src/HOL/Number_Theory/Euclidean_Algorithm.thy
 author haftmann Fri Jun 12 08:53:23 2015 +0200 (2015-06-12) changeset 60432 68d75cff8809 parent 60431 db9c67b760f1 child 60433 720f210c5b1d permissions -rw-r--r--
given up trivial definition
     1 (* Author: Manuel Eberl *)

     2

     3 section {* Abstract euclidean algorithm *}

     4

     5 theory Euclidean_Algorithm

     6 imports Complex_Main

     7 begin

     8

     9 context semiring_div

    10 begin

    11

    12 abbreviation is_unit :: "'a \<Rightarrow> bool"

    13 where

    14   "is_unit a \<equiv> a dvd 1"

    15

    16 definition associated :: "'a \<Rightarrow> 'a \<Rightarrow> bool"

    17 where

    18   "associated a b \<longleftrightarrow> a dvd b \<and> b dvd a"

    19

    20 lemma unit_prod [intro]:

    21   "is_unit a \<Longrightarrow> is_unit b \<Longrightarrow> is_unit (a * b)"

    22   by (subst mult_1_left [of 1, symmetric], rule mult_dvd_mono)

    23

    24 lemma unit_divide_1:

    25   "is_unit b \<Longrightarrow> a div b = a * divide 1 b"

    26   by (simp add: div_mult_swap)

    27

    28 lemma unit_divide_1_divide_1 [simp]:

    29   "is_unit a \<Longrightarrow> divide 1 (divide 1 a) = a"

    30   by (metis div_mult_mult1_if div_mult_self1_is_id dvd_mult_div_cancel mult_1_right)

    31

    32 lemma inv_imp_eq_divide_1:

    33   "a * b = 1 \<Longrightarrow> divide 1 a = b"

    34   by (metis dvd_mult_div_cancel dvd_mult_right mult_1_right mult.left_commute one_dvd)

    35

    36 lemma unit_divide_1_unit [simp, intro]:

    37   assumes "is_unit a"

    38   shows "is_unit (divide 1 a)"

    39 proof -

    40   from assms have "1 = divide 1 a * a" by simp

    41   then show "is_unit (divide 1 a)" by (rule dvdI)

    42 qed

    43

    44 lemma mult_unit_dvd_iff:

    45   "is_unit b \<Longrightarrow> a * b dvd c \<longleftrightarrow> a dvd c"

    46 proof

    47   assume "is_unit b" "a * b dvd c"

    48   then show "a dvd c" by (simp add: dvd_mult_left)

    49 next

    50   assume "is_unit b" "a dvd c"

    51   then obtain k where "c = a * k" unfolding dvd_def by blast

    52   with is_unit b have "c = (a * b) * (divide 1 b * k)"

    53       by (simp add: mult_ac)

    54   then show "a * b dvd c" by (rule dvdI)

    55 qed

    56

    57 lemma div_unit_dvd_iff:

    58   "is_unit b \<Longrightarrow> a div b dvd c \<longleftrightarrow> a dvd c"

    59   by (subst unit_divide_1) (assumption, simp add: mult_unit_dvd_iff)

    60

    61 lemma dvd_mult_unit_iff:

    62   "is_unit b \<Longrightarrow> a dvd c * b \<longleftrightarrow> a dvd c"

    63 proof

    64   assume "is_unit b" and "a dvd c * b"

    65   have "c * b dvd c * (b * divide 1 b)" by (subst mult_assoc [symmetric]) simp

    66   also from is_unit b have "b * divide 1 b = 1" by simp

    67   finally have "c * b dvd c" by simp

    68   with a dvd c * b show "a dvd c" by (rule dvd_trans)

    69 next

    70   assume "a dvd c"

    71   then show "a dvd c * b" by simp

    72 qed

    73

    74 lemma dvd_div_unit_iff:

    75   "is_unit b \<Longrightarrow> a dvd c div b \<longleftrightarrow> a dvd c"

    76   by (subst unit_divide_1) (assumption, simp add: dvd_mult_unit_iff)

    77

    78 lemmas unit_dvd_iff = mult_unit_dvd_iff div_unit_dvd_iff dvd_mult_unit_iff dvd_div_unit_iff

    79

    80 lemma unit_div [intro]:

    81   "is_unit a \<Longrightarrow> is_unit b \<Longrightarrow> is_unit (a div b)"

    82   by (subst unit_divide_1) (assumption, rule unit_prod, simp_all)

    83

    84 lemma unit_div_mult_swap:

    85   "is_unit c \<Longrightarrow> a * (b div c) = a * b div c"

    86   by (simp only: unit_divide_1 [of _ b] unit_divide_1 [of _ "a*b"] ac_simps)

    87

    88 lemma unit_div_commute:

    89   "is_unit b \<Longrightarrow> a div b * c = a * c div b"

    90   by (simp only: unit_divide_1 [of _ a] unit_divide_1 [of _ "a*c"] ac_simps)

    91

    92 lemma unit_imp_dvd [dest]:

    93   "is_unit b \<Longrightarrow> b dvd a"

    94   by (rule dvd_trans [of _ 1]) simp_all

    95

    96 lemma dvd_unit_imp_unit:

    97   "is_unit b \<Longrightarrow> a dvd b \<Longrightarrow> is_unit a"

    98   by (rule dvd_trans)

    99

   100 lemma unit_divide_1'1:

   101   assumes "is_unit b"

   102   shows "a div (b * c) = a * divide 1 b div c"

   103 proof -

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

   105     by simp

   106   also have "... = b * (a * divide 1 b) div (b * c)"

   107     by (simp only: mult_ac)

   108   also have "... = a * divide 1 b div c"

   109     by (cases "b = 0", simp, rule div_mult_mult1)

   110   finally show ?thesis .

   111 qed

   112

   113 lemma associated_comm:

   114   "associated a b \<Longrightarrow> associated b a"

   115   by (simp add: associated_def)

   116

   117 lemma associated_0 [simp]:

   118   "associated 0 b \<longleftrightarrow> b = 0"

   119   "associated a 0 \<longleftrightarrow> a = 0"

   120   unfolding associated_def by simp_all

   121

   122 lemma associated_unit:

   123   "is_unit a \<Longrightarrow> associated a b \<Longrightarrow> is_unit b"

   124   unfolding associated_def using dvd_unit_imp_unit by auto

   125

   126 lemma is_unit_1 [simp]:

   127   "is_unit 1"

   128   by simp

   129

   130 lemma not_is_unit_0 [simp]:

   131   "\<not> is_unit 0"

   132   by auto

   133

   134 lemma unit_mult_left_cancel:

   135   assumes "is_unit a"

   136   shows "(a * b) = (a * c) \<longleftrightarrow> b = c"

   137 proof -

   138   from assms have "a \<noteq> 0" by auto

   139   then show ?thesis by (metis div_mult_self1_is_id)

   140 qed

   141

   142 lemma unit_mult_right_cancel:

   143   "is_unit a \<Longrightarrow> (b * a) = (c * a) \<longleftrightarrow> b = c"

   144   by (simp add: ac_simps unit_mult_left_cancel)

   145

   146 lemma unit_div_cancel:

   147   "is_unit a \<Longrightarrow> (b div a) = (c div a) \<longleftrightarrow> b = c"

   148   apply (subst unit_divide_1[of _ b], assumption)

   149   apply (subst unit_divide_1[of _ c], assumption)

   150   apply (rule unit_mult_right_cancel, erule unit_divide_1_unit)

   151   done

   152

   153 lemma unit_eq_div1:

   154   "is_unit b \<Longrightarrow> a div b = c \<longleftrightarrow> a = c * b"

   155   apply (subst unit_divide_1, assumption)

   156   apply (subst unit_mult_right_cancel[symmetric], assumption)

   157   apply (subst mult_assoc, subst dvd_div_mult_self, assumption, simp)

   158   done

   159

   160 lemma unit_eq_div2:

   161   "is_unit b \<Longrightarrow> a = c div b \<longleftrightarrow> a * b = c"

   162   by (subst (1 2) eq_commute, simp add: unit_eq_div1, subst eq_commute, rule refl)

   163

   164 lemma associated_iff_div_unit:

   165   "associated a b \<longleftrightarrow> (\<exists>c. is_unit c \<and> a = c * b)"

   166 proof

   167   assume "associated a b"

   168   show "\<exists>c. is_unit c \<and> a = c * b"

   169   proof (cases "a = 0")

   170     assume "a = 0"

   171     then show "\<exists>c. is_unit c \<and> a = c * b" using associated a b

   172         by (intro exI[of _ 1], simp add: associated_def)

   173   next

   174     assume [simp]: "a \<noteq> 0"

   175     hence [simp]: "a dvd b" "b dvd a" using associated a b

   176         unfolding associated_def by simp_all

   177     hence "1 = a div b * (b div a)"

   178       by (simp add: div_mult_swap)

   179     hence "is_unit (a div b)" ..

   180     moreover have "a = (a div b) * b" by simp

   181     ultimately show ?thesis by blast

   182   qed

   183 next

   184   assume "\<exists>c. is_unit c \<and> a = c * b"

   185   then obtain c where "is_unit c" and "a = c * b" by blast

   186   hence "b = a * divide 1 c" by (simp add: algebra_simps)

   187   hence "a dvd b" by simp

   188   moreover from a = c * b have "b dvd a" by simp

   189   ultimately show "associated a b" unfolding associated_def by simp

   190 qed

   191

   192 lemmas unit_simps = mult_unit_dvd_iff div_unit_dvd_iff dvd_mult_unit_iff

   193   dvd_div_unit_iff unit_div_mult_swap unit_div_commute

   194   unit_mult_left_cancel unit_mult_right_cancel unit_div_cancel

   195   unit_eq_div1 unit_eq_div2

   196

   197 end

   198

   199 context ring_div

   200 begin

   201

   202 lemma is_unit_neg [simp]:

   203   "is_unit (- a) \<Longrightarrow> is_unit a"

   204   by simp

   205

   206 lemma is_unit_neg_1 [simp]:

   207   "is_unit (-1)"

   208   by simp

   209

   210 end

   211

   212 lemma is_unit_nat [simp]:

   213   "is_unit (a::nat) \<longleftrightarrow> a = 1"

   214   by simp

   215

   216 lemma is_unit_int:

   217   "is_unit (a::int) \<longleftrightarrow> a = 1 \<or> a = -1"

   218   by auto

   219

   220 text {*

   221   A Euclidean semiring is a semiring upon which the Euclidean algorithm can be

   222   implemented. It must provide:

   223   \begin{itemize}

   224   \item division with remainder

   225   \item a size function such that @{term "size (a mod b) < size b"}

   226         for any @{term "b \<noteq> 0"}

   227   \item a normalisation factor such that two associated numbers are equal iff

   228         they are the same when divided by their normalisation factors.

   229   \end{itemize}

   230   The existence of these functions makes it possible to derive gcd and lcm functions

   231   for any Euclidean semiring.

