Quelle Cong.thy Sprache: unbekannt

(*  Title:      HOL/Number_Theory/Cong.thy
    Author:     Christophe Tabacznyj
    Author:     Lawrence C. Paulson
    Author:     Amine Chaieb
    Author:     Thomas M. Rasmussen
    Author:     Jeremy Avigad

Defines congruence (notation: [x = y] (mod z)) for natural numbers and
integers.

This file combines and revises a number of prior developments.

The original theories "GCD" and "Primes" were by Christophe Tabacznyj
and Lawrence C. Paulson, based on @{cite davenport92}. They introduced
gcd, lcm, and prime for the natural numbers.

The original theory "IntPrimes" was by Thomas M. Rasmussen, and
extended gcd, lcm, primes to the integers. Amine Chaieb provided
another extension of the notions to the integers, and added a number
of results to "Primes" and "GCD".

The original theory, "IntPrimes", by Thomas M. Rasmussen, defined and
developed the congruence relations on the integers. The notion was
extended to the natural numbers by Chaieb. Jeremy Avigad combined
these, revised and tidied them, made the development uniform for the
natural numbers and the integers, and added a number of new theorems.
*)

section \<open>Congruence\<close>

theory Cong
  imports "HOL-Computational_Algebra.Primes"
begin

subsection \<open>Generic congruences\<close>

context unique_euclidean_semiring
begin

definition cong :: "'a \ 'a \ 'a \ bool"
    (\<open>(\<open>indent=1 notation=\<open>mixfix cong\<close>\<close>[_ = _] '(' mod _'))\<close>)
  where "[b = c] (mod a) \ b mod a = c mod a"

abbreviation notcong :: "'a \ 'a \ 'a \ bool"
    (\<open>(\<open>indent=1 notation=\<open>mixfix notcong\<close>\<close>[_ \<noteq> _] '(' mod _'))\<close>)
  where "[b \ c] (mod a) \ \ cong b c a"

lemma cong_refl [simp]:
  "[b = b] (mod a)"
  by (simp add: cong_def)

lemma cong_sym:
  "[b = c] (mod a) \ [c = b] (mod a)"
  by (simp add: cong_def)

lemma cong_sym_eq:
  "[b = c] (mod a) \ [c = b] (mod a)"
  by (auto simp add: cong_def)

lemma cong_trans [trans]:
  "[b = c] (mod a) \ [c = d] (mod a) \ [b = d] (mod a)"
  by (simp add: cong_def)

lemma cong_mult_self_right:
  "[b * a = 0] (mod a)"
  by (simp add: cong_def)

lemma cong_mult_self_left:
  "[a * b = 0] (mod a)"
  by (simp add: cong_def)

lemma cong_mod_left [simp]:
  "[b mod a = c] (mod a) \ [b = c] (mod a)"
  by (simp add: cong_def)

lemma cong_mod_right [simp]:
  "[b = c mod a] (mod a) \ [b = c] (mod a)"
  by (simp add: cong_def)

lemma cong_0 [simp, presburger]:
  "[b = c] (mod 0) \ b = c"
  by (simp add: cong_def)

lemma cong_1 [simp, presburger]:
  "[b = c] (mod 1)"
  by (simp add: cong_def)

lemma cong_dvd_iff:
  "a dvd b \ a dvd c" if "[b = c] (mod a)"
  using that by (auto simp: cong_def dvd_eq_mod_eq_0)

lemma cong_0_iff: "[b = 0] (mod a) \ a dvd b"
  by (simp add: cong_def dvd_eq_mod_eq_0)

lemma cong_add:
  "[b = c] (mod a) \ [d = e] (mod a) \ [b + d = c + e] (mod a)"
  by (auto simp add: cong_def intro: mod_add_cong)

lemma cong_mult:
  "[b = c] (mod a) \ [d = e] (mod a) \ [b * d = c * e] (mod a)"
  by (auto simp add: cong_def intro: mod_mult_cong)

lemma cong_scalar_right:
  "[b = c] (mod a) \ [b * d = c * d] (mod a)"
  by (simp add: cong_mult)

lemma cong_scalar_left:
  "[b = c] (mod a) \ [d * b = d * c] (mod a)"
  by (simp add: cong_mult)

lemma cong_pow:
  "[b = c] (mod a) \ [b ^ n = c ^ n] (mod a)"
  by (simp add: cong_def power_mod [symmetric, of b n a] power_mod [symmetric, of c n a])

lemma cong_sum:
  "[sum f A = sum g A] (mod a)" if "\x. x \ A \ [f x = g x] (mod a)"
  using that by (induct A rule: infinite_finite_induct) (auto intro: cong_add)

lemma cong_prod:
  "[prod f A = prod g A] (mod a)" if "(\x. x \ A \ [f x = g x] (mod a))"
  using that by (induct A rule: infinite_finite_induct) (auto intro: cong_mult)

