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Quellcode-Bibliothek
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(* Title: HOL/Euclidean_Division.thy
Author: Manuel Eberl, TU Muenchen
Author: Florian Haftmann, TU Muenchen
*)
section \<open>Division in euclidean (semi)rings\<close>
theory Euclidean_Division
imports Int Lattices_Big
begin
subsection \<open>Euclidean (semi)rings with explicit division and remainder\<close>
class euclidean_semiring = semidom_modulo +
fixes euclidean_size :: "'a \ 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 \<open>euclidean_size b = 0\<close> 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 \<open>b dvd a\<close> have "b dvd b mod a" by (simp add: dvd_mod_iff)
then obtain c where "b mod a = b * c" unfolding dvd_def by blast
with \<open>b mod a \<noteq> 0\<close> have "c \<noteq> 0" by auto
with \<open>b mod a = b * c\<close> have "euclidean_size (b mod a) \<ge> euclidean_size b"
using size_mult_mono by force
moreover from \<open>\<not> a dvd b\<close> and \<open>a \<noteq> 0\<close>
have "euclidean_size (b mod a) < euclidean_size a"
using mod_size_less by blast
ultimately show False using \<open>euclidean_size a = euclidean_size b\<close>
by simp
qed
lemma euclidean_size_times_unit:
assumes "is_unit a"
shows "euclidean_size (a * b) = euclidean_size b"
proof (rule antisym)
from assms have [simp]: "a \ 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"
by (simp add: algebra_simps unit_div_mult_swap)
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 \<open>b \<noteq> 0\<close> have "is_unit a" by (subst (asm) dvd_times_right_cancel_iff)
with assms(3) show False by contradiction
qed
lemma dvd_imp_size_le:
assumes "a dvd b" "b \ 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 mod_eq_self_iff_div_eq_0:
"a mod b = a \ a div b = 0" (is "?P \ ?Q")
proof
assume ?P
with div_mult_mod_eq [of a b] show ?Q
by auto
next
assume ?Q
with div_mult_mod_eq [of a b] show ?P
by simp
qed
lemma coprime_mod_left_iff [simp]:
"coprime (a mod b) b \ coprime a b" if "b \ 0"
by (rule; rule coprimeI)
(use that in \<open>auto dest!: dvd_mod_imp_dvd coprime_common_divisor simp add: dvd_mod_iff\<close>)
lemma coprime_mod_right_iff [simp]:
"coprime a (b mod a) \ 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 \<open>Euclidean (semi)rings with cancel rules\<close>
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"
using assms by (simp add: add.commute)
lemma div_mult_self4 [simp]:
assumes "b \ 0"
shows "(b * c + a) div b = c + a div b"
using assms by (simp add: add.commute)
lemma mod_mult_self1 [simp]: "(a + c * b) mod b = a mod b"
proof (cases "b = 0")
case True then show ?thesis by simp
next
case False
have "a + c * b = (a + c * b) div b * b + (a + c * b) mod b"
by (simp add: div_mult_mod_eq)
also from False div_mult_self1 [of b a c] have
"\ = (c + a div b) * b + (a + c * b) mod b"
by (simp add: algebra_simps)
finally have "a = a div b * b + (a + c * b) mod b"
by (simp add: add.commute [of a] add.assoc distrib_right)
then have "a div b * b + (a + c * b) mod b = a div b * b + a mod b"
by (simp add: div_mult_mod_eq)
then show ?thesis by simp
qed
lemma mod_mult_self2 [simp]:
"(a + b * c) mod b = a mod b"
by (simp add: mult.commute [of b])
lemma mod_mult_self3 [simp]:
"(c * b + a) mod b = a mod b"
by (simp add: add.commute)
lemma mod_mult_self4 [simp]:
"(b * c + a) mod b = a mod b"
by (simp add: add.commute)
lemma mod_mult_self1_is_0 [simp]:
"b * a mod b = 0"
using mod_mult_self2 [of 0 b a] by simp
lemma mod_mult_self2_is_0 [simp]:
"a * b mod b = 0"
using mod_mult_self1 [of 0 a b] by simp
lemma div_add_self1:
assumes "b \ 0"
shows "(b + a) div b = a div b + 1"
using assms div_mult_self1 [of b a 1] by (simp add: add.commute)
lemma div_add_self2:
assumes "b \ 0"
shows "(a + b) div b = a div b + 1"
using assms div_add_self1 [of b a] by (simp add: add.commute)
lemma mod_add_self1 [simp]:
"(b + a) mod b = a mod b"
using mod_mult_self1 [of a 1 b] by (simp add: add.commute)
lemma mod_add_self2 [simp]:
"(a + b) mod b = a mod b"
using mod_mult_self1 [of a 1 b] by simp
lemma mod_div_trivial [simp]:
"a mod b div b = 0"
proof (cases "b = 0")
assume "b = 0"
thus ?thesis by simp
next
assume "b \ 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
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 "\ = a mod b"
by (simp only: mod_div_mult_eq)
finally show ?thesis .
qed
lemma mod_mod_cancel:
assumes "c dvd b"
shows "a mod b mod c = a mod c"
proof -
from \<open>c dvd b\<close> obtain k where "b = c * k"
by (rule dvdE)
have "a mod b mod c = a mod (c * k) mod c"
by (simp only: \<open>b = c * k\<close>)
also have "\ = (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]
by (simp add: ac_simps)
lemma sum_div_partition:
\<open>(\<Sum>a\<in>A. f a) div b = (\<Sum>a\<in>A \<inter> {a. b dvd f a}. f a div b) + (\<Sum>a\<in>A \<inter> {a. \<not> b dvd f a}. f a) div b\<close>
if \<open>finite A\<close>
proof -
have \<open>A = A \<inter> {a. b dvd f a} \<union> A \<inter> {a. \<not> b dvd f a}\<close>
by auto
then have \<open>(\<Sum>a\<in>A. f a) = (\<Sum>a\<in>A \<inter> {a. b dvd f a} \<union> A \<inter> {a. \<not> b dvd f a}. f a)\<close>
by simp
also have \<open>\<dots> = (\<Sum>a\<in>A \<inter> {a. b dvd f a}. f a) + (\<Sum>a\<in>A \<inter> {a. \<not> b dvd f a}. f a)\<close>
using \<open>finite A\<close> by (auto intro: sum.union_inter_neutral)
finally have *: \<open>sum f A = sum f (A \<inter> {a. b dvd f a}) + sum f (A \<inter> {a. \<not> b dvd f a})\<close> .