   232 *}

   233 class euclidean_semiring = semiring_div +

   234   fixes euclidean_size :: "'a \<Rightarrow> nat"

   235   fixes normalisation_factor :: "'a \<Rightarrow> 'a"

   236   assumes mod_size_less [simp]:

   237     "b \<noteq> 0 \<Longrightarrow> euclidean_size (a mod b) < euclidean_size b"

   238   assumes size_mult_mono:

   239     "b \<noteq> 0 \<Longrightarrow> euclidean_size (a * b) \<ge> euclidean_size a"

   240   assumes normalisation_factor_is_unit [intro,simp]:

   241     "a \<noteq> 0 \<Longrightarrow> is_unit (normalisation_factor a)"

   242   assumes normalisation_factor_mult: "normalisation_factor (a * b) =

   243     normalisation_factor a * normalisation_factor b"

   244   assumes normalisation_factor_unit: "is_unit a \<Longrightarrow> normalisation_factor a = a"

   245   assumes normalisation_factor_0 [simp]: "normalisation_factor 0 = 0"

   246 begin

   247

   248 lemma normalisation_factor_dvd [simp]:

   249   "a \<noteq> 0 \<Longrightarrow> normalisation_factor a dvd b"

   250   by (rule unit_imp_dvd, simp)

   251

   252 lemma normalisation_factor_1 [simp]:

   253   "normalisation_factor 1 = 1"

   254   by (simp add: normalisation_factor_unit)

   255

   256 lemma normalisation_factor_0_iff [simp]:

   257   "normalisation_factor a = 0 \<longleftrightarrow> a = 0"

   258 proof

   259   assume "normalisation_factor a = 0"

   260   hence "\<not> is_unit (normalisation_factor a)"

   261     by (metis not_is_unit_0)

   262   then show "a = 0" by force

   263 next

   264   assume "a = 0"

   265   then show "normalisation_factor a = 0" by simp

   266 qed

   267

   268 lemma normalisation_factor_pow:

   269   "normalisation_factor (a ^ n) = normalisation_factor a ^ n"

   270   by (induct n) (simp_all add: normalisation_factor_mult power_Suc2)

   271

   272 lemma normalisation_correct [simp]:

   273   "normalisation_factor (a div normalisation_factor a) = (if a = 0 then 0 else 1)"

   274 proof (cases "a = 0", simp)

   275   assume "a \<noteq> 0"

   276   let ?nf = "normalisation_factor"

   277   from normalisation_factor_is_unit[OF a \<noteq> 0] have "?nf a \<noteq> 0"

   278     by (metis not_is_unit_0)

   279   have "?nf (a div ?nf a) * ?nf (?nf a) = ?nf (a div ?nf a * ?nf a)"

   280     by (simp add: normalisation_factor_mult)

   281   also have "a div ?nf a * ?nf a = a" using a \<noteq> 0

   282     by simp

   283   also have "?nf (?nf a) = ?nf a" using a \<noteq> 0

   284     normalisation_factor_is_unit normalisation_factor_unit by simp

   285   finally show ?thesis using a \<noteq> 0 and ?nf a \<noteq> 0

   286     by (metis div_mult_self2_is_id div_self)

   287 qed

   288

   289 lemma normalisation_0_iff [simp]:

   290   "a div normalisation_factor a = 0 \<longleftrightarrow> a = 0"

   291   by (cases "a = 0", simp, subst unit_eq_div1, blast, simp)

   292

   293 lemma associated_iff_normed_eq:

   294   "associated a b \<longleftrightarrow> a div normalisation_factor a = b div normalisation_factor b"

   295 proof (cases "b = 0", simp, cases "a = 0", metis associated_0(1) normalisation_0_iff, rule iffI)

   296   let ?nf = normalisation_factor

   297   assume "a \<noteq> 0" "b \<noteq> 0" "a div ?nf a = b div ?nf b"

   298   hence "a = b * (?nf a div ?nf b)"

   299     apply (subst (asm) unit_eq_div1, blast, subst (asm) unit_div_commute, blast)

   300     apply (subst div_mult_swap, simp, simp)

   301     done

   302   with a \<noteq> 0 b \<noteq> 0 have "\<exists>c. is_unit c \<and> a = c * b"

   303     by (intro exI[of _ "?nf a div ?nf b"], force simp: mult_ac)

   304   with associated_iff_div_unit show "associated a b" by simp

   305 next

   306   let ?nf = normalisation_factor

   307   assume "a \<noteq> 0" "b \<noteq> 0" "associated a b"

   308   with associated_iff_div_unit obtain c where "is_unit c" and "a = c * b" by blast

   309   then show "a div ?nf a = b div ?nf b"

   310     apply (simp only: a = c * b normalisation_factor_mult normalisation_factor_unit)

   311     apply (rule div_mult_mult1, force)

   312     done

   313   qed

   314

   315 lemma normed_associated_imp_eq:

   316   "associated a b \<Longrightarrow> normalisation_factor a \<in> {0, 1} \<Longrightarrow> normalisation_factor b \<in> {0, 1} \<Longrightarrow> a = b"

   317   by (simp add: associated_iff_normed_eq, elim disjE, simp_all)

   318

   319 lemmas normalisation_factor_dvd_iff [simp] =

   320   unit_dvd_iff [OF normalisation_factor_is_unit]

   321

   322 lemma euclidean_division:

   323   fixes a :: 'a and b :: 'a

   324   assumes "b \<noteq> 0"

   325   obtains s and t where "a = s * b + t"

   326     and "euclidean_size t < euclidean_size b"

   327 proof -

   328   from div_mod_equality[of a b 0]

   329      have "a = a div b * b + a mod b" by simp

   330   with that and assms show ?thesis by force

   331 qed

   332

   333 lemma dvd_euclidean_size_eq_imp_dvd:

   334   assumes "a \<noteq> 0" and b_dvd_a: "b dvd a" and size_eq: "euclidean_size a = euclidean_size b"

   335   shows "a dvd b"

   336 proof (subst dvd_eq_mod_eq_0, rule ccontr)

   337   assume "b mod a \<noteq> 0"

   338   from b_dvd_a have b_dvd_mod: "b dvd b mod a" by (simp add: dvd_mod_iff)

   339   from b_dvd_mod obtain c where "b mod a = b * c" unfolding dvd_def by blast

   340     with b mod a \<noteq> 0 have "c \<noteq> 0" by auto

   341   with b mod a = b * c have "euclidean_size (b mod a) \<ge> euclidean_size b"

   342       using size_mult_mono by force

   343   moreover from a \<noteq> 0 have "euclidean_size (b mod a) < euclidean_size a"

   344       using mod_size_less by blast

   345   ultimately show False using size_eq by simp

   346 qed

   347

   348 function gcd_eucl :: "'a \<Rightarrow> 'a \<Rightarrow> 'a"

   349 where

   350   "gcd_eucl a b = (if b = 0 then a div normalisation_factor a else gcd_eucl b (a mod b))"

   351   by (pat_completeness, simp)

   352 termination by (relation "measure (euclidean_size \<circ> snd)", simp_all)

   353

   354 declare gcd_eucl.simps [simp del]

   355

   356 lemma gcd_induct: "\<lbrakk>\<And>b. P b 0; \<And>a b. 0 \<noteq> b \<Longrightarrow> P b (a mod b) \<Longrightarrow> P a b\<rbrakk> \<Longrightarrow> P a b"

   357 proof (induct a b rule: gcd_eucl.induct)

   358   case ("1" m n)

   359     then show ?case by (cases "n = 0") auto

   360 qed

   361

   362 definition lcm_eucl :: "'a \<Rightarrow> 'a \<Rightarrow> 'a"

   363 where

   364   "lcm_eucl a b = a * b div (gcd_eucl a b * normalisation_factor (a * b))"

   365

   366   (* Somewhat complicated definition of Lcm that has the advantage of working

   367      for infinite sets as well *)

   368

   369 definition Lcm_eucl :: "'a set \<Rightarrow> 'a"

   370 where

   371   "Lcm_eucl A = (if \<exists>l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l) then

   372      let l = SOME l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l) \<and> euclidean_size l =

   373        (LEAST n. \<exists>l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l) \<and> euclidean_size l = n)

   374        in l div normalisation_factor l

   375       else 0)"

   376

   377 definition Gcd_eucl :: "'a set \<Rightarrow> 'a"

   378 where

   379   "Gcd_eucl A = Lcm_eucl {d. \<forall>a\<in>A. d dvd a}"

   380

   381 end

   382

   383 class euclidean_semiring_gcd = euclidean_semiring + gcd + Gcd +

   384   assumes gcd_gcd_eucl: "gcd = gcd_eucl" and lcm_lcm_eucl: "lcm = lcm_eucl"

   385   assumes Gcd_Gcd_eucl: "Gcd = Gcd_eucl" and Lcm_Lcm_eucl: "Lcm = Lcm_eucl"

   386 begin

   387

   388 lemma gcd_red:

   389   "gcd a b = gcd b (a mod b)"

   390   by (metis gcd_eucl.simps mod_0 mod_by_0 gcd_gcd_eucl)

   391

   392 lemma gcd_non_0:

   393   "b \<noteq> 0 \<Longrightarrow> gcd a b = gcd b (a mod b)"

   394   by (rule gcd_red)

   395

   396 lemma gcd_0_left:

   397   "gcd 0 a = a div normalisation_factor a"

   398    by (simp only: gcd_gcd_eucl, subst gcd_eucl.simps, subst gcd_eucl.simps, simp add: Let_def)

   399

   400 lemma gcd_0:

   401   "gcd a 0 = a div normalisation_factor a"

   402   by (simp only: gcd_gcd_eucl, subst gcd_eucl.simps, simp add: Let_def)

   403

   404 lemma gcd_dvd1 [iff]: "gcd a b dvd a"

   405   and gcd_dvd2 [iff]: "gcd a b dvd b"

   406 proof (induct a b rule: gcd_eucl.induct)

   407   fix a b :: 'a

   408   assume IH1: "b \<noteq> 0 \<Longrightarrow> gcd b (a mod b) dvd b"

   409   assume IH2: "b \<noteq> 0 \<Longrightarrow> gcd b (a mod b) dvd (a mod b)"

   410

   411   have "gcd a b dvd a \<and> gcd a b dvd b"

   412   proof (cases "b = 0")

   413     case True

   414       then show ?thesis by (cases "a = 0", simp_all add: gcd_0)

   415   next

   416     case False

   417       with IH1 and IH2 show ?thesis by (simp add: gcd_non_0 dvd_mod_iff)

   418   qed

   419   then show "gcd a b dvd a" "gcd a b dvd b" by simp_all

   420 qed

   421

   422 lemma dvd_gcd_D1: "k dvd gcd m n \<Longrightarrow> k dvd m"

   423   by (rule dvd_trans, assumption, rule gcd_dvd1)

   424

   425 lemma dvd_gcd_D2: "k dvd gcd m n \<Longrightarrow> k dvd n"

   426   by (rule dvd_trans, assumption, rule gcd_dvd2)

   427

   428 lemma gcd_greatest:

   429   fixes k a b :: 'a

   430   shows "k dvd a \<Longrightarrow> k dvd b \<Longrightarrow> k dvd gcd a b"

   431 proof (induct a b rule: gcd_eucl.induct)

   432   case (1 a b)

   433   show ?case

   434     proof (cases "b = 0")

   435       assume "b = 0"

   436       with 1 show ?thesis by (cases "a = 0", simp_all add: gcd_0)

   437     next

   438       assume "b \<noteq> 0"

   439       with 1 show ?thesis by (simp add: gcd_non_0 dvd_mod_iff)

   440     qed

   441 qed

   442

   443 lemma dvd_gcd_iff:

   444   "k dvd gcd a b \<longleftrightarrow> k dvd a \<and> k dvd b"

   445   by (blast intro!: gcd_greatest intro: dvd_trans)

   446

   447 lemmas gcd_greatest_iff = dvd_gcd_iff

   448

   449 lemma gcd_zero [simp]:

   450   "gcd a b = 0 \<longleftrightarrow> a = 0 \<and> b = 0"

   451   by (metis dvd_0_left dvd_refl gcd_dvd1 gcd_dvd2 gcd_greatest)+

   452

   453 lemma normalisation_factor_gcd [simp]:

   454   "normalisation_factor (gcd a b) = (if a = 0 \<and> b = 0 then 0 else 1)" (is "?f a b = ?g a b")

   455 proof (induct a b rule: gcd_eucl.induct)

   456   fix a b :: 'a

   457   assume IH: "b \<noteq> 0 \<Longrightarrow> ?f b (a mod b) = ?g b (a mod b)"

   458   then show "?f a b = ?g a b" by (cases "b = 0", auto simp: gcd_non_0 gcd_0)

   459 qed

   460

   461 lemma gcdI:

   462   "k dvd a \<Longrightarrow> k dvd b \<Longrightarrow> (\<And>l. l dvd a \<Longrightarrow> l dvd b \<Longrightarrow> l dvd k)

   463     \<Longrightarrow> normalisation_factor k = (if k = 0 then 0 else 1) \<Longrightarrow> k = gcd a b"

   464   by (intro normed_associated_imp_eq) (auto simp: associated_def intro: gcd_greatest)

   465

   466 sublocale gcd!: abel_semigroup gcd

   467 proof

   468   fix a b c

   469   show "gcd (gcd a b) c = gcd a (gcd b c)"

   470   proof (rule gcdI)

   471     have "gcd (gcd a b) c dvd gcd a b" "gcd a b dvd a" by simp_all

   472     then show "gcd (gcd a b) c dvd a" by (rule dvd_trans)