lemma mod_mult_cong_right:
  "[c mod (a * b) = d] (mod a) \ [c = d] (mod a)"
  by (simp add: cong_def mod_mod_cancel mod_add_left_eq)

lemma mod_mult_cong_left:
  "[c mod (b * a) = d] (mod a) \ [c = d] (mod a)"
  using mod_mult_cong_right [of c a b d] by (simp add: ac_simps)

lemma cong_mod_leftI [simp]:
  "[b = c] (mod a) \ [b mod a = c] (mod a)"
  by (simp add: cong_def)

lemma cong_mod_rightI [simp]:
  "[b = c] (mod a) \ [b = c mod a] (mod a)"
  by (simp add: cong_def)

lemma cong_cmult_leftI: "[a = b] (mod m) \ [c * a = c * b] (mod (c * m))"
  by (metis cong_def local.mult_mod_right)

lemma cong_cmult_rightI: "[a = b] (mod m) \ [a * c = b * c] (mod (m * c))"
  using cong_cmult_leftI[of a b m c] by (simp add: mult.commute)

lemma cong_dvd_mono_modulus:
  assumes "[a = b] (mod m)" "m' dvd m"
  shows   "[a = b] (mod m')"
  using assms by (metis cong_def local.mod_mod_cancel)

lemma coprime_cong_transfer_left:
  assumes "coprime a b" "[a = a'] (mod b)"
  shows   "coprime a' b"
  using assms by (metis cong_0 cong_def local.coprime_mod_left_iff)

lemma coprime_cong_transfer_right:
  assumes "coprime a b" "[b = b'] (mod a)"
  shows   "coprime a b'"
  using coprime_cong_transfer_left[of b a b'] assms
  by (simp add: coprime_commute)

lemma coprime_cong_cong_left:
  assumes "[a = a'] (mod b)"
  shows   "coprime a b \ coprime a' b"
  using assms cong_sym_eq coprime_cong_transfer_left by blast

lemma coprime_cong_cong_right:
  assumes "[b = b'] (mod a)"
  shows   "coprime a b \ coprime a b'"
  using coprime_cong_cong_left[OF assms] by (simp add: coprime_commute)

end

context unique_euclidean_ring
begin

lemma cong_diff:
  "[b = c] (mod a) \ [d = e] (mod a) \ [b - d = c - e] (mod a)"
  by (auto simp add: cong_def intro: mod_diff_cong)

lemma cong_diff_iff_cong_0:
  "[b - c = 0] (mod a) \ [b = c] (mod a)" (is "?P \ ?Q")
proof
  assume ?P
  then have "[b - c + c = 0 + c] (mod a)"
    by (rule cong_add) simp
  then show ?Q
    by simp
next
  assume ?Q
  with cong_diff [of b c a c c] show ?P
    by simp
qed

lemma cong_minus_minus_iff:
  "[- b = - c] (mod a) \ [b = c] (mod a)"
  using cong_diff_iff_cong_0 [of b c a] cong_diff_iff_cong_0 [of "- b" "- c" a]
  by (simp add: cong_0_iff dvd_diff_commute)

lemma cong_modulus_minus_iff [iff]:
  "[b = c] (mod - a) \ [b = c] (mod a)"
  using cong_diff_iff_cong_0 [of b c a] cong_diff_iff_cong_0 [of b c " -a"]
  by (simp add: cong_0_iff)

lemma cong_iff_dvd_diff:
  "[a = b] (mod m) \ m dvd (a - b)"
  by (simp add: cong_0_iff [symmetric] cong_diff_iff_cong_0)

lemma cong_iff_lin:
  "[a = b] (mod m) \ (\k. b = a + m * k)" (is "?P \ ?Q")
proof -
  have "?P \ m dvd b - a"
    by (simp add: cong_iff_dvd_diff dvd_diff_commute)
  also have "\ \ ?Q"
    by (auto simp add: algebra_simps elim!: dvdE)
  finally show ?thesis
    by simp
qed

lemma cong_add_lcancel:
  "[a + x = a + y] (mod n) \ [x = y] (mod n)"
  by (simp add: cong_iff_lin algebra_simps)

lemma cong_add_rcancel:
  "[x + a = y + a] (mod n) \ [x = y] (mod n)"
  by (simp add: cong_iff_lin algebra_simps)

lemma cong_add_lcancel_0:
  "[a + x = a] (mod n) \ [x = 0] (mod n)"
  using cong_add_lcancel [of a x 0 n] by simp

lemma cong_add_rcancel_0:
  "[x + a = a] (mod n) \ [x = 0] (mod n)"
  using cong_add_rcancel [of x a 0 n] by simp

lemma cong_dvd_modulus:
  "[x = y] (mod n)" if "[x = y] (mod m)" and "n dvd m"
  using that by (auto intro: dvd_trans simp add: cong_iff_dvd_diff)

lemma cong_modulus_mult:
  "[x = y] (mod m)" if "[x = y] (mod m * n)"
  using that by (simp add: cong_iff_dvd_diff) (rule dvd_mult_left)

lemma cong_uminus: "[x = y] (mod m) \ [-x = -y] (mod m)"
  unfolding cong_minus_minus_iff .