define B where B: \<open>B = A \<inter> {a. b dvd f a}\<close>
with \<open>finite A\<close> have \<open>finite B\<close> and \<open>a \<in> B \<Longrightarrow> b dvd f a\<close> for a
by simp_all
then have \<open>(\<Sum>a\<in>B. f a) div b = (\<Sum>a\<in>B. f a div b)\<close> and \<open>b dvd (\<Sum>a\<in>B. f a)\<close>
by induction (simp_all add: div_plus_div_distrib_dvd_left)
then show ?thesis using *
by (simp add: B div_plus_div_distrib_dvd_left)
qed
named_theorems mod_simps
text \<open>Addition respects modular equivalence.\<close>
lemma mod_add_left_eq [mod_simps]:
"(a mod c + b) mod c = (a + b) mod c"
proof -
have "(a + b) mod c = (a div c * c + a mod c + b) mod c"
by (simp only: div_mult_mod_eq)
also have "\ = (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
lemma mod_add_right_eq [mod_simps]:
"(a + b mod c) mod c = (a + b) mod c"
using mod_add_left_eq [of b c a] by (simp add: ac_simps)
lemma mod_add_eq:
"(a mod c + b mod c) mod c = (a + b) mod c"
by (simp add: mod_add_left_eq mod_add_right_eq)
lemma mod_sum_eq [mod_simps]:
"(\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 + (\<Sum>i\<in>A. f i mod a)) mod a"
by simp
also have "\ = (f i + (\i\A. f i mod a) mod a) mod a"
by (simp add: mod_simps)
also have "\ = (f i + (\i\A. f i) mod a) mod a"
by (simp add: insert.hyps)
finally show ?case
by (simp add: insert.hyps mod_simps)
qed simp_all
lemma mod_add_cong:
assumes "a mod c = a' mod c"
assumes "b mod c = b' mod c"
shows "(a + b) mod c = (a' + b') mod c"
proof -
have "(a mod c + b mod c) mod c = (a' mod c + b' mod c) mod c"
unfolding assms ..
then show ?thesis
by (simp add: mod_add_eq)
qed
text \<open>Multiplication respects modular equivalence.\<close>
lemma mod_mult_left_eq [mod_simps]:
"((a mod c) * b) mod c = (a * b) mod c"
proof -
have "(a * b) mod c = ((a div c * c + a mod c) * b) mod c"
by (simp only: div_mult_mod_eq)
also have "\ = (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"
by (simp add: mod_mult_left_eq mod_mult_right_eq)
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 * (\<Prod>i\<in>A. f i mod a)) mod a"
by simp
also have "\ = (f i * ((\i\A. f i mod a) mod a)) mod a"
by (simp add: mod_simps)
also have "\ = (f i * ((\i\A. f i) mod a)) mod a"
by (simp add: insert.hyps)
finally show ?case
by (simp add: insert.hyps mod_simps)
qed simp_all
lemma mod_mult_cong:
assumes "a mod c = a' mod c"
assumes "b mod c = b' mod c"
shows "(a * b) mod c = (a' * b') mod c"
proof -
have "(a mod c * (b mod c)) mod c = (a' mod c * (b' mod c)) mod c"
unfolding assms ..
then show ?thesis
by (simp add: mod_mult_eq)
qed
text \<open>Exponentiation respects modular equivalence.\<close>
lemma power_mod [mod_simps]:
"((a mod b) ^ n) mod b = (a ^ n) mod b"
proof (induct n)
case 0
then show ?case by simp
next
case (Suc n)
have "(a mod b) ^ Suc n mod b = (a mod b) * ((a mod b) ^ n mod b) mod b"
by (simp add: mod_mult_right_eq)
with Suc show ?case
by (simp add: mod_mult_left_eq mod_mult_right_eq)
qed
lemma power_diff_power_eq:
\<open>a ^ m div a ^ n = (if n \<le> m then a ^ (m - n) else 1 div a ^ (n - m))\<close>
if \<open>a \<noteq> 0\<close>
proof (cases \<open>n \<le> m\<close>)
case True
with that power_diff [symmetric, of a n m] show ?thesis by simp
next
case False
then obtain q where n: \<open>n = m + Suc q\<close>
by (auto simp add: not_le dest: less_imp_Suc_add)
then have \<open>a ^ m div a ^ n = (a ^ m * 1) div (a ^ m * a ^ Suc q)\<close>
by (simp add: power_add ac_simps)
moreover from that have \<open>a ^ m \<noteq> 0\<close>
by simp
ultimately have \<open>a ^ m div a ^ n = 1 div a ^ Suc q\<close>
by (subst (asm) div_mult_mult1) simp
with False n show ?thesis
by simp
qed
end
class euclidean_ring_cancel = euclidean_ring + euclidean_semiring_cancel
begin
subclass idom_divide ..
