diff --git a/src/HOL/Parity.thy b/src/HOL/Parity.thy
--- a/src/HOL/Parity.thy
+++ b/src/HOL/Parity.thy
@@ -1,1381 +1,1380 @@
 (*  Title:      HOL/Parity.thy
     Author:     Jeremy Avigad
     Author:     Jacques D. Fleuriot
 *)
 
 section \<open>Parity in rings and semirings\<close>
 
 theory Parity
   imports Euclidean_Rings
 begin
 
 subsection \<open>Ring structures with parity and \<open>even\<close>/\<open>odd\<close> predicates\<close>
 
 class semiring_parity = comm_semiring_1 + semiring_modulo +
   assumes even_iff_mod_2_eq_zero: "2 dvd a \<longleftrightarrow> a mod 2 = 0"
     and odd_iff_mod_2_eq_one: "\<not> 2 dvd a \<longleftrightarrow> a mod 2 = 1"
     and odd_one [simp]: "\<not> 2 dvd 1"
 begin
 
 abbreviation even :: "'a \<Rightarrow> bool"
   where "even a \<equiv> 2 dvd a"
 
 abbreviation odd :: "'a \<Rightarrow> bool"
   where "odd a \<equiv> \<not> 2 dvd a"
 
 end
 
 class ring_parity = ring + semiring_parity
 begin
 
 subclass comm_ring_1 ..
 
 end
 
 instance nat :: semiring_parity
   by standard (simp_all add: dvd_eq_mod_eq_0)
 
 instance int :: ring_parity
   by standard (auto simp add: dvd_eq_mod_eq_0)
 
 context semiring_parity
 begin
 
 lemma parity_cases [case_names even odd]:
   assumes "even a \<Longrightarrow> a mod 2 = 0 \<Longrightarrow> P"
   assumes "odd a \<Longrightarrow> a mod 2 = 1 \<Longrightarrow> P"
   shows P
   using assms by (cases "even a")
     (simp_all add: even_iff_mod_2_eq_zero [symmetric] odd_iff_mod_2_eq_one [symmetric])
 
 lemma odd_of_bool_self [simp]:
   \<open>odd (of_bool p) \<longleftrightarrow> p\<close>
   by (cases p) simp_all
 
 lemma not_mod_2_eq_0_eq_1 [simp]:
   "a mod 2 \<noteq> 0 \<longleftrightarrow> a mod 2 = 1"
   by (cases a rule: parity_cases) simp_all
 
 lemma not_mod_2_eq_1_eq_0 [simp]:
   "a mod 2 \<noteq> 1 \<longleftrightarrow> a mod 2 = 0"
   by (cases a rule: parity_cases) simp_all
 
 lemma evenE [elim?]:
   assumes "even a"
   obtains b where "a = 2 * b"
   using assms by (rule dvdE)
 
 lemma oddE [elim?]:
   assumes "odd a"
   obtains b where "a = 2 * b + 1"
 proof -
   have "a = 2 * (a div 2) + a mod 2"
     by (simp add: mult_div_mod_eq)
   with assms have "a = 2 * (a div 2) + 1"
     by (simp add: odd_iff_mod_2_eq_one)
   then show ?thesis ..
 qed
 
 lemma mod_2_eq_odd:
   "a mod 2 = of_bool (odd a)"
   by (auto elim: oddE simp add: even_iff_mod_2_eq_zero)
 
 lemma of_bool_odd_eq_mod_2:
   "of_bool (odd a) = a mod 2"
   by (simp add: mod_2_eq_odd)
 
 lemma even_mod_2_iff [simp]:
   \<open>even (a mod 2) \<longleftrightarrow> even a\<close>
   by (simp add: mod_2_eq_odd)
 
 lemma mod2_eq_if:
   "a mod 2 = (if even a then 0 else 1)"
   by (simp add: mod_2_eq_odd)
 
 lemma even_zero [simp]:
   "even 0"
   by (fact dvd_0_right)
 
