# Theory Euclidean_Division

theory Euclidean_Division
imports Int Lattices_Big
```(*  Title:      HOL/Euclidean_Division.thy
Author:     Manuel Eberl, TU Muenchen
Author:     Florian Haftmann, TU Muenchen
*)

section ‹Division in euclidean (semi)rings›

theory Euclidean_Division
imports Int Lattices_Big
begin

subsection ‹Euclidean (semi)rings with explicit division and remainder›

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

lemma euclidean_size_eq_0_iff [simp]:
"euclidean_size b = 0 ⟷ b = 0"
proof
assume "b = 0"
then show "euclidean_size b = 0"
by simp
next
assume "euclidean_size b = 0"
show "b = 0"
proof (rule ccontr)
assume "b ≠ 0"
with mod_size_less have "euclidean_size (b mod b) < euclidean_size b" .
with ‹euclidean_size b = 0› show False
by simp
qed
qed

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

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

lemma dvd_euclidean_size_eq_imp_dvd:
assumes "a ≠ 0" and "euclidean_size a = euclidean_size b"
and "b dvd a"
shows "a dvd b"
proof (rule ccontr)
assume "¬ a dvd b"
hence "b mod a ≠ 0" using mod_0_imp_dvd [of b a] by blast
then have "b mod a ≠ 0" by (simp add: mod_eq_0_iff_dvd)
from ‹b dvd a› have "b dvd b mod a" by (simp add: dvd_mod_iff)
then obtain c where "b mod a = b * c" unfolding dvd_def by blast
with ‹b mod a ≠ 0› have "c ≠ 0" by auto
with ‹b mod a = b * c› have "euclidean_size (b mod a) ≥ euclidean_size b"
using size_mult_mono by force
moreover from ‹¬ a dvd b› and ‹a ≠ 0›
have "euclidean_size (b mod a) < euclidean_size a"
using mod_size_less by blast
ultimately show False using ‹euclidean_size a = euclidean_size b›
by simp
qed

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

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

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

lemma euclidean_size_times_nonunit:
assumes "a ≠ 0" "b ≠ 0" "¬ is_unit a"
shows   "euclidean_size b < euclidean_size (a * b)"
proof (rule ccontr)
assume "¬euclidean_size b < euclidean_size (a * b)"
with size_mult_mono'[OF assms(1), of b]
have eq: "euclidean_size (a * b) = euclidean_size b" by simp
have "a * b dvd b"
by (rule dvd_euclidean_size_eq_imp_dvd [OF _ eq]) (insert assms, simp_all)
hence "a * b dvd 1 * b" by simp
with ‹b ≠ 0› have "is_unit a" by (subst (asm) dvd_times_right_cancel_iff)
with assms(3) show False by contradiction
qed

lemma dvd_imp_size_le:
assumes "a dvd b" "b ≠ 0"
shows   "euclidean_size a ≤ euclidean_size b"
using assms by (auto elim!: dvdE simp: size_mult_mono)

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

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

lemma coprime_mod_left_iff [simp]:
"coprime (a mod b) b ⟷ coprime a b" if "b ≠ 0"
by (rule; rule coprimeI)
(use that in ‹auto dest!: dvd_mod_imp_dvd coprime_common_divisor simp add: dvd_mod_iff›)

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

end

class euclidean_ring = idom_modulo + euclidean_semiring
begin

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

end

subsection ‹Euclidean (semi)rings with cancel rules›

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

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

lemma div_mult_self3 [simp]:
assumes "b ≠ 0"
shows "(c * b + a) div b = c + a div b"

lemma div_mult_self4 [simp]:
assumes "b ≠ 0"
shows "(b * c + a) div b = c + a div b"

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"
also from False div_mult_self1 [of b a c] have
"… = (c + a div b) * b + (a + c * b) mod b"
finally have "a = a div b * b + (a + c * b) mod b"
then have "a div b * b + (a + c * b) mod b = a div b * b + a mod b"
then show ?thesis by simp
qed

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

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

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

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

assumes "b ≠ 0"
shows "(b + a) div b = a div b + 1"

assumes "b ≠ 0"
shows "(a + b) div b = a div b + 1"

"(b + a) mod b = a mod b"

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

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

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

lemma mod_mod_cancel:
assumes "c dvd b"
shows "a mod b mod c = a mod c"
proof -
from ‹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 "… = (a mod (c * k) + a div (c * k) * k * c) mod c"
by (simp only: mod_mult_self1)
also have "… = (a div (c * k) * (c * k) + a mod (c * k)) mod c"
by (simp only: ac_simps)
also have "… = a mod c"
by (simp only: div_mult_mod_eq)
finally show ?thesis .