   473     have "gcd (gcd a b) c dvd gcd a b" "gcd a b dvd b" by simp_all

   474     hence "gcd (gcd a b) c dvd b" by (rule dvd_trans)

   475     moreover have "gcd (gcd a b) c dvd c" by simp

   476     ultimately show "gcd (gcd a b) c dvd gcd b c"

   477       by (rule gcd_greatest)

   478     show "normalisation_factor (gcd (gcd a b) c) =  (if gcd (gcd a b) c = 0 then 0 else 1)"

   479       by auto

   480     fix l assume "l dvd a" and "l dvd gcd b c"

   481     with dvd_trans[OF _ gcd_dvd1] and dvd_trans[OF _ gcd_dvd2]

   482       have "l dvd b" and "l dvd c" by blast+

   483     with l dvd a show "l dvd gcd (gcd a b) c"

   484       by (intro gcd_greatest)

   485   qed

   486 next

   487   fix a b

   488   show "gcd a b = gcd b a"

   489     by (rule gcdI) (simp_all add: gcd_greatest)

   490 qed

   491

   492 lemma gcd_unique: "d dvd a \<and> d dvd b \<and>

   493     normalisation_factor d = (if d = 0 then 0 else 1) \<and>

   494     (\<forall>e. e dvd a \<and> e dvd b \<longrightarrow> e dvd d) \<longleftrightarrow> d = gcd a b"

   495   by (rule, auto intro: gcdI simp: gcd_greatest)

   496

   497 lemma gcd_dvd_prod: "gcd a b dvd k * b"

   498   using mult_dvd_mono [of 1] by auto

   499

   500 lemma gcd_1_left [simp]: "gcd 1 a = 1"

   501   by (rule sym, rule gcdI, simp_all)

   502

   503 lemma gcd_1 [simp]: "gcd a 1 = 1"

   504   by (rule sym, rule gcdI, simp_all)

   505

   506 lemma gcd_proj2_if_dvd:

   507   "b dvd a \<Longrightarrow> gcd a b = b div normalisation_factor b"

   508   by (cases "b = 0", simp_all add: dvd_eq_mod_eq_0 gcd_non_0 gcd_0)

   509

   510 lemma gcd_proj1_if_dvd:

   511   "a dvd b \<Longrightarrow> gcd a b = a div normalisation_factor a"

   512   by (subst gcd.commute, simp add: gcd_proj2_if_dvd)

   513

   514 lemma gcd_proj1_iff: "gcd m n = m div normalisation_factor m \<longleftrightarrow> m dvd n"

   515 proof

   516   assume A: "gcd m n = m div normalisation_factor m"

   517   show "m dvd n"

   518   proof (cases "m = 0")

   519     assume [simp]: "m \<noteq> 0"

   520     from A have B: "m = gcd m n * normalisation_factor m"

   521       by (simp add: unit_eq_div2)

   522     show ?thesis by (subst B, simp add: mult_unit_dvd_iff)

   523   qed (insert A, simp)

   524 next

   525   assume "m dvd n"

   526   then show "gcd m n = m div normalisation_factor m" by (rule gcd_proj1_if_dvd)

   527 qed

   528

   529 lemma gcd_proj2_iff: "gcd m n = n div normalisation_factor n \<longleftrightarrow> n dvd m"

   530   by (subst gcd.commute, simp add: gcd_proj1_iff)

   531

   532 lemma gcd_mod1 [simp]:

   533   "gcd (a mod b) b = gcd a b"

   534   by (rule gcdI, metis dvd_mod_iff gcd_dvd1 gcd_dvd2, simp_all add: gcd_greatest dvd_mod_iff)

   535

   536 lemma gcd_mod2 [simp]:

   537   "gcd a (b mod a) = gcd a b"

   538   by (rule gcdI, simp, metis dvd_mod_iff gcd_dvd1 gcd_dvd2, simp_all add: gcd_greatest dvd_mod_iff)

   539

   540 lemma normalisation_factor_dvd' [simp]:

   541   "normalisation_factor a dvd a"

   542   by (cases "a = 0", simp_all)

   543

   544 lemma gcd_mult_distrib':

   545   "k div normalisation_factor k * gcd a b = gcd (k*a) (k*b)"

   546 proof (induct a b rule: gcd_eucl.induct)

   547   case (1 a b)

   548   show ?case

   549   proof (cases "b = 0")

   550     case True

   551     then show ?thesis by (simp add: normalisation_factor_mult gcd_0 algebra_simps div_mult_div_if_dvd)

   552   next

   553     case False

   554     hence "k div normalisation_factor k * gcd a b =  gcd (k * b) (k * (a mod b))"

   555       using 1 by (subst gcd_red, simp)

   556     also have "... = gcd (k * a) (k * b)"

   557       by (simp add: mult_mod_right gcd.commute)

   558     finally show ?thesis .

   559   qed

   560 qed

   561

   562 lemma gcd_mult_distrib:

   563   "k * gcd a b = gcd (k*a) (k*b) * normalisation_factor k"

   564 proof-

   565   let ?nf = "normalisation_factor"

   566   from gcd_mult_distrib'

   567     have "gcd (k*a) (k*b) = k div ?nf k * gcd a b" ..

   568   also have "... = k * gcd a b div ?nf k"

   569     by (metis dvd_div_mult dvd_eq_mod_eq_0 mod_0 normalisation_factor_dvd)

   570   finally show ?thesis

   571     by simp

   572 qed

   573

   574 lemma euclidean_size_gcd_le1 [simp]:

   575   assumes "a \<noteq> 0"

   576   shows "euclidean_size (gcd a b) \<le> euclidean_size a"

   577 proof -

   578    have "gcd a b dvd a" by (rule gcd_dvd1)

   579    then obtain c where A: "a = gcd a b * c" unfolding dvd_def by blast

   580    with a \<noteq> 0 show ?thesis by (subst (2) A, intro size_mult_mono) auto

   581 qed

   582

   583 lemma euclidean_size_gcd_le2 [simp]:

   584   "b \<noteq> 0 \<Longrightarrow> euclidean_size (gcd a b) \<le> euclidean_size b"

   585   by (subst gcd.commute, rule euclidean_size_gcd_le1)

   586

   587 lemma euclidean_size_gcd_less1:

   588   assumes "a \<noteq> 0" and "\<not>a dvd b"

   589   shows "euclidean_size (gcd a b) < euclidean_size a"

   590 proof (rule ccontr)

   591   assume "\<not>euclidean_size (gcd a b) < euclidean_size a"

   592   with a \<noteq> 0 have "euclidean_size (gcd a b) = euclidean_size a"

   593     by (intro le_antisym, simp_all)

   594   with assms have "a dvd gcd a b" by (auto intro: dvd_euclidean_size_eq_imp_dvd)

   595   hence "a dvd b" using dvd_gcd_D2 by blast

   596   with \<not>a dvd b show False by contradiction

   597 qed

   598

   599 lemma euclidean_size_gcd_less2:

   600   assumes "b \<noteq> 0" and "\<not>b dvd a"

   601   shows "euclidean_size (gcd a b) < euclidean_size b"

   602   using assms by (subst gcd.commute, rule euclidean_size_gcd_less1)

   603

   604 lemma gcd_mult_unit1: "is_unit a \<Longrightarrow> gcd (b * a) c = gcd b c"

   605   apply (rule gcdI)

   606   apply (rule dvd_trans, rule gcd_dvd1, simp add: unit_simps)

   607   apply (rule gcd_dvd2)

   608   apply (rule gcd_greatest, simp add: unit_simps, assumption)

   609   apply (subst normalisation_factor_gcd, simp add: gcd_0)

   610   done

   611

   612 lemma gcd_mult_unit2: "is_unit a \<Longrightarrow> gcd b (c * a) = gcd b c"

   613   by (subst gcd.commute, subst gcd_mult_unit1, assumption, rule gcd.commute)

   614

   615 lemma gcd_div_unit1: "is_unit a \<Longrightarrow> gcd (b div a) c = gcd b c"

   616   by (subst unit_divide_1) (simp_all add: gcd_mult_unit1)

   617

   618 lemma gcd_div_unit2: "is_unit a \<Longrightarrow> gcd b (c div a) = gcd b c"

   619   by (subst unit_divide_1) (simp_all add: gcd_mult_unit2)

   620

   621 lemma gcd_idem: "gcd a a = a div normalisation_factor a"

   622   by (cases "a = 0") (simp add: gcd_0_left, rule sym, rule gcdI, simp_all)

   623

   624 lemma gcd_right_idem: "gcd (gcd a b) b = gcd a b"

   625   apply (rule gcdI)

   626   apply (simp add: ac_simps)

   627   apply (rule gcd_dvd2)

   628   apply (rule gcd_greatest, erule (1) gcd_greatest, assumption)

   629   apply simp

   630   done

   631

   632 lemma gcd_left_idem: "gcd a (gcd a b) = gcd a b"

   633   apply (rule gcdI)

   634   apply simp

   635   apply (rule dvd_trans, rule gcd_dvd2, rule gcd_dvd2)

   636   apply (rule gcd_greatest, assumption, erule gcd_greatest, assumption)

   637   apply simp

   638   done

   639

   640 lemma comp_fun_idem_gcd: "comp_fun_idem gcd"

   641 proof

   642   fix a b show "gcd a \<circ> gcd b = gcd b \<circ> gcd a"

   643     by (simp add: fun_eq_iff ac_simps)

   644 next

   645   fix a show "gcd a \<circ> gcd a = gcd a"

   646     by (simp add: fun_eq_iff gcd_left_idem)

   647 qed

   648

   649 lemma coprime_dvd_mult:

   650   assumes "gcd c b = 1" and "c dvd a * b"

   651   shows "c dvd a"

   652 proof -

   653   let ?nf = "normalisation_factor"

   654   from assms gcd_mult_distrib [of a c b]

   655     have A: "a = gcd (a * c) (a * b) * ?nf a" by simp

   656   from c dvd a * b show ?thesis by (subst A, simp_all add: gcd_greatest)

   657 qed

   658

   659 lemma coprime_dvd_mult_iff:

   660   "gcd c b = 1 \<Longrightarrow> (c dvd a * b) = (c dvd a)"

   661   by (rule, rule coprime_dvd_mult, simp_all)

   662

   663 lemma gcd_dvd_antisym:

   664   "gcd a b dvd gcd c d \<Longrightarrow> gcd c d dvd gcd a b \<Longrightarrow> gcd a b = gcd c d"

   665 proof (rule gcdI)

   666   assume A: "gcd a b dvd gcd c d" and B: "gcd c d dvd gcd a b"

   667   have "gcd c d dvd c" by simp

   668   with A show "gcd a b dvd c" by (rule dvd_trans)

   669   have "gcd c d dvd d" by simp

   670   with A show "gcd a b dvd d" by (rule dvd_trans)

   671   show "normalisation_factor (gcd a b) = (if gcd a b = 0 then 0 else 1)"

   672     by simp

   673   fix l assume "l dvd c" and "l dvd d"

   674   hence "l dvd gcd c d" by (rule gcd_greatest)

   675   from this and B show "l dvd gcd a b" by (rule dvd_trans)

   676 qed

   677

   678 lemma gcd_mult_cancel:

   679   assumes "gcd k n = 1"

   680   shows "gcd (k * m) n = gcd m n"

   681 proof (rule gcd_dvd_antisym)

   682   have "gcd (gcd (k * m) n) k = gcd (gcd k n) (k * m)" by (simp add: ac_simps)

   683   also note gcd k n = 1

   684   finally have "gcd (gcd (k * m) n) k = 1" by simp

   685   hence "gcd (k * m) n dvd m" by (rule coprime_dvd_mult, simp add: ac_simps)

   686   moreover have "gcd (k * m) n dvd n" by simp

   687   ultimately show "gcd (k * m) n dvd gcd m n" by (rule gcd_greatest)

   688   have "gcd m n dvd (k * m)" and "gcd m n dvd n" by simp_all

   689   then show "gcd m n dvd gcd (k * m) n" by (rule gcd_greatest)

   690 qed

   691

   692 lemma coprime_crossproduct:

   693   assumes [simp]: "gcd a d = 1" "gcd b c = 1"