 lemma odd_even_add:
   "even (a + b)" if "odd a" and "odd b"
 proof -
   from that obtain c d where "a = 2 * c + 1" and "b = 2 * d + 1"
     by (blast elim: oddE)
   then have "a + b = 2 * c + 2 * d + (1 + 1)"
     by (simp only: ac_simps)
   also have "\<dots> = 2 * (c + d + 1)"
     by (simp add: algebra_simps)
   finally show ?thesis ..
 qed
 
 lemma even_add [simp]:
   "even (a + b) \<longleftrightarrow> (even a \<longleftrightarrow> even b)"
   by (auto simp add: dvd_add_right_iff dvd_add_left_iff odd_even_add)
 
 lemma odd_add [simp]:
   "odd (a + b) \<longleftrightarrow> \<not> (odd a \<longleftrightarrow> odd b)"
   by simp
 
 lemma even_plus_one_iff [simp]:
   "even (a + 1) \<longleftrightarrow> odd a"
   by (auto simp add: dvd_add_right_iff intro: odd_even_add)
 
 lemma even_mult_iff [simp]:
   "even (a * b) \<longleftrightarrow> even a \<or> even b" (is "?P \<longleftrightarrow> ?Q")
 proof
   assume ?Q
   then show ?P
     by auto
 next
   assume ?P
   show ?Q
   proof (rule ccontr)
     assume "\<not> (even a \<or> even b)"
     then have "odd a" and "odd b"
       by auto
     then obtain r s where "a = 2 * r + 1" and "b = 2 * s + 1"
       by (blast elim: oddE)
     then have "a * b = (2 * r + 1) * (2 * s + 1)"
       by simp
     also have "\<dots> = 2 * (2 * r * s + r + s) + 1"
       by (simp add: algebra_simps)
     finally have "odd (a * b)"
       by simp
     with \<open>?P\<close> show False
       by auto
   qed
 qed
 
 lemma even_numeral [simp]: "even (numeral (Num.Bit0 n))"
 proof -
   have "even (2 * numeral n)"
     unfolding even_mult_iff by simp
   then have "even (numeral n + numeral n)"
     unfolding mult_2 .
   then show ?thesis
     unfolding numeral.simps .
 qed
 
 lemma odd_numeral [simp]: "odd (numeral (Num.Bit1 n))"
 proof
   assume "even (numeral (num.Bit1 n))"
   then have "even (numeral n + numeral n + 1)"
     unfolding numeral.simps .
   then have "even (2 * numeral n + 1)"
     unfolding mult_2 .
   then have "2 dvd numeral n * 2 + 1"
     by (simp add: ac_simps)
   then have "2 dvd 1"
     using dvd_add_times_triv_left_iff [of 2 "numeral n" 1] by simp
   then show False by simp
 qed
 
 lemma odd_numeral_BitM [simp]:
   \<open>odd (numeral (Num.BitM w))\<close>
   by (cases w) simp_all
 
 lemma even_power [simp]: "even (a ^ n) \<longleftrightarrow> even a \<and> n > 0"
   by (induct n) auto
 
 lemma even_prod_iff:
   \<open>even (prod f A) \<longleftrightarrow> (\<exists>a\<in>A. even (f a))\<close> if \<open>finite A\<close>
   using that by (induction A) simp_all
 
 lemma mask_eq_sum_exp:
   \<open>2 ^ n - 1 = (\<Sum>m\<in>{q. q < n}. 2 ^ m)\<close>
 proof -
   have *: \<open>{q. q < Suc m} = insert m {q. q < m}\<close> for m
     by auto
   have \<open>2 ^ n = (\<Sum>m\<in>{q. q < n}. 2 ^ m) + 1\<close>
     by (induction n) (simp_all add: ac_simps mult_2 *)
   then have \<open>2 ^ n - 1 = (\<Sum>m\<in>{q. q < n}. 2 ^ m) + 1 - 1\<close>
     by simp
   then show ?thesis
     by simp
 qed
 