qed

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

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

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

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

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

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

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

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

lemma div_plus_div_distrib_dvd_right:
"c dvd b ⟹ (a + b) div c = a div c + b div c"
using div_plus_div_distrib_dvd_left [of c b a]

named_theorems mod_simps

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

"(a + b mod c) mod c = (a + b) mod c"

"(a mod c + b mod c) mod c = (a + b) mod c"

lemma mod_sum_eq [mod_simps]:
"(∑i∈A. f i mod a) mod a = sum f A mod a"
proof (induct A rule: infinite_finite_induct)
case (insert i A)
then have "(∑i∈insert i A. f i mod a) mod a
= (f i mod a + (∑i∈A. f i mod a)) mod a"
by simp
also have "… = (f i + (∑i∈A. f i mod a) mod a) mod a"
also have "… = (f i + (∑i∈A. f i) mod a) mod a"
finally show ?case
qed simp_all

assumes "a mod c = a' mod c"
assumes "b mod c = b' mod c"
shows "(a + b) mod c = (a' + b') mod c"
proof -
have "(a mod c + b mod c) mod c = (a' mod c + b' mod c) mod c"
unfolding assms ..
then show ?thesis
qed

text ‹Multiplication respects modular equivalence.›

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

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

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

lemma mod_prod_eq [mod_simps]:
"(∏i∈A. f i mod a) mod a = prod f A mod a"
proof (induct A rule: infinite_finite_induct)
case (insert i A)
then have "(∏i∈insert i A. f i mod a) mod a
= (f i mod a * (∏i∈A. f i mod a)) mod a"
by simp
also have "… = (f i * ((∏i∈A. f i mod a) mod a)) mod a"
also have "… = (f i * ((∏i∈A. f i) mod a)) mod a"
finally show ?case
qed simp_all

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

text ‹Exponentiation respects modular equivalence.›

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

end

class euclidean_ring_cancel = euclidean_ring + euclidean_semiring_cancel
begin

subclass idom_divide ..

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

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

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

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

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

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

text ‹Negation respects modular equivalence.›

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

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

text ‹Subtraction respects modular equivalence.›

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

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

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

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

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

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

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

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

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

end

subsection ‹Uniquely determined division›

class unique_euclidean_semiring = euclidean_semiring +
assumes euclidean_size_mult: "euclidean_size (a * b) = euclidean_size a * euclidean_size b"
fixes division_segment :: "'a ⇒ 'a"
assumes is_unit_division_segment [simp]: "is_unit (division_segment a)"
and division_segment_mult:
"a ≠ 0 ⟹ b ≠ 0 ⟹ division_segment (a * b) = division_segment a * division_segment b"
and division_segment_mod:
"b ≠ 0 ⟹ ¬ b dvd a ⟹ division_segment (a mod b) = division_segment b"
assumes div_bounded:
"b ≠ 0 ⟹ division_segment r = division_segment b
⟹ euclidean_size r < euclidean_size b
⟹ (q * b + r) div b = q"
begin

lemma division_segment_not_0 [simp]:
"division_segment a ≠ 0"
using is_unit_division_segment [of a] is_unitE [of "division_segment a"] by blast

lemma divmod_cases [case_names divides remainder by0]:
obtains
(divides) q where "b ≠ 0"
and "a div b = q"
and "a mod b = 0"
and "a = q * b"
| (remainder) q r where "b ≠ 0"
and "division_segment r = division_segment b"
and "euclidean_size r < euclidean_size b"
and "r ≠ 0"
and "a div b = q"
and "a mod b = r"
and "a = q * b + r"
| (by0) "b = 0"
proof (cases "b = 0")
case True
then show thesis
by (rule by0)
next
case False
show thesis
proof (cases "b dvd a")
case True
then obtain q where "a = b * q" ..