lemma div_minus_minus [simp]: "(- a) div (- b) = a div b"
using div_mult_mult1 [of "- 1" a b] by simp
lemma mod_minus_minus [simp]: "(- a) mod (- b) = - (a mod b)"
using mod_mult_mult1 [of "- 1" a b] by simp
lemma div_minus_right: "a div (- b) = (- a) div b"
using div_minus_minus [of "- a" b] by simp
lemma mod_minus_right: "a mod (- b) = - ((- a) mod b)"
using mod_minus_minus [of "- a" b] by simp
lemma div_minus1_right [simp]: "a div (- 1) = - a"
using div_minus_right [of a 1] by simp
lemma mod_minus1_right [simp]: "a mod (- 1) = 0"
using mod_minus_right [of a 1] by simp
text \<open>Negation respects modular equivalence.\<close>
lemma mod_minus_eq [mod_simps]:
"(- (a mod b)) mod b = (- a) mod b"
proof -
have "(- a) mod b = (- (a div b * b + a mod b)) mod b"
by (simp only: div_mult_mod_eq)
also have "\ = (- (a mod b) + - (a div b) * b) mod b"
by (simp add: ac_simps)
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
by (simp add: mod_minus_eq)
qed
text \<open>Subtraction respects modular equivalence.\<close>
lemma mod_diff_left_eq [mod_simps]:
"(a mod c - b) mod c = (a - b) mod c"
using mod_add_cong [of a c "a mod c" "- b" "- b"]
by simp
lemma mod_diff_right_eq [mod_simps]:
"(a - b mod c) mod c = (a - b) mod c"
using mod_add_cong [of a c a "- b" "- (b mod c)"] mod_minus_cong [of "b mod c" c b]
by simp
lemma mod_diff_eq:
"(a mod c - b mod c) mod c = (a - b) mod c"
using mod_add_cong [of a c "a mod c" "- b" "- (b mod c)"] mod_minus_cong [of "b mod c" c b]
by simp
lemma mod_diff_cong:
assumes "a mod c = a' mod c"
assumes "b mod c = b' mod c"
shows "(a - b) mod c = (a' - b') mod c"
using assms mod_add_cong [of a c a' "- b" "- b'"] mod_minus_cong [of b c "b'"]
by simp
lemma minus_mod_self2 [simp]:
"(a - b) mod b = a mod b"
using mod_diff_right_eq [of a b b]
by (simp add: mod_diff_right_eq)
lemma minus_mod_self1 [simp]:
"(b - a) mod b = - a mod b"
using mod_add_self2 [of "- a" b] by simp
lemma mod_eq_dvd_iff:
"a mod c = b mod c \ 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
by (simp add: dvd_eq_mod_eq_0 mod_simps)
next
assume ?Q
then obtain d where d: "a - b = c * d" ..
then have "a = c * d + b"
by (simp add: algebra_simps)
then show ?P by simp
qed
lemma mod_eqE:
assumes "a mod c = b mod c"
obtains d where "b = a + c * d"
proof -
from assms have "c dvd a - b"
by (simp add: mod_eq_dvd_iff)
then obtain d where "a - b = c * d" ..
then have "b = a + c * - d"
by (simp add: algebra_simps)
with that show thesis .
qed
lemma invertible_coprime:
"coprime a c" if "a * b mod c = 1"
by (rule coprimeI) (use that dvd_mod_iff [of _ c "a * b"] in auto)
end
subsection \<open>Uniquely determined division\<close>
class unique_euclidean_semiring = euclidean_semiring +
assumes euclidean_size_mult: "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
\<Longrightarrow> euclidean_size r < euclidean_size b
\<Longrightarrow> (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 \<open>b \<noteq> 0\<close> divides
show thesis
by (simp add: ac_simps)
next
case False
then have "a mod b \ 0"
by (simp add: mod_eq_0_iff_dvd)
moreover from \<open>b \<noteq> 0\<close> \<open>\<not> b dvd a\<close> 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 \<open>b \<noteq> 0\<close> by (rule mod_size_less)
moreover have "a = a div b * b + a mod b"
by (simp add: div_mult_mod_eq)
ultimately show thesis
using \<open>b \<noteq> 0\<close> 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 \<open>q * b + r = a\<close> 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 \<open>b \<noteq> 0\<close> show ?thesis
by simp
next
case (divides q)
then show ?thesis
by (simp add: ac_simps)
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 \<open>c \<noteq> 0\<close> show ?thesis
by (simp add: mult.left_commute [of c])
next
case (remainder q r)
from \<open>b \<noteq> 0\<close> \<open>c \<noteq> 0\<close> have "b * c \<noteq> 0"
by simp
from remainder \<open>c \<noteq> 0\<close>
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 \<open>b * c \<noteq> 0\<close> 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)"
by (simp add: div_mult_mod_eq)
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)"
by (simp add: div_mult_mod_eq)
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
lemma div_add1_eq:
"(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)"
by (simp add: algebra_simps)
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"
by (simp add: algebra_simps)
also have "\ = a + b + (a mod c + b mod c)"
by (simp add: div_mult_mod_eq remainder) (simp add: ac_simps)
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 \<open>Euclidean division on \<^typ>\<open>nat\<close>\<close>
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"
by (simp_all add: unit_factor_nat_def)
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"
by (simp add: modulo_nat_def)
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"
by (simp add: not_less)
moreover from \<open>n \<noteq> 0\<close> have "Max {k. k * n \<le> m} < Suc (m div n)"
by (simp add: divide_nat_def)
with \<open>n \<noteq> 0\<close> ex fin have "\<And>k. k * n \<le> m \<Longrightarrow> k < Suc (m div n)"
by auto
ultimately have "Suc (m div n) < Suc (m div n)"
by blast
then show False
by simp
qed
with \<open>n \<noteq> 0\<close> show ?thesis
by (simp add: modulo_nat_def)
qed
show "euclidean_size m \ euclidean_size (m * n)"
using \<open>n \<noteq> 0\<close> by (cases n) simp_all
fix q r :: nat
show "(q * n + r) div n = q" if "euclidean_size r < euclidean_size n"
proof -
from that have "r < n"
by simp
have "k \ 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)"
by (auto simp add: less_iff_Suc_add)
with \<open>r < n\<close> that show False
by (simp add: algebra_simps)
qed
with \<open>n \<noteq> 0\<close> ex fin show ?thesis
by (auto simp add: divide_nat_def Max_eq_iff)
qed
qed simp_all
end