   694   shows "associated (a * c) (b * d) \<longleftrightarrow> associated a b \<and> associated c d" (is "?lhs \<longleftrightarrow> ?rhs")

   695 proof

   696   assume ?rhs then show ?lhs unfolding associated_def by (fast intro: mult_dvd_mono)

   697 next

   698   assume ?lhs

   699   from ?lhs have "a dvd b * d" unfolding associated_def by (metis dvd_mult_left)

   700   hence "a dvd b" by (simp add: coprime_dvd_mult_iff)

   701   moreover from ?lhs have "b dvd a * c" unfolding associated_def by (metis dvd_mult_left)

   702   hence "b dvd a" by (simp add: coprime_dvd_mult_iff)

   703   moreover from ?lhs have "c dvd d * b"

   704     unfolding associated_def by (auto dest: dvd_mult_right simp add: ac_simps)

   705   hence "c dvd d" by (simp add: coprime_dvd_mult_iff gcd.commute)

   706   moreover from ?lhs have "d dvd c * a"

   707     unfolding associated_def by (auto dest: dvd_mult_right simp add: ac_simps)

   708   hence "d dvd c" by (simp add: coprime_dvd_mult_iff gcd.commute)

   709   ultimately show ?rhs unfolding associated_def by simp

   710 qed

   711

   712 lemma gcd_add1 [simp]:

   713   "gcd (m + n) n = gcd m n"

   714   by (cases "n = 0", simp_all add: gcd_non_0)

   715

   716 lemma gcd_add2 [simp]:

   717   "gcd m (m + n) = gcd m n"

   718   using gcd_add1 [of n m] by (simp add: ac_simps)

   719

   720 lemma gcd_add_mult: "gcd m (k * m + n) = gcd m n"

   721   by (subst gcd.commute, subst gcd_red, simp)

   722

   723 lemma coprimeI: "(\<And>l. \<lbrakk>l dvd a; l dvd b\<rbrakk> \<Longrightarrow> l dvd 1) \<Longrightarrow> gcd a b = 1"

   724   by (rule sym, rule gcdI, simp_all)

   725

   726 lemma coprime: "gcd a b = 1 \<longleftrightarrow> (\<forall>d. d dvd a \<and> d dvd b \<longleftrightarrow> is_unit d)"

   727   by (auto intro: coprimeI gcd_greatest dvd_gcd_D1 dvd_gcd_D2)

   728

   729 lemma div_gcd_coprime:

   730   assumes nz: "a \<noteq> 0 \<or> b \<noteq> 0"

   731   defines [simp]: "d \<equiv> gcd a b"

   732   defines [simp]: "a' \<equiv> a div d" and [simp]: "b' \<equiv> b div d"

   733   shows "gcd a' b' = 1"

   734 proof (rule coprimeI)

   735   fix l assume "l dvd a'" "l dvd b'"

   736   then obtain s t where "a' = l * s" "b' = l * t" unfolding dvd_def by blast

   737   moreover have "a = a' * d" "b = b' * d" by simp_all

   738   ultimately have "a = (l * d) * s" "b = (l * d) * t"

   739     by (simp_all only: ac_simps)

   740   hence "l*d dvd a" and "l*d dvd b" by (simp_all only: dvd_triv_left)

   741   hence "l*d dvd d" by (simp add: gcd_greatest)

   742   then obtain u where "d = l * d * u" ..

   743   then have "d * (l * u) = d" by (simp add: ac_simps)

   744   moreover from nz have "d \<noteq> 0" by simp

   745   with div_mult_self1_is_id have "d * (l * u) div d = l * u" .

   746   ultimately have "1 = l * u"

   747     using d \<noteq> 0 by simp

   748   then show "l dvd 1" ..

   749 qed

   750

   751 lemma coprime_mult:

   752   assumes da: "gcd d a = 1" and db: "gcd d b = 1"

   753   shows "gcd d (a * b) = 1"

   754   apply (subst gcd.commute)

   755   using da apply (subst gcd_mult_cancel)

   756   apply (subst gcd.commute, assumption)

   757   apply (subst gcd.commute, rule db)

   758   done

   759

   760 lemma coprime_lmult:

   761   assumes dab: "gcd d (a * b) = 1"

   762   shows "gcd d a = 1"

   763 proof (rule coprimeI)

   764   fix l assume "l dvd d" and "l dvd a"

   765   hence "l dvd a * b" by simp

   766   with l dvd d and dab show "l dvd 1" by (auto intro: gcd_greatest)

   767 qed

   768

   769 lemma coprime_rmult:

   770   assumes dab: "gcd d (a * b) = 1"

   771   shows "gcd d b = 1"

   772 proof (rule coprimeI)

   773   fix l assume "l dvd d" and "l dvd b"

   774   hence "l dvd a * b" by simp

   775   with l dvd d and dab show "l dvd 1" by (auto intro: gcd_greatest)

   776 qed

   777

   778 lemma coprime_mul_eq: "gcd d (a * b) = 1 \<longleftrightarrow> gcd d a = 1 \<and> gcd d b = 1"

   779   using coprime_rmult[of d a b] coprime_lmult[of d a b] coprime_mult[of d a b] by blast

   780

   781 lemma gcd_coprime:

   782   assumes c: "gcd a b \<noteq> 0" and a: "a = a' * gcd a b" and b: "b = b' * gcd a b"

   783   shows "gcd a' b' = 1"

   784 proof -

   785   from c have "a \<noteq> 0 \<or> b \<noteq> 0" by simp

   786   with div_gcd_coprime have "gcd (a div gcd a b) (b div gcd a b) = 1" .

   787   also from assms have "a div gcd a b = a'" by (metis div_mult_self2_is_id)+

   788   also from assms have "b div gcd a b = b'" by (metis div_mult_self2_is_id)+

   789   finally show ?thesis .

   790 qed

   791

   792 lemma coprime_power:

   793   assumes "0 < n"

   794   shows "gcd a (b ^ n) = 1 \<longleftrightarrow> gcd a b = 1"

   795 using assms proof (induct n)

   796   case (Suc n) then show ?case

   797     by (cases n) (simp_all add: coprime_mul_eq)

   798 qed simp

   799

   800 lemma gcd_coprime_exists:

   801   assumes nz: "gcd a b \<noteq> 0"

   802   shows "\<exists>a' b'. a = a' * gcd a b \<and> b = b' * gcd a b \<and> gcd a' b' = 1"

   803   apply (rule_tac x = "a div gcd a b" in exI)

   804   apply (rule_tac x = "b div gcd a b" in exI)

   805   apply (insert nz, auto intro: div_gcd_coprime)

   806   done

   807

   808 lemma coprime_exp:

   809   "gcd d a = 1 \<Longrightarrow> gcd d (a^n) = 1"

   810   by (induct n, simp_all add: coprime_mult)

   811

   812 lemma coprime_exp2 [intro]:

   813   "gcd a b = 1 \<Longrightarrow> gcd (a^n) (b^m) = 1"

   814   apply (rule coprime_exp)

   815   apply (subst gcd.commute)

   816   apply (rule coprime_exp)

   817   apply (subst gcd.commute)

   818   apply assumption

   819   done

   820

   821 lemma gcd_exp:

   822   "gcd (a^n) (b^n) = (gcd a b) ^ n"

   823 proof (cases "a = 0 \<and> b = 0")

   824   assume "a = 0 \<and> b = 0"

   825   then show ?thesis by (cases n, simp_all add: gcd_0_left)

   826 next

   827   assume A: "\<not>(a = 0 \<and> b = 0)"

   828   hence "1 = gcd ((a div gcd a b)^n) ((b div gcd a b)^n)"

   829     using div_gcd_coprime by (subst sym, auto simp: div_gcd_coprime)

   830   hence "(gcd a b) ^ n = (gcd a b) ^ n * ..." by simp

   831   also note gcd_mult_distrib

   832   also have "normalisation_factor ((gcd a b)^n) = 1"

   833     by (simp add: normalisation_factor_pow A)

   834   also have "(gcd a b)^n * (a div gcd a b)^n = a^n"

   835     by (subst ac_simps, subst div_power, simp, rule dvd_div_mult_self, rule dvd_power_same, simp)

   836   also have "(gcd a b)^n * (b div gcd a b)^n = b^n"

   837     by (subst ac_simps, subst div_power, simp, rule dvd_div_mult_self, rule dvd_power_same, simp)

   838   finally show ?thesis by simp

   839 qed

   840

   841 lemma coprime_common_divisor:

   842   "gcd a b = 1 \<Longrightarrow> a dvd a \<Longrightarrow> a dvd b \<Longrightarrow> is_unit a"

   843   apply (subgoal_tac "a dvd gcd a b")

   844   apply simp

   845   apply (erule (1) gcd_greatest)

   846   done

   847

   848 lemma division_decomp:

   849   assumes dc: "a dvd b * c"

   850   shows "\<exists>b' c'. a = b' * c' \<and> b' dvd b \<and> c' dvd c"

   851 proof (cases "gcd a b = 0")

   852   assume "gcd a b = 0"

   853   hence "a = 0 \<and> b = 0" by simp

   854   hence "a = 0 * c \<and> 0 dvd b \<and> c dvd c" by simp

   855   then show ?thesis by blast

   856 next

   857   let ?d = "gcd a b"

   858   assume "?d \<noteq> 0"

   859   from gcd_coprime_exists[OF this]

   860     obtain a' b' where ab': "a = a' * ?d" "b = b' * ?d" "gcd a' b' = 1"

   861     by blast

   862   from ab'(1) have "a' dvd a" unfolding dvd_def by blast

   863   with dc have "a' dvd b*c" using dvd_trans[of a' a "b*c"] by simp

   864   from dc ab'(1,2) have "a'*?d dvd (b'*?d) * c" by simp

   865   hence "?d * a' dvd ?d * (b' * c)" by (simp add: mult_ac)

   866   with ?d \<noteq> 0 have "a' dvd b' * c" by simp

   867   with coprime_dvd_mult[OF ab'(3)]

   868     have "a' dvd c" by (subst (asm) ac_simps, blast)

   869   with ab'(1) have "a = ?d * a' \<and> ?d dvd b \<and> a' dvd c" by (simp add: mult_ac)

   870   then show ?thesis by blast

   871 qed

   872

   873 lemma pow_divides_pow:

   874   assumes ab: "a ^ n dvd b ^ n" and n: "n \<noteq> 0"

   875   shows "a dvd b"

   876 proof (cases "gcd a b = 0")

   877   assume "gcd a b = 0"

   878   then show ?thesis by simp

   879 next

   880   let ?d = "gcd a b"

   881   assume "?d \<noteq> 0"

   882   from n obtain m where m: "n = Suc m" by (cases n, simp_all)

   883   from ?d \<noteq> 0 have zn: "?d ^ n \<noteq> 0" by (rule power_not_zero)

   884   from gcd_coprime_exists[OF ?d \<noteq> 0]

   885     obtain a' b' where ab': "a = a' * ?d" "b = b' * ?d" "gcd a' b' = 1"

   886     by blast

   887   from ab have "(a' * ?d) ^ n dvd (b' * ?d) ^ n"

   888     by (simp add: ab'(1,2)[symmetric])

   889   hence "?d^n * a'^n dvd ?d^n * b'^n"

   890     by (simp only: power_mult_distrib ac_simps)

   891   with zn have "a'^n dvd b'^n" by simp

   892   hence "a' dvd b'^n" using dvd_trans[of a' "a'^n" "b'^n"] by (simp add: m)

   893   hence "a' dvd b'^m * b'" by (simp add: m ac_simps)

   894   with coprime_dvd_mult[OF coprime_exp[OF ab'(3), of m]]

   895     have "a' dvd b'" by (subst (asm) ac_simps, blast)

   896   hence "a'*?d dvd b'*?d" by (rule mult_dvd_mono, simp)

   897   with ab'(1,2) show ?thesis by simp

   898 qed

   899

   900 lemma pow_divides_eq [simp]:

   901   "n \<noteq> 0 \<Longrightarrow> a ^ n dvd b ^ n \<longleftrightarrow> a dvd b"

   902   by (auto intro: pow_divides_pow dvd_power_same)

   903

   904 lemma divides_mult:

   905   assumes mr: "m dvd r" and nr: "n dvd r" and mn: "gcd m n = 1"

   906   shows "m * n dvd r"