 lemma (in -) mask_eq_sum_exp_nat:
   \<open>2 ^ n - Suc 0 = (\<Sum>m\<in>{q. q < n}. 2 ^ m)\<close>
   using mask_eq_sum_exp [where ?'a = nat] by simp
 
 end
 
 context ring_parity
 begin
 
 lemma even_minus:
   "even (- a) \<longleftrightarrow> even a"
   by (fact dvd_minus_iff)
 
 lemma even_diff [simp]:
   "even (a - b) \<longleftrightarrow> even (a + b)"
   using even_add [of a "- b"] by simp
 
 end
 
 
 subsection \<open>Instance for \<^typ>\<open>nat\<close>\<close>
 
 lemma even_Suc_Suc_iff [simp]:
   "even (Suc (Suc n)) \<longleftrightarrow> even n"
   using dvd_add_triv_right_iff [of 2 n] by simp
 
 lemma even_Suc [simp]: "even (Suc n) \<longleftrightarrow> odd n"
   using even_plus_one_iff [of n] by simp
 
 lemma even_diff_nat [simp]:
   "even (m - n) \<longleftrightarrow> m < n \<or> even (m + n)" for m n :: nat
 proof (cases "n \<le> m")
   case True
   then have "m - n + n * 2 = m + n" by (simp add: mult_2_right)
   moreover have "even (m - n) \<longleftrightarrow> even (m - n + n * 2)" by simp
   ultimately have "even (m - n) \<longleftrightarrow> even (m + n)" by (simp only:)
   then show ?thesis by auto
 next
   case False
   then show ?thesis by simp
 qed
 
 lemma odd_pos:
   "odd n \<Longrightarrow> 0 < n" for n :: nat
   by (auto elim: oddE)
 
 lemma Suc_double_not_eq_double:
   "Suc (2 * m) \<noteq> 2 * n"
 proof
   assume "Suc (2 * m) = 2 * n"
   moreover have "odd (Suc (2 * m))" and "even (2 * n)"
     by simp_all
   ultimately show False by simp
 qed
 
 lemma double_not_eq_Suc_double:
   "2 * m \<noteq> Suc (2 * n)"
   using Suc_double_not_eq_double [of n m] by simp
 
 lemma odd_Suc_minus_one [simp]: "odd n \<Longrightarrow> Suc (n - Suc 0) = n"
   by (auto elim: oddE)
 
 lemma even_Suc_div_two [simp]:
   "even n \<Longrightarrow> Suc n div 2 = n div 2"
   by auto
 
 lemma odd_Suc_div_two [simp]:
   "odd n \<Longrightarrow> Suc n div 2 = Suc (n div 2)"
   by (auto elim: oddE)
 
 lemma odd_two_times_div_two_nat [simp]:
   assumes "odd n"
   shows "2 * (n div 2) = n - (1 :: nat)"
 proof -
   from assms have "2 * (n div 2) + 1 = n"
     by (auto elim: oddE)
   then have "Suc (2 * (n div 2)) - 1 = n - 1"
     by simp
   then show ?thesis
     by simp
 qed
 
 lemma not_mod2_eq_Suc_0_eq_0 [simp]:
   "n mod 2 \<noteq> Suc 0 \<longleftrightarrow> n mod 2 = 0"
   using not_mod_2_eq_1_eq_0 [of n] by simp
 
 lemma odd_card_imp_not_empty:
   \<open>A \<noteq> {}\<close> if \<open>odd (card A)\<close>
   using that by auto
 
 lemma nat_induct2 [case_names 0 1 step]:
   assumes "P 0" "P 1" and step: "\<And>n::nat. P n \<Longrightarrow> P (n + 2)"
   shows "P n"
 proof (induct n rule: less_induct)
   case (less n)
   show ?case
   proof (cases "n < Suc (Suc 0)")
     case True
     then show ?thesis
       using assms by (auto simp: less_Suc_eq)
   next
     case False
     then obtain k where k: "n = Suc (Suc k)"
       by (force simp: not_less nat_le_iff_add)
     then have "k<n"
       by simp
     with less assms have "P (k+2)"
       by blast
     then show ?thesis
       by (simp add: k)
   qed
 qed
 