text \<open>Tool support\<close>
ML \<open>
structure Cancel_Div_Mod_Nat = Cancel_Div_Mod
(
val div_name = \<^const_name>\<open>divide\<close>;
val mod_name = \<^const_name>\<open>modulo\<close>;
val mk_binop = HOLogic.mk_binop;
val dest_plus = HOLogic.dest_bin \<^const_name>\<open>Groups.plus\<close> HOLogic.natT;
val mk_sum = Arith_Data.mk_sum;
fun dest_sum tm =
if HOLogic.is_zero tm then []
else
(case try HOLogic.dest_Suc tm of
SOME t => HOLogic.Suc_zero :: dest_sum t
| NONE =>
(case try dest_plus tm of
SOME (t, u) => dest_sum t @ dest_sum u
| NONE => [tm]));
val div_mod_eqs = map mk_meta_eq @{thms cancel_div_mod_rules};
val prove_eq_sums = Arith_Data.prove_conv2 all_tac
(Arith_Data.simp_all_tac @{thms add_0_left add_0_right ac_simps})
)
\<close>
simproc_setup cancel_div_mod_nat ("(m::nat) + n") =
\<open>K Cancel_Div_Mod_Nat.proc\<close>
lemma div_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 \<open>simp_all add: algebra_simps\<close>)
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 \<open>simp_all add: algebra_simps\<close>)
lemma div_mult_self_is_m [simp]:
"m * n div n = m" if "n > 0" for m n :: nat
using that by simp
lemma div_mult_self1_is_m [simp]:
"n * m div n = m" if "n > 0" for m n :: nat
using that by simp
lemma mod_less_divisor [simp]:
"m mod n < n" if "n > 0" for m n :: nat
using mod_size_less [of n m] that by simp
lemma mod_le_divisor [simp]:
"m mod n \ 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
by (simp add: minus_mod_eq_div_mult [symmetric])
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]
by (simp add: ac_simps)
lemma dividend_less_div_times:
"m < n + (m div n) * n" if "0 < n" for m n :: nat
proof -
from that have "m mod n < n"
by simp
then show ?thesis
by (simp add: minus_mod_eq_div_mult [symmetric])
qed
lemma dividend_less_times_div:
"m < n + n * (m div n)" if "0 < n" for m n :: nat
using dividend_less_div_times [of n m] that
by (simp add: ac_simps)
lemma mod_Suc_le_divisor [simp]:
"m mod Suc n \ 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
proof (rule add_leD2)
from div_mult_mod_eq have "m div n * n + m mod n = m" .
then show "m div n * n + m mod n \ 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 \<open>n \<le> m\<close> obtain q where "m = n + q"
by (auto simp add: le_iff_add)
with \<open>0 < n\<close> show ?thesis
by (simp add: div_add_self1)
qed
lemma le_mod_geq:
"m mod n = (m - n) mod n" if "n \ m" for m n :: nat
proof -
from \<open>n \<le> m\<close> obtain q where "m = n + q"
by (auto simp add: le_iff_add)
then show ?thesis
by simp
qed
lemma div_if:
"m div n = (if m < n \ n = 0 then 0 else Suc ((m - n) div n))"
by (simp add: le_div_geq)
lemma mod_if:
"m mod n = (if m < n then m else (m - n) mod n)" for m n :: nat
by (simp add: le_mod_geq)
lemma div_eq_0_iff:
"m div n = 0 \ m < n \ n = 0" for m n :: nat
by (simp add: div_eq_0_iff)
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
by (simp add: dvd_eq_mod_eq_0)
lemma div_by_Suc_0 [simp]:
"m div Suc 0 = m"
using div_by_1 [of m] by simp
lemma mod_by_Suc_0 [simp]:
"m mod Suc 0 = 0"
using mod_by_1 [of m] by simp
lemma div2_Suc_Suc [simp]:
"Suc (Suc m) div 2 = Suc (m div 2)"
by (simp add: numeral_2_eq_2 le_div_geq)
lemma Suc_n_div_2_gt_zero [simp]:
"0 < Suc n div 2" if "n > 0" for n :: nat
using that by (cases n) simp_all
lemma div_2_gt_zero [simp]:
"0 < n div 2" if "Suc 0 < n" for n :: nat
using that Suc_n_div_2_gt_zero [of "n - 1"] by simp
lemma mod2_Suc_Suc [simp]:
"Suc (Suc m) mod 2 = m mod 2"
by (simp add: numeral_2_eq_2 le_mod_geq)
lemma add_self_div_2 [simp]:
"(m + m) div 2 = m" for m :: nat
by (simp add: mult_2 [symmetric])
lemma add_self_mod_2 [simp]:
"(m + m) mod 2 = 0" for m :: nat
by (simp add: mult_2 [symmetric])
lemma mod2_gr_0 [simp]:
"0 < m mod 2 \ 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 \<open>n > 0\<close> have "m mod n < n"
by (rule mod_less_divisor)
from \<open>q > 0\<close> have "m div n mod q < q"
by (rule mod_less_divisor)
then obtain s where "q = Suc (m div n mod q + s)"
by (blast dest: less_imp_Suc_add)
moreover have "m mod n + n * (m div n mod q) < n * Suc (m div n mod q + s)"
using \<open>m mod n < n\<close> by (simp add: add_mult_distrib2)
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"]
simp add: algebra_simps add_mult_distrib2 [symmetric])
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"]
simp add: algebra_simps add_mult_distrib2 [symmetric])
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"
by (auto simp add: le_iff_add)
then show ?thesis
by (simp add: div_add1_eq [of m q k])
qed
text \<open>Antimonotonicity of \<^const>\<open>divide\<close> in second argument\<close>
lemma div_le_mono2:
"k div n \ 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 \<open>n \<le> k\<close> less.prems less.hyps [of "k - m" m]
by simp
finally show ?thesis
using \<open>n \<le> k\<close> less.prems
by (simp add: le_div_geq)
qed
qed
lemma div_le_dividend [simp]:
"m div n \ m" for m n :: nat
using div_le_mono2 [of 1 n m] by (cases "n = 0") simp_all
lemma div_less_dividend [simp]:
"m div n < m" if "1 < n" and "0 < m" for m n :: nat
using that proof (induct m rule: less_induct)
case (less m)
show ?case
proof (cases "n < m")
case False
with less show ?thesis
by (cases "n = m") simp_all
next
case True
then show ?thesis
using less.hyps [of "m - n"] less.prems
by (simp add: le_div_geq)
qed
qed
lemma div_eq_dividend_iff:
"m div n = m \ 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 \<open>n > 0\<close> show ?thesis
by simp
qed
text \<open>A fact for the mutilated chess board\<close>
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"
by (simp add: mod_simps)
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"
by (simp add: div_le_mono)
text \<open>These lemmas collapse some needless occurrences of Suc:
at least three Sucs, since two and fewer are rewritten back to Suc again!