with ‹b ≠ 0› divides
show thesis
next
case False
then have "a mod b ≠ 0"
moreover from ‹b ≠ 0› ‹¬ b dvd a› have "division_segment (a mod b) = division_segment b"
by (rule division_segment_mod)
moreover have "euclidean_size (a mod b) < euclidean_size b"
using ‹b ≠ 0› by (rule mod_size_less)
moreover have "a = a div b * b + a mod b"
ultimately show thesis
using ‹b ≠ 0› by (blast intro!: remainder)
qed
qed

lemma div_eqI:
"a div b = q" if "b ≠ 0" "division_segment r = division_segment b"
"euclidean_size r < euclidean_size b" "q * b + r = a"
proof -
from that have "(q * b + r) div b = q"
by (auto intro: div_bounded)
with that show ?thesis
by simp
qed

lemma mod_eqI:
"a mod b = r" if "b ≠ 0" "division_segment r = division_segment b"
"euclidean_size r < euclidean_size b" "q * b + r = a"
proof -
from that have "a div b = q"
by (rule div_eqI)
moreover have "a div b * b + a mod b = a"
by (fact div_mult_mod_eq)
ultimately have "a div b * b + a mod b = a div b * b + r"
using ‹q * b + r = a› by simp
then show ?thesis
by simp
qed

subclass euclidean_semiring_cancel
proof
show "(a + c * b) div b = c + a div b" if "b ≠ 0" for a b c
proof (cases a b rule: divmod_cases)
case by0
with ‹b ≠ 0› show ?thesis
by simp
next
case (divides q)
then show ?thesis
next
case (remainder q r)
then show ?thesis
by (auto intro: div_eqI simp add: algebra_simps)
qed
next
show"(c * a) div (c * b) = a div b" if "c ≠ 0" for a b c
proof (cases a b rule: divmod_cases)
case by0
then show ?thesis
by simp
next
case (divides q)
with ‹c ≠ 0› show ?thesis
by (simp add: mult.left_commute [of c])
next
case (remainder q r)
from ‹b ≠ 0› ‹c ≠ 0› have "b * c ≠ 0"
by simp
from remainder ‹c ≠ 0›
have "division_segment (r * c) = division_segment (b * c)"
and "euclidean_size (r * c) < euclidean_size (b * c)"
by (simp_all add: division_segment_mult division_segment_mod euclidean_size_mult)
with remainder show ?thesis
by (auto intro!: div_eqI [of _ "c * (a mod b)"] simp add: algebra_simps)
(use ‹b * c ≠ 0› in simp)
qed
qed

lemma div_mult1_eq:
"(a * b) div c = a * (b div c) + a * (b mod c) div c"
proof (cases "a * (b mod c)" c rule: divmod_cases)
case (divides q)
have "a * b = a * (b div c * c + b mod c)"
also have "… = (a * (b div c) + q) * c"
using divides by (simp add: algebra_simps)
finally have "(a * b) div c = … div c"
by simp
with divides show ?thesis
by simp
next
case (remainder q r)
from remainder(1-3) show ?thesis
proof (rule div_eqI)
have "a * b = a * (b div c * c + b mod c)"
also have "… = a * c * (b div c) + q * c + r"
using remainder by (simp add: algebra_simps)
finally show "(a * (b div c) + a * (b mod c) div c) * c + r = a * b"
using remainder(5-7) by (simp add: algebra_simps)
qed
next
case by0
then show ?thesis
by simp
qed

"(a + b) div c = a div c + b div c + (a mod c + b mod c) div c"
proof (cases "a mod c + b mod c" c rule: divmod_cases)
case (divides q)
have "a + b = (a div c * c + a mod c) + (b div c * c + b mod c)"
using mod_mult_div_eq [of a c] mod_mult_div_eq [of b c] by (simp add: ac_simps)
also have "… = (a div c + b div c) * c + (a mod c + b mod c)"
also have "… = (a div c + b div c + q) * c"
using divides by (simp add: algebra_simps)
finally have "(a + b) div c = (a div c + b div c + q) * c div c"
by simp
with divides show ?thesis
by simp
next