   907 proof -

   908   from mr nr obtain m' n' where m': "r = m*m'" and n': "r = n*n'"

   909     unfolding dvd_def by blast

   910   from mr n' have "m dvd n'*n" by (simp add: ac_simps)

   911   hence "m dvd n'" using coprime_dvd_mult_iff[OF mn] by simp

   912   then obtain k where k: "n' = m*k" unfolding dvd_def by blast

   913   with n' have "r = m * n * k" by (simp add: mult_ac)

   914   then show ?thesis unfolding dvd_def by blast

   915 qed

   916

   917 lemma coprime_plus_one [simp]: "gcd (n + 1) n = 1"

   918   by (subst add_commute, simp)

   919

   920 lemma setprod_coprime [rule_format]:

   921   "(\<forall>i\<in>A. gcd (f i) a = 1) \<longrightarrow> gcd (\<Prod>i\<in>A. f i) a = 1"

   922   apply (cases "finite A")

   923   apply (induct set: finite)

   924   apply (auto simp add: gcd_mult_cancel)

   925   done

   926

   927 lemma coprime_divisors:

   928   assumes "d dvd a" "e dvd b" "gcd a b = 1"

   929   shows "gcd d e = 1"

   930 proof -

   931   from assms obtain k l where "a = d * k" "b = e * l"

   932     unfolding dvd_def by blast

   933   with assms have "gcd (d * k) (e * l) = 1" by simp

   934   hence "gcd (d * k) e = 1" by (rule coprime_lmult)

   935   also have "gcd (d * k) e = gcd e (d * k)" by (simp add: ac_simps)

   936   finally have "gcd e d = 1" by (rule coprime_lmult)

   937   then show ?thesis by (simp add: ac_simps)

   938 qed

   939

   940 lemma invertible_coprime:

   941   assumes "a * b mod m = 1"

   942   shows "coprime a m"

   943 proof -

   944   from assms have "coprime m (a * b mod m)"

   945     by simp

   946   then have "coprime m (a * b)"

   947     by simp

   948   then have "coprime m a"

   949     by (rule coprime_lmult)

   950   then show ?thesis

   951     by (simp add: ac_simps)

   952 qed

   953

   954 lemma lcm_gcd:

   955   "lcm a b = a * b div (gcd a b * normalisation_factor (a*b))"

   956   by (simp only: lcm_lcm_eucl gcd_gcd_eucl lcm_eucl_def)

   957

   958 lemma lcm_gcd_prod:

   959   "lcm a b * gcd a b = a * b div normalisation_factor (a*b)"

   960 proof (cases "a * b = 0")

   961   let ?nf = normalisation_factor

   962   assume "a * b \<noteq> 0"

   963   hence "gcd a b \<noteq> 0" by simp

   964   from lcm_gcd have "lcm a b * gcd a b = gcd a b * (a * b div (?nf (a*b) * gcd a b))"

   965     by (simp add: mult_ac)

   966   also from a * b \<noteq> 0 have "... = a * b div ?nf (a*b)"

   967     by (simp add: div_mult_swap mult.commute)

   968   finally show ?thesis .

   969 qed (auto simp add: lcm_gcd)

   970

   971 lemma lcm_dvd1 [iff]:

   972   "a dvd lcm a b"

   973 proof (cases "a*b = 0")

   974   assume "a * b \<noteq> 0"

   975   hence "gcd a b \<noteq> 0" by simp

   976   let ?c = "divide 1 (normalisation_factor (a*b))"

   977   from a * b \<noteq> 0 have [simp]: "is_unit (normalisation_factor (a*b))" by simp

   978   from lcm_gcd_prod[of a b] have "lcm a b * gcd a b = a * ?c * b"

   979     by (simp add: div_mult_swap unit_div_commute)

   980   hence "lcm a b * gcd a b div gcd a b = a * ?c * b div gcd a b" by simp

   981   with gcd a b \<noteq> 0 have "lcm a b = a * ?c * b div gcd a b"

   982     by (subst (asm) div_mult_self2_is_id, simp_all)

   983   also have "... = a * (?c * b div gcd a b)"

   984     by (metis div_mult_swap gcd_dvd2 mult_assoc)

   985   finally show ?thesis by (rule dvdI)

   986 qed (auto simp add: lcm_gcd)

   987

   988 lemma lcm_least:

   989   "\<lbrakk>a dvd k; b dvd k\<rbrakk> \<Longrightarrow> lcm a b dvd k"

   990 proof (cases "k = 0")

   991   let ?nf = normalisation_factor

   992   assume "k \<noteq> 0"

   993   hence "is_unit (?nf k)" by simp

   994   hence "?nf k \<noteq> 0" by (metis not_is_unit_0)

   995   assume A: "a dvd k" "b dvd k"

   996   hence "gcd a b \<noteq> 0" using k \<noteq> 0 by auto

   997   from A obtain r s where ar: "k = a * r" and bs: "k = b * s"

   998     unfolding dvd_def by blast

   999   with k \<noteq> 0 have "r * s \<noteq> 0"

  1000     by auto (drule sym [of 0], simp)

  1001   hence "is_unit (?nf (r * s))" by simp

  1002   let ?c = "?nf k div ?nf (r*s)"

  1003   from is_unit (?nf k) and is_unit (?nf (r * s)) have "is_unit ?c" by (rule unit_div)

  1004   hence "?c \<noteq> 0" using not_is_unit_0 by fast

  1005   from ar bs have "k * k * gcd s r = ?nf k * k * gcd (k * s) (k * r)"

  1006     by (subst mult_assoc, subst gcd_mult_distrib[of k s r], simp only: ac_simps)

  1007   also have "... = ?nf k * k * gcd ((r*s) * a) ((r*s) * b)"

  1008     by (subst (3) k = a * r, subst (3) k = b * s, simp add: algebra_simps)

  1009   also have "... = ?c * r*s * k * gcd a b" using r * s \<noteq> 0

  1010     by (subst gcd_mult_distrib'[symmetric], simp add: algebra_simps unit_simps)

  1011   finally have "(a*r) * (b*s) * gcd s r = ?c * k * r * s * gcd a b"

  1012     by (subst ar[symmetric], subst bs[symmetric], simp add: mult_ac)

  1013   hence "a * b * gcd s r * (r * s) = ?c * k * gcd a b * (r * s)"

  1014     by (simp add: algebra_simps)

  1015   hence "?c * k * gcd a b = a * b * gcd s r" using r * s \<noteq> 0

  1016     by (metis div_mult_self2_is_id)

  1017   also have "... = lcm a b * gcd a b * gcd s r * ?nf (a*b)"

  1018     by (subst lcm_gcd_prod[of a b], metis gcd_mult_distrib gcd_mult_distrib')

  1019   also have "... = lcm a b * gcd s r * ?nf (a*b) * gcd a b"

  1020     by (simp add: algebra_simps)

  1021   finally have "k * ?c = lcm a b * gcd s r * ?nf (a*b)" using gcd a b \<noteq> 0

  1022     by (metis mult.commute div_mult_self2_is_id)

  1023   hence "k = lcm a b * (gcd s r * ?nf (a*b)) div ?c" using ?c \<noteq> 0

  1024     by (metis div_mult_self2_is_id mult_assoc)

  1025   also have "... = lcm a b * (gcd s r * ?nf (a*b) div ?c)" using is_unit ?c

  1026     by (simp add: unit_simps)

  1027   finally show ?thesis by (rule dvdI)

  1028 qed simp

  1029

  1030 lemma lcm_zero:

  1031   "lcm a b = 0 \<longleftrightarrow> a = 0 \<or> b = 0"

  1032 proof -

  1033   let ?nf = normalisation_factor

  1034   {

  1035     assume "a \<noteq> 0" "b \<noteq> 0"

  1036     hence "a * b div ?nf (a * b) \<noteq> 0" by (simp add: no_zero_divisors)

  1037     moreover from a \<noteq> 0 and b \<noteq> 0 have "gcd a b \<noteq> 0" by simp

  1038     ultimately have "lcm a b \<noteq> 0" using lcm_gcd_prod[of a b] by (intro notI, simp)

  1039   } moreover {

  1040     assume "a = 0 \<or> b = 0"

  1041     hence "lcm a b = 0" by (elim disjE, simp_all add: lcm_gcd)

  1042   }

  1043   ultimately show ?thesis by blast

  1044 qed

  1045

  1046 lemmas lcm_0_iff = lcm_zero

  1047

  1048 lemma gcd_lcm:

  1049   assumes "lcm a b \<noteq> 0"

  1050   shows "gcd a b = a * b div (lcm a b * normalisation_factor (a * b))"

  1051 proof-

  1052   from assms have "gcd a b \<noteq> 0" by (simp add: lcm_zero)

  1053   let ?c = "normalisation_factor (a*b)"

  1054   from lcm a b \<noteq> 0 have "?c \<noteq> 0" by (intro notI, simp add: lcm_zero no_zero_divisors)

  1055   hence "is_unit ?c" by simp

  1056   from lcm_gcd_prod [of a b] have "gcd a b = a * b div ?c div lcm a b"

  1057     by (subst (2) div_mult_self2_is_id[OF lcm a b \<noteq> 0, symmetric], simp add: mult_ac)

  1058   also from is_unit ?c have "... = a * b div (?c * lcm a b)"

  1059     by (metis local.unit_divide_1 local.unit_divide_1'1)

  1060   finally show ?thesis by (simp only: ac_simps)

  1061 qed

  1062

  1063 lemma normalisation_factor_lcm [simp]:

  1064   "normalisation_factor (lcm a b) = (if a = 0 \<or> b = 0 then 0 else 1)"

  1065 proof (cases "a = 0 \<or> b = 0")

  1066   case True then show ?thesis

  1067     by (auto simp add: lcm_gcd)

  1068 next

  1069   case False

  1070   let ?nf = normalisation_factor

  1071   from lcm_gcd_prod[of a b]

  1072     have "?nf (lcm a b) * ?nf (gcd a b) = ?nf (a*b) div ?nf (a*b)"

  1073     by (metis div_by_0 div_self normalisation_correct normalisation_factor_0 normalisation_factor_mult)

  1074   also have "... = (if a*b = 0 then 0 else 1)"

  1075     by simp

  1076   finally show ?thesis using False by simp

  1077 qed

  1078

  1079 lemma lcm_dvd2 [iff]: "b dvd lcm a b"

  1080   using lcm_dvd1 [of b a] by (simp add: lcm_gcd ac_simps)

  1081

  1082 lemma lcmI:

  1083   "\<lbrakk>a dvd k; b dvd k; \<And>l. a dvd l \<Longrightarrow> b dvd l \<Longrightarrow> k dvd l;

  1084     normalisation_factor k = (if k = 0 then 0 else 1)\<rbrakk> \<Longrightarrow> k = lcm a b"

  1085   by (intro normed_associated_imp_eq) (auto simp: associated_def intro: lcm_least)

  1086

  1087 sublocale lcm!: abel_semigroup lcm

  1088 proof

  1089   fix a b c

  1090   show "lcm (lcm a b) c = lcm a (lcm b c)"

  1091   proof (rule lcmI)

  1092     have "a dvd lcm a b" and "lcm a b dvd lcm (lcm a b) c" by simp_all

  1093     then show "a dvd lcm (lcm a b) c" by (rule dvd_trans)

  1094

  1095     have "b dvd lcm a b" and "lcm a b dvd lcm (lcm a b) c" by simp_all

  1096     hence "b dvd lcm (lcm a b) c" by (rule dvd_trans)

  1097     moreover have "c dvd lcm (lcm a b) c" by simp

  1098     ultimately show "lcm b c dvd lcm (lcm a b) c" by (rule lcm_least)

  1099

  1100     fix l assume "a dvd l" and "lcm b c dvd l"

  1101     have "b dvd lcm b c" by simp

  1102     from this and lcm b c dvd l have "b dvd l" by (rule dvd_trans)

  1103     have "c dvd lcm b c" by simp

  1104     from this and lcm b c dvd l have "c dvd l" by (rule dvd_trans)

  1105     from a dvd l and b dvd l have "lcm a b dvd l" by (rule lcm_least)

  1106     from this and c dvd l show "lcm (lcm a b) c dvd l" by (rule lcm_least)

  1107   qed (simp add: lcm_zero)

  1108 next

  1109   fix a b

  1110   show "lcm a b = lcm b a"

  1111     by (simp add: lcm_gcd ac_simps)

  1112 qed

  1113

  1114 lemma dvd_lcm_D1:

  1115   "lcm m n dvd k \<Longrightarrow> m dvd k"

  1116   by (rule dvd_trans, rule lcm_dvd1, assumption)

  1117

  1118 lemma dvd_lcm_D2:

  1119   "lcm m n dvd k \<Longrightarrow> n dvd k"

  1120   by (rule dvd_trans, rule lcm_dvd2, assumption)

  1121

  1122 lemma gcd_dvd_lcm [simp]:

  1123   "gcd a b dvd lcm a b"

  1124   by (metis dvd_trans gcd_dvd2 lcm_dvd2)

  1125

  1126 lemma lcm_1_iff:

  1127   "lcm a b = 1 \<longleftrightarrow> is_unit a \<and> is_unit b"

  1128 proof

  1129   assume "lcm a b = 1"

  1130   then show "is_unit a \<and> is_unit b" by auto

  1131 next

  1132   assume "is_unit a \<and> is_unit b"

  1133   hence "a dvd 1" and "b dvd 1" by simp_all

  1134   hence "is_unit (lcm a b)" by (rule lcm_least)

  1135   hence "lcm a b = normalisation_factor (lcm a b)"

  1136     by (subst normalisation_factor_unit, simp_all)

  1137   also have "\<dots> = 1" using is_unit a \<and> is_unit b

  1138     by auto

  1139   finally show "lcm a b = 1" .