We already have some rules to simplify operands smaller than 3.\<close>
lemma div_Suc_eq_div_add3 [simp]:
"m div Suc (Suc (Suc n)) = m div (3 + n)"
by (simp add: Suc3_eq_add_3)
lemma mod_Suc_eq_mod_add3 [simp]:
"m mod Suc (Suc (Suc n)) = m mod (3 + n)"
by (simp add: Suc3_eq_add_3)
lemma Suc_div_eq_add3_div:
"Suc (Suc (Suc m)) div n = (3 + m) div n"
by (simp add: Suc3_eq_add_3)
lemma Suc_mod_eq_add3_mod:
"Suc (Suc (Suc m)) mod n = (3 + m) mod n"
by (simp add: Suc3_eq_add_3)
lemmas Suc_div_eq_add3_div_numeral [simp] =
Suc_div_eq_add3_div [of _ "numeral v"] for v
lemmas Suc_mod_eq_add3_mod_numeral [simp] =
Suc_mod_eq_add3_mod [of _ "numeral v"] for v
lemma (in field_char_0) of_nat_div:
"of_nat (m div n) = ((of_nat m - of_nat (m mod n)) / of_nat n)"
proof -
have "of_nat (m div n) = ((of_nat (m div n * n + m mod n) - of_nat (m mod n)) / of_nat n :: 'a)"
unfolding of_nat_add by (cases "n = 0") simp_all
then show ?thesis
by simp
qed
text \<open>An ``induction'' law for modulus arithmetic.\<close>
lemma mod_induct [consumes 3, case_names step]:
"P m" if "P n" and "n < p" and "m < p"
and step: "\n. n < p \ P n \ P (Suc n mod p)"
using \<open>m < p\<close> proof (induct m)
case 0
show ?case
proof (rule ccontr)
assume "\ P 0"
from \<open>n < p\<close> have "0 < p"
by simp
from \<open>n < p\<close> obtain m where "0 < m" and "p = n + m"
by (blast dest: less_imp_add_positive)
with \<open>P n\<close> have "P (p - m)"
by simp
moreover have "\ P (p - m)"
using \<open>0 < m\<close> proof (induct m)
case 0
then show ?case
by simp
next
case (Suc m)
show ?case
proof
assume P: "P (p - Suc m)"
with \<open>\<not> P 0\<close> have "Suc m < p"
by (auto intro: ccontr)
then have "Suc (p - Suc m) = p - m"
by arith
moreover from \<open>0 < p\<close> have "p - Suc m < p"
by arith
with P step have "P ((Suc (p - Suc m)) mod p)"
by blast
ultimately show False
using \<open>\<not> P 0\<close> Suc.hyps by (cases "m = 0") simp_all
qed
qed
ultimately show False
by blast
qed
next
case (Suc m)
then have "m < p" and mod: "Suc m mod p = Suc m"
by simp_all
from \<open>m < p\<close> have "P m"
by (rule Suc.hyps)
with \<open>m < p\<close> have "P (Suc m mod p)"
by (rule step)
with mod show ?case
by simp
qed
lemma split_div:
"P (m div n) \ (n = 0 \ P 0) \ (n \ 0 \
(\<forall>i j. j < n \<longrightarrow> m = n * i + j \<longrightarrow> 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 \
(\<forall>i j. j < n \<longrightarrow> m = n * i + j \<longrightarrow> 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 \<open>Euclidean division on \<^typ>\<open>int\<close>\<close>
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 \<bar>k\<bar> div nat \<bar>l\<bar>)
else - int (nat \<bar>k\<bar> div nat \<bar>l\<bar> + of_bool (\<not> l dvd k)))"
lemma divide_int_unfold:
"(sgn k * int m) div (sgn l * int n) =
(if sgn l = 0 \<or> sgn k = 0 \<or> n = 0 then 0
else if sgn k = sgn l
then int (m div n)
else - int (m div n + of_bool (\<not> 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"
by (simp add: divide_int_def)
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)"
by (simp add: ac_simps sgn_mult)
with k l \<open>l \<noteq> 0\<close> show "k * l div l = k"
by (simp only: divide_int_unfold)
(auto simp add: algebra_simps sgn_mult sgn_1_pos sgn_0_0)
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 \<open>?P\<close> have "is_unit (int q)"
by (rule coprime_common_divisor)
then show "is_unit q"
by simp
qed
next
assume ?Q
show ?P
proof (rule coprimeI)
fix k
assume "k dvd int m" "k dvd int n"
then have "nat \k\ dvd m" "nat \k\ dvd n"
by simp_all
with \<open>?Q\<close> have "is_unit (nat \<bar>k\<bar>)"
by (rule coprime_common_divisor)
then show "is_unit k"
by simp
qed
qed
lemma coprime_abs_left_iff [simp]:
"coprime \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 \<bar>k\<bar> mod nat \<bar>l\<bar>)
else sgn l * (\<bar>l\<bar> * of_bool (\<not> l dvd k) - int (nat \<bar>k\<bar> mod nat \<bar>l\<bar>)))"
lemma modulo_int_unfold:
"(sgn k * int m) mod (sgn l * int n) =
(if sgn l = 0 \<or> sgn k = 0 \<or> 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 (\<not> 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)
(simp_all add: of_nat_mult [symmetric] of_nat_add [symmetric]
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
by (simp add: division_segment_int_def)
