assume \<open>0 \<le> a div b\<close>
   show \<open>0 \<le> a\<close>
   proof (rule ccontr)
     assume \<open>\<not> 0 \<le> a\<close>
     then have \<open>a < 0\<close>
       by simp
     then have \<open>a div b < 0\<close>
       using that by (rule div_neg_pos_less0)
     with \<open>0 \<le> a div b\<close> show False
       by simp
   qed
 next
   assume "0 \<le> a"
   then have "0 div b \<le> a div b"
     using zdiv_mono1 that by blast
   then show "0 \<le> a div b"
     by auto
 qed
 
 lemma neg_imp_zdiv_nonneg_iff:
   \<open>0 \<le> a div b \<longleftrightarrow> a \<le> 0\<close> if \<open>b < 0\<close> for a b :: int
   using that pos_imp_zdiv_nonneg_iff [of \<open>- b\<close> \<open>- a\<close>] by simp
 
 lemma pos_imp_zdiv_pos_iff:
   \<open>0 < (i::int) div k \<longleftrightarrow> k \<le> i\<close> if \<open>0 < k\<close> for i k :: int
   using that pos_imp_zdiv_nonneg_iff [of k i] zdiv_eq_0_iff [of i k] by arith
 
 lemma pos_imp_zdiv_neg_iff:
   \<open>a div b < 0 \<longleftrightarrow> a < 0\<close> if \<open>0 < b\<close> for a b :: int
     \<comment> \<open>But not \<^prop>\<open>a div b \<le> 0 \<longleftrightarrow> a \<le> 0\<close>; consider \<^prop>\<open>a = 1\<close>, \<^prop>\<open>b = 2\<close> when \<^prop>\<open>a div b = 0\<close>.\<close>
   using that by (simp add: pos_imp_zdiv_nonneg_iff flip: linorder_not_le)
 
 lemma neg_imp_zdiv_neg_iff:
     \<comment> \<open>But not \<^prop>\<open>a div b \<le> 0 \<longleftrightarrow> 0 \<le> a\<close>; consider \<^prop>\<open>a = - 1\<close>, \<^prop>\<open>b = - 2\<close> when \<^prop>\<open>a div b = 0\<close>.\<close>
   \<open>a div b < 0 \<longleftrightarrow> 0 < a\<close> if \<open>b < 0\<close> for a b :: int
   using that by (simp add: neg_imp_zdiv_nonneg_iff flip: linorder_not_le)
 
 lemma nonneg1_imp_zdiv_pos_iff:
   \<open>a div b > 0 \<longleftrightarrow> a \<ge> b \<and> b > 0\<close> if \<open>0 \<le> a\<close> for a b :: int
 proof -
   have "0 < a div b \<Longrightarrow> b \<le> a"
     using div_pos_pos_trivial[of a b] that by arith
   moreover have "0 < a div b \<Longrightarrow> b > 0"
     using that div_nonneg_neg_le0[of a b] by (cases "b=0"; force)
   moreover have "b \<le> a \<and> 0 < b \<Longrightarrow> 0 < a div b"
     using int_one_le_iff_zero_less[of "a div b"] zdiv_mono1[of b a b] by simp
   ultimately show ?thesis
     by blast
 qed
 
 lemma zmod_le_nonneg_dividend:
   \<open>m mod k \<le> m\<close> if \<open>(m::int) \<ge> 0\<close> for m k :: int
 proof -
   from that have \<open>m > 0 \<or> m = 0\<close>
     by auto
   then show ?thesis proof
     assume \<open>m = 0\<close> then show ?thesis
       by simp
   next
     assume \<open>m > 0\<close> then show ?thesis
     proof (cases k \<open>0::int\<close> rule: linorder_cases)
       case less
       moreover define l where \<open>l = - k\<close>
       ultimately have \<open>l > 0\<close>
         by simp
       with \<open>m > 0\<close> have \<open>int (nat m mod nat l) \<le> m\<close>
         by (simp flip: le_nat_iff)
       then have \<open>int (nat m mod nat l) - l \<le> m\<close>
         using \<open>l > 0\<close> by simp
       with \<open>m > 0\<close> \<open>l > 0\<close> show ?thesis
         by (simp add: modulo_int_def l_def flip: le_nat_iff)
     qed (simp_all add: modulo_int_def flip: le_nat_iff)
   qed
 qed
 