end

lemma cong_abs [simp]:
  "[x = y] (mod \m\) \ [x = y] (mod m)"
  for x y :: "'a :: {unique_euclidean_ring, linordered_idom}"
  by (simp add: cong_iff_dvd_diff)

lemma cong_square:
  "prime p \ 0 < a \ [a * a = 1] (mod p) \ [a = 1] (mod p) \ [a = - 1] (mod p)"
  for a p :: "'a :: {normalization_semidom, linordered_idom, unique_euclidean_ring}"
  by (auto simp add: cong_iff_dvd_diff square_diff_one_factored dest: prime_dvd_multD)

lemma cong_mult_rcancel:
  "[a * k = b * k] (mod m) \ [a = b] (mod m)"
  if "coprime k m" for a k m :: "'a::{unique_euclidean_ring, ring_gcd}"
  using that by (auto simp add: cong_iff_dvd_diff left_diff_distrib [symmetric] ac_simps coprime_dvd_mult_right_iff)

lemma cong_mult_lcancel:
  "[k * a = k * b] (mod m) = [a = b] (mod m)"
  if "coprime k m" for a k m :: "'a::{unique_euclidean_ring, ring_gcd}"
  using that cong_mult_rcancel [of k m a b] by (simp add: ac_simps)

lemma coprime_cong_mult:
  "[a = b] (mod m) \ [a = b] (mod n) \ coprime m n \ [a = b] (mod m * n)"
  for a b :: "'a :: {unique_euclidean_ring, semiring_gcd}"
  by (simp add: cong_iff_dvd_diff divides_mult)

lemma cong_gcd_eq:
  "gcd a m = gcd b m" if "[a = b] (mod m)"
  for a b :: "'a :: {unique_euclidean_semiring, euclidean_semiring_gcd}"
proof (cases "m = 0")
  case True
  with that show ?thesis
    by simp
next
  case False
  moreover have "gcd (a mod m) m = gcd (b mod m) m"
    using that by (simp add: cong_def)
  ultimately show ?thesis
    by simp
qed

lemma cong_imp_coprime:
  "[a = b] (mod m) \ coprime a m \ coprime b m"
  for a b :: "'a :: {unique_euclidean_semiring, euclidean_semiring_gcd}"
  by (auto simp add: coprime_iff_gcd_eq_1 dest: cong_gcd_eq)

lemma cong_cong_prod_coprime:
  "[x = y] (mod (\i\A. m i))" if
    "(\i\A. [x = y] (mod m i))"
    "(\i\A. (\j\A. i \ j \ coprime (m i) (m j)))"
  for x y :: "'a :: {unique_euclidean_ring, semiring_gcd}"
  using that by (induct A rule: infinite_finite_induct)
    (auto intro!: coprime_cong_mult prod_coprime_right)

subsection \<open>Congruences on \<^typ>\<open>nat\<close> and \<^typ>\<open>int\<close>\<close>

lemma cong_int_iff:
  "[int m = int q] (mod int n) \ [m = q] (mod n)"
  by (simp add: cong_def of_nat_mod [symmetric])

lemma cong_Suc_0 [simp, presburger]:
  "[m = n] (mod Suc 0)"
  using cong_1 [of m n] by simp

lemma cong_diff_nat:
  "[a - c = b - d] (mod m)" if "[a = b] (mod m)" "[c = d] (mod m)"
    and "a \ c" "b \ d" for a b c d m :: nat
proof -
  have "[c + (a - c) = d + (b - d)] (mod m)"
    using that by simp
  with \<open>[c = d] (mod m)\<close> have "[c + (a - c) = c + (b - d)] (mod m)"
    using mod_add_cong by (auto simp add: cong_def) fastforce
  then show ?thesis
    by (simp add: cong_def nat_mod_eq_iff)
qed

lemma cong_diff_iff_cong_0_nat:
  "[a - b = 0] (mod m) \ [a = b] (mod m)" if "a \ b" for a b :: nat
  using that by (simp add: cong_0_iff) (simp add: cong_def mod_eq_dvd_iff_nat)