 context semiring_parity
 begin
 
 lemma even_sum_iff:
   \<open>even (sum f A) \<longleftrightarrow> even (card {a\<in>A. odd (f a)})\<close> if \<open>finite A\<close>
 using that proof (induction A)
   case empty
   then show ?case
     by simp
 next
   case (insert a A)
   moreover have \<open>{b \<in> insert a A. odd (f b)} = (if odd (f a) then {a} else {}) \<union> {b \<in> A. odd (f b)}\<close>
     by auto
   ultimately show ?case
     by simp
 qed
 
 lemma even_mask_iff [simp]:
   \<open>even (2 ^ n - 1) \<longleftrightarrow> n = 0\<close>
 proof (cases \<open>n = 0\<close>)
   case True
   then show ?thesis
     by simp
 next
   case False
   then have \<open>{a. a = 0 \<and> a < n} = {0}\<close>
     by auto
   then show ?thesis
     by (auto simp add: mask_eq_sum_exp even_sum_iff)
 qed
 
 lemma even_of_nat_iff [simp]:
   "even (of_nat n) \<longleftrightarrow> even n"
   by (induction n) simp_all
 
 end
 
 
 subsection \<open>Parity and powers\<close>
 
 context ring_1
 begin
 
 lemma power_minus_even [simp]: "even n \<Longrightarrow> (- a) ^ n = a ^ n"
   by (auto elim: evenE)
 
 lemma power_minus_odd [simp]: "odd n \<Longrightarrow> (- a) ^ n = - (a ^ n)"
   by (auto elim: oddE)
 
 lemma uminus_power_if:
   "(- a) ^ n = (if even n then a ^ n else - (a ^ n))"
   by auto
 
 lemma neg_one_even_power [simp]: "even n \<Longrightarrow> (- 1) ^ n = 1"
   by simp
 
 lemma neg_one_odd_power [simp]: "odd n \<Longrightarrow> (- 1) ^ n = - 1"
   by simp
 
 lemma neg_one_power_add_eq_neg_one_power_diff: "k \<le> n \<Longrightarrow> (- 1) ^ (n + k) = (- 1) ^ (n - k)"
   by (cases "even (n + k)") auto
 
 lemma minus_one_power_iff: "(- 1) ^ n = (if even n then 1 else - 1)"
   by (induct n) auto
 
 end
 
 context linordered_idom
 begin
 
 lemma zero_le_even_power: "even n \<Longrightarrow> 0 \<le> a ^ n"
   by (auto elim: evenE)
 
 lemma zero_le_odd_power: "odd n \<Longrightarrow> 0 \<le> a ^ n \<longleftrightarrow> 0 \<le> a"
   by (auto simp add: power_even_eq zero_le_mult_iff elim: oddE)
 
 lemma zero_le_power_eq: "0 \<le> a ^ n \<longleftrightarrow> even n \<or> odd n \<and> 0 \<le> a"
   by (auto simp add: zero_le_even_power zero_le_odd_power)
 
 lemma zero_less_power_eq: "0 < a ^ n \<longleftrightarrow> n = 0 \<or> even n \<and> a \<noteq> 0 \<or> odd n \<and> 0 < a"
 proof -
   have [simp]: "0 = a ^ n \<longleftrightarrow> a = 0 \<and> n > 0"
     unfolding power_eq_0_iff [of a n, symmetric] by blast
   show ?thesis
     unfolding less_le zero_le_power_eq by auto
 qed
 
 lemma power_less_zero_eq [simp]: "a ^ n < 0 \<longleftrightarrow> odd n \<and> a < 0"
   unfolding not_le [symmetric] zero_le_power_eq by auto
 
 lemma power_le_zero_eq: "a ^ n \<le> 0 \<longleftrightarrow> n > 0 \<and> (odd n \<and> a \<le> 0 \<or> even n \<and> a = 0)"
   unfolding not_less [symmetric] zero_less_power_eq by auto
 
