src/HOL/Library/Legacy_GCD.thy
changeset 32479 521cc9bf2958
parent 32478 87201c60ae7d
child 32480 6c19da8e661a
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     1 (*  Title:      HOL/GCD.thy
       
     2     Author:     Christophe Tabacznyj and Lawrence C Paulson
       
     3     Copyright   1996  University of Cambridge
       
     4 *)
       
     5 
       
     6 header {* The Greatest Common Divisor *}
       
     7 
       
     8 theory Legacy_GCD
       
     9 imports Main
       
    10 begin
       
    11 
       
    12 text {*
       
    13   See \cite{davenport92}. \bigskip
       
    14 *}
       
    15 
       
    16 subsection {* Specification of GCD on nats *}
       
    17 
       
    18 definition
       
    19   is_gcd :: "nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> bool" where -- {* @{term gcd} as a relation *}
       
    20   [code del]: "is_gcd m n p \<longleftrightarrow> p dvd m \<and> p dvd n \<and>
       
    21     (\<forall>d. d dvd m \<longrightarrow> d dvd n \<longrightarrow> d dvd p)"
       
    22 
       
    23 text {* Uniqueness *}
       
    24 
       
    25 lemma is_gcd_unique: "is_gcd a b m \<Longrightarrow> is_gcd a b n \<Longrightarrow> m = n"
       
    26   by (simp add: is_gcd_def) (blast intro: dvd_anti_sym)
       
    27 
       
    28 text {* Connection to divides relation *}
       
    29 
       
    30 lemma is_gcd_dvd: "is_gcd a b m \<Longrightarrow> k dvd a \<Longrightarrow> k dvd b \<Longrightarrow> k dvd m"
       
    31   by (auto simp add: is_gcd_def)
       
    32 
       
    33 text {* Commutativity *}
       
    34 
       
    35 lemma is_gcd_commute: "is_gcd m n k = is_gcd n m k"
       
    36   by (auto simp add: is_gcd_def)
       
    37 
       
    38 
       
    39 subsection {* GCD on nat by Euclid's algorithm *}
       
    40 
       
    41 fun
       
    42   gcd  :: "nat => nat => nat"
       
    43 where
       
    44   "gcd m n = (if n = 0 then m else gcd n (m mod n))"
       
    45 lemma gcd_induct [case_names "0" rec]:
       
    46   fixes m n :: nat
       
    47   assumes "\<And>m. P m 0"
       
    48     and "\<And>m n. 0 < n \<Longrightarrow> P n (m mod n) \<Longrightarrow> P m n"
       
    49   shows "P m n"
       
    50 proof (induct m n rule: gcd.induct)
       
    51   case (1 m n) with assms show ?case by (cases "n = 0") simp_all
       
    52 qed
       
    53 
       
    54 lemma gcd_0 [simp, algebra]: "gcd m 0 = m"
       
    55   by simp
       
    56 
       
    57 lemma gcd_0_left [simp,algebra]: "gcd 0 m = m"
       
    58   by simp
       
    59 
       
    60 lemma gcd_non_0: "n > 0 \<Longrightarrow> gcd m n = gcd n (m mod n)"
       
    61   by simp
       
    62 
       
    63 lemma gcd_1 [simp, algebra]: "gcd m (Suc 0) = Suc 0"
       
    64   by simp
       
    65 
       
    66 lemma nat_gcd_1_right [simp, algebra]: "gcd m 1 = 1"
       
    67   unfolding One_nat_def by (rule gcd_1)
       
    68 
       
    69 declare gcd.simps [simp del]
       
    70 
       
    71 text {*
       
    72   \medskip @{term "gcd m n"} divides @{text m} and @{text n}.  The
       
    73   conjunctions don't seem provable separately.
       
    74 *}
       
    75 
       
    76 lemma gcd_dvd1 [iff, algebra]: "gcd m n dvd m"
       
    77   and gcd_dvd2 [iff, algebra]: "gcd m n dvd n"
       
    78   apply (induct m n rule: gcd_induct)
       
    79      apply (simp_all add: gcd_non_0)
       
    80   apply (blast dest: dvd_mod_imp_dvd)
       
    81   done
       
    82 
       
    83 text {*
       
    84   \medskip Maximality: for all @{term m}, @{term n}, @{term k}
       
    85   naturals, if @{term k} divides @{term m} and @{term k} divides
       
    86   @{term n} then @{term k} divides @{term "gcd m n"}.
       