 lemma sgn_div_eq_sgn_mult:
   \<open>sgn (k div l) = of_bool (k div l \<noteq> 0) * sgn (k * l)\<close>
   for k l :: int
 proof (cases \<open>k div l = 0\<close>)
   case True
   then show ?thesis
     by simp
 next
   case False
   have \<open>0 \<le> \<bar>k\<bar> div \<bar>l\<bar>\<close>
     by (cases \<open>l = 0\<close>) (simp_all add: pos_imp_zdiv_nonneg_iff)
   then have \<open>\<bar>k\<bar> div \<bar>l\<bar> \<noteq> 0 \<longleftrightarrow> 0 < \<bar>k\<bar> div \<bar>l\<bar>\<close>
     by (simp add: less_le)
   also have \<open>\<dots> \<longleftrightarrow> \<bar>k\<bar> \<ge> \<bar>l\<bar>\<close>
     using False nonneg1_imp_zdiv_pos_iff by auto
   finally have *: \<open>\<bar>k\<bar> div \<bar>l\<bar> \<noteq> 0 \<longleftrightarrow> \<bar>l\<bar> \<le> \<bar>k\<bar>\<close> .
   show ?thesis
     using \<open>0 \<le> \<bar>k\<bar> div \<bar>l\<bar>\<close> False
   by (auto simp add: div_eq_div_abs [of k l] div_eq_sgn_abs [of k l]
     sgn_mult sgn_1_pos sgn_1_neg sgn_eq_0_iff nonneg1_imp_zdiv_pos_iff * dest: sgn_not_eq_imp)
 qed
 
 
 subsubsection \<open>Further properties\<close>
 
 lemma div_int_pos_iff:
   "k div l \<ge> 0 \<longleftrightarrow> k = 0 \<or> l = 0 \<or> k \<ge> 0 \<and> l \<ge> 0
     \<or> k < 0 \<and> l < 0"
   for k l :: int
 proof (cases "k = 0 \<or> l = 0")
   case False
   then have *: "k \<noteq> 0" "l \<noteq> 0"
     by auto
   then have "0 \<le> k div l \<Longrightarrow> \<not> k < 0 \<Longrightarrow> 0 \<le> l"
     by (meson neg_imp_zdiv_neg_iff not_le not_less_iff_gr_or_eq)
   then show ?thesis
    using * by (auto simp add: pos_imp_zdiv_nonneg_iff neg_imp_zdiv_nonneg_iff)
 qed auto
 
 lemma mod_int_pos_iff:
   \<open>k mod l \<ge> 0 \<longleftrightarrow> l dvd k \<or> l = 0 \<and> k \<ge> 0 \<or> l > 0\<close>
   for k l :: int
 proof (cases "l > 0")
   case False
   then show ?thesis
     by (simp add: dvd_eq_mod_eq_0) (use neg_mod_sign [of l k] in \<open>auto simp add: le_less not_less\<close>)
 qed auto
 
 lemma abs_div:
   \<open>\<bar>x div y\<bar> = \<bar>x\<bar> div \<bar>y\<bar>\<close> if \<open>y dvd x\<close> for x y :: int
   using that by (cases \<open>y = 0\<close>) (auto simp add: abs_mult)
 
 lemma int_power_div_base: \<^marker>\<open>contributor \<open>Matthias Daum\<close>\<close>
   \<open>k ^ m div k = k ^ (m - Suc 0)\<close> if \<open>0 < m\<close> \<open>0 < k\<close> for k :: int
   using that by (cases m) simp_all
 
 lemma int_div_less_self: \<^marker>\<open>contributor \<open>Matthias Daum\<close>\<close>
   \<open>x div k < x\<close> if \<open>0 < x\<close> \<open>1 < k\<close> for x k :: int
 proof -
   from that have \<open>nat (x div k) = nat x div nat k\<close>
     by (simp add: nat_div_distrib)
   also from that have \<open>nat x div nat k < nat x\<close>
     by simp
   finally show ?thesis
     by simp
 qed
 
 
 subsubsection \<open>Computing \<open>div\<close> and \<open>mod\<close> by shifting\<close>
 
 lemma div_pos_geq:
   \<open>k div l = (k - l) div l + 1\<close> if \<open>0 < l\<close> \<open>l \<le> k\<close> for k l :: int
 proof -
   have "k = (k - l) + l" by simp
   then obtain j where k: "k = j + l" ..
   with that show ?thesis by (simp add: div_add_self2)
 qed
 
 lemma mod_pos_geq:
   \<open>k mod l = (k - l) mod l\<close>  if \<open>0 < l\<close> \<open>l \<le> k\<close> for k l :: int
 proof -
   have "k = (k - l) + l" by simp
   then obtain j where k: "k = j + l" ..
   with that show ?thesis by simp
 qed
 