case (remainder q r)
from remainder(1-3) show ?thesis
proof (rule div_eqI)
have "(a div c + b div c + q) * c + r + (a mod c + b mod c) =
(a div c * c + a mod c) + (b div c * c + b mod c) + q * c + r"
also have "… = a + b + (a mod c + b mod c)"
finally show "(a div c + b div c + (a mod c + b mod c) div c) * c + r = a + b"
using remainder by simp
qed
next
case by0
then show ?thesis
by simp
qed

lemma div_eq_0_iff:
"a div b = 0 ⟷ euclidean_size a < euclidean_size b ∨ b = 0" (is "_ ⟷ ?P")
if "division_segment a = division_segment b"
proof
assume ?P
with that show "a div b = 0"
by (cases "b = 0") (auto intro: div_eqI)
next
assume "a div b = 0"
then have "a mod b = a"
using div_mult_mod_eq [of a b] by simp
with mod_size_less [of b a] show ?P
by auto
qed

end

class unique_euclidean_ring = euclidean_ring + unique_euclidean_semiring
begin

subclass euclidean_ring_cancel ..

end

subsection ‹Euclidean division on @{typ nat}›

instantiation nat :: normalization_semidom
begin

definition normalize_nat :: "nat ⇒ nat"
where [simp]: "normalize = (id :: nat ⇒ nat)"

definition unit_factor_nat :: "nat ⇒ nat"
where "unit_factor n = (if n = 0 then 0 else 1 :: nat)"

lemma unit_factor_simps [simp]:
"unit_factor 0 = (0::nat)"
"unit_factor (Suc n) = 1"

definition divide_nat :: "nat ⇒ nat ⇒ nat"
where "m div n = (if n = 0 then 0 else Max {k::nat. k * n ≤ m})"

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

end

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

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

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

instantiation nat :: unique_euclidean_semiring
begin

definition euclidean_size_nat :: "nat ⇒ nat"
where [simp]: "euclidean_size_nat = id"

definition division_segment_nat :: "nat ⇒ nat"
where [simp]: "division_segment_nat n = 1"

definition modulo_nat :: "nat ⇒ nat ⇒ nat"
where "m mod n = m - (m div n * (n::nat))"

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

end

text ‹Tool support›

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

val div_mod_eqs = map mk_meta_eq @{thms cancel_div_mod_rules};

val prove_eq_sums = Arith_Data.prove_conv2 all_tac
)
›

simproc_setup cancel_div_mod_nat ("(m::nat) + n") =
‹K Cancel_Div_Mod_Nat.proc›

lemma div_nat_eqI:
"m div n = q" if "n * q ≤ m" and "m < n * Suc q" for m n q :: nat
by (rule div_eqI [of _ "m - n * q"]) (use that in ‹simp_all add: algebra_simps›)

lemma mod_nat_eqI:
"m mod n = r" if "r < n" and "r ≤ m" and "n dvd m - r" for m n r :: nat
by (rule mod_eqI [of _ _ "(m - r) div n"]) (use that in ‹simp_all add: algebra_simps›)

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

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

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

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

lemma div_times_less_eq_dividend [simp]:
"m div n * n ≤ m" for m n :: nat

lemma times_div_less_eq_dividend [simp]:
"n * (m div n) ≤ m" for m n :: nat
using div_times_less_eq_dividend [of m n]

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

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

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

lemma mod_less_eq_dividend [simp]:
"m mod n ≤ m" for m n :: nat
from div_mult_mod_eq have "m div n * n + m mod n = m" .