    87 *}
       
    88 
       
    89 lemma gcd_greatest: "k dvd m \<Longrightarrow> k dvd n \<Longrightarrow> k dvd gcd m n"
       
    90   by (induct m n rule: gcd_induct) (simp_all add: gcd_non_0 dvd_mod)
       
    91 
       
    92 text {*
       
    93   \medskip Function gcd yields the Greatest Common Divisor.
       
    94 *}
       
    95 
       
    96 lemma is_gcd: "is_gcd m n (gcd m n) "
       
    97   by (simp add: is_gcd_def gcd_greatest)
       
    98 
       
    99 
       
   100 subsection {* Derived laws for GCD *}
       
   101 
       
   102 lemma gcd_greatest_iff [iff, algebra]: "k dvd gcd m n \<longleftrightarrow> k dvd m \<and> k dvd n"
       
   103   by (blast intro!: gcd_greatest intro: dvd_trans)
       
   104 
       
   105 lemma gcd_zero[algebra]: "gcd m n = 0 \<longleftrightarrow> m = 0 \<and> n = 0"
       
   106   by (simp only: dvd_0_left_iff [symmetric] gcd_greatest_iff)
       
   107 
       
   108 lemma gcd_commute: "gcd m n = gcd n m"
       
   109   apply (rule is_gcd_unique)
       
   110    apply (rule is_gcd)
       
   111   apply (subst is_gcd_commute)
       
   112   apply (simp add: is_gcd)
       
   113   done
       
   114 
       
   115 lemma gcd_assoc: "gcd (gcd k m) n = gcd k (gcd m n)"
       
   116   apply (rule is_gcd_unique)
       
   117    apply (rule is_gcd)
       
   118   apply (simp add: is_gcd_def)
       
   119   apply (blast intro: dvd_trans)
       
   120   done
       
   121 
       
   122 lemma gcd_1_left [simp, algebra]: "gcd (Suc 0) m = Suc 0"
       
   123   by (simp add: gcd_commute)
       
   124 
       
   125 lemma nat_gcd_1_left [simp, algebra]: "gcd 1 m = 1"
       
   126   unfolding One_nat_def by (rule gcd_1_left)
       
   127 
       
   128 text {*
       
   129   \medskip Multiplication laws
       
   130 *}
       
   131 
       
   132 lemma gcd_mult_distrib2: "k * gcd m n = gcd (k * m) (k * n)"
       
   133     -- {* \cite[page 27]{davenport92} *}
       
   134   apply (induct m n rule: gcd_induct)
       
   135    apply simp
       
   136   apply (case_tac "k = 0")
       
   137    apply (simp_all add: mod_geq gcd_non_0 mod_mult_distrib2)
       
   138   done
       
   139 
       
   140 lemma gcd_mult [simp, algebra]: "gcd k (k * n) = k"
       
   141   apply (rule gcd_mult_distrib2 [of k 1 n, simplified, symmetric])
       
   142   done
       
   143 
       
   144 lemma gcd_self [simp, algebra]: "gcd k k = k"
       
   145   apply (rule gcd_mult [of k 1, simplified])
       
   146   done
       
   147 
       
   148 lemma relprime_dvd_mult: "gcd k n = 1 ==> k dvd m * n ==> k dvd m"
       
   149   apply (insert gcd_mult_distrib2 [of m k n])
       
   150   apply simp
       
   151   apply (erule_tac t = m in ssubst)
       
   152   apply simp
       
   153   done
       
   154 
       
   155 lemma relprime_dvd_mult_iff: "gcd k n = 1 ==> (k dvd m * n) = (k dvd m)"
       
   156   by (auto intro: relprime_dvd_mult dvd_mult2)
       
   157 
       
   158 lemma gcd_mult_cancel: "gcd k n = 1 ==> gcd (k * m) n = gcd m n"
       
   159   apply (rule dvd_anti_sym)
       
   160    apply (rule gcd_greatest)
       
   161     apply (rule_tac n = k in relprime_dvd_mult)
       
   162      apply (simp add: gcd_assoc)
       
   163      apply (simp add: gcd_commute)
       
   164     apply (simp_all add: mult_commute)
       