 lemma pos_zdiv_mult_2: \<open>(1 + 2 * b) div (2 * a) = b div a\<close> (is ?Q)
   and pos_zmod_mult_2: \<open>(1 + 2 * b) mod (2 * a) = 1 + 2 * (b mod a)\<close> (is ?R)
   if \<open>0 \<le> a\<close> for a b :: int
 proof -
   have \<open>((1 + 2 * b) div (2 * a), (1 + 2 * b) mod (2 * a)) = (b div a, 1 + 2 * (b mod a))\<close>
   proof (induction rule: euclidean_relation_intI)
     case by0
     then show ?case
       by simp
   next
     case divides
     have \<open>2 dvd (2 * a)\<close>
       by simp
     then have \<open>2 dvd (1 + 2 * b)\<close>
       using \<open>2 * a dvd 1 + 2 * b\<close> by (rule dvd_trans)
     then have \<open>2 dvd (1 + b * 2)\<close>
       by (simp add: ac_simps)
     then have \<open>is_unit (2 :: int)\<close>
       by simp
     then show ?case
       by simp
   next
     case euclidean_relation
     with that have \<open>a > 0\<close>
       by simp
     moreover have \<open>b mod a < a\<close>
       using \<open>a > 0\<close> by simp
     then have \<open>1 + 2 * (b mod a) < 2 * a\<close>
       by simp
     moreover have \<open>2 * (b mod a) + a * (2 * (b div a)) = 2 * (b div a * a + b mod a)\<close>
       by (simp only: algebra_simps)
     moreover have \<open>0 \<le> 2 * (b mod a)\<close>
       using \<open>a > 0\<close> by simp
     ultimately show ?case
       by (simp add: algebra_simps)
   qed
   then show ?Q and ?R
     by simp_all
 qed
 
 lemma neg_zdiv_mult_2: \<open>(1 + 2 * b) div (2 * a) = (b + 1) div a\<close> (is ?Q)
   and neg_zmod_mult_2: \<open>(1 + 2 * b) mod (2 * a) = 2 * ((b + 1) mod a) - 1\<close> (is ?R)
   if \<open>a \<le> 0\<close> for a b :: int
 proof -
   have \<open>((1 + 2 * b) div (2 * a), (1 + 2 * b) mod (2 * a)) = ((b + 1) div a, 2 * ((b + 1) mod a) - 1)\<close>
   proof (induction rule: euclidean_relation_intI)
     case by0
     then show ?case
       by simp
   next
     case divides
     have \<open>2 dvd (2 * a)\<close>
       by simp
     then have \<open>2 dvd (1 + 2 * b)\<close>
       using \<open>2 * a dvd 1 + 2 * b\<close> by (rule dvd_trans)
     then have \<open>2 dvd (1 + b * 2)\<close>
       by (simp add: ac_simps)
     then have \<open>is_unit (2 :: int)\<close>
       by simp
     then show ?case
       by simp
   next
     case euclidean_relation
     with that have \<open>a < 0\<close>
       by simp
     moreover have \<open>(b + 1) mod a > a\<close>
       using \<open>a < 0\<close> by simp
     then have \<open>2 * ((b + 1) mod a) > 1 + 2 * a\<close>
       by simp
     moreover have \<open>((1 + b) mod a) \<le> 0\<close>
       using \<open>a < 0\<close> by simp
     then have \<open>2 * ((1 + b) mod a) \<le> 0\<close>
       by simp
     moreover have \<open>2 * ((1 + b) mod a) + a * (2 * ((1 + b) div a)) =
       2 * ((1 + b) div a * a + (1 + b) mod a)\<close>
       by (simp only: algebra_simps)
     ultimately show ?case
       by (simp add: algebra_simps sgn_mult abs_mult)
   qed
   then show ?Q and ?R
     by simp_all
 qed
 
 lemma zdiv_numeral_Bit0 [simp]:
   \<open>numeral (Num.Bit0 v) div numeral (Num.Bit0 w) =
     numeral v div (numeral w :: int)\<close>
   unfolding numeral.simps unfolding mult_2 [symmetric]
   by (rule div_mult_mult1) simp
 
 lemma zdiv_numeral_Bit1 [simp]:
   \<open>numeral (Num.Bit1 v) div numeral (Num.Bit0 w) =
     (numeral v div (numeral w :: int))\<close>
   unfolding numeral.simps
   unfolding mult_2 [symmetric] add.commute [of _ 1]
   by (rule pos_zdiv_mult_2) simp
 