lemma cong_diff_iff_cong_0_nat':
  "[nat \int a - int b\ = 0] (mod m) \ [a = b] (mod m)"
proof (cases "b \ a")
  case True
  then show ?thesis
    by (simp add: nat_diff_distrib' cong_diff_iff_cong_0_nat [of b a m])
next
  case False
  then have "a \ b"
    by simp
  then show ?thesis
    by (simp add: nat_diff_distrib' cong_diff_iff_cong_0_nat [of a b m])
      (auto simp add: cong_def)
qed

lemma cong_altdef_nat:
  "a \ b \ [a = b] (mod m) \ m dvd (a - b)"
  for a b :: nat
  by (simp add: cong_0_iff [symmetric] cong_diff_iff_cong_0_nat)

lemma cong_altdef_nat':
  "[a = b] (mod m) \ m dvd nat \int a - int b\"
  using cong_diff_iff_cong_0_nat' [of a b m]
  by (simp only: cong_0_iff [symmetric])

lemma cong_mult_rcancel_nat:
  "[a * k = b * k] (mod m) \ [a = b] (mod m)"
  if "coprime k m" for a k m :: nat
proof -
  have "[a * k = b * k] (mod m) \ m dvd nat \int (a * k) - int (b * k)\"
    by (simp add: cong_altdef_nat')
  also have "\ \ m dvd nat \(int a - int b) * int k\"
    by (simp add: algebra_simps)
  also have "\ \ m dvd nat \int a - int b\ * k"
    by (simp add: abs_mult nat_times_as_int)
  also have "\ \ m dvd nat \int a - int b\"
    by (rule coprime_dvd_mult_left_iff) (use \<open>coprime k m\<close> in \<open>simp add: ac_simps\<close>)
  also have "\ \ [a = b] (mod m)"
    by (simp add: cong_altdef_nat')
  finally show ?thesis .
qed

lemma cong_mult_lcancel_nat:
  "[k * a = k * b] (mod m) = [a = b] (mod m)"
  if "coprime k m" for a k m :: nat
  using that by (simp add: cong_mult_rcancel_nat ac_simps)

lemma coprime_cong_mult_nat:
  "[a = b] (mod m) \ [a = b] (mod n) \ coprime m n \ [a = b] (mod m * n)"
  for a b :: nat
  by (simp add: cong_altdef_nat' divides_mult)

lemma cong_less_imp_eq_nat: "0 \ a \ a < m \ 0 \ b \ b < m \ [a = b] (mod m) \ a = b"
  for a b :: nat
  by (auto simp add: cong_def)

lemma cong_less_imp_eq_int: "0 \ a \ a < m \ 0 \ b \ b < m \ [a = b] (mod m) \ a = b"
  for a b :: int
  by (auto simp add: cong_def)

lemma cong_less_unique_nat: "0 < m \ (\!b. 0 \ b \ b < m \ [a = b] (mod m))"
  for a m :: nat
  by (auto simp: cong_def) (metis mod_mod_trivial mod_less_divisor)

lemma cong_less_unique_int: "0 < m \ (\!b. 0 \ b \ b < m \ [a = b] (mod m))"
  for a m :: int
  by (auto simp add: cong_def) (metis mod_mod_trivial pos_mod_bound pos_mod_sign)

lemma cong_iff_lin_nat: "[a = b] (mod m) \ (\k1 k2. b + k1 * m = a + k2 * m)"
  for a b :: nat
  apply (auto simp add: cong_def nat_mod_eq_iff)
   apply (metis mult.commute)
  apply (metis mult.commute)
  done

lemma cong_cong_mod_nat: "[a = b] (mod m) \ [a mod m = b mod m] (mod m)"
  for a b :: nat
  by simp

lemma cong_cong_mod_int: "[a = b] (mod m) \ [a mod m = b mod m] (mod m)"
  for a b :: int
  by simp

lemma cong_add_lcancel_nat: "[a + x = a + y] (mod n) \ [x = y] (mod n)"
  for a x y :: nat
  by (simp add: cong_iff_lin_nat)

lemma cong_add_rcancel_nat: "[x + a = y + a] (mod n) \ [x = y] (mod n)"
  for a x y :: nat
  by (simp add: cong_iff_lin_nat)

lemma cong_add_lcancel_0_nat: "[a + x = a] (mod n) \ [x = 0] (mod n)"
  for a x :: nat
  using cong_add_lcancel_nat [of a x 0 n] by simp

lemma cong_add_rcancel_0_nat: "[x + a = a] (mod n) \ [x = 0] (mod n)"
  for a x :: nat
  using cong_add_rcancel_nat [of x a 0 n] by simp

lemma cong_dvd_modulus_nat: "[x = y] (mod m) \ n dvd m \ [x = y] (mod n)"
  for x y :: nat
  by (auto simp add: cong_altdef_nat')

lemma cong_to_1_nat:
  fixes a :: nat
  assumes "[a = 1] (mod n)"
  shows "n dvd (a - 1)"
proof (cases "a = 0")
  case True
  then show ?thesis by force
next
  case False
  with assms show ?thesis by (metis cong_altdef_nat leI less_one)
qed