   165   done
       
   166 
       
   167 
       
   168 text {* \medskip Addition laws *}
       
   169 
       
   170 lemma gcd_add1 [simp, algebra]: "gcd (m + n) n = gcd m n"
       
   171   by (cases "n = 0") (auto simp add: gcd_non_0)
       
   172 
       
   173 lemma gcd_add2 [simp, algebra]: "gcd m (m + n) = gcd m n"
       
   174 proof -
       
   175   have "gcd m (m + n) = gcd (m + n) m" by (rule gcd_commute)
       
   176   also have "... = gcd (n + m) m" by (simp add: add_commute)
       
   177   also have "... = gcd n m" by simp
       
   178   also have  "... = gcd m n" by (rule gcd_commute)
       
   179   finally show ?thesis .
       
   180 qed
       
   181 
       
   182 lemma gcd_add2' [simp, algebra]: "gcd m (n + m) = gcd m n"
       
   183   apply (subst add_commute)
       
   184   apply (rule gcd_add2)
       
   185   done
       
   186 
       
   187 lemma gcd_add_mult[algebra]: "gcd m (k * m + n) = gcd m n"
       
   188   by (induct k) (simp_all add: add_assoc)
       
   189 
       
   190 lemma gcd_dvd_prod: "gcd m n dvd m * n" 
       
   191   using mult_dvd_mono [of 1] by auto
       
   192 
       
   193 text {*
       
   194   \medskip Division by gcd yields rrelatively primes.
       
   195 *}
       
   196 
       
   197 lemma div_gcd_relprime:
       
   198   assumes nz: "a \<noteq> 0 \<or> b \<noteq> 0"
       
   199   shows "gcd (a div gcd a b) (b div gcd a b) = 1"
       
   200 proof -
       
   201   let ?g = "gcd a b"
       
   202   let ?a' = "a div ?g"
       
   203   let ?b' = "b div ?g"
       
   204   let ?g' = "gcd ?a' ?b'"
       
   205   have dvdg: "?g dvd a" "?g dvd b" by simp_all
       
   206   have dvdg': "?g' dvd ?a'" "?g' dvd ?b'" by simp_all
       
   207   from dvdg dvdg' obtain ka kb ka' kb' where
       
   208       kab: "a = ?g * ka" "b = ?g * kb" "?a' = ?g' * ka'" "?b' = ?g' * kb'"
       
   209     unfolding dvd_def by blast
       
   210   then have "?g * ?a' = (?g * ?g') * ka'" "?g * ?b' = (?g * ?g') * kb'" by simp_all
       
   211   then have dvdgg':"?g * ?g' dvd a" "?g* ?g' dvd b"
       
   212     by (auto simp add: dvd_mult_div_cancel [OF dvdg(1)]
       
   213       dvd_mult_div_cancel [OF dvdg(2)] dvd_def)
       
   214   have "?g \<noteq> 0" using nz by (simp add: gcd_zero)
       
   215   then have gp: "?g > 0" by simp
       
   216   from gcd_greatest [OF dvdgg'] have "?g * ?g' dvd ?g" .
       
   217   with dvd_mult_cancel1 [OF gp] show "?g' = 1" by simp
       
   218 qed
       
   219 
       
   220 
       
   221 lemma gcd_unique: "d dvd a\<and>d dvd b \<and> (\<forall>e. e dvd a \<and> e dvd b \<longrightarrow> e dvd d) \<longleftrightarrow> d = gcd a b"
       
   222 proof(auto)
       
   223   assume H: "d dvd a" "d dvd b" "\<forall>e. e dvd a \<and> e dvd b \<longrightarrow> e dvd d"
       
   224   from H(3)[rule_format] gcd_dvd1[of a b] gcd_dvd2[of a b] 
       
   225   have th: "gcd a b dvd d" by blast
       
   226   from dvd_anti_sym[OF th gcd_greatest[OF H(1,2)]]  show "d = gcd a b" by blast 
       
   227 qed
       
   228 
       
   229 lemma gcd_eq: assumes H: "\<forall>d. d dvd x \<and> d dvd y \<longleftrightarrow> d dvd u \<and> d dvd v"
       
   230   shows "gcd x y = gcd u v"
       
   231 proof-
       
   232   from H have "\<forall>d. d dvd x \<and> d dvd y \<longleftrightarrow> d dvd gcd u v" by simp
       
   233   with gcd_unique[of "gcd u v" x y]  show ?thesis by auto