  1140 qed

  1141

  1142 lemma lcm_0_left [simp]:

  1143   "lcm 0 a = 0"

  1144   by (rule sym, rule lcmI, simp_all)

  1145

  1146 lemma lcm_0 [simp]:

  1147   "lcm a 0 = 0"

  1148   by (rule sym, rule lcmI, simp_all)

  1149

  1150 lemma lcm_unique:

  1151   "a dvd d \<and> b dvd d \<and>

  1152   normalisation_factor d = (if d = 0 then 0 else 1) \<and>

  1153   (\<forall>e. a dvd e \<and> b dvd e \<longrightarrow> d dvd e) \<longleftrightarrow> d = lcm a b"

  1154   by (rule, auto intro: lcmI simp: lcm_least lcm_zero)

  1155

  1156 lemma dvd_lcm_I1 [simp]:

  1157   "k dvd m \<Longrightarrow> k dvd lcm m n"

  1158   by (metis lcm_dvd1 dvd_trans)

  1159

  1160 lemma dvd_lcm_I2 [simp]:

  1161   "k dvd n \<Longrightarrow> k dvd lcm m n"

  1162   by (metis lcm_dvd2 dvd_trans)

  1163

  1164 lemma lcm_1_left [simp]:

  1165   "lcm 1 a = a div normalisation_factor a"

  1166   by (cases "a = 0") (simp, rule sym, rule lcmI, simp_all)

  1167

  1168 lemma lcm_1_right [simp]:

  1169   "lcm a 1 = a div normalisation_factor a"

  1170   using lcm_1_left [of a] by (simp add: ac_simps)

  1171

  1172 lemma lcm_coprime:

  1173   "gcd a b = 1 \<Longrightarrow> lcm a b = a * b div normalisation_factor (a*b)"

  1174   by (subst lcm_gcd) simp

  1175

  1176 lemma lcm_proj1_if_dvd:

  1177   "b dvd a \<Longrightarrow> lcm a b = a div normalisation_factor a"

  1178   by (cases "a = 0") (simp, rule sym, rule lcmI, simp_all)

  1179

  1180 lemma lcm_proj2_if_dvd:

  1181   "a dvd b \<Longrightarrow> lcm a b = b div normalisation_factor b"

  1182   using lcm_proj1_if_dvd [of a b] by (simp add: ac_simps)

  1183

  1184 lemma lcm_proj1_iff:

  1185   "lcm m n = m div normalisation_factor m \<longleftrightarrow> n dvd m"

  1186 proof

  1187   assume A: "lcm m n = m div normalisation_factor m"

  1188   show "n dvd m"

  1189   proof (cases "m = 0")

  1190     assume [simp]: "m \<noteq> 0"

  1191     from A have B: "m = lcm m n * normalisation_factor m"

  1192       by (simp add: unit_eq_div2)

  1193     show ?thesis by (subst B, simp)

  1194   qed simp

  1195 next

  1196   assume "n dvd m"

  1197   then show "lcm m n = m div normalisation_factor m" by (rule lcm_proj1_if_dvd)

  1198 qed

  1199

  1200 lemma lcm_proj2_iff:

  1201   "lcm m n = n div normalisation_factor n \<longleftrightarrow> m dvd n"

  1202   using lcm_proj1_iff [of n m] by (simp add: ac_simps)

  1203

  1204 lemma euclidean_size_lcm_le1:

  1205   assumes "a \<noteq> 0" and "b \<noteq> 0"

  1206   shows "euclidean_size a \<le> euclidean_size (lcm a b)"

  1207 proof -

  1208   have "a dvd lcm a b" by (rule lcm_dvd1)

  1209   then obtain c where A: "lcm a b = a * c" unfolding dvd_def by blast

  1210   with a \<noteq> 0 and b \<noteq> 0 have "c \<noteq> 0" by (auto simp: lcm_zero)

  1211   then show ?thesis by (subst A, intro size_mult_mono)

  1212 qed

  1213

  1214 lemma euclidean_size_lcm_le2:

  1215   "a \<noteq> 0 \<Longrightarrow> b \<noteq> 0 \<Longrightarrow> euclidean_size b \<le> euclidean_size (lcm a b)"

  1216   using euclidean_size_lcm_le1 [of b a] by (simp add: ac_simps)

  1217

  1218 lemma euclidean_size_lcm_less1:

  1219   assumes "b \<noteq> 0" and "\<not>b dvd a"

  1220   shows "euclidean_size a < euclidean_size (lcm a b)"

  1221 proof (rule ccontr)

  1222   from assms have "a \<noteq> 0" by auto

  1223   assume "\<not>euclidean_size a < euclidean_size (lcm a b)"

  1224   with a \<noteq> 0 and b \<noteq> 0 have "euclidean_size (lcm a b) = euclidean_size a"

  1225     by (intro le_antisym, simp, intro euclidean_size_lcm_le1)

  1226   with assms have "lcm a b dvd a"

  1227     by (rule_tac dvd_euclidean_size_eq_imp_dvd) (auto simp: lcm_zero)

  1228   hence "b dvd a" by (rule dvd_lcm_D2)

  1229   with \<not>b dvd a show False by contradiction

  1230 qed

  1231

  1232 lemma euclidean_size_lcm_less2:

  1233   assumes "a \<noteq> 0" and "\<not>a dvd b"

  1234   shows "euclidean_size b < euclidean_size (lcm a b)"

  1235   using assms euclidean_size_lcm_less1 [of a b] by (simp add: ac_simps)

  1236

  1237 lemma lcm_mult_unit1:

  1238   "is_unit a \<Longrightarrow> lcm (b * a) c = lcm b c"

  1239   apply (rule lcmI)

  1240   apply (rule dvd_trans[of _ "b * a"], simp, rule lcm_dvd1)

  1241   apply (rule lcm_dvd2)

  1242   apply (rule lcm_least, simp add: unit_simps, assumption)

  1243   apply (subst normalisation_factor_lcm, simp add: lcm_zero)

  1244   done

  1245

  1246 lemma lcm_mult_unit2:

  1247   "is_unit a \<Longrightarrow> lcm b (c * a) = lcm b c"

  1248   using lcm_mult_unit1 [of a c b] by (simp add: ac_simps)

  1249

  1250 lemma lcm_div_unit1:

  1251   "is_unit a \<Longrightarrow> lcm (b div a) c = lcm b c"

  1252   by (metis lcm_mult_unit1 local.unit_divide_1 local.unit_divide_1_unit)

  1253

  1254 lemma lcm_div_unit2:

  1255   "is_unit a \<Longrightarrow> lcm b (c div a) = lcm b c"

  1256   by (metis lcm_mult_unit2 local.unit_divide_1 local.unit_divide_1_unit)

  1257

  1258 lemma lcm_left_idem:

  1259   "lcm a (lcm a b) = lcm a b"

  1260   apply (rule lcmI)

  1261   apply simp

  1262   apply (subst lcm.assoc [symmetric], rule lcm_dvd2)

  1263   apply (rule lcm_least, assumption)

  1264   apply (erule (1) lcm_least)

  1265   apply (auto simp: lcm_zero)

  1266   done

  1267

  1268 lemma lcm_right_idem:

  1269   "lcm (lcm a b) b = lcm a b"

  1270   apply (rule lcmI)

  1271   apply (subst lcm.assoc, rule lcm_dvd1)

  1272   apply (rule lcm_dvd2)

  1273   apply (rule lcm_least, erule (1) lcm_least, assumption)

  1274   apply (auto simp: lcm_zero)

  1275   done

  1276

  1277 lemma comp_fun_idem_lcm: "comp_fun_idem lcm"

  1278 proof

  1279   fix a b show "lcm a \<circ> lcm b = lcm b \<circ> lcm a"

  1280     by (simp add: fun_eq_iff ac_simps)

  1281 next

  1282   fix a show "lcm a \<circ> lcm a = lcm a" unfolding o_def

  1283     by (intro ext, simp add: lcm_left_idem)

  1284 qed

  1285

  1286 lemma dvd_Lcm [simp]: "a \<in> A \<Longrightarrow> a dvd Lcm A"

  1287   and Lcm_dvd [simp]: "(\<forall>a\<in>A. a dvd l') \<Longrightarrow> Lcm A dvd l'"

  1288   and normalisation_factor_Lcm [simp]:

  1289           "normalisation_factor (Lcm A) = (if Lcm A = 0 then 0 else 1)"

  1290 proof -

  1291   have "(\<forall>a\<in>A. a dvd Lcm A) \<and> (\<forall>l'. (\<forall>a\<in>A. a dvd l') \<longrightarrow> Lcm A dvd l') \<and>

  1292     normalisation_factor (Lcm A) = (if Lcm A = 0 then 0 else 1)" (is ?thesis)

  1293   proof (cases "\<exists>l. l \<noteq>  0 \<and> (\<forall>a\<in>A. a dvd l)")

  1294     case False

  1295     hence "Lcm A = 0" by (auto simp: Lcm_Lcm_eucl Lcm_eucl_def)

  1296     with False show ?thesis by auto

  1297   next

  1298     case True

  1299     then obtain l\<^sub>0 where l\<^sub>0_props: "l\<^sub>0 \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l\<^sub>0)" by blast

  1300     def n \<equiv> "LEAST n. \<exists>l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l) \<and> euclidean_size l = n"

  1301     def l \<equiv> "SOME l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l) \<and> euclidean_size l = n"

  1302     have "\<exists>l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l) \<and> euclidean_size l = n"

  1303       apply (subst n_def)

  1304       apply (rule LeastI[of _ "euclidean_size l\<^sub>0"])

  1305       apply (rule exI[of _ l\<^sub>0])

  1306       apply (simp add: l\<^sub>0_props)

  1307       done

  1308     from someI_ex[OF this] have "l \<noteq> 0" and "\<forall>a\<in>A. a dvd l" and "euclidean_size l = n"

  1309       unfolding l_def by simp_all

  1310     {

  1311       fix l' assume "\<forall>a\<in>A. a dvd l'"

  1312       with \<forall>a\<in>A. a dvd l have "\<forall>a\<in>A. a dvd gcd l l'" by (auto intro: gcd_greatest)

  1313       moreover from l \<noteq> 0 have "gcd l l' \<noteq> 0" by simp

  1314       ultimately have "\<exists>b. b \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd b) \<and> euclidean_size b = euclidean_size (gcd l l')"

  1315         by (intro exI[of _ "gcd l l'"], auto)

  1316       hence "euclidean_size (gcd l l') \<ge> n" by (subst n_def) (rule Least_le)

  1317       moreover have "euclidean_size (gcd l l') \<le> n"

  1318       proof -

  1319         have "gcd l l' dvd l" by simp

  1320         then obtain a where "l = gcd l l' * a" unfolding dvd_def by blast

  1321         with l \<noteq> 0 have "a \<noteq> 0" by auto

  1322         hence "euclidean_size (gcd l l') \<le> euclidean_size (gcd l l' * a)"

  1323           by (rule size_mult_mono)

  1324         also have "gcd l l' * a = l" using l = gcd l l' * a ..