then show "m div n * n + m mod n ≤ m" by auto
qed

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

lemma le_div_geq:
"m div n = Suc ((m - n) div n)" if "0 < n" and "n ≤ m" for m n :: nat
proof -
from ‹n ≤ m› obtain q where "m = n + q"
with ‹0 < n› show ?thesis
qed

lemma le_mod_geq:
"m mod n = (m - n) mod n" if "n ≤ m" for m n :: nat
proof -
from ‹n ≤ m› obtain q where "m = n + q"
then show ?thesis
by simp
qed

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

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

lemma div_eq_0_iff:
"m div n = 0 ⟷ m < n ∨ n = 0" for m n :: nat

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

lemma mod_greater_zero_iff_not_dvd:
"m mod n > 0 ⟷ ¬ n dvd m" for m n :: nat

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

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

lemma div2_Suc_Suc [simp]:
"Suc (Suc m) div 2 = Suc (m div 2)"

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

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

lemma mod2_Suc_Suc [simp]:
"Suc (Suc m) mod 2 = m mod 2"

"(m + m) div 2 = m" for m :: nat

"(m + m) mod 2 = 0" for m :: nat

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

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

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

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

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

context
fixes m n q :: nat
begin

private lemma eucl_rel_mult2:
"m mod n + n * (m div n mod q) < n * q"
if "n > 0" and "q > 0"
proof -
from ‹n > 0› have "m mod n < n"
by (rule mod_less_divisor)
from ‹q > 0› have "m div n mod q < q"
by (rule mod_less_divisor)
then obtain s where "q = Suc (m div n mod q + s)"
moreover have "m mod n + n * (m div n mod q) < n * Suc (m div n mod q + s)"
ultimately show ?thesis
by simp
qed

lemma div_mult2_eq:
"m div (n * q) = (m div n) div q"
proof (cases "n = 0 ∨ q = 0")
case True
then show ?thesis
by auto
next
case False
with eucl_rel_mult2 show ?thesis
by (auto intro: div_eqI [of _ "n * (m div n mod q) + m mod n"]
qed

lemma mod_mult2_eq:
"m mod (n * q) = n * (m div n mod q) + m mod n"
proof (cases "n = 0 ∨ q = 0")
case True
then show ?thesis
by auto
next
case False
with eucl_rel_mult2 show ?thesis
by (auto intro: mod_eqI [of _ _ "(m div n) div q"]
qed

end

lemma div_le_mono:
"m div k ≤ n div k" if "m ≤ n" for m n k :: nat
proof -
from that obtain q where "n = m + q"
then show ?thesis
qed

text ‹Antimonotonicity of @{const divide} in second argument›

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

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

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

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

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

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))" (is "_ = ?rhs")
proof (cases "n = 0")
case True
then show ?thesis
by simp
next
case False
have "Suc m mod n = Suc (m mod n) mod n"
also have "… = ?rhs"
using False by (auto intro!: mod_nat_eqI intro: neq_le_trans simp add: Suc_le_eq)
finally show ?thesis .
