(*  Title:      HOL/Divides.thy

    Author:     Lawrence C Paulson, Cambridge University Computer Laboratory

    Copyright   1999  University of Cambridge

*)

header {* The division operators div and mod *}

theory Divides

imports Nat_Numeral Nat_Transfer

uses "~~/src/Provers/Arith/cancel_div_mod.ML"

begin

subsection {* Syntactic division operations *}

class div = dvd +

  fixes div :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixl "div" 70)

    and mod :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixl "mod" 70)

subsection {* Abstract division in commutative semirings. *}

class semiring_div = comm_semiring_1_cancel + no_zero_divisors + div +

  assumes mod_div_equality: "a div b * b + a mod b = a"

    and div_by_0 [simp]: "a div 0 = 0"

    and div_0 [simp]: "0 div a = 0"

    and 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

text {* @{const div} and @{const mod} *}

lemma mod_div_equality2: "b * (a div b) + a mod b = a"

  unfolding mult_commute [of b]

  by (rule mod_div_equality)

lemma mod_div_equality': "a mod b + a div b * b = a"

  using mod_div_equality [of a b]

  by (simp only: add_ac)

lemma div_mod_equality: "((a div b) * b + a mod b) + c = a + c"

  by (simp add: mod_div_equality)

lemma div_mod_equality2: "(b * (a div b) + a mod b) + c = a + c"

  by (simp add: mod_div_equality2)

lemma mod_by_0 [simp]: "a mod 0 = a"

  using mod_div_equality [of a zero] by simp

lemma mod_0 [simp]: "0 mod a = 0"

  using mod_div_equality [of zero a] div_0 by simp

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 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: mod_div_equality)

  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 left_distrib)

  then have "a div b * b + (a + c * b) mod b = a div b * b + a mod b"

    by (simp add: mod_div_equality)

  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 div_mult_self1_is_id [simp]: "b \<noteq> 0 \<Longrightarrow> b * a div b = a"

  using div_mult_self2 [of b 0 a] by simp

lemma div_mult_self2_is_id [simp]: "b \<noteq> 0 \<Longrightarrow> a * b div b = a"

  using div_mult_self1 [of b 0 a] by simp

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_by_1 [simp]: "a div 1 = a"

  using div_mult_self2_is_id [of 1 a] zero_neq_one by simp

lemma mod_by_1 [simp]: "a mod 1 = 0"

proof -

  from mod_div_equality [of a one] div_by_1 have "a + a mod 1 = a" by simp

  then have "a + a mod 1 = a + 0" by simp

  then show ?thesis by (rule add_left_imp_eq)

qed

lemma mod_self [simp]: "a mod a = 0"

  using mod_mult_self2_is_0 [of 1] by simp

lemma div_self [simp]: "a \<noteq> 0 \<Longrightarrow> a div a = 1"

  using div_mult_self2_is_id [of _ 1] by simp

lemma div_add_self1 [simp]:

  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 [simp]:

  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_decomp:

  fixes a b

  obtains q r where "q = a div b" and "r = a mod b"

    and "a = q * b + r"

proof -

  from mod_div_equality have "a = a div b * b + a mod b" by simp

  moreover have "a div b = a div b" ..

  moreover have "a mod b = a mod b" ..

  note that ultimately show thesis by blast

qed

lemma dvd_eq_mod_eq_0 [code]: "a dvd b \<longleftrightarrow> b mod a = 0"

proof

  assume "b mod a = 0"

  with mod_div_equality [of b a] have "b div a * a = b" by simp

  then have "b = a * (b div a)" unfolding mult_commute ..

  then have "\<exists>c. b = a * c" ..

  then show "a dvd b" unfolding dvd_def .

next

  assume "a dvd b"

  then have "\<exists>c. b = a * c" unfolding dvd_def .

  then obtain c where "b = a * c" ..

  then have "b mod a = a * c mod a" by simp

  then have "b mod a = c * a mod a" by (simp add: mult_commute)

  then show "b mod a = 0" by simp

qed

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_equality')

  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_equality')

  finally show ?thesis .