 lemma zmod_numeral_Bit0 [simp]:
   \<open>numeral (Num.Bit0 v) mod numeral (Num.Bit0 w) =
     (2::int) * (numeral v mod numeral w)\<close>
   unfolding numeral_Bit0 [of v] numeral_Bit0 [of w]
   unfolding mult_2 [symmetric] by (rule mod_mult_mult1)
 
 lemma zmod_numeral_Bit1 [simp]:
   \<open>numeral (Num.Bit1 v) mod numeral (Num.Bit0 w) =
     2 * (numeral v mod numeral w) + (1::int)\<close>
   unfolding numeral_Bit1 [of v] numeral_Bit0 [of w]
   unfolding mult_2 [symmetric] add.commute [of _ 1]
   by (rule pos_zmod_mult_2) simp
 
 
 subsection \<open>Generic symbolic computations\<close>
 
 text \<open>
   The following type class contains everything necessary to formulate
   a division algorithm in ring structures with numerals, restricted
   to its positive segments.
 \<close>
 
 class unique_euclidean_semiring_with_nat_division = unique_euclidean_semiring_with_nat +
   fixes divmod :: \<open>num \<Rightarrow> num \<Rightarrow> 'a \<times> 'a\<close>
     and divmod_step :: \<open>'a \<Rightarrow> 'a \<times> 'a \<Rightarrow> 'a \<times> 'a\<close> \<comment> \<open>
       These are conceptually definitions but force generated code
       to be monomorphic wrt. particular instances of this class which
       yields a significant speedup.\<close>
   assumes divmod_def: \<open>divmod m n = (numeral m div numeral n, numeral m mod numeral n)\<close>
     and divmod_step_def [simp]: \<open>divmod_step l (q, r) =
       (if euclidean_size l \<le> euclidean_size r then (2 * q + 1, r - l)
        else (2 * q, r))\<close> \<comment> \<open>
          This is a formulation of one step (referring to one digit position)
          in school-method division: compare the dividend at the current
          digit position with the remainder from previous division steps
          and evaluate accordingly.\<close>
 begin
 
 lemma fst_divmod:
   \<open>fst (divmod m n) = numeral m div numeral n\<close>
   by (simp add: divmod_def)
 
 lemma snd_divmod:
   \<open>snd (divmod m n) = numeral m mod numeral n\<close>
   by (simp add: divmod_def)
 
 text \<open>
   Following a formulation of school-method division.
   If the divisor is smaller than the dividend, terminate.
   If not, shift the dividend to the right until termination
   occurs and then reiterate single division steps in the
   opposite direction.
 \<close>
 
 lemma divmod_divmod_step:
   \<open>divmod m n = (if m < n then (0, numeral m)
     else divmod_step (numeral n) (divmod m (Num.Bit0 n)))\<close>
 proof (cases \<open>m < n\<close>)
   case True
   then show ?thesis
     by (simp add: prod_eq_iff fst_divmod snd_divmod flip: of_nat_numeral of_nat_div of_nat_mod)
 next
   case False
   define r s t where \<open>r = (numeral m :: nat)\<close> \<open>s = (numeral n :: nat)\<close> \<open>t = 2 * s\<close>
   then have *: \<open>numeral m = of_nat r\<close> \<open>numeral n = of_nat s\<close> \<open>numeral (num.Bit0 n) = of_nat t\<close>
     and \<open>\<not> s \<le> r mod s\<close>
     by (simp_all add: not_le)
   have t: \<open>2 * (r div t) = r div s - r div s mod 2\<close>
     \<open>r mod t = s * (r div s mod 2) + r mod s\<close>
     by (simp add: Rings.minus_mod_eq_mult_div Groups.mult.commute [of 2] Euclidean_Division.div_mult2_eq \<open>t = 2 * s\<close>)
       (use mod_mult2_eq [of r s 2] in \<open>simp add: ac_simps \<open>t = 2 * s\<close>\<close>)
   have rs: \<open>r div s mod 2 = 0 \<or> r div s mod 2 = Suc 0\<close>
     by auto
   from \<open>\<not> s \<le> r mod s\<close> have \<open>s \<le> r mod t \<Longrightarrow>
      r div s = Suc (2 * (r div t)) \<and>
      r mod s = r mod t - s\<close>
     using rs
     by (auto simp add: t)
   moreover have \<open>r mod t < s \<Longrightarrow>
      r div s = 2 * (r div t) \<and>
      r mod s = r mod t\<close>
     using rs
     by (auto simp add: t)
   ultimately show ?thesis
     by (simp add: divmod_def prod_eq_iff split_def Let_def
         not_less mod_eq_0_iff_dvd Rings.mod_eq_0_iff_dvd False not_le *)
     (simp add: flip: of_nat_numeral of_nat_mult add.commute [of 1] of_nat_div of_nat_mod of_nat_Suc of_nat_diff)
 qed
 