 lemma power_even_abs: "even n \<Longrightarrow> \<bar>a\<bar> ^ n = a ^ n"
   using power_abs [of a n] by (simp add: zero_le_even_power)
 
 lemma power_mono_even:
   assumes "even n" and "\<bar>a\<bar> \<le> \<bar>b\<bar>"
   shows "a ^ n \<le> b ^ n"
 proof -
   have "0 \<le> \<bar>a\<bar>" by auto
   with \<open>\<bar>a\<bar> \<le> \<bar>b\<bar>\<close> have "\<bar>a\<bar> ^ n \<le> \<bar>b\<bar> ^ n"
     by (rule power_mono)
   with \<open>even n\<close> show ?thesis
     by (simp add: power_even_abs)
 qed
 
 lemma power_mono_odd:
   assumes "odd n" and "a \<le> b"
   shows "a ^ n \<le> b ^ n"
 proof (cases "b < 0")
   case True
   with \<open>a \<le> b\<close> have "- b \<le> - a" and "0 \<le> - b" by auto
   then have "(- b) ^ n \<le> (- a) ^ n" by (rule power_mono)
   with \<open>odd n\<close> show ?thesis by simp
 next
   case False
   then have "0 \<le> b" by auto
   show ?thesis
   proof (cases "a < 0")
     case True
     then have "n \<noteq> 0" and "a \<le> 0" using \<open>odd n\<close> [THEN odd_pos] by auto
     then have "a ^ n \<le> 0" unfolding power_le_zero_eq using \<open>odd n\<close> by auto
     moreover from \<open>0 \<le> b\<close> have "0 \<le> b ^ n" by auto
     ultimately show ?thesis by auto
   next
     case False
     then have "0 \<le> a" by auto
     with \<open>a \<le> b\<close> show ?thesis
       using power_mono by auto
   qed
 qed
 
 text \<open>Simplify, when the exponent is a numeral\<close>
 
 lemma zero_le_power_eq_numeral [simp]:
   "0 \<le> a ^ numeral w \<longleftrightarrow> even (numeral w :: nat) \<or> odd (numeral w :: nat) \<and> 0 \<le> a"
   by (fact zero_le_power_eq)
 
 lemma zero_less_power_eq_numeral [simp]:
   "0 < a ^ numeral w \<longleftrightarrow>
     numeral w = (0 :: nat) \<or>
     even (numeral w :: nat) \<and> a \<noteq> 0 \<or>
     odd (numeral w :: nat) \<and> 0 < a"
   by (fact zero_less_power_eq)
 
 lemma power_le_zero_eq_numeral [simp]:
   "a ^ numeral w \<le> 0 \<longleftrightarrow>
     (0 :: nat) < numeral w \<and>
     (odd (numeral w :: nat) \<and> a \<le> 0 \<or> even (numeral w :: nat) \<and> a = 0)"
   by (fact power_le_zero_eq)
 
 lemma power_less_zero_eq_numeral [simp]:
   "a ^ numeral w < 0 \<longleftrightarrow> odd (numeral w :: nat) \<and> a < 0"
   by (fact power_less_zero_eq)
 
 lemma power_even_abs_numeral [simp]:
   "even (numeral w :: nat) \<Longrightarrow> \<bar>a\<bar> ^ numeral w = a ^ numeral w"
   by (fact power_even_abs)
 
 end
 
 
 subsection \<open>Instance for \<^typ>\<open>int\<close>\<close>
   
 lemma even_diff_iff:
   "even (k - l) \<longleftrightarrow> even (k + l)" for k l :: int
   by (fact even_diff)
 
 lemma even_abs_add_iff:
   "even (\<bar>k\<bar> + l) \<longleftrightarrow> even (k + l)" for k l :: int
   by simp
 
 lemma even_add_abs_iff:
   "even (k + \<bar>l\<bar>) \<longleftrightarrow> even (k + l)" for k l :: int
   by simp
 