lemma cong_0_1_nat': "[0 = Suc 0] (mod n) \ n = Suc 0"
  by (auto simp: cong_def)

lemma cong_0_1_nat: "[0 = 1] (mod n) \ n = 1"
  for n :: nat
  by (auto simp: cong_def)

lemma cong_0_1_int: "[0 = 1] (mod n) \ n = 1 \ n = - 1"
  for n :: int
  by (auto simp: cong_def zmult_eq_1_iff)

lemma cong_to_1'_nat: "[a = 1] (mod n) \ a = 0 \ n = 1 \ (\m. a = 1 + m * n)"
  for a :: nat
  by (metis add.right_neutral cong_0_1_nat cong_iff_lin_nat cong_to_1_nat
      dvd_div_mult_self leI le_add_diff_inverse less_one mult_eq_if)

lemma cong_le_nat: "y \ x \ [x = y] (mod n) \ (\q. x = q * n + y)"
  for x y :: nat
  by (auto simp add: cong_altdef_nat le_imp_diff_is_add)

lemma cong_solve_nat:
  fixes a :: nat
  shows "\x. [a * x = gcd a n] (mod n)"
proof (cases "a = 0 \ n = 0")
  case True
  then show ?thesis
    by (force simp add: cong_0_iff cong_sym)
next
  case False
  then show ?thesis
    using bezout_nat [of a n]
    by auto (metis cong_add_rcancel_0_nat cong_mult_self_left)
qed

lemma cong_solve_int:
  fixes a :: int
  shows "\x. [a * x = gcd a n] (mod n)"
    by (metis bezout_int cong_iff_lin mult.commute)

lemma cong_solve_dvd_nat:
  fixes a :: nat
  assumes "gcd a n dvd d"
  shows "\x. [a * x = d] (mod n)"
proof -
  from cong_solve_nat [of a] obtain x where "[a * x = gcd a n](mod n)"
    by auto
  then have "[(d div gcd a n) * (a * x) = (d div gcd a n) * gcd a n] (mod n)"
    using cong_scalar_left by blast
  also from assms have "(d div gcd a n) * gcd a n = d"
    by (rule dvd_div_mult_self)
  also have "(d div gcd a n) * (a * x) = a * (d div gcd a n * x)"
    by auto
  finally show ?thesis
    by auto
qed

lemma cong_solve_dvd_int:
  fixes a::int
  assumes b: "gcd a n dvd d"
  shows "\x. [a * x = d] (mod n)"
proof -
  from cong_solve_int [of a] obtain x where "[a * x = gcd a n](mod n)"
    by auto
  then have "[(d div gcd a n) * (a * x) = (d div gcd a n) * gcd a n] (mod n)"
    using cong_scalar_left by blast
  also from b have "(d div gcd a n) * gcd a n = d"
    by (rule dvd_div_mult_self)
  also have "(d div gcd a n) * (a * x) = a * (d div gcd a n * x)"
    by auto
  finally show ?thesis
    by auto
qed

lemma cong_solve_coprime_nat:
  "\x. [a * x = Suc 0] (mod n)" if "coprime a n"
  using that cong_solve_nat [of a n] by auto

lemma cong_solve_coprime_int:
  "\x. [a * x = 1] (mod n)" if "coprime a n" for a n x :: int
  using that cong_solve_int [of a n] by (auto simp add: zabs_def split: if_splits)

lemma coprime_iff_invertible_nat:
  "coprime a m \ (\x. [a * x = Suc 0] (mod m))" (is "?P \ ?Q")
proof
  assume ?P then show ?Q
    by (auto dest!: cong_solve_coprime_nat)
next
  assume ?Q
  then obtain b where "[a * b = Suc 0] (mod m)"
    by blast
  with coprime_mod_left_iff [of m "a * b"] show ?P
    by (cases "m = 0 \ m = 1")
      (unfold cong_def, auto simp add: cong_def)
qed

lemma coprime_iff_invertible_int:
  "coprime a m \ (\x. [a * x = 1] (mod m))" (is "?P \ ?Q") for m :: int
proof
  assume ?P then show ?Q
    by (auto dest: cong_solve_coprime_int)
next
  assume ?Q
  then obtain b where "[a * b = 1] (mod m)"
    by blast
  with coprime_mod_left_iff [of m "a * b"] show ?P
    by (cases "m = 0 \ m = 1")
      (unfold cong_def, auto simp add: zmult_eq_1_iff)
qed

lemma coprime_iff_invertible'_nat:
  assumes "m > 0"
  shows "coprime a m \ (\x. 0 \ x \ x < m \ [a * x = Suc 0] (mod m))"
proof -
  have "\b. \0 < m; [a * b = Suc 0] (mod m)\ \ \b'
    by (metis cong_def mod_less_divisor [OF assms] mod_mult_right_eq)
  then show ?thesis
    using assms coprime_iff_invertible_nat by auto
qed