qed

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

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

lemma Suc_div_le_mono [simp]:
"m div n ≤ Suc m div n"

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

"m div Suc (Suc (Suc n)) = m div (3 + n)"

"m mod Suc (Suc (Suc n)) = m mod (3 + n)"

"Suc (Suc (Suc m)) div n = (3 + m) div n"

"Suc (Suc (Suc m)) mod n = (3 + m) mod n"

Suc_div_eq_add3_div [of _ "numeral v"] for v

Suc_mod_eq_add3_mod [of _ "numeral v"] for v

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

text ‹An ``induction'' law for modulus arithmetic.›

lemma mod_induct [consumes 3, case_names step]:
"P m" if "P n" and "n < p" and "m < p"
and step: "⋀n. n < p ⟹ P n ⟹ P (Suc n mod p)"
using ‹m < p› proof (induct m)
case 0
show ?case
proof (rule ccontr)
assume "¬ P 0"
from ‹n < p› have "0 < p"
by simp
from ‹n < p› obtain m where "0 < m" and "p = n + m"
with ‹P n› have "P (p - m)"
by simp
moreover have "¬ P (p - m)"
using ‹0 < m› proof (induct m)
case 0
then show ?case
by simp
next
case (Suc m)
show ?case
proof
assume P: "P (p - Suc m)"
with ‹¬ P 0› have "Suc m < p"
by (auto intro: ccontr)
then have "Suc (p - Suc m) = p - m"
by arith
moreover from ‹0 < p› have "p - Suc m < p"
by arith
with P step have "P ((Suc (p - Suc m)) mod p)"
by blast
ultimately show False
using ‹¬ P 0› Suc.hyps by (cases "m = 0") simp_all
qed
qed
ultimately show False
by blast
qed
next
case (Suc m)
then have "m < p" and mod: "Suc m mod p = Suc m"
by simp_all
from ‹m < p› have "P m"
by (rule Suc.hyps)
with ‹m < p› have "P (Suc m mod p)"
by (rule step)
with mod show ?case
by simp
qed

lemma split_div:
"P (m div n) ⟷ (n = 0 ⟶ P 0) ∧ (n ≠ 0 ⟶
(∀i j. j < n ⟶ m = n * i + j ⟶ P i))"
(is "?P = ?Q") for m n :: nat
proof (cases "n = 0")
case True
then show ?thesis
by simp
next
case False
show ?thesis
proof
assume ?P
with False show ?Q
by auto
next
assume ?Q
with False have *: "⋀i j. j < n ⟹ m = n * i + j ⟹ P i"
by simp
with False show ?P
by (auto intro: * [of "m mod n"])
qed
qed

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

lemma split_mod:
"P (m mod n) ⟷ (n = 0 ⟶ P m) ∧ (n ≠ 0 ⟶
(∀i j. j < n ⟶ m = n * i + j ⟶ P j))"
(is "?P ⟷ ?Q") for m n :: nat
proof (cases "n = 0")
case True
then show ?thesis
by simp
next
case False
show ?thesis
proof
assume ?P
with False show ?Q
by auto
next
assume ?Q
with False have *: "⋀i j. j < n ⟹ m = n * i + j ⟹ P j"
by simp
with False show ?P
by (auto intro: * [of _ "m div n"])
qed
qed

subsection ‹Euclidean division on @{typ int}›

instantiation int :: normalization_semidom
begin

definition normalize_int :: "int ⇒ int"
where [simp]: "normalize = (abs :: int ⇒ int)"

definition unit_factor_int :: "int ⇒ int"
where [simp]: "unit_factor = (sgn :: int ⇒ int)"

definition divide_int :: "int ⇒ int ⇒ int"
where "k div l = (if l = 0 then 0
else if sgn k = sgn l
then int (nat ¦k¦ div nat ¦l¦)
else - int (nat ¦k¦ div nat ¦l¦ + of_bool (¬ l dvd k)))"

lemma divide_int_unfold:
"(sgn k * int m) div (sgn l * int n) =
(if sgn l = 0 ∨ sgn k = 0 ∨ n = 0 then 0
else if sgn k = sgn l
then int (m div n)
else - int (m div n + of_bool (¬ n dvd m)))"
by (auto simp add: divide_int_def sgn_0_0 sgn_1_pos sgn_mult abs_mult
nat_mult_distrib)

instance proof
fix k :: int show "k div 0 = 0"
next
fix k l :: int
assume "l ≠ 0"
obtain n m and s t where k: "k = sgn s * int n" and l: "l = sgn t * int m"
by (blast intro: int_sgnE elim: that)
then have "k * l = sgn (s * t) * int (n * m)"
with k l ‹l ≠ 0› show "k * l div l = k"
by (simp only: divide_int_unfold)
(auto simp add: algebra_simps sgn_mult sgn_1_pos sgn_0_0)