qed

lemma dvd_imp_mod_0: "a dvd b \<Longrightarrow> b mod a = 0"

by (rule dvd_eq_mod_eq_0[THEN iffD1])

lemma dvd_div_mult_self: "a dvd b \<Longrightarrow> (b div a) * a = b"

by (subst (2) mod_div_equality [of b a, symmetric]) (simp add:dvd_imp_mod_0)

lemma dvd_mult_div_cancel: "a dvd b \<Longrightarrow> a * (b div a) = b"

by (drule dvd_div_mult_self) (simp add: mult_commute)

lemma dvd_div_mult: "a dvd b \<Longrightarrow> (b div a) * c = b * c div a"

apply (cases "a = 0")

 apply simp

apply (auto simp: dvd_def mult_assoc)

done

lemma div_dvd_div[simp]:

  "a dvd b \<Longrightarrow> a dvd c \<Longrightarrow> (b div a dvd c div a) = (b dvd c)"

apply (cases "a = 0")

 apply simp

apply (unfold dvd_def)

apply auto

 apply(blast intro:mult_assoc[symmetric])

apply(fastforce simp add: mult_assoc)

done

lemma dvd_mod_imp_dvd: "[| k dvd m mod n;  k dvd n |] ==> k dvd m"

  apply (subgoal_tac "k dvd (m div n) *n + m mod n")

   apply (simp add: mod_div_equality)

  apply (simp only: dvd_add dvd_mult)

  done

text {* Addition respects modular equivalence. *}

lemma mod_add_left_eq: "(a + b) mod c = (a mod c + b) mod c"

proof -

  have "(a + b) mod c = (a div c * c + a mod c + b) mod c"

    by (simp only: mod_div_equality)

  also have "\<dots> = (a mod c + b + a div c * c) mod c"

    by (simp only: add_ac)

  also have "\<dots> = (a mod c + b) mod c"

    by (rule mod_mult_self1)

  finally show ?thesis .

qed

lemma mod_add_right_eq: "(a + b) mod c = (a + b mod c) mod c"

proof -

  have "(a + b) mod c = (a + (b div c * c + b mod c)) mod c"

    by (simp only: mod_div_equality)

  also have "\<dots> = (a + b mod c + b div c * c) mod c"

    by (simp only: add_ac)

  also have "\<dots> = (a + b mod c) mod c"

    by (rule mod_mult_self1)

  finally show ?thesis .

qed

lemma mod_add_eq: "(a + b) mod c = (a mod c + b mod c) mod c"

by (rule trans [OF mod_add_left_eq mod_add_right_eq])

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 ..

  thus ?thesis

    by (simp only: mod_add_eq [symmetric])

qed

lemma div_add [simp]: "z dvd x \<Longrightarrow> z dvd y

  \<Longrightarrow> (x + y) div z = x div z + y div z"

by (cases "z = 0", simp, unfold dvd_def, auto simp add: algebra_simps)

text {* Multiplication respects modular equivalence. *}

lemma mod_mult_left_eq: "(a * b) mod c = ((a mod c) * b) mod c"

proof -

  have "(a * b) mod c = ((a div c * c + a mod c) * b) mod c"

    by (simp only: mod_div_equality)

  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 .

qed

lemma mod_mult_right_eq: "(a * b) mod c = (a * (b mod c)) mod c"

proof -

  have "(a * b) mod c = (a * (b div c * c + b mod c)) mod c"

    by (simp only: mod_div_equality)

  also have "\<dots> = (a * (b mod c) + a * (b div c) * c) mod c"

    by (simp only: algebra_simps)

  also have "\<dots> = (a * (b mod c)) mod c"

    by (rule mod_mult_self1)

  finally show ?thesis .

qed

lemma mod_mult_eq: "(a * b) mod c = ((a mod c) * (b mod c)) mod c"

by (rule trans [OF mod_mult_left_eq mod_mult_right_eq])

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 ..

  thus ?thesis

    by (simp only: mod_mult_eq [symmetric])

qed

text {* Exponentiation respects modular equivalence. *}

lemma power_mod: "(a mod b)^n mod b = a^n mod b"

apply (induct n, simp_all)

apply (rule mod_mult_right_eq [THEN trans])

apply (simp (no_asm_simp))

apply (rule mod_mult_eq [symmetric])

done

lemma mod_mod_cancel:

  assumes "c dvd b"

  shows "a mod b mod c = a mod c"

proof -

  from `c dvd b` obtain k where "b = c * k"

    by (rule dvdE)

  have "a mod b mod c = a mod (c * k) mod c"

    by (simp only: `b = c * k`)