  1325         also note euclidean_size l = n

  1326         finally show "euclidean_size (gcd l l') \<le> n" .

  1327       qed

  1328       ultimately have "euclidean_size l = euclidean_size (gcd l l')"

  1329         by (intro le_antisym, simp_all add: euclidean_size l = n)

  1330       with l \<noteq> 0 have "l dvd gcd l l'" by (blast intro: dvd_euclidean_size_eq_imp_dvd)

  1331       hence "l dvd l'" by (blast dest: dvd_gcd_D2)

  1332     }

  1333

  1334     with (\<forall>a\<in>A. a dvd l) and normalisation_factor_is_unit[OF l \<noteq> 0] and l \<noteq> 0

  1335       have "(\<forall>a\<in>A. a dvd l div normalisation_factor l) \<and>

  1336         (\<forall>l'. (\<forall>a\<in>A. a dvd l') \<longrightarrow> l div normalisation_factor l dvd l') \<and>

  1337         normalisation_factor (l div normalisation_factor l) =

  1338         (if l div normalisation_factor l = 0 then 0 else 1)"

  1339       by (auto simp: unit_simps)

  1340     also from True have "l div normalisation_factor l = Lcm A"

  1341       by (simp add: Lcm_Lcm_eucl Lcm_eucl_def Let_def n_def l_def)

  1342     finally show ?thesis .

  1343   qed

  1344   note A = this

  1345

  1346   {fix a assume "a \<in> A" then show "a dvd Lcm A" using A by blast}

  1347   {fix l' assume "\<forall>a\<in>A. a dvd l'" then show "Lcm A dvd l'" using A by blast}

  1348   from A show "normalisation_factor (Lcm A) = (if Lcm A = 0 then 0 else 1)" by blast

  1349 qed

  1350

  1351 lemma LcmI:

  1352   "(\<And>a. a\<in>A \<Longrightarrow> a dvd l) \<Longrightarrow> (\<And>l'. (\<forall>a\<in>A. a dvd l') \<Longrightarrow> l dvd l') \<Longrightarrow>

  1353       normalisation_factor l = (if l = 0 then 0 else 1) \<Longrightarrow> l = Lcm A"

  1354   by (intro normed_associated_imp_eq)

  1355     (auto intro: Lcm_dvd dvd_Lcm simp: associated_def)

  1356

  1357 lemma Lcm_subset:

  1358   "A \<subseteq> B \<Longrightarrow> Lcm A dvd Lcm B"

  1359   by (blast intro: Lcm_dvd dvd_Lcm)

  1360

  1361 lemma Lcm_Un:

  1362   "Lcm (A \<union> B) = lcm (Lcm A) (Lcm B)"

  1363   apply (rule lcmI)

  1364   apply (blast intro: Lcm_subset)

  1365   apply (blast intro: Lcm_subset)

  1366   apply (intro Lcm_dvd ballI, elim UnE)

  1367   apply (rule dvd_trans, erule dvd_Lcm, assumption)

  1368   apply (rule dvd_trans, erule dvd_Lcm, assumption)

  1369   apply simp

  1370   done

  1371

  1372 lemma Lcm_1_iff:

  1373   "Lcm A = 1 \<longleftrightarrow> (\<forall>a\<in>A. is_unit a)"

  1374 proof

  1375   assume "Lcm A = 1"

  1376   then show "\<forall>a\<in>A. is_unit a" by auto

  1377 qed (rule LcmI [symmetric], auto)

  1378

  1379 lemma Lcm_no_units:

  1380   "Lcm A = Lcm (A - {a. is_unit a})"

  1381 proof -

  1382   have "(A - {a. is_unit a}) \<union> {a\<in>A. is_unit a} = A" by blast

  1383   hence "Lcm A = lcm (Lcm (A - {a. is_unit a})) (Lcm {a\<in>A. is_unit a})"

  1384     by (simp add: Lcm_Un[symmetric])

  1385   also have "Lcm {a\<in>A. is_unit a} = 1" by (simp add: Lcm_1_iff)

  1386   finally show ?thesis by simp

  1387 qed

  1388

  1389 lemma Lcm_empty [simp]:

  1390   "Lcm {} = 1"

  1391   by (simp add: Lcm_1_iff)

  1392

  1393 lemma Lcm_eq_0 [simp]:

  1394   "0 \<in> A \<Longrightarrow> Lcm A = 0"

  1395   by (drule dvd_Lcm) simp

  1396

  1397 lemma Lcm0_iff':

  1398   "Lcm A = 0 \<longleftrightarrow> \<not>(\<exists>l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l))"

  1399 proof

  1400   assume "Lcm A = 0"

  1401   show "\<not>(\<exists>l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l))"

  1402   proof

  1403     assume ex: "\<exists>l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l)"

  1404     then obtain l\<^sub>0 where l\<^sub>0_props: "l\<^sub>0 \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l\<^sub>0)" by blast

  1405     def n \<equiv> "LEAST n. \<exists>l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l) \<and> euclidean_size l = n"

  1406     def l \<equiv> "SOME l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l) \<and> euclidean_size l = n"

  1407     have "\<exists>l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l) \<and> euclidean_size l = n"

  1408       apply (subst n_def)

  1409       apply (rule LeastI[of _ "euclidean_size l\<^sub>0"])

  1410       apply (rule exI[of _ l\<^sub>0])

  1411       apply (simp add: l\<^sub>0_props)

  1412       done

  1413     from someI_ex[OF this] have "l \<noteq> 0" unfolding l_def by simp_all

  1414     hence "l div normalisation_factor l \<noteq> 0" by simp

  1415     also from ex have "l div normalisation_factor l = Lcm A"

  1416        by (simp only: Lcm_Lcm_eucl Lcm_eucl_def n_def l_def if_True Let_def)

  1417     finally show False using Lcm A = 0 by contradiction

  1418   qed

  1419 qed (simp only: Lcm_Lcm_eucl Lcm_eucl_def if_False)

  1420

  1421 lemma Lcm0_iff [simp]:

  1422   "finite A \<Longrightarrow> Lcm A = 0 \<longleftrightarrow> 0 \<in> A"

  1423 proof -

  1424   assume "finite A"

  1425   have "0 \<in> A \<Longrightarrow> Lcm A = 0"  by (intro dvd_0_left dvd_Lcm)

  1426   moreover {

  1427     assume "0 \<notin> A"

  1428     hence "\<Prod>A \<noteq> 0"

  1429       apply (induct rule: finite_induct[OF finite A])

  1430       apply simp

  1431       apply (subst setprod.insert, assumption, assumption)

  1432       apply (rule no_zero_divisors)

  1433       apply blast+

  1434       done

  1435     moreover from finite A have "\<forall>a\<in>A. a dvd \<Prod>A" by blast

  1436     ultimately have "\<exists>l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l)" by blast

  1437     with Lcm0_iff' have "Lcm A \<noteq> 0" by simp

  1438   }

  1439   ultimately show "Lcm A = 0 \<longleftrightarrow> 0 \<in> A" by blast

  1440 qed

  1441

  1442 lemma Lcm_no_multiple:

  1443   "(\<forall>m. m \<noteq> 0 \<longrightarrow> (\<exists>a\<in>A. \<not>a dvd m)) \<Longrightarrow> Lcm A = 0"

  1444 proof -

  1445   assume "\<forall>m. m \<noteq> 0 \<longrightarrow> (\<exists>a\<in>A. \<not>a dvd m)"

  1446   hence "\<not>(\<exists>l. l \<noteq> 0 \<and> (\<forall>a\<in>A. a dvd l))" by blast

  1447   then show "Lcm A = 0" by (simp only: Lcm_Lcm_eucl Lcm_eucl_def if_False)

  1448 qed

  1449

  1450 lemma Lcm_insert [simp]:

  1451   "Lcm (insert a A) = lcm a (Lcm A)"

  1452 proof (rule lcmI)

  1453   fix l assume "a dvd l" and "Lcm A dvd l"

  1454   hence "\<forall>a\<in>A. a dvd l" by (blast intro: dvd_trans dvd_Lcm)

  1455   with a dvd l show "Lcm (insert a A) dvd l" by (force intro: Lcm_dvd)

  1456 qed (auto intro: Lcm_dvd dvd_Lcm)

  1457

  1458 lemma Lcm_finite:

  1459   assumes "finite A"

  1460   shows "Lcm A = Finite_Set.fold lcm 1 A"

  1461   by (induct rule: finite.induct[OF finite A])

  1462     (simp_all add: comp_fun_idem.fold_insert_idem[OF comp_fun_idem_lcm])

  1463

  1464 lemma Lcm_set [code_unfold]:

  1465   "Lcm (set xs) = fold lcm xs 1"

  1466   using comp_fun_idem.fold_set_fold[OF comp_fun_idem_lcm] Lcm_finite by (simp add: ac_simps)

  1467

  1468 lemma Lcm_singleton [simp]:

  1469   "Lcm {a} = a div normalisation_factor a"

  1470   by simp

  1471

  1472 lemma Lcm_2 [simp]:

  1473   "Lcm {a,b} = lcm a b"

  1474   by (simp only: Lcm_insert Lcm_empty lcm_1_right)

  1475     (cases "b = 0", simp, rule lcm_div_unit2, simp)

  1476

  1477 lemma Lcm_coprime:

  1478   assumes "finite A" and "A \<noteq> {}"

  1479   assumes "\<And>a b. a \<in> A \<Longrightarrow> b \<in> A \<Longrightarrow> a \<noteq> b \<Longrightarrow> gcd a b = 1"

  1480   shows "Lcm A = \<Prod>A div normalisation_factor (\<Prod>A)"

  1481 using assms proof (induct rule: finite_ne_induct)

  1482   case (insert a A)

  1483   have "Lcm (insert a A) = lcm a (Lcm A)" by simp

  1484   also from insert have "Lcm A = \<Prod>A div normalisation_factor (\<Prod>A)" by blast

  1485   also have "lcm a \<dots> = lcm a (\<Prod>A)" by (cases "\<Prod>A = 0") (simp_all add: lcm_div_unit2)

  1486   also from insert have "gcd a (\<Prod>A) = 1" by (subst gcd.commute, intro setprod_coprime) auto

  1487   with insert have "lcm a (\<Prod>A) = \<Prod>(insert a A) div normalisation_factor (\<Prod>(insert a A))"

  1488     by (simp add: lcm_coprime)

  1489   finally show ?case .