 lemma even_nat_iff: "0 \<le> k \<Longrightarrow> even (nat k) \<longleftrightarrow> even k"
   by (simp add: even_of_nat_iff [of "nat k", where ?'a = int, symmetric])
 
 context
   assumes "SORT_CONSTRAINT('a::division_ring)"
 begin
 
 lemma power_int_minus_left:
   "power_int (-a :: 'a) n = (if even n then power_int a n else -power_int a n)"
   by (auto simp: power_int_def minus_one_power_iff even_nat_iff)
 
 lemma power_int_minus_left_even [simp]: "even n \<Longrightarrow> power_int (-a :: 'a) n = power_int a n"
   by (simp add: power_int_minus_left)
 
 lemma power_int_minus_left_odd [simp]: "odd n \<Longrightarrow> power_int (-a :: 'a) n = -power_int a n"
   by (simp add: power_int_minus_left)
 
 lemma power_int_minus_left_distrib:
   "NO_MATCH (-1) x \<Longrightarrow> power_int (-a :: 'a) n = power_int (-1) n * power_int a n"
   by (simp add: power_int_minus_left)
 
 lemma power_int_minus_one_minus: "power_int (-1 :: 'a) (-n) = power_int (-1) n"
   by (simp add: power_int_minus_left)
 
 lemma power_int_minus_one_diff_commute: "power_int (-1 :: 'a) (a - b) = power_int (-1) (b - a)"
   by (subst power_int_minus_one_minus [symmetric]) auto
 
 lemma power_int_minus_one_mult_self [simp]:
   "power_int (-1 :: 'a) m * power_int (-1) m = 1"
   by (simp add: power_int_minus_left)
 
 lemma power_int_minus_one_mult_self' [simp]:
   "power_int (-1 :: 'a) m * (power_int (-1) m * b) = b"
   by (simp add: power_int_minus_left)
 
 end
 
 
 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)
 
 
 context unique_euclidean_semiring_with_nat
 begin
 
 subclass semiring_parity
 proof
   show "2 dvd a \<longleftrightarrow> a mod 2 = 0" for a
     by (fact dvd_eq_mod_eq_0)
   show "\<not> 2 dvd a \<longleftrightarrow> a mod 2 = 1" for a
   proof
     assume "a mod 2 = 1"
     then show "\<not> 2 dvd a"
       by auto
   next
     assume "\<not> 2 dvd a"
     have eucl: "euclidean_size (a mod 2) = 1"
     proof (rule order_antisym)
       show "euclidean_size (a mod 2) \<le> 1"
         using mod_size_less [of 2 a] by simp
       show "1 \<le> euclidean_size (a mod 2)"
         using \<open>\<not> 2 dvd a\<close> by (simp add: Suc_le_eq dvd_eq_mod_eq_0)
     qed 
     from \<open>\<not> 2 dvd a\<close> have "\<not> of_nat 2 dvd division_segment a * of_nat (euclidean_size a)"
       by simp
     then have "\<not> of_nat 2 dvd of_nat (euclidean_size a)"
       by (auto simp only: dvd_mult_unit_iff' is_unit_division_segment)
     then have "\<not> 2 dvd euclidean_size a"
       using of_nat_dvd_iff [of 2] by simp
     then have "euclidean_size a mod 2 = 1"
       by (simp add: semidom_modulo_class.dvd_eq_mod_eq_0)
     then have "of_nat (euclidean_size a mod 2) = of_nat 1"
       by simp
     then have "of_nat (euclidean_size a) mod 2 = 1"
       by (simp add: of_nat_mod)
     from \<open>\<not> 2 dvd a\<close> eucl
     show "a mod 2 = 1"
       by (auto intro: division_segment_eq_iff simp add: division_segment_mod)
   qed
   show "\<not> is_unit 2"
   proof (rule notI)
     assume "is_unit 2"
     then have "of_nat 2 dvd of_nat 1"
       by simp
     then have "is_unit (2::nat)"
       by (simp only: of_nat_dvd_iff)
     then show False
       by simp
   qed
 qed
 