  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: add_ac mult_ac)

  also have "\<dots> = a mod c"

    by (simp only: mod_div_equality)

  finally show ?thesis .

qed

lemma div_mult_div_if_dvd:

  "y dvd x \<Longrightarrow> z dvd w \<Longrightarrow> (x div y) * (w div z) = (x * w) div (y * z)"

  apply (cases "y = 0", simp)

  apply (cases "z = 0", simp)

  apply (auto elim!: dvdE simp add: algebra_simps)

  apply (subst mult_assoc [symmetric])

  apply (simp add: no_zero_divisors)

  done

lemma div_mult_swap:

  assumes "c dvd b"

  shows "a * (b div c) = (a * b) div c"

proof -

  from assms have "b div c * (a div 1) = b * a div (c * 1)"

    by (simp only: div_mult_div_if_dvd one_dvd)

  then show ?thesis by (simp add: mult_commute)

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 mod_div_equality

  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 mod_div_equality 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 dvd_mod_iff: "k dvd n \<Longrightarrow> k dvd (m mod n) \<longleftrightarrow> k dvd m"

by (blast intro: dvd_mod_imp_dvd dvd_mod)

lemma div_power:

  "y dvd x \<Longrightarrow> (x div y) ^ n = x ^ n div y ^ n"

apply (induct n)

 apply simp

apply(simp add: div_mult_div_if_dvd dvd_power_same)

done

lemma dvd_div_eq_mult:

  assumes "a \<noteq> 0" and "a dvd b"  

  shows "b div a = c \<longleftrightarrow> b = c * a"

proof

  assume "b = c * a"

  then show "b div a = c" by (simp add: assms)

next

  assume "b div a = c"

  then have "b div a * a = c * a" by simp

  moreover from `a dvd b` have "b div a * a = b" by (simp add: dvd_div_mult_self)

  ultimately show "b = c * a" by simp

qed

lemma dvd_div_div_eq_mult:

  assumes "a \<noteq> 0" "c \<noteq> 0" and "a dvd b" "c dvd d"

  shows "b div a = d div c \<longleftrightarrow> b * c = a * d"

  using assms by (auto simp add: mult_commute [of _ a] dvd_div_mult_self dvd_div_eq_mult div_mult_swap intro: sym)

end

class ring_div = semiring_div + comm_ring_1

begin

subclass ring_1_no_zero_divisors ..

text {* Negation respects modular equivalence. *}

lemma mod_minus_eq: "(- a) mod b = (- (a mod b)) mod b"

proof -

  have "(- a) mod b = (- (a div b * b + a mod b)) mod b"

    by (simp only: mod_div_equality)

  also have "\<dots> = (- (a mod b) + - (a div b) * b) mod b"

    by (simp only: minus_add_distrib minus_mult_left add_ac)

  also have "\<dots> = (- (a mod b)) mod b"

    by (rule mod_mult_self1)

  finally show ?thesis .

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 ..

  thus ?thesis

    by (simp only: mod_minus_eq [symmetric])

qed

text {* Subtraction respects modular equivalence. *}

lemma mod_diff_left_eq: "(a - b) mod c = (a mod c - b) mod c"

  unfolding diff_minus

  by (intro mod_add_cong mod_minus_cong) simp_all

lemma mod_diff_right_eq: "(a - b) mod c = (a - b mod c) mod c"

  unfolding diff_minus

  by (intro mod_add_cong mod_minus_cong) simp_all

lemma mod_diff_eq: "(a - b) mod c = (a mod c - b mod c) mod c"

  unfolding diff_minus

  by (intro mod_add_cong mod_minus_cong) simp_all

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"

  unfolding diff_minus using assms

  by (intro mod_add_cong mod_minus_cong)

lemma dvd_neg_div: "y dvd x \<Longrightarrow> -x div y = - (x div y)"

apply (case_tac "y = 0") apply simp

apply (auto simp add: dvd_def)

apply (subgoal_tac "-(y * k) = y * - k")

 apply (erule ssubst)

 apply (erule div_mult_self1_is_id)

apply simp

done

lemma dvd_div_neg: "y dvd x \<Longrightarrow> x div -y = - (x div y)"

apply (case_tac "y = 0") apply simp

apply (auto simp add: dvd_def)