 text \<open>The division rewrite proper -- first, trivial results involving \<open>1\<close>\<close>
 
 lemma divmod_trivial [simp]:
   "divmod m Num.One = (numeral m, 0)"
   "divmod num.One (num.Bit0 n) = (0, Numeral1)"
   "divmod num.One (num.Bit1 n) = (0, Numeral1)"
   using divmod_divmod_step [of "Num.One"] by (simp_all add: divmod_def)
 
 text \<open>Division by an even number is a right-shift\<close>
 
 lemma divmod_cancel [simp]:
   \<open>divmod (Num.Bit0 m) (Num.Bit0 n) = (case divmod m n of (q, r) \<Rightarrow> (q, 2 * r))\<close> (is ?P)
   \<open>divmod (Num.Bit1 m) (Num.Bit0 n) = (case divmod m n of (q, r) \<Rightarrow> (q, 2 * r + 1))\<close> (is ?Q)
 proof -
   define r s where \<open>r = (numeral m :: nat)\<close> \<open>s = (numeral n :: nat)\<close>
   then have *: \<open>numeral m = of_nat r\<close> \<open>numeral n = of_nat s\<close>
     \<open>numeral (num.Bit0 m) = of_nat (2 * r)\<close> \<open>numeral (num.Bit0 n) = of_nat (2 * s)\<close>
     \<open>numeral (num.Bit1 m) = of_nat (Suc (2 * r))\<close>
     by simp_all
   have **: \<open>Suc (2 * r) div 2 = r\<close>
     by simp
   show ?P and ?Q
     by (simp_all add: divmod_def *)
       (simp_all flip: of_nat_numeral of_nat_div of_nat_mod of_nat_mult add.commute [of 1] of_nat_Suc
        add: Euclidean_Division.mod_mult_mult1 div_mult2_eq [of _ 2] mod_mult2_eq [of _ 2] **)
 qed
 
 text \<open>The really hard work\<close>
 
 lemma divmod_steps [simp]:
   "divmod (num.Bit0 m) (num.Bit1 n) =
       (if m \<le> n then (0, numeral (num.Bit0 m))
        else divmod_step (numeral (num.Bit1 n))
              (divmod (num.Bit0 m)
                (num.Bit0 (num.Bit1 n))))"
   "divmod (num.Bit1 m) (num.Bit1 n) =
       (if m < n then (0, numeral (num.Bit1 m))
        else divmod_step (numeral (num.Bit1 n))
              (divmod (num.Bit1 m)
                (num.Bit0 (num.Bit1 n))))"
   by (simp_all add: divmod_divmod_step)
 
 lemmas divmod_algorithm_code = divmod_trivial divmod_cancel divmod_steps
 
 text \<open>Special case: divisibility\<close>
 
 definition divides_aux :: "'a \<times> 'a \<Rightarrow> bool"
 where
   "divides_aux qr \<longleftrightarrow> snd qr = 0"
 
 lemma divides_aux_eq [simp]:
   "divides_aux (q, r) \<longleftrightarrow> r = 0"
   by (simp add: divides_aux_def)
 
 lemma dvd_numeral_simp [simp]:
   "numeral m dvd numeral n \<longleftrightarrow> divides_aux (divmod n m)"
   by (simp add: divmod_def mod_eq_0_iff_dvd)
 
 text \<open>Generic computation of quotient and remainder\<close>
 
 lemma numeral_div_numeral [simp]:
   "numeral k div numeral l = fst (divmod k l)"
   by (simp add: fst_divmod)
 
 lemma numeral_mod_numeral [simp]:
   "numeral k mod numeral l = snd (divmod k l)"
   by (simp add: snd_divmod)
 
 lemma one_div_numeral [simp]:
   "1 div numeral n = fst (divmod num.One n)"
   by (simp add: fst_divmod)
 
 lemma one_mod_numeral [simp]:
   "1 mod numeral n = snd (divmod num.One n)"
   by (simp add: snd_divmod)
 
 end
 
 instantiation nat :: unique_euclidean_semiring_with_nat_division
 begin
 
 definition divmod_nat :: "num \<Rightarrow> num \<Rightarrow> nat \<times> nat"
 where
   divmod'_nat_def: "divmod_nat m n = (numeral m div numeral n, numeral m mod numeral n)"
 