 lemma even_succ_div_two [simp]:
   "even a \<Longrightarrow> (a + 1) div 2 = a div 2"
   by (cases "a = 0") (auto elim!: evenE dest: mult_not_zero)
 
 lemma odd_succ_div_two [simp]:
   "odd a \<Longrightarrow> (a + 1) div 2 = a div 2 + 1"
   by (auto elim!: oddE simp add: add.assoc)
 
 lemma even_two_times_div_two:
   "even a \<Longrightarrow> 2 * (a div 2) = a"
   by (fact dvd_mult_div_cancel)
 
 lemma odd_two_times_div_two_succ [simp]:
   "odd a \<Longrightarrow> 2 * (a div 2) + 1 = a"
   using mult_div_mod_eq [of 2 a]
   by (simp add: even_iff_mod_2_eq_zero)
 
 lemma coprime_left_2_iff_odd [simp]:
   "coprime 2 a \<longleftrightarrow> odd a"
 proof
   assume "odd a"
   show "coprime 2 a"
   proof (rule coprimeI)
     fix b
     assume "b dvd 2" "b dvd a"
     then have "b dvd a mod 2"
       by (auto intro: dvd_mod)
     with \<open>odd a\<close> show "is_unit b"
       by (simp add: mod_2_eq_odd)
   qed
 next
   assume "coprime 2 a"
   show "odd a"
   proof (rule notI)
     assume "even a"
     then obtain b where "a = 2 * b" ..
     with \<open>coprime 2 a\<close> have "coprime 2 (2 * b)"
       by simp
     moreover have "\<not> coprime 2 (2 * b)"
       by (rule not_coprimeI [of 2]) simp_all
     ultimately show False
       by blast
   qed
 qed
 
 lemma coprime_right_2_iff_odd [simp]:
   "coprime a 2 \<longleftrightarrow> odd a"
   using coprime_left_2_iff_odd [of a] by (simp add: ac_simps)
 
 end
 
 context unique_euclidean_ring_with_nat
 begin
 
 subclass ring_parity ..
 
 lemma minus_1_mod_2_eq [simp]:
   "- 1 mod 2 = 1"
   by (simp add: mod_2_eq_odd)
 
 lemma minus_1_div_2_eq [simp]:
   "- 1 div 2 = - 1"
 proof -
   from div_mult_mod_eq [of "- 1" 2]
   have "- 1 div 2 * 2 = - 1 * 2"
     using add_implies_diff by fastforce
   then show ?thesis
     using mult_right_cancel [of 2 "- 1 div 2" "- 1"] by simp
 qed
 
 end
 
 context unique_euclidean_semiring_with_nat
 begin
 
 lemma even_mask_div_iff':
   \<open>even ((2 ^ m - 1) div 2 ^ n) \<longleftrightarrow> m \<le> n\<close>
 proof -
   have \<open>even ((2 ^ m - 1) div 2 ^ n) \<longleftrightarrow> even (of_nat ((2 ^ m - Suc 0) div 2 ^ n))\<close>
     by (simp only: of_nat_div) (simp add: of_nat_diff)
   also have \<open>\<dots> \<longleftrightarrow> even ((2 ^ m - Suc 0) div 2 ^ n)\<close>
     by simp
   also have \<open>\<dots> \<longleftrightarrow> m \<le> n\<close>
   proof (cases \<open>m \<le> n\<close>)
     case True
     then show ?thesis
       by (simp add: Suc_le_lessD)
   next
     case False
     then obtain r where r: \<open>m = n + Suc r\<close>
       using less_imp_Suc_add by fastforce
     from r have \<open>{q. q < m} \<inter> {q. 2 ^ n dvd (2::nat) ^ q} = {q. n \<le> q \<and> q < m}\<close>
       by (auto simp add: dvd_power_iff_le)
     moreover from r have \<open>{q. q < m} \<inter> {q. \<not> 2 ^ n dvd (2::nat) ^ q} = {q. q < n}\<close>
       by (auto simp add: dvd_power_iff_le)
     moreover from False have \<open>{q. n \<le> q \<and> q < m \<and> q \<le> n} = {n}\<close>
       by auto
     then have \<open>odd ((\<Sum>a\<in>{q. n \<le> q \<and> q < m}. 2 ^ a div (2::nat) ^ n) + sum ((^) 2) {q. q < n} div 2 ^ n)\<close>
       by (simp_all add: euclidean_semiring_cancel_class.power_diff_power_eq semiring_parity_class.even_sum_iff not_less mask_eq_sum_exp_nat [symmetric])
     ultimately have \<open>odd (sum ((^) (2::nat)) {q. q < m} div 2 ^ n)\<close>
       by (subst euclidean_semiring_cancel_class.sum_div_partition) simp_all
     with False show ?thesis
       by (simp add: mask_eq_sum_exp_nat)
   qed
   finally show ?thesis .
 qed
 