apply (subgoal_tac "y * k = -y * -k")

 apply (erule ssubst)

 apply (rule div_mult_self1_is_id)

 apply simp

apply simp

done

lemma div_minus_minus [simp]: "(-a) div (-b) = a div b"

  using div_mult_mult1 [of "- 1" a b]

  unfolding neg_equal_0_iff_equal 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

end

subsection {* Division on @{typ nat} *}

text {*

  We define @{const div} and @{const mod} on @{typ nat} by means

  of a characteristic relation with two input arguments

  @{term "m\<Colon>nat"}, @{term "n\<Colon>nat"} and two output arguments

  @{term "q\<Colon>nat"}(uotient) and @{term "r\<Colon>nat"}(emainder).

*}

definition divmod_nat_rel :: "nat \<Rightarrow> nat \<Rightarrow> nat \<times> nat \<Rightarrow> bool" where

  "divmod_nat_rel m n qr \<longleftrightarrow>

    m = fst qr * n + snd qr \<and>

      (if n = 0 then fst qr = 0 else if n > 0 then 0 \<le> snd qr \<and> snd qr < n else n < snd qr \<and> snd qr \<le> 0)"

text {* @{const divmod_nat_rel} is total: *}

lemma divmod_nat_rel_ex:

  obtains q r where "divmod_nat_rel m n (q, r)"

proof (cases "n = 0")

  case True  with that show thesis

    by (auto simp add: divmod_nat_rel_def)

next

  case False

  have "\<exists>q r. m = q * n + r \<and> r < n"

  proof (induct m)

    case 0 with `n \<noteq> 0`

    have "(0\<Colon>nat) = 0 * n + 0 \<and> 0 < n" by simp

    then show ?case by blast

  next

    case (Suc m) then obtain q' r'

      where m: "m = q' * n + r'" and n: "r' < n" by auto

    then show ?case proof (cases "Suc r' < n")

      case True

      from m n have "Suc m = q' * n + Suc r'" by simp

      with True show ?thesis by blast

    next

      case False then have "n \<le> Suc r'" by auto

      moreover from n have "Suc r' \<le> n" by auto

      ultimately have "n = Suc r'" by auto

      with m have "Suc m = Suc q' * n + 0" by simp

      with `n \<noteq> 0` show ?thesis by blast

    qed

  qed

  with that show thesis

    using `n \<noteq> 0` by (auto simp add: divmod_nat_rel_def)

qed

text {* @{const divmod_nat_rel} is injective: *}

lemma divmod_nat_rel_unique:

  assumes "divmod_nat_rel m n qr"

    and "divmod_nat_rel m n qr'"

  shows "qr = qr'"

proof (cases "n = 0")

  case True with assms show ?thesis

    by (cases qr, cases qr')

      (simp add: divmod_nat_rel_def)

next

  case False

  have aux: "\<And>q r q' r'. q' * n + r' = q * n + r \<Longrightarrow> r < n \<Longrightarrow> q' \<le> (q\<Colon>nat)"

  apply (rule leI)

  apply (subst less_iff_Suc_add)

  apply (auto simp add: add_mult_distrib)

  done

  from `n \<noteq> 0` assms have "fst qr = fst qr'"

    by (auto simp add: divmod_nat_rel_def intro: order_antisym dest: aux sym)

  moreover from this assms have "snd qr = snd qr'"

    by (simp add: divmod_nat_rel_def)

  ultimately show ?thesis by (cases qr, cases qr') simp

qed

text {*

  We instantiate divisibility on the natural numbers by

  means of @{const divmod_nat_rel}:

*}

definition divmod_nat :: "nat \<Rightarrow> nat \<Rightarrow> nat \<times> nat" where

  "divmod_nat m n = (THE qr. divmod_nat_rel m n qr)"

lemma divmod_nat_rel_divmod_nat:

  "divmod_nat_rel m n (divmod_nat m n)"

proof -

  from divmod_nat_rel_ex

    obtain qr where rel: "divmod_nat_rel m n qr" .

  then show ?thesis

  by (auto simp add: divmod_nat_def intro: theI elim: divmod_nat_rel_unique)

qed

lemma divmod_nat_unique:

  assumes "divmod_nat_rel m n qr" 

  shows "divmod_nat m n = qr"

  using assms by (auto intro: divmod_nat_rel_unique divmod_nat_rel_divmod_nat)

instantiation nat :: semiring_div

begin

definition div_nat where

  "m div n = fst (divmod_nat m n)"

lemma fst_divmod_nat [simp]:

  "fst (divmod_nat m n) = m div n"

  by (simp add: div_nat_def)

definition mod_nat where

  "m mod n = snd (divmod_nat m n)"

lemma snd_divmod_nat [simp]:

  "snd (divmod_nat m n) = m mod n"

  by (simp add: mod_nat_def)

lemma divmod_nat_div_mod:

  "divmod_nat m n = (m div n, m mod n)"

  by (simp add: prod_eq_iff)

lemma div_nat_unique:

  assumes "divmod_nat_rel m n (q, r)" 

  shows "m div n = q"

  using assms by (auto dest!: divmod_nat_unique simp add: prod_eq_iff)

lemma mod_nat_unique:

  assumes "divmod_nat_rel m n (q, r)" 

  shows "m mod n = r"

  using assms by (auto dest!: divmod_nat_unique simp add: prod_eq_iff)

lemma divmod_nat_rel: "divmod_nat_rel m n (m div n, m mod n)"

  using divmod_nat_rel_divmod_nat by (simp add: divmod_nat_div_mod)

lemma divmod_nat_zero: "divmod_nat m 0 = (0, m)"

  by (simp add: divmod_nat_unique divmod_nat_rel_def)

lemma divmod_nat_zero_left: "divmod_nat 0 n = (0, 0)"

  by (simp add: divmod_nat_unique divmod_nat_rel_def)

lemma divmod_nat_base: "m < n \<Longrightarrow> divmod_nat m n = (0, m)"

  by (simp add: divmod_nat_unique divmod_nat_rel_def)

lemma divmod_nat_step:

  assumes "0 < n" and "n \<le> m"

  shows "divmod_nat m n = (Suc ((m - n) div n), (m - n) mod n)"

proof (rule divmod_nat_unique)

  have "divmod_nat_rel (m - n) n ((m - n) div n, (m - n) mod n)"

    by (rule divmod_nat_rel)

  thus "divmod_nat_rel m n (Suc ((m - n) div n), (m - n) mod n)"

    unfolding divmod_nat_rel_def using assms by auto

qed

text {* The ''recursion'' equations for @{const div} and @{const mod} *}

lemma div_less [simp]:

  fixes m n :: nat

  assumes "m < n"

  shows "m div n = 0"

  using assms divmod_nat_base by (simp add: prod_eq_iff)

lemma le_div_geq:

  fixes m n :: nat

  assumes "0 < n" and "n \<le> m"

  shows "m div n = Suc ((m - n) div n)"

  using assms divmod_nat_step by (simp add: prod_eq_iff)

lemma mod_less [simp]:

  fixes m n :: nat

  assumes "m < n"

  shows "m mod n = m"

  using assms divmod_nat_base by (simp add: prod_eq_iff)

lemma le_mod_geq:

  fixes m n :: nat

  assumes "n \<le> m"

  shows "m mod n = (m - n) mod n"

  using assms divmod_nat_step by (cases "n = 0") (simp_all add: prod_eq_iff)

instance proof

  fix m n :: nat

  show "m div n * n + m mod n = m"

    using divmod_nat_rel [of m n] by (simp add: divmod_nat_rel_def)

next

  fix m n q :: nat

  assume "n \<noteq> 0"

  then show "(q + m * n) div n = m + q div n"

    by (induct m) (simp_all add: le_div_geq)

next

  fix m n q :: nat

  assume "m \<noteq> 0"

  hence "\<And>a b. divmod_nat_rel n q (a, b) \<Longrightarrow> divmod_nat_rel (m * n) (m * q) (a, m * b)"

    unfolding divmod_nat_rel_def

    by (auto split: split_if_asm, simp_all add: algebra_simps)

  moreover from divmod_nat_rel have "divmod_nat_rel n q (n div q, n mod q)" .

  ultimately have "divmod_nat_rel (m * n) (m * q) (n div q, m * (n mod q))" .