lemma coprime_iff_invertible'_int:
  fixes m :: int
  assumes "m > 0"
  shows "coprime a m \ (\x. 0 \ x \ x < m \ [a * x = 1] (mod m))"
  using assms by (simp add: coprime_iff_invertible_int)
    (metis assms cong_mod_left mod_mult_right_eq pos_mod_bound pos_mod_sign)

lemma cong_cong_lcm_nat: "[x = y] (mod a) \ [x = y] (mod b) \ [x = y] (mod lcm a b)"
  for x y :: nat
  by (meson cong_altdef_nat' lcm_least)

lemma cong_cong_lcm_int: "[x = y] (mod a) \ [x = y] (mod b) \ [x = y] (mod lcm a b)"
  for x y :: int
  by (auto simp add: cong_iff_dvd_diff lcm_least)

lemma cong_cong_prod_coprime_nat:
  "[x = y] (mod (\i\A. m i))" if
    "(\i\A. [x = y] (mod m i))"
    "(\i\A. (\j\A. i \ j \ coprime (m i) (m j)))"
  for x y :: nat
  using that by (induct A rule: infinite_finite_induct)
    (auto intro!: coprime_cong_mult_nat prod_coprime_right)

lemma binary_chinese_remainder_nat:
  fixes m1 m2 :: nat
  assumes a: "coprime m1 m2"
  shows "\x. [x = u1] (mod m1) \ [x = u2] (mod m2)"
proof -
  have "\b1 b2. [b1 = 1] (mod m1) \ [b1 = 0] (mod m2) \ [b2 = 0] (mod m1) \ [b2 = 1] (mod m2)"
  proof -
    from cong_solve_coprime_nat [OF a] obtain x1 where 1: "[m1 * x1 = 1] (mod m2)"
      by auto
    from a have b: "coprime m2 m1"
      by (simp add: ac_simps)
    from cong_solve_coprime_nat [OF b] obtain x2 where 2: "[m2 * x2 = 1] (mod m1)"
      by auto
    have "[m1 * x1 = 0] (mod m1)"
      by (simp add: cong_mult_self_left)
    moreover have "[m2 * x2 = 0] (mod m2)"
      by (simp add: cong_mult_self_left)
    ultimately show ?thesis
      using 1 2 by blast
  qed
  then obtain b1 b2
    where "[b1 = 1] (mod m1)" and "[b1 = 0] (mod m2)"
      and "[b2 = 0] (mod m1)" and "[b2 = 1] (mod m2)"
    by blast
  let ?x = "u1 * b1 + u2 * b2"
  have "[?x = u1 * 1 + u2 * 0] (mod m1)"
    using \<open>[b1 = 1] (mod m1)\<close> \<open>[b2 = 0] (mod m1)\<close> cong_add cong_scalar_left by blast
  then have "[?x = u1] (mod m1)" by simp
  have "[?x = u1 * 0 + u2 * 1] (mod m2)"
    using \<open>[b1 = 0] (mod m2)\<close> \<open>[b2 = 1] (mod m2)\<close> cong_add cong_scalar_left by blast
  then have "[?x = u2] (mod m2)"
    by simp
  with \<open>[?x = u1] (mod m1)\<close> show ?thesis
    by blast
qed

lemma binary_chinese_remainder_int:
  fixes m1 m2 :: int
  assumes a: "coprime m1 m2"
  shows "\x. [x = u1] (mod m1) \ [x = u2] (mod m2)"
proof -
  have "\b1 b2. [b1 = 1] (mod m1) \ [b1 = 0] (mod m2) \ [b2 = 0] (mod m1) \ [b2 = 1] (mod m2)"
  proof -
    from cong_solve_coprime_int [OF a] obtain x1 where 1: "[m1 * x1 = 1] (mod m2)"
      by auto
    from a have b: "coprime m2 m1"
      by (simp add: ac_simps)
    from cong_solve_coprime_int [OF b] obtain x2 where 2: "[m2 * x2 = 1] (mod m1)"
      by auto
    have "[m1 * x1 = 0] (mod m1)"
     by (simp add: cong_mult_self_left)
    moreover have "[m2 * x2 = 0] (mod m2)"
      by (simp add: cong_mult_self_left)
    ultimately show ?thesis
      using 1 2 by blast
  qed
  then obtain b1 b2
    where "[b1 = 1] (mod m1)" and "[b1 = 0] (mod m2)"
      and "[b2 = 0] (mod m1)" and "[b2 = 1] (mod m2)"
    by blast
  let ?x = "u1 * b1 + u2 * b2"
  have "[?x = u1 * 1 + u2 * 0] (mod m1)"
    using \<open>[b1 = 1] (mod m1)\<close> \<open>[b2 = 0] (mod m1)\<close> cong_add cong_scalar_left by blast
  then have "[?x = u1] (mod m1)" by simp
  have "[?x = u1 * 0 + u2 * 1] (mod m2)"
    using \<open>[b1 = 0] (mod m2)\<close> \<open>[b2 = 1] (mod m2)\<close> cong_add cong_scalar_left by blast
  then have "[?x = u2] (mod m2)" by simp
  with \<open>[?x = u1] (mod m1)\<close> show ?thesis
    by blast
qed