 definition divmod_step_nat :: "nat \<Rightarrow> nat \<times> nat \<Rightarrow> nat \<times> nat"
 where
   "divmod_step_nat l qr = (let (q, r) = qr
     in if r \<ge> l then (2 * q + 1, r - l)
     else (2 * q, r))"
 
 instance
   by standard (simp_all add: divmod'_nat_def divmod_step_nat_def)
 
 end
 
 declare divmod_algorithm_code [where ?'a = nat, code]
 
 lemma Suc_0_div_numeral [simp]:
   \<open>Suc 0 div numeral Num.One = 1\<close>
   \<open>Suc 0 div numeral (Num.Bit0 n) = 0\<close>
   \<open>Suc 0 div numeral (Num.Bit1 n) = 0\<close>
   by simp_all
 
 lemma Suc_0_mod_numeral [simp]:
   \<open>Suc 0 mod numeral Num.One = 0\<close>
   \<open>Suc 0 mod numeral (Num.Bit0 n) = 1\<close>
   \<open>Suc 0 mod numeral (Num.Bit1 n) = 1\<close>
   by simp_all
 
 instantiation int :: unique_euclidean_semiring_with_nat_division
 begin
 
 definition divmod_int :: "num \<Rightarrow> num \<Rightarrow> int \<times> int"
 where
   "divmod_int m n = (numeral m div numeral n, numeral m mod numeral n)"
 
 definition divmod_step_int :: "int \<Rightarrow> int \<times> int \<Rightarrow> int \<times> int"
 where
   "divmod_step_int l qr = (let (q, r) = qr
     in if \<bar>l\<bar> \<le> \<bar>r\<bar> then (2 * q + 1, r - l)
     else (2 * q, r))"
 
 instance
   by standard (auto simp add: divmod_int_def divmod_step_int_def)
 
 end
 
 declare divmod_algorithm_code [where ?'a = int, code]
 
 context
 begin
 
 qualified definition adjust_div :: "int \<times> int \<Rightarrow> int"
 where
   "adjust_div qr = (let (q, r) = qr in q + of_bool (r \<noteq> 0))"
 
 qualified lemma adjust_div_eq [simp, code]:
   "adjust_div (q, r) = q + of_bool (r \<noteq> 0)"
   by (simp add: adjust_div_def)
 
 qualified definition adjust_mod :: "num \<Rightarrow> int \<Rightarrow> int"
 where
   [simp]: "adjust_mod l r = (if r = 0 then 0 else numeral l - r)"
 
 lemma minus_numeral_div_numeral [simp]:
   "- numeral m div numeral n = - (adjust_div (divmod m n) :: int)"
 proof -
   have "int (fst (divmod m n)) = fst (divmod m n)"
     by (simp only: fst_divmod divide_int_def) auto
   then show ?thesis
     by (auto simp add: split_def Let_def adjust_div_def divides_aux_def divide_int_def)
 qed
 
 lemma minus_numeral_mod_numeral [simp]:
   "- numeral m mod numeral n = adjust_mod n (snd (divmod m n) :: int)"
 proof (cases "snd (divmod m n) = (0::int)")
   case True
   then show ?thesis
     by (simp add: mod_eq_0_iff_dvd divides_aux_def)
 next
   case False
   then have "int (snd (divmod m n)) = snd (divmod m n)" if "snd (divmod m n) \<noteq> (0::int)"
     by (simp only: snd_divmod modulo_int_def) auto
   then show ?thesis
     by (simp add: divides_aux_def adjust_div_def)
       (simp add: divides_aux_def modulo_int_def)
 qed
 
 lemma numeral_div_minus_numeral [simp]:
   "numeral m div - numeral n = - (adjust_div (divmod m n) :: int)"
 proof -
   have "int (fst (divmod m n)) = fst (divmod m n)"
     by (simp only: fst_divmod divide_int_def) auto
   then show ?thesis
     by (auto simp add: split_def Let_def adjust_div_def divides_aux_def divide_int_def)
 qed
 
 lemma numeral_mod_minus_numeral [simp]:
   "numeral m mod - numeral n = - adjust_mod n (snd (divmod m n) :: int)"
 proof (cases "snd (divmod m n) = (0::int)")
   case True
   then show ?thesis
     by (simp add: mod_eq_0_iff_dvd divides_aux_def)
 next
   case False
   then have "int (snd (divmod m n)) = snd (divmod m n)" if "snd (divmod m n) \<noteq> (0::int)"
     by (simp only: snd_divmod modulo_int_def) auto
   then show ?thesis
     by (simp add: divides_aux_def adjust_div_def)
       (simp add: divides_aux_def modulo_int_def)
 qed
 