  1490 qed simp

  1491

  1492 lemma Lcm_coprime':

  1493   "card A \<noteq> 0 \<Longrightarrow> (\<And>a b. a \<in> A \<Longrightarrow> b \<in> A \<Longrightarrow> a \<noteq> b \<Longrightarrow> gcd a b = 1)

  1494     \<Longrightarrow> Lcm A = \<Prod>A div normalisation_factor (\<Prod>A)"

  1495   by (rule Lcm_coprime) (simp_all add: card_eq_0_iff)

  1496

  1497 lemma Gcd_Lcm:

  1498   "Gcd A = Lcm {d. \<forall>a\<in>A. d dvd a}"

  1499   by (simp add: Gcd_Gcd_eucl Lcm_Lcm_eucl Gcd_eucl_def)

  1500

  1501 lemma Gcd_dvd [simp]: "a \<in> A \<Longrightarrow> Gcd A dvd a"

  1502   and dvd_Gcd [simp]: "(\<forall>a\<in>A. g' dvd a) \<Longrightarrow> g' dvd Gcd A"

  1503   and normalisation_factor_Gcd [simp]:

  1504     "normalisation_factor (Gcd A) = (if Gcd A = 0 then 0 else 1)"

  1505 proof -

  1506   fix a assume "a \<in> A"

  1507   hence "Lcm {d. \<forall>a\<in>A. d dvd a} dvd a" by (intro Lcm_dvd) blast

  1508   then show "Gcd A dvd a" by (simp add: Gcd_Lcm)

  1509 next

  1510   fix g' assume "\<forall>a\<in>A. g' dvd a"

  1511   hence "g' dvd Lcm {d. \<forall>a\<in>A. d dvd a}" by (intro dvd_Lcm) blast

  1512   then show "g' dvd Gcd A" by (simp add: Gcd_Lcm)

  1513 next

  1514   show "normalisation_factor (Gcd A) = (if Gcd A = 0 then 0 else 1)"

  1515     by (simp add: Gcd_Lcm)

  1516 qed

  1517

  1518 lemma GcdI:

  1519   "(\<And>a. a\<in>A \<Longrightarrow> l dvd a) \<Longrightarrow> (\<And>l'. (\<forall>a\<in>A. l' dvd a) \<Longrightarrow> l' dvd l) \<Longrightarrow>

  1520     normalisation_factor l = (if l = 0 then 0 else 1) \<Longrightarrow> l = Gcd A"

  1521   by (intro normed_associated_imp_eq)

  1522     (auto intro: Gcd_dvd dvd_Gcd simp: associated_def)

  1523

  1524 lemma Lcm_Gcd:

  1525   "Lcm A = Gcd {m. \<forall>a\<in>A. a dvd m}"

  1526   by (rule LcmI[symmetric]) (auto intro: dvd_Gcd Gcd_dvd)

  1527

  1528 lemma Gcd_0_iff:

  1529   "Gcd A = 0 \<longleftrightarrow> A \<subseteq> {0}"

  1530   apply (rule iffI)

  1531   apply (rule subsetI, drule Gcd_dvd, simp)

  1532   apply (auto intro: GcdI[symmetric])

  1533   done

  1534

  1535 lemma Gcd_empty [simp]:

  1536   "Gcd {} = 0"

  1537   by (simp add: Gcd_0_iff)

  1538

  1539 lemma Gcd_1:

  1540   "1 \<in> A \<Longrightarrow> Gcd A = 1"

  1541   by (intro GcdI[symmetric]) (auto intro: Gcd_dvd dvd_Gcd)

  1542

  1543 lemma Gcd_insert [simp]:

  1544   "Gcd (insert a A) = gcd a (Gcd A)"

  1545 proof (rule gcdI)

  1546   fix l assume "l dvd a" and "l dvd Gcd A"

  1547   hence "\<forall>a\<in>A. l dvd a" by (blast intro: dvd_trans Gcd_dvd)

  1548   with l dvd a show "l dvd Gcd (insert a A)" by (force intro: Gcd_dvd)

  1549 qed auto

  1550

  1551 lemma Gcd_finite:

  1552   assumes "finite A"

  1553   shows "Gcd A = Finite_Set.fold gcd 0 A"

  1554   by (induct rule: finite.induct[OF finite A])

  1555     (simp_all add: comp_fun_idem.fold_insert_idem[OF comp_fun_idem_gcd])

  1556

  1557 lemma Gcd_set [code_unfold]:

  1558   "Gcd (set xs) = fold gcd xs 0"

  1559   using comp_fun_idem.fold_set_fold[OF comp_fun_idem_gcd] Gcd_finite by (simp add: ac_simps)

  1560

  1561 lemma Gcd_singleton [simp]: "Gcd {a} = a div normalisation_factor a"

  1562   by (simp add: gcd_0)

  1563

  1564 lemma Gcd_2 [simp]: "Gcd {a,b} = gcd a b"

  1565   by (simp only: Gcd_insert Gcd_empty gcd_0) (cases "b = 0", simp, rule gcd_div_unit2, simp)

  1566

  1567 end

  1568

  1569 text {*

  1570   A Euclidean ring is a Euclidean semiring with additive inverses. It provides a

  1571   few more lemmas; in particular, Bezout's lemma holds for any Euclidean ring.

  1572 *}

  1573

  1574 class euclidean_ring = euclidean_semiring + idom

  1575

  1576 class euclidean_ring_gcd = euclidean_semiring_gcd + idom

  1577 begin

  1578

  1579 subclass euclidean_ring ..

  1580

  1581 lemma gcd_neg1 [simp]:

  1582   "gcd (-a) b = gcd a b"

  1583   by (rule sym, rule gcdI, simp_all add: gcd_greatest)

  1584

  1585 lemma gcd_neg2 [simp]:

  1586   "gcd a (-b) = gcd a b"

  1587   by (rule sym, rule gcdI, simp_all add: gcd_greatest)

  1588

  1589 lemma gcd_neg_numeral_1 [simp]:

  1590   "gcd (- numeral n) a = gcd (numeral n) a"

  1591   by (fact gcd_neg1)

  1592

  1593 lemma gcd_neg_numeral_2 [simp]:

  1594   "gcd a (- numeral n) = gcd a (numeral n)"

  1595   by (fact gcd_neg2)

  1596

  1597 lemma gcd_diff1: "gcd (m - n) n = gcd m n"

  1598   by (subst diff_conv_add_uminus, subst gcd_neg2[symmetric],  subst gcd_add1, simp)

  1599

  1600 lemma gcd_diff2: "gcd (n - m) n = gcd m n"

  1601   by (subst gcd_neg1[symmetric], simp only: minus_diff_eq gcd_diff1)

  1602

  1603 lemma coprime_minus_one [simp]: "gcd (n - 1) n = 1"

  1604 proof -

  1605   have "gcd (n - 1) n = gcd n (n - 1)" by (fact gcd.commute)

  1606   also have "\<dots> = gcd ((n - 1) + 1) (n - 1)" by simp

  1607   also have "\<dots> = 1" by (rule coprime_plus_one)

  1608   finally show ?thesis .

  1609 qed

  1610

  1611 lemma lcm_neg1 [simp]: "lcm (-a) b = lcm a b"

  1612   by (rule sym, rule lcmI, simp_all add: lcm_least lcm_zero)

  1613

  1614 lemma lcm_neg2 [simp]: "lcm a (-b) = lcm a b"

  1615   by (rule sym, rule lcmI, simp_all add: lcm_least lcm_zero)

  1616

  1617 lemma lcm_neg_numeral_1 [simp]: "lcm (- numeral n) a = lcm (numeral n) a"

  1618   by (fact lcm_neg1)

  1619

  1620 lemma lcm_neg_numeral_2 [simp]: "lcm a (- numeral n) = lcm a (numeral n)"

  1621   by (fact lcm_neg2)

  1622

  1623 function euclid_ext :: "'a \<Rightarrow> 'a \<Rightarrow> 'a \<times> 'a \<times> 'a" where

  1624   "euclid_ext a b =

  1625      (if b = 0 then

  1626         let c = divide 1 (normalisation_factor a) in (c, 0, a * c)

  1627       else

  1628         case euclid_ext b (a mod b) of

  1629             (s,t,c) \<Rightarrow> (t, s - t * (a div b), c))"

  1630   by (pat_completeness, simp)

  1631   termination by (relation "measure (euclidean_size \<circ> snd)", simp_all)

  1632

  1633 declare euclid_ext.simps [simp del]

  1634

  1635 lemma euclid_ext_0:

  1636   "euclid_ext a 0 = (divide 1 (normalisation_factor a), 0, a * divide 1 (normalisation_factor a))"

  1637   by (subst euclid_ext.simps, simp add: Let_def)

  1638

  1639 lemma euclid_ext_non_0:

  1640   "b \<noteq> 0 \<Longrightarrow> euclid_ext a b = (case euclid_ext b (a mod b) of

  1641     (s,t,c) \<Rightarrow> (t, s - t * (a div b), c))"

  1642   by (subst euclid_ext.simps, simp)

  1643

  1644 definition euclid_ext' :: "'a \<Rightarrow> 'a \<Rightarrow> 'a \<times> 'a"

  1645 where

  1646   "euclid_ext' a b = (case euclid_ext a b of (s, t, _) \<Rightarrow> (s, t))"

  1647

  1648 lemma euclid_ext_gcd [simp]:

  1649   "(case euclid_ext a b of (_,_,t) \<Rightarrow> t) = gcd a b"

  1650 proof (induct a b rule: euclid_ext.induct)

  1651   case (1 a b)

  1652   then show ?case

  1653   proof (cases "b = 0")

  1654     case True

  1655       then show ?thesis by (cases "a = 0")

  1656         (simp_all add: euclid_ext_0 unit_div mult_ac unit_simps gcd_0)

  1657     next

  1658     case False with 1 show ?thesis

  1659       by (simp add: euclid_ext_non_0 ac_simps split: prod.split prod.split_asm)

  1660     qed

  1661 qed

  1662

  1663 lemma euclid_ext_gcd' [simp]:

  1664   "euclid_ext a b = (r, s, t) \<Longrightarrow> t = gcd a b"

  1665   by (insert euclid_ext_gcd[of a b], drule (1) subst, simp)

  1666

  1667 lemma euclid_ext_correct:

  1668   "case euclid_ext a b of (s,t,c) \<Rightarrow> s*a + t*b = c"

  1669 proof (induct a b rule: euclid_ext.induct)

  1670   case (1 a b)

  1671   show ?case

  1672   proof (cases "b = 0")

  1673     case True

  1674     then show ?thesis by (simp add: euclid_ext_0 mult_ac)

  1675   next

  1676     case False

  1677     obtain s t c where stc: "euclid_ext b (a mod b) = (s,t,c)"

  1678       by (cases "euclid_ext b (a mod b)", blast)

  1679     from 1 have "c = s * b + t * (a mod b)" by (simp add: stc False)

  1680     also have "... = t*((a div b)*b + a mod b) + (s - t * (a div b))*b"

  1681       by (simp add: algebra_simps)

  1682     also have "(a div b)*b + a mod b = a" using mod_div_equality .

  1683     finally show ?thesis

  1684       by (subst euclid_ext.simps, simp add: False stc)

  1685     qed

  1686 qed

  1687

  1688 lemma euclid_ext'_correct:

  1689   "fst (euclid_ext' a b) * a + snd (euclid_ext' a b) * b = gcd a b"

  1690 proof-

  1691   obtain s t c where "euclid_ext a b = (s,t,c)"

  1692     by (cases "euclid_ext a b", blast)

  1693   with euclid_ext_correct[of a b] euclid_ext_gcd[of a b]

  1694     show ?thesis unfolding euclid_ext'_def by simp

  1695 qed

  1696

  1697 lemma bezout: "\<exists>s t. s * a + t * b = gcd a b"

  1698   using euclid_ext'_correct by blast

  1699

  1700 lemma euclid_ext'_0 [simp]: "euclid_ext' a 0 = (divide 1 (normalisation_factor a), 0)"

  1701   by (simp add: bezw_def euclid_ext'_def euclid_ext_0)

  1702

  1703 lemma euclid_ext'_non_0: "b \<noteq> 0 \<Longrightarrow> euclid_ext' a b = (snd (euclid_ext' b (a mod b)),

  1704   fst (euclid_ext' b (a mod b)) - snd (euclid_ext' b (a mod b)) * (a div b))"

  1705   by (cases "euclid_ext b (a mod b)")

  1706     (simp add: euclid_ext'_def euclid_ext_non_0)

  1707

  1708 end

  1709

  1710 instantiation nat :: euclidean_semiring

  1711 begin

  1712

  1713 definition [simp]:

  1714   "euclidean_size_nat = (id :: nat \<Rightarrow> nat)"

  1715

  1716 definition [simp]:

  1717   "normalisation_factor_nat (n::nat) = (if n = 0 then 0 else 1 :: nat)"

  1718

  1719 instance proof

  1720 qed simp_all

  1721

  1722 end

  1723

  1724 instantiation int :: euclidean_ring

  1725 begin

  1726

  1727 definition [simp]:

  1728   "euclidean_size_int = (nat \<circ> abs :: int \<Rightarrow> nat)"

  1729

  1730 definition [simp]:

  1731   "normalisation_factor_int = (sgn :: int \<Rightarrow> int)"

  1732

  1733 instance proof

  1734   case goal2 then show ?case by (auto simp add: abs_mult nat_mult_distrib)

  1735 next

  1736   case goal3 then show ?case by (simp add: zsgn_def)

  1737 next

  1738   case goal5 then show ?case by (auto simp: zsgn_def)

  1739 next

  1740   case goal6 then show ?case by (auto split: abs_split simp: zsgn_def)

  1741 qed (auto simp: sgn_times split: abs_split)

  1742

  1743 end

  1744

  1745 end