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 \<open>l \<noteq> 0\<close>
with l have "s \ 0" and "n > 0"
by (simp_all add: sgn_0_0)
assume "division_segment r = division_segment l"
moreover have "r = sgn r * \r\"
by (simp add: sgn_mult_abs)
moreover define u where "u = nat \r\"
ultimately have "r = sgn l * int u"
using division_segment_eq_sgn \<open>l \<noteq> 0\<close> by (cases "r = 0") simp_all
with l \<open>n > 0\<close> have r: "r = sgn s * int u"
by (simp add: sgn_mult)
assume "euclidean_size r < euclidean_size l"
with l r \<open>s \<noteq> 0\<close> have "u < n"
by (simp add: abs_mult)
show "(q * l + r) div l = q"
proof (cases "q = 0 \ r = 0")
case True
then show ?thesis
proof
assume "q = 0"
then show ?thesis
using l r \<open>u < n\<close> by (simp add: divide_int_unfold)
next
assume "r = 0"
from \<open>r = 0\<close> have *: "q * l + r = sgn (t * s) * int (n * m)"
using q l by (simp add: ac_simps sgn_mult)
from \<open>s \<noteq> 0\<close> \<open>n > 0\<close> show ?thesis
by (simp only: *, simp only: q l divide_int_unfold)
(auto simp add: sgn_mult 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"
by (simp_all add: sgn_0_0)
moreover from \<open>0 < m\<close> \<open>u < n\<close> have "u \<le> m * n"
using mult_le_less_imp_less [of 1 m u n] by simp
ultimately have *: "q * l + r = sgn (s * t)
* int (if t < 0 then m * n - u else m * n + u)"
using l q r
by (simp add: sgn_mult algebra_simps of_nat_diff)
have "(m * n - u) div n = m - 1" if "u > 0"
using \<open>0 < m\<close> \<open>u < n\<close> that
by (auto intro: div_nat_eqI simp add: algebra_simps)
moreover have "n dvd m * n - u \ n dvd u"
using \<open>u \<le> m * n\<close> dvd_diffD1 [of n "m * n" u]
by auto
ultimately show ?thesis
using \<open>s \<noteq> 0\<close> \<open>m > 0\<close> \<open>u > 0\<close> \<open>u < n\<close> \<open>u \<le> m * n\<close>
by (simp only: *, simp only: l q divide_int_unfold)
(auto simp add: sgn_mult sgn_0_0 sgn_1_pos algebra_simps dest: dvd_imp_le)
qed
qed (use mult_le_mono2 [of 1] in \<open>auto simp add: division_segment_int_def not_le zero_less_mult_iff mult_less_0_iff abs_mult sgn_mult abs_mod_less sgn_mod nat_mult_distrib\<close>)
end
lemma pos_mod_bound [simp]:
"k mod l < l" if "l > 0" for k l :: int
proof -
obtain m and s where "k = sgn s * int m"
by (rule int_sgnE)
moreover from that obtain n where "l = sgn 1 * int n"
by (cases l) simp_all
moreover from this that have "n > 0"
by simp
ultimately show ?thesis
by (simp only: modulo_int_unfold)
(simp add: mod_greater_zero_iff_not_dvd)
qed
lemma neg_mod_bound [simp]:
"l < k mod l" if "l < 0" for k l :: int
proof -
obtain m and s where "k = sgn s * int m"
by (rule int_sgnE)
moreover from that obtain q where "l = sgn (- 1) * int (Suc q)"
by (cases l) simp_all
moreover define n where "n = Suc q"
then have "Suc q = n"
by simp
ultimately show ?thesis
by (simp only: modulo_int_unfold)
(simp add: mod_greater_zero_iff_not_dvd)
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 (rule int_sgnE)
moreover from that obtain n where "l = sgn 1 * int n"
by (cases l) auto
moreover from this that have "n > 0"
by simp
ultimately show ?thesis
by (simp only: modulo_int_unfold) simp
qed
lemma neg_mod_sign [simp]:
"k mod l \ 0" if "l < 0" for k l :: int
proof -
obtain m and s where "k = sgn s * int m"
by (rule int_sgnE)
moreover from that obtain q where "l = sgn (- 1) * int (Suc q)"
by (cases l) simp_all
moreover define n where "n = Suc q"
then have "Suc q = n"
by simp
ultimately show ?thesis
by (simp only: modulo_int_unfold) simp
qed
lemma div_pos_pos_trivial [simp]:
"k div l = 0" if "k \ 0" and "k < l" for k l :: int
using that by (simp add: unique_euclidean_semiring_class.div_eq_0_iff division_segment_int_def)
lemma mod_pos_pos_trivial [simp]:
"k mod l = k" if "k \ 0" and "k < l" for k l :: int
using that by (simp add: mod_eq_self_iff_div_eq_0)
lemma div_neg_neg_trivial [simp]:
"k div l = 0" if "k \ 0" and "l < k" for k l :: int
using that by (cases "k = 0") (simp, simp add: unique_euclidean_semiring_class.div_eq_0_iff division_segment_int_def)
lemma mod_neg_neg_trivial [simp]:
"k mod l = k" if "k \ 0" and "l < k" for k l :: int
using that by (simp add: mod_eq_self_iff_div_eq_0)
lemma div_pos_neg_trivial:
"k div l = - 1" if "0 < k" and "k + l \ 0" for k l :: int
proof (cases \<open>l = - k\<close>)
case True