qed (auto simp add: sgn_mult mult_sgn_abs abs_eq_iff')

end

lemma coprime_int_iff [simp]:
"coprime (int m) (int n) ⟷ coprime m n" (is "?P ⟷ ?Q")
proof
assume ?P
show ?Q
proof (rule coprimeI)
fix q
assume "q dvd m" "q dvd n"
then have "int q dvd int m" "int q dvd int n"
by simp_all
with ‹?P› have "is_unit (int q)"
by (rule coprime_common_divisor)
then show "is_unit q"
by simp
qed
next
assume ?Q
show ?P
proof (rule coprimeI)
fix k
assume "k dvd int m" "k dvd int n"
then have "nat ¦k¦ dvd m" "nat ¦k¦ dvd n"
by simp_all
with ‹?Q› have "is_unit (nat ¦k¦)"
by (rule coprime_common_divisor)
then show "is_unit k"
by simp
qed
qed

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

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

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

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

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

instantiation int :: idom_modulo
begin

definition modulo_int :: "int ⇒ int ⇒ int"
where "k mod l = (if l = 0 then k
else if sgn k = sgn l
then sgn l * int (nat ¦k¦ mod nat ¦l¦)
else sgn l * (¦l¦ * of_bool (¬ l dvd k) - int (nat ¦k¦ mod nat ¦l¦)))"

lemma modulo_int_unfold:
"(sgn k * int m) mod (sgn l * int n) =
(if sgn l = 0 ∨ sgn k = 0 ∨ n = 0 then sgn k * int m
else if sgn k = sgn l
then sgn l * int (m mod n)
else sgn l * (int (n * of_bool (¬ n dvd m)) - int (m mod n)))"
by (auto simp add: modulo_int_def sgn_0_0 sgn_1_pos sgn_mult abs_mult
nat_mult_distrib)

instance proof
fix k l :: int
obtain n m and s t where "k = sgn s * int n" and "l = sgn t * int m"
by (blast intro: int_sgnE elim: that)
then show "k div l * l + k mod l = k"
by (auto simp add: divide_int_unfold modulo_int_unfold algebra_simps dest!: sgn_not_eq_imp)
distrib_left [symmetric] minus_mult_right
del: of_nat_mult minus_mult_right [symmetric])
qed

end

instantiation int :: unique_euclidean_ring
begin

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

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

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

lemma abs_division_segment [simp]:
"¦division_segment k¦ = 1" for k :: int

lemma abs_mod_less:
"¦k mod l¦ < ¦l¦" if "l ≠ 0" for k l :: int
proof -
obtain n m and s t where "k = sgn s * int n" and "l = sgn t * int m"
by (blast intro: int_sgnE elim: that)
with that show ?thesis
by (simp add: modulo_int_unfold sgn_0_0 sgn_1_pos sgn_1_neg
abs_mult mod_greater_zero_iff_not_dvd)
qed

lemma sgn_mod:
"sgn (k mod l) = sgn l" if "l ≠ 0" "¬ l dvd k" for k l :: int
proof -
obtain n m and s t where "k = sgn s * int n" and "l = sgn t * int m"
by (blast intro: int_sgnE elim: that)
with that show ?thesis
by (simp add: modulo_int_unfold sgn_0_0 sgn_1_pos sgn_1_neg
sgn_mult mod_eq_0_iff_dvd)
qed

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

end

lemma pos_mod_bound [simp]:
"k mod l < l" if "l > 0" for k l :: int
proof -
obtain m and s where "k = sgn s * int m"
by (blast intro: int_sgnE elim: that)
moreover from that obtain n where "l = sgn 1 * int n"
by (cases l) auto
ultimately show ?thesis
using that by (simp only: modulo_int_unfold)
qed

lemma pos_mod_sign [simp]:
"0 ≤ k mod l" if "l > 0" for k l :: int
proof -
obtain m and s where "k = sgn s * int m"
by (blast intro: int_sgnE elim: that)
moreover from that obtain n where "l = sgn 1 * int n"
by (cases l) auto
ultimately show ?thesis
using that by (simp only: modulo_int_unfold) simp
qed

subsection ‹Code generation›

code_identifier
code_module Euclidean_Division ⇀ (SML) Arith and (OCaml) Arith and (Haskell) Arith

end
```