  thus "(m * n) div (m * q) = n div q" by (rule div_nat_unique)

next

  fix n :: nat show "n div 0 = 0"

    by (simp add: div_nat_def divmod_nat_zero)

next

  fix n :: nat show "0 div n = 0"

    by (simp add: div_nat_def divmod_nat_zero_left)

qed

end

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: prod_eq_iff prod_case_beta not_less le_div_geq le_mod_geq)

text {* Simproc for cancelling @{const div} and @{const mod} *}

ML {*

structure Cancel_Div_Mod_Nat = Cancel_Div_Mod

(

  val div_name = @{const_name div};

  val mod_name = @{const_name mod};

  val mk_binop = HOLogic.mk_binop;

  val mk_sum = Nat_Arith.mk_sum;

  val dest_sum = Nat_Arith.dest_sum;

  val div_mod_eqs = map mk_meta_eq [@{thm div_mod_equality}, @{thm div_mod_equality2}];

  val prove_eq_sums = Arith_Data.prove_conv2 all_tac (Arith_Data.simp_all_tac

    (@{thm add_0_left} :: @{thm add_0_right} :: @{thms add_ac}))

)

*}

simproc_setup cancel_div_mod_nat ("(m::nat) + n") = {* K Cancel_Div_Mod_Nat.proc *}

subsubsection {* Quotient *}

lemma div_geq: "0 < n \<Longrightarrow>  \<not> m < n \<Longrightarrow> m div n = Suc ((m - n) div n)"

by (simp add: le_div_geq linorder_not_less)

lemma div_if: "0 < n \<Longrightarrow> m div n = (if m < n then 0 else Suc ((m - n) div n))"

by (simp add: div_geq)

lemma div_mult_self_is_m [simp]: "0<n ==> (m*n) div n = (m::nat)"

by simp

lemma div_mult_self1_is_m [simp]: "0<n ==> (n*m) div n = (m::nat)"

by simp

subsubsection {* Remainder *}

lemma mod_less_divisor [simp]:

  fixes m n :: nat

  assumes "n > 0"

  shows "m mod n < (n::nat)"

  using assms divmod_nat_rel [of m n] unfolding divmod_nat_rel_def by auto

lemma mod_less_eq_dividend [simp]:

  fixes m n :: nat

  shows "m mod n \<le> m"

proof (rule add_leD2)

  from mod_div_equality have "m div n * n + m mod n = m" .

  then show "m div n * n + m mod n \<le> m" by auto

qed

lemma mod_geq: "\<not> m < (n\<Colon>nat) \<Longrightarrow> m mod n = (m - n) mod n"

by (simp add: le_mod_geq linorder_not_less)

lemma mod_if: "m mod (n\<Colon>nat) = (if m < n then m else (m - n) mod n)"

by (simp add: le_mod_geq)

lemma mod_1 [simp]: "m mod Suc 0 = 0"

by (induct m) (simp_all add: mod_geq)

(* a simple rearrangement of mod_div_equality: *)

lemma mult_div_cancel: "(n::nat) * (m div n) = m - (m mod n)"

  using mod_div_equality2 [of n m] by arith

lemma mod_le_divisor[simp]: "0 < n \<Longrightarrow> m mod n \<le> (n::nat)"

  apply (drule mod_less_divisor [where m = m])

  apply simp

  done

subsubsection {* Quotient and Remainder *}

lemma divmod_nat_rel_mult1_eq:

  "divmod_nat_rel b c (q, r)

   \<Longrightarrow> divmod_nat_rel (a * b) c (a * q + a * r div c, a * r mod c)"

by (auto simp add: split_ifs divmod_nat_rel_def algebra_simps)

lemma div_mult1_eq:

  "(a * b) div c = a * (b div c) + a * (b mod c) div (c::nat)"

by (blast intro: divmod_nat_rel_mult1_eq [THEN div_nat_unique] divmod_nat_rel)

lemma divmod_nat_rel_add1_eq:

  "divmod_nat_rel a c (aq, ar) \<Longrightarrow> divmod_nat_rel b c (bq, br)

   \<Longrightarrow> divmod_nat_rel (a + b) c (aq + bq + (ar + br) div c, (ar + br) mod c)"

by (auto simp add: split_ifs divmod_nat_rel_def algebra_simps)

(*NOT suitable for rewriting: the RHS has an instance of the LHS*)

lemma div_add1_eq:

  "(a+b) div (c::nat) = a div c + b div c + ((a mod c + b mod c) div c)"

by (blast intro: divmod_nat_rel_add1_eq [THEN div_nat_unique] divmod_nat_rel)

lemma mod_lemma: "[| (0::nat) < c; r < b |] ==> b * (q mod c) + r < b * c"

  apply (cut_tac m = q and n = c in mod_less_divisor)

  apply (drule_tac [2] m = "q mod c" in less_imp_Suc_add, auto)

  apply (erule_tac P = "%x. ?lhs < ?rhs x" in ssubst)

  apply (simp add: add_mult_distrib2)

  done

lemma divmod_nat_rel_mult2_eq:

  "divmod_nat_rel a b (q, r)

   \<Longrightarrow> divmod_nat_rel a (b * c) (q div c, b *(q mod c) + r)"

by (auto simp add: mult_ac divmod_nat_rel_def add_mult_distrib2 [symmetric] mod_lemma)

lemma div_mult2_eq: "a div (b*c) = (a div b) div (c::nat)"

by (force simp add: divmod_nat_rel [THEN divmod_nat_rel_mult2_eq, THEN div_nat_unique])

lemma mod_mult2_eq: "a mod (b*c) = b*(a div b mod c) + a mod (b::nat)"

by (auto simp add: mult_commute divmod_nat_rel [THEN divmod_nat_rel_mult2_eq, THEN mod_nat_unique])

subsubsection {* Further Facts about Quotient and Remainder *}

lemma div_1 [simp]: "m div Suc 0 = m"

by (induct m) (simp_all add: div_geq)

(* Monotonicity of div in first argument *)

lemma div_le_mono [rule_format (no_asm)]:

    "\<forall>m::nat. m \<le> n --> (m div k) \<le> (n div k)"

apply (case_tac "k=0", simp)

apply (induct "n" rule: nat_less_induct, clarify)

apply (case_tac "n<k")

(* 1  case n<k *)

apply simp

(* 2  case n >= k *)

apply (case_tac "m<k")

(* 2.1  case m<k *)

apply simp

(* 2.2  case m>=k *)

apply (simp add: div_geq diff_le_mono)

done

(* Antimonotonicity of div in second argument *)

lemma div_le_mono2: "!!m::nat. [| 0<m; m\<le>n |] ==> (k div n) \<le> (k div m)"

apply (subgoal_tac "0<n")

 prefer 2 apply simp

apply (induct_tac k rule: nat_less_induct)

apply (rename_tac "k")

apply (case_tac "k<n", simp)

apply (subgoal_tac "~ (k<m) ")

 prefer 2 apply simp

apply (simp add: div_geq)

apply (subgoal_tac "(k-n) div n \<le> (k-m) div n")

 prefer 2

 apply (blast intro: div_le_mono diff_le_mono2)

apply (rule le_trans, simp)

apply (simp)

done

lemma div_le_dividend [simp]: "m div n \<le> (m::nat)"

apply (case_tac "n=0", simp)

apply (subgoal_tac "m div n \<le> m div 1", simp)

apply (rule div_le_mono2)

apply (simp_all (no_asm_simp))

done

(* Similar for "less than" *)

lemma div_less_dividend [simp]:

  "\<lbrakk>(1::nat) < n; 0 < m\<rbrakk> \<Longrightarrow> m div n < m"

apply (induct m rule: nat_less_induct)

apply (rename_tac "m")

apply (case_tac "m<n", simp)

apply (subgoal_tac "0<n")

 prefer 2 apply simp

apply (simp add: div_geq)

apply (case_tac "n<m")

 apply (subgoal_tac "(m-n) div n < (m-n) ")

  apply (rule impI less_trans_Suc)+

apply assumption

  apply (simp_all)

done

text{*A fact for the mutilated chess board*}

lemma mod_Suc: "Suc(m) mod n = (if Suc(m mod n) = n then 0 else Suc(m mod n))"

apply (case_tac "n=0", simp)

apply (induct "m" rule: nat_less_induct)

apply (case_tac "Suc (na) <n")

(* case Suc(na) < n *)

apply (frule lessI [THEN less_trans], simp add: less_not_refl3)