with that show ?thesis
by (simp add: divide_int_def)
next
case False
show ?thesis
apply (rule div_eqI [of _ "k + l"])
using False that apply (simp_all add: division_segment_int_def)
done
qed
lemma mod_pos_neg_trivial:
"k mod l = k + l" if "0 < k" and "k + l \ 0" for k l :: int
proof (cases \<open>l = - k\<close>)
case True
with that show ?thesis
by (simp add: divide_int_def)
next
case False
show ?thesis
apply (rule mod_eqI [of _ _ \<open>- 1\<close>])
using False that apply (simp_all add: division_segment_int_def)
done
qed
text \<open>There is neither \<open>div_neg_pos_trivial\<close> nor \<open>mod_neg_pos_trivial\<close>
because \<^term>\<open>0 div l = 0\<close> would supersede it.\<close>
text \<open>Distributive laws for function \<open>nat\<close>.\<close>
lemma nat_div_distrib:
\<open>nat (x div y) = nat x div nat y\<close> if \<open>0 \<le> x\<close>
using that by (simp add: divide_int_def sgn_if)
lemma nat_div_distrib':
\<open>nat (x div y) = nat x div nat y\<close> if \<open>0 \<le> y\<close>
using that by (simp add: divide_int_def sgn_if)
lemma nat_mod_distrib: \<comment> \<open>Fails if y<0: the LHS collapses to (nat z) but the RHS doesn't\<close>
\<open>nat (x mod y) = nat x mod nat y\<close> if \<open>0 \<le> x\<close> \<open>0 \<le> y\<close>
using that by (simp add: modulo_int_def sgn_if)
subsection \<open>Special case: euclidean rings containing the natural numbers\<close>
class unique_euclidean_semiring_with_nat = semidom + semiring_char_0 + unique_euclidean_semiring +
assumes of_nat_div: "of_nat (m div n) = of_nat m div of_nat n"
and division_segment_of_nat [simp]: "division_segment (of_nat n) = 1"
and division_segment_euclidean_size [simp]: "division_segment a * of_nat (euclidean_size a) = a"
begin
lemma division_segment_eq_iff:
"a = b" if "division_segment a = division_segment b"
and "euclidean_size a = euclidean_size b"
using that division_segment_euclidean_size [of a] by simp
lemma euclidean_size_of_nat [simp]:
"euclidean_size (of_nat n) = n"
proof -
have "division_segment (of_nat n) * of_nat (euclidean_size (of_nat n)) = of_nat n"
by (fact division_segment_euclidean_size)
then show ?thesis by simp
qed
lemma of_nat_euclidean_size:
"of_nat (euclidean_size a) = a div division_segment a"
proof -
have "of_nat (euclidean_size a) = division_segment a * of_nat (euclidean_size a) div division_segment a"
by (subst nonzero_mult_div_cancel_left) simp_all
also have "\ = a div division_segment a"
by simp
finally show ?thesis .
qed
lemma division_segment_1 [simp]:
"division_segment 1 = 1"
using division_segment_of_nat [of 1] by simp
lemma division_segment_numeral [simp]:
"division_segment (numeral k) = 1"
using division_segment_of_nat [of "numeral k"] by simp
lemma euclidean_size_1 [simp]:
"euclidean_size 1 = 1"
using euclidean_size_of_nat [of 1] by simp
lemma euclidean_size_numeral [simp]:
"euclidean_size (numeral k) = numeral k"
using euclidean_size_of_nat [of "numeral k"] by simp
lemma of_nat_dvd_iff:
"of_nat m dvd of_nat n \ m dvd n" (is "?P \ ?Q")
proof (cases "m = 0")
case True
then show ?thesis
by simp
next
case False
show ?thesis
proof
assume ?Q
then show ?P
by auto
next
assume ?P
with False have "of_nat n = of_nat n div of_nat m * of_nat m"
by simp
then have "of_nat n = of_nat (n div m * m)"
by (simp add: of_nat_div)
then have "n = n div m * m"
by (simp only: of_nat_eq_iff)
then have "n = m * (n div m)"
by (simp add: ac_simps)
then show ?Q ..
qed
qed
lemma of_nat_mod:
"of_nat (m mod n) = of_nat m mod of_nat n"
proof -
have "of_nat m div of_nat n * of_nat n + of_nat m mod of_nat n = of_nat m"
by (simp add: div_mult_mod_eq)
also have "of_nat m = of_nat (m div n * n + m mod n)"
by simp
finally show ?thesis
by (simp only: of_nat_div of_nat_mult of_nat_add) simp
qed
lemma one_div_two_eq_zero [simp]:
"1 div 2 = 0"
proof -
from of_nat_div [symmetric] have "of_nat 1 div of_nat 2 = of_nat 0"
by (simp only:) simp
then show ?thesis
by simp
qed
lemma one_mod_two_eq_one [simp]:
"1 mod 2 = 1"
proof -
from of_nat_mod [symmetric] have "of_nat 1 mod of_nat 2 = of_nat 1"
by (simp only:) simp
then show ?thesis
by simp
qed
lemma one_mod_2_pow_eq [simp]:
"1 mod (2 ^ n) = of_bool (n > 0)"
proof -
have "1 mod (2 ^ n) = of_nat (1 mod (2 ^ n))"
using of_nat_mod [of 1 "2 ^ n"] by simp
also have "\ = of_bool (n > 0)"
by simp
finally show ?thesis .