 lemma minus_one_div_numeral [simp]:
   "- 1 div numeral n = - (adjust_div (divmod Num.One n) :: int)"
   using minus_numeral_div_numeral [of Num.One n] by simp
 
 lemma minus_one_mod_numeral [simp]:
   "- 1 mod numeral n = adjust_mod n (snd (divmod Num.One n) :: int)"
   using minus_numeral_mod_numeral [of Num.One n] by simp
 
 lemma one_div_minus_numeral [simp]:
   "1 div - numeral n = - (adjust_div (divmod Num.One n) :: int)"
   using numeral_div_minus_numeral [of Num.One n] by simp
 
 lemma one_mod_minus_numeral [simp]:
   "1 mod - numeral n = - adjust_mod n (snd (divmod Num.One n) :: int)"
   using numeral_mod_minus_numeral [of Num.One n] by simp
 
 lemma [code]:
   fixes k :: int
   shows
     "k div 0 = 0"
     "k mod 0 = k"
     "0 div k = 0"
     "0 mod k = 0"
     "k div Int.Pos Num.One = k"
     "k mod Int.Pos Num.One = 0"
     "k div Int.Neg Num.One = - k"
     "k mod Int.Neg Num.One = 0"
     "Int.Pos m div Int.Pos n = (fst (divmod m n) :: int)"
     "Int.Pos m mod Int.Pos n = (snd (divmod m n) :: int)"
     "Int.Neg m div Int.Pos n = - (adjust_div (divmod m n) :: int)"
     "Int.Neg m mod Int.Pos n = adjust_mod n (snd (divmod m n) :: int)"
     "Int.Pos m div Int.Neg n = - (adjust_div (divmod m n) :: int)"
     "Int.Pos m mod Int.Neg n = - adjust_mod n (snd (divmod m n) :: int)"
     "Int.Neg m div Int.Neg n = (fst (divmod m n) :: int)"
     "Int.Neg m mod Int.Neg n = - (snd (divmod m n) :: int)"
   by simp_all
 
 end
 
 lemma divmod_BitM_2_eq [simp]:
   \<open>divmod (Num.BitM m) (Num.Bit0 Num.One) = (numeral m - 1, (1 :: int))\<close>
   by (cases m) simp_all
 
 
 subsubsection \<open>Computation by simplification\<close>
 
 lemma euclidean_size_nat_less_eq_iff:
   \<open>euclidean_size m \<le> euclidean_size n \<longleftrightarrow> m \<le> n\<close> for m n :: nat
   by simp
 
 lemma euclidean_size_int_less_eq_iff:
   \<open>euclidean_size k \<le> euclidean_size l \<longleftrightarrow> \<bar>k\<bar> \<le> \<bar>l\<bar>\<close> for k l :: int
   by auto
 