(* case n \<le> Suc(na) *)

apply (simp add: linorder_not_less le_Suc_eq mod_geq)

apply (auto simp add: Suc_diff_le le_mod_geq)

done

lemma mod_eq_0_iff: "(m mod d = 0) = (\<exists>q::nat. m = d*q)"

by (auto simp add: dvd_eq_mod_eq_0 [symmetric] dvd_def)

lemmas mod_eq_0D [dest!] = mod_eq_0_iff [THEN iffD1]

(*Loses information, namely we also have r<d provided d is nonzero*)

lemma mod_eqD: "(m mod d = r) ==> \<exists>q::nat. m = r + q*d"

  apply (cut_tac a = m in mod_div_equality)

  apply (simp only: add_ac)

  apply (blast intro: sym)

  done

lemma split_div:

 "P(n div k :: nat) =

 ((k = 0 \<longrightarrow> P 0) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P i)))"

 (is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")

proof

  assume P: ?P

  show ?Q

  proof (cases)

    assume "k = 0"

    with P show ?Q by simp

  next

    assume not0: "k \<noteq> 0"

    thus ?Q

    proof (simp, intro allI impI)

      fix i j

      assume n: "n = k*i + j" and j: "j < k"

      show "P i"

      proof (cases)

        assume "i = 0"

        with n j P show "P i" by simp

      next

        assume "i \<noteq> 0"

        with not0 n j P show "P i" by(simp add:add_ac)

      qed

    qed

  qed

next

  assume Q: ?Q

  show ?P

  proof (cases)

    assume "k = 0"

    with Q show ?P by simp

  next

    assume not0: "k \<noteq> 0"

    with Q have R: ?R by simp

    from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]

    show ?P by simp

  qed

qed

lemma split_div_lemma:

  assumes "0 < n"

  shows "n * q \<le> m \<and> m < n * Suc q \<longleftrightarrow> q = ((m\<Colon>nat) div n)" (is "?lhs \<longleftrightarrow> ?rhs")

proof

  assume ?rhs

  with mult_div_cancel have nq: "n * q = m - (m mod n)" by simp

  then have A: "n * q \<le> m" by simp

  have "n - (m mod n) > 0" using mod_less_divisor assms by auto

  then have "m < m + (n - (m mod n))" by simp

  then have "m < n + (m - (m mod n))" by simp

  with nq have "m < n + n * q" by simp

  then have B: "m < n * Suc q" by simp

  from A B show ?lhs ..

next

  assume P: ?lhs

  then have "divmod_nat_rel m n (q, m - n * q)"

    unfolding divmod_nat_rel_def by (auto simp add: mult_ac)

  with divmod_nat_rel_unique divmod_nat_rel [of m n]

  have "(q, m - n * q) = (m div n, m mod n)" by auto

  then show ?rhs by simp

qed

theorem split_div':

  "P ((m::nat) div n) = ((n = 0 \<and> P 0) \<or>

   (\<exists>q. (n * q \<le> m \<and> m < n * (Suc q)) \<and> P q))"

  apply (case_tac "0 < n")

  apply (simp only: add: split_div_lemma)

  apply simp_all

  done

lemma split_mod:

 "P(n mod k :: nat) =

 ((k = 0 \<longrightarrow> P n) \<and> (k \<noteq> 0 \<longrightarrow> (!i. !j<k. n = k*i + j \<longrightarrow> P j)))"

 (is "?P = ?Q" is "_ = (_ \<and> (_ \<longrightarrow> ?R))")

proof

  assume P: ?P

  show ?Q

  proof (cases)

    assume "k = 0"

    with P show ?Q by simp

  next

    assume not0: "k \<noteq> 0"

    thus ?Q

    proof (simp, intro allI impI)

      fix i j

      assume "n = k*i + j" "j < k"

      thus "P j" using not0 P by(simp add:add_ac mult_ac)

    qed

  qed

next

  assume Q: ?Q

  show ?P

  proof (cases)

    assume "k = 0"

    with Q show ?P by simp

  next

    assume not0: "k \<noteq> 0"

    with Q have R: ?R by simp

    from not0 R[THEN spec,of "n div k",THEN spec, of "n mod k"]

    show ?P by simp

  qed

qed

theorem mod_div_equality': "(m::nat) mod n = m - (m div n) * n"

  using mod_div_equality [of m n] by arith

lemma div_mod_equality': "(m::nat) div n * n = m - m mod n"

  using mod_div_equality [of m n] by arith

(* FIXME: very similar to mult_div_cancel *)