qed
lemma one_div_2_pow_eq [simp]:
"1 div (2 ^ n) = of_bool (n = 0)"
using div_mult_mod_eq [of 1 "2 ^ n"] by auto
lemma div_mult2_eq':
"a div (of_nat m * of_nat n) = a div of_nat m div of_nat n"
proof (cases a "of_nat m * of_nat n" rule: divmod_cases)
case (divides q)
then show ?thesis
using nonzero_mult_div_cancel_right [of "of_nat m" "q * of_nat n"]
by (simp add: ac_simps)
next
case (remainder q r)
then have "division_segment r = 1"
using division_segment_of_nat [of "m * n"] by simp
with division_segment_euclidean_size [of r]
have "of_nat (euclidean_size r) = r"
by simp
have "a mod (of_nat m * of_nat n) div (of_nat m * of_nat n) = 0"
by simp
with remainder(6) have "r div (of_nat m * of_nat n) = 0"
by simp
with \<open>of_nat (euclidean_size r) = r\<close>
have "of_nat (euclidean_size r) div (of_nat m * of_nat n) = 0"
by simp
then have "of_nat (euclidean_size r div (m * n)) = 0"
by (simp add: of_nat_div)
then have "of_nat (euclidean_size r div m div n) = 0"
by (simp add: div_mult2_eq)
with \<open>of_nat (euclidean_size r) = r\<close> have "r div of_nat m div of_nat n = 0"
by (simp add: of_nat_div)
with remainder(1)
have "q = (r div of_nat m + q * of_nat n * of_nat m div of_nat m) div of_nat n"
by simp
with remainder(5) remainder(7) show ?thesis
using div_plus_div_distrib_dvd_right [of "of_nat m" "q * (of_nat m * of_nat n)" r]
by (simp add: ac_simps)
next
case by0
then show ?thesis
by auto
qed
lemma mod_mult2_eq':
"a mod (of_nat m * of_nat n) = of_nat m * (a div of_nat m mod of_nat n) + a mod of_nat m"
proof -
have "a div (of_nat m * of_nat n) * (of_nat m * of_nat n) + a mod (of_nat m * of_nat n) = a div of_nat m div of_nat n * of_nat n * of_nat m + (a div of_nat m mod of_nat n * of_nat m + a mod of_nat m)"
by (simp add: combine_common_factor div_mult_mod_eq)
moreover have "a div of_nat m div of_nat n * of_nat n * of_nat m = of_nat n * of_nat m * (a div of_nat m div of_nat n)"
by (simp add: ac_simps)
ultimately show ?thesis
by (simp add: div_mult2_eq' mult_commute)
qed
lemma div_mult2_numeral_eq:
"a div numeral k div numeral l = a div numeral (k * l)" (is "?A = ?B")
proof -
have "?A = a div of_nat (numeral k) div of_nat (numeral l)"
by simp
also have "\ = a div (of_nat (numeral k) * of_nat (numeral l))"
by (fact div_mult2_eq' [symmetric])
also have "\ = ?B"
by simp
finally show ?thesis .
qed
lemma numeral_Bit0_div_2:
"numeral (num.Bit0 n) div 2 = numeral n"
proof -
have "numeral (num.Bit0 n) = numeral n + numeral n"
by (simp only: numeral.simps)
also have "\ = numeral n * 2"
by (simp add: mult_2_right)
finally have "numeral (num.Bit0 n) div 2 = numeral n * 2 div 2"
by simp
also have "\ = numeral n"
by (rule nonzero_mult_div_cancel_right) simp
finally show ?thesis .
qed
lemma numeral_Bit1_div_2:
"numeral (num.Bit1 n) div 2 = numeral n"
proof -
have "numeral (num.Bit1 n) = numeral n + numeral n + 1"
by (simp only: numeral.simps)
also have "\ = numeral n * 2 + 1"
by (simp add: mult_2_right)
finally have "numeral (num.Bit1 n) div 2 = (numeral n * 2 + 1) div 2"
by simp
also have "\ = numeral n * 2 div 2 + 1 div 2"
using dvd_triv_right by (rule div_plus_div_distrib_dvd_left)
also have "\ = numeral n * 2 div 2"
by simp
also have "\ = numeral n"
by (rule nonzero_mult_div_cancel_right) simp
finally show ?thesis .
qed
lemma exp_mod_exp:
\<open>2 ^ m mod 2 ^ n = of_bool (m < n) * 2 ^ m\<close>
proof -
have \<open>(2::nat) ^ m mod 2 ^ n = of_bool (m < n) * 2 ^ m\<close> (is \<open>?lhs = ?rhs\<close>)
by (auto simp add: not_less monoid_mult_class.power_add dest!: le_Suc_ex)
then have \<open>of_nat ?lhs = of_nat ?rhs\<close>
by simp
then show ?thesis
by (simp add: of_nat_mod)
qed
lemma mask_mod_exp:
\<open>(2 ^ n - 1) mod 2 ^ m = 2 ^ min m n - 1\<close>
proof -
have \<open>(2 ^ n - 1) mod 2 ^ m = 2 ^ min m n - (1::nat)\<close> (is \<open>?lhs = ?rhs\<close>)
proof (cases \<open>n \<le> m\<close>)
case True
then show ?thesis
by (simp add: Suc_le_lessD min.absorb2)
next
case False
then have \<open>m < n\<close>
by simp
then obtain q where n: \<open>n = Suc q + m\<close>
by (auto dest: less_imp_Suc_add)
then have \<open>min m n = m\<close>
by simp
moreover have \<open>(2::nat) ^ m \<le> 2 * 2 ^ q * 2 ^ m\<close>
using mult_le_mono1 [of 1 \<open>2 * 2 ^ q\<close> \<open>2 ^ m\<close>] by simp
with n have \<open>2 ^ n - 1 = (2 ^ Suc q - 1) * 2 ^ m + (2 ^ m - (1::nat))\<close>
--> --------------------
--> maximum size reached
--> --------------------
¤ Dauer der Verarbeitung: 0.86 Sekunden
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