 simproc_setup numeral_divmod
   ("0 div 0 :: 'a :: unique_euclidean_semiring_with_nat_division" | "0 mod 0 :: 'a :: unique_euclidean_semiring_with_nat_division" |
    "0 div 1 :: 'a :: unique_euclidean_semiring_with_nat_division" | "0 mod 1 :: 'a :: unique_euclidean_semiring_with_nat_division" |
    "0 div - 1 :: int" | "0 mod - 1 :: int" |
    "0 div numeral b :: 'a :: unique_euclidean_semiring_with_nat_division" | "0 mod numeral b :: 'a :: unique_euclidean_semiring_with_nat_division" |
    "0 div - numeral b :: int" | "0 mod - numeral b :: int" |
    "1 div 0 :: 'a :: unique_euclidean_semiring_with_nat_division" | "1 mod 0 :: 'a :: unique_euclidean_semiring_with_nat_division" |
    "1 div 1 :: 'a :: unique_euclidean_semiring_with_nat_division" | "1 mod 1 :: 'a :: unique_euclidean_semiring_with_nat_division" |
    "1 div - 1 :: int" | "1 mod - 1 :: int" |
    "1 div numeral b :: 'a :: unique_euclidean_semiring_with_nat_division" | "1 mod numeral b :: 'a :: unique_euclidean_semiring_with_nat_division" |
    "1 div - numeral b :: int" |"1 mod - numeral b :: int" |
    "- 1 div 0 :: int" | "- 1 mod 0 :: int" | "- 1 div 1 :: int" | "- 1 mod 1 :: int" |
    "- 1 div - 1 :: int" | "- 1 mod - 1 :: int" | "- 1 div numeral b :: int" | "- 1 mod numeral b :: int" |
    "- 1 div - numeral b :: int" | "- 1 mod - numeral b :: int" |
    "numeral a div 0 :: 'a :: unique_euclidean_semiring_with_nat_division" | "numeral a mod 0 :: 'a :: unique_euclidean_semiring_with_nat_division" |
    "numeral a div 1 :: 'a :: unique_euclidean_semiring_with_nat_division" | "numeral a mod 1 :: 'a :: unique_euclidean_semiring_with_nat_division" |
    "numeral a div - 1 :: int" | "numeral a mod - 1 :: int" |
    "numeral a div numeral b :: 'a :: unique_euclidean_semiring_with_nat_division" | "numeral a mod numeral b :: 'a :: unique_euclidean_semiring_with_nat_division" |
    "numeral a div - numeral b :: int" | "numeral a mod - numeral b :: int" |
    "- numeral a div 0 :: int" | "- numeral a mod 0 :: int" |
    "- numeral a div 1 :: int" | "- numeral a mod 1 :: int" |
    "- numeral a div - 1 :: int" | "- numeral a mod - 1 :: int" |
    "- numeral a div numeral b :: int" | "- numeral a mod numeral b :: int" |
    "- numeral a div - numeral b :: int" | "- numeral a mod - numeral b :: int") = \<open>
   let
     val if_cong = the (Code.get_case_cong \<^theory> \<^const_name>\<open>If\<close>);
     fun successful_rewrite ctxt ct =
       let
         val thm = Simplifier.rewrite ctxt ct
       in if Thm.is_reflexive thm then NONE else SOME thm end;
   in fn phi =>
     let
       val simps = Morphism.fact phi (@{thms div_0 mod_0 div_by_0 mod_by_0 div_by_1 mod_by_1
         one_div_numeral one_mod_numeral minus_one_div_numeral minus_one_mod_numeral
         one_div_minus_numeral one_mod_minus_numeral
         numeral_div_numeral numeral_mod_numeral minus_numeral_div_numeral minus_numeral_mod_numeral
         numeral_div_minus_numeral numeral_mod_minus_numeral
         div_minus_minus mod_minus_minus Euclidean_Division.adjust_div_eq of_bool_eq one_neq_zero
         numeral_neq_zero neg_equal_0_iff_equal arith_simps arith_special divmod_trivial
         divmod_cancel divmod_steps divmod_step_def fst_conv snd_conv numeral_One
         case_prod_beta rel_simps Euclidean_Division.adjust_mod_def div_minus1_right mod_minus1_right
         minus_minus numeral_times_numeral mult_zero_right mult_1_right
         euclidean_size_nat_less_eq_iff euclidean_size_int_less_eq_iff diff_nat_numeral nat_numeral}
         @ [@{lemma "0 = 0 \<longleftrightarrow> True" by simp}]);
       fun prepare_simpset ctxt = HOL_ss |> Simplifier.simpset_map ctxt
         (Simplifier.add_cong if_cong #> fold Simplifier.add_simp simps)
     in fn ctxt => successful_rewrite (Simplifier.put_simpset (prepare_simpset ctxt) ctxt) end
   end
 \<close> \<comment> \<open>
   There is space for improvement here: the calculation itself
   could be carried out outside the logic, and a generic simproc
   (simplifier setup) for generic calculation would be helpful.
 \<close>
 
 
 subsubsection \<open>Code generation\<close>
 
 context
 begin
 
 qualified definition divmod_nat :: "nat \<Rightarrow> nat \<Rightarrow> nat \<times> nat"
   where "divmod_nat m n = (m div n, m mod n)"
 
 qualified lemma divmod_nat_if [code]:
   "divmod_nat m n = (if n = 0 \<or> m < n then (0, m) else
     let (q, r) = divmod_nat (m - n) n in (Suc q, r))"
   by (simp add: divmod_nat_def prod_eq_iff case_prod_beta not_less le_div_geq le_mod_geq)
 
 qualified lemma [code]:
   "m div n = fst (divmod_nat m n)"
   "m mod n = snd (divmod_nat m n)"
   by (simp_all add: divmod_nat_def)
 
 end
 
 code_identifier
   code_module Euclidean_Division \<rightharpoonup> (SML) Arith and (OCaml) Arith and (Haskell) Arith
 
 end