lemma cong_modulus_mult_nat: "[x = y] (mod m * n) \ [x = y] (mod m)"
  for x y :: nat
  by (metis cong_def mod_mult_cong_right)

lemma cong_less_modulus_unique_nat: "[x = y] (mod m) \ x < m \ y < m \ x = y"
  for x y :: nat
  by (simp add: cong_def)

lemma binary_chinese_remainder_unique_nat:
  fixes m1 m2 :: nat
  assumes a: "coprime m1 m2"
    and nz: "m1 \ 0" "m2 \ 0"
  shows "\!x. x < m1 * m2 \ [x = u1] (mod m1) \ [x = u2] (mod m2)"
proof -
  obtain y where y1: "[y = u1] (mod m1)" and y2: "[y = u2] (mod m2)"
    using binary_chinese_remainder_nat [OF a] by blast
  let ?x = "y mod (m1 * m2)"
  from nz have less: "?x < m1 * m2"
    by auto
  have 1: "[?x = u1] (mod m1)"
    using y1 mod_mult_cong_right by blast
  have 2: "[?x = u2] (mod m2)"
    using y2 mod_mult_cong_left by blast
  have "z = ?x" if "z < m1 * m2" "[z = u1] (mod m1)"  "[z = u2] (mod m2)" for z
  proof -
    have "[?x = z] (mod m1)"
      by (metis "1" cong_def that(2))
    moreover have "[?x = z] (mod m2)"
      by (metis "2" cong_def that(3))
    ultimately have "[?x = z] (mod m1 * m2)"
      using a by (auto intro: coprime_cong_mult_nat simp add: mod_mult_cong_left mod_mult_cong_right)
    with \<open>z < m1 * m2\<close> \<open>?x < m1 * m2\<close> show "z = ?x"
      by (auto simp add: cong_def)
  qed
  with less 1 2 show ?thesis
    by blast
qed

lemma chinese_remainder_nat:
  fixes A :: "'a set"
    and m :: "'a \ nat"
    and u :: "'a \ nat"
  assumes fin: "finite A"
    and cop: "\i \ A. \j \ A. i \ j \ coprime (m i) (m j)"
  shows "\x. \i \ A. [x = u i] (mod m i)"
proof -
  have "\b. (\i \ A. [b i = 1] (mod m i) \ [b i = 0] (mod (\j \ A - {i}. m j)))"
  proof (rule finite_set_choice, rule fin, rule ballI)
    fix i
    assume "i \ A"
    with cop have "coprime (\j \ A - {i}. m j) (m i)"
      by (intro prod_coprime_left) auto
    then have "\x. [(\j \ A - {i}. m j) * x = Suc 0] (mod m i)"
      by (elim cong_solve_coprime_nat)
    then obtain x where "[(\j \ A - {i}. m j) * x = 1] (mod m i)"
      by auto
    moreover have "[(\j \ A - {i}. m j) * x = 0] (mod (\j \ A - {i}. m j))"
      by (simp add: cong_0_iff)
    ultimately show "\a. [a = 1] (mod m i) \ [a = 0] (mod prod m (A - {i}))"
      by blast
  qed
  then obtain b where b: "\i. i \ A \ [b i = 1] (mod m i) \ [b i = 0] (mod (\j \ A - {i}. m j))"
    by blast
  let ?x = "\i\A. (u i) * (b i)"
  show ?thesis
  proof (rule exI, clarify)
    fix i
    assume a: "i \ A"
    show "[?x = u i] (mod m i)"
    proof -
      from fin a have "?x = (\j \ {i}. u j * b j) + (\j \ A - {i}. u j * b j)"
        by (subst sum.union_disjoint [symmetric]) (auto intro: sum.cong)
      then have "[?x = u i * b i + (\j \ A - {i}. u j * b j)] (mod m i)"
        by auto
      also have "[u i * b i + (\j \ A - {i}. u j * b j) =
                  u i * 1 + (\<Sum>j \<in> A - {i}. u j * 0)] (mod m i)"
      proof (intro cong_add cong_scalar_left cong_sum)
        show "[b i = 1] (mod m i)"
          using a b by blast
        show "[b x = 0] (mod m i)" if "x \ A - {i}" for x
        proof -
          have "x \ A" "x \ i"
            using that by auto
          then show ?thesis
            using a b [OF \<open>x \<in> A\<close>] cong_dvd_modulus_nat fin by blast
        qed
      qed
      finally show ?thesis
        by simp
    qed
  qed
qed