 end
 
 
 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_Rings.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_Rings.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 Parity.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 Parity.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
+    val simps = @{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 Parity.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 Parity.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
+    K (fn ctxt => successful_rewrite (Simplifier.put_simpset (prepare_simpset ctxt) ctxt))
   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>
 
 
 subsection \<open>Computing congruences modulo \<open>2 ^ q\<close>\<close>
 
 context unique_euclidean_semiring_with_nat_division
 begin
 
 lemma cong_exp_iff_simps:
   "numeral n mod numeral Num.One = 0
     \<longleftrightarrow> True"
   "numeral (Num.Bit0 n) mod numeral (Num.Bit0 q) = 0
     \<longleftrightarrow> numeral n mod numeral q = 0"
   "numeral (Num.Bit1 n) mod numeral (Num.Bit0 q) = 0
     \<longleftrightarrow> False"
   "numeral m mod numeral Num.One = (numeral n mod numeral Num.One)
     \<longleftrightarrow> True"
   "numeral Num.One mod numeral (Num.Bit0 q) = (numeral Num.One mod numeral (Num.Bit0 q))
     \<longleftrightarrow> True"
   "numeral Num.One mod numeral (Num.Bit0 q) = (numeral (Num.Bit0 n) mod numeral (Num.Bit0 q))
     \<longleftrightarrow> False"
   "numeral Num.One mod numeral (Num.Bit0 q) = (numeral (Num.Bit1 n) mod numeral (Num.Bit0 q))
     \<longleftrightarrow> (numeral n mod numeral q) = 0"
   "numeral (Num.Bit0 m) mod numeral (Num.Bit0 q) = (numeral Num.One mod numeral (Num.Bit0 q))
     \<longleftrightarrow> False"
   "numeral (Num.Bit0 m) mod numeral (Num.Bit0 q) = (numeral (Num.Bit0 n) mod numeral (Num.Bit0 q))
     \<longleftrightarrow> numeral m mod numeral q = (numeral n mod numeral q)"
   "numeral (Num.Bit0 m) mod numeral (Num.Bit0 q) = (numeral (Num.Bit1 n) mod numeral (Num.Bit0 q))
     \<longleftrightarrow> False"
   "numeral (Num.Bit1 m) mod numeral (Num.Bit0 q) = (numeral Num.One mod numeral (Num.Bit0 q))
     \<longleftrightarrow> (numeral m mod numeral q) = 0"
   "numeral (Num.Bit1 m) mod numeral (Num.Bit0 q) = (numeral (Num.Bit0 n) mod numeral (Num.Bit0 q))
     \<longleftrightarrow> False"
   "numeral (Num.Bit1 m) mod numeral (Num.Bit0 q) = (numeral (Num.Bit1 n) mod numeral (Num.Bit0 q))
     \<longleftrightarrow> numeral m mod numeral q = (numeral n mod numeral q)"
   by (auto simp add: case_prod_beta dest: arg_cong [of _ _ even])
 
 end
 
 
 code_identifier
   code_module Parity \<rightharpoonup> (SML) Arith and (OCaml) Arith and (Haskell) Arith
 
 lemmas even_of_nat = even_of_nat_iff
 
 end