lemma coprime_cong_prod_nat: "[x = y] (mod (\i\A. m i))"
  if "\i j. \i \ A; j \ A; i \ j\ \ coprime (m i) (m j)"
    and "\i. i \ A \ [x = y] (mod m i)" for x y :: nat
  using that
proof (induct A rule: infinite_finite_induct)
  case (insert x A)
  then show ?case
    by simp (metis coprime_cong_mult_nat prod_coprime_right)
qed auto

lemma chinese_remainder_unique_nat:
  fixes A :: "'a set"
    and m :: "'a \ nat"
    and u :: "'a \ nat"
  assumes fin: "finite A"
    and nz: "\i\A. m i \ 0"
    and cop: "\i\A. \j\A. i \ j \ coprime (m i) (m j)"
  shows "\!x. x < (\i\A. m i) \ (\i\A. [x = u i] (mod m i))"
proof -
  from chinese_remainder_nat [OF fin cop]
  obtain y where one: "(\i\A. [y = u i] (mod m i))"
    by blast
  let ?x = "y mod (\i\A. m i)"
  from fin nz have prodnz: "(\i\A. m i) \ 0"
    by auto
  then have less: "?x < (\i\A. m i)"
    by auto
  have cong: "\i\A. [?x = u i] (mod m i)"
    using fin one
    by (auto simp add: cong_def dvd_prod_eqI mod_mod_cancel)
  have unique: "\z. z < (\i\A. m i) \ (\i\A. [z = u i] (mod m i)) \ z = ?x"
  proof clarify
    fix z
    assume zless: "z < (\i\A. m i)"
    assume zcong: "(\i\A. [z = u i] (mod m i))"
    have "\i\A. [?x = z] (mod m i)"
      using cong zcong by (auto simp add: cong_def)
    with fin cop have "[?x = z] (mod (\i\A. m i))"
      by (intro coprime_cong_prod_nat) auto
    with zless less show "z = ?x"
      by (auto simp add: cong_def)
  qed
  from less cong unique show ?thesis
    by blast
qed

lemma (in semiring_1_cancel) of_nat_eq_iff_cong_CHAR:
  "of_nat x = (of_nat y :: 'a) \ [x = y] (mod CHAR('a))"
proof (induction x y rule: linorder_wlog)
  case (le x y)
  define z where "z = y - x"
  have [simp]: "y = x + z"
    using le by (auto simp: z_def)
  have "(CHAR('a) dvd z) = [x = x + z] (mod CHAR('a))"
    by (metis \<open>y = x + z\<close> cong_def le mod_eq_dvd_iff_nat z_def)
  thus ?case
    by (simp add: of_nat_eq_0_iff_char_dvd)
qed (simp add: eq_commute cong_sym_eq)

lemma (in ring_1) of_int_eq_iff_cong_CHAR:
  "of_int x = (of_int y :: 'a) \ [x = y] (mod int CHAR('a))"
proof -
  have "of_int x = (of_int y :: 'a) \ of_int (x - y) = (0 :: 'a)"
    by auto
  also have "\ \ (int CHAR('a) dvd x - y)"
    by (rule of_int_eq_0_iff_char_dvd)
  also have "\ \ [x = y] (mod int CHAR('a))"
    by (simp add: cong_iff_dvd_diff)
  finally show ?thesis .
qed

text \<open>Thanks to Manuel Eberl\<close>
lemma prime_cong_4_nat_cases [consumes 1, case_names 2 cong_1 cong_3]:
  assumes "prime (p :: nat)"
  obtains "p = 2" | "[p = 1] (mod 4)" | "[p = 3] (mod 4)"
proof -
  have "[p = 2] (mod 4) \ p = 2"
  proof
    assume "[p = 2] (mod 4)"
    hence "p mod 4 = 2"
      by (auto simp: cong_def)
    hence "even p"
      by (simp add: even_even_mod_4_iff)
    with assms show "p = 2"
      unfolding prime_nat_iff by force
  qed auto
  moreover have "[p \ 0] (mod 4)"
  proof
    assume "[p = 0] (mod 4)"
    hence "4 dvd p"
      by (auto simp: cong_0_iff)
    with assms have "p = 4"
      by (subst (asm) prime_nat_iff) auto
    thus False
      using assms by simp
  qed
  ultimately consider "[p = 3] (mod 4)" | "[p = 1] (mod 4)" | "p = 2"
    by (fastforce simp: cong_def)
  thus ?thesis
    using that by metis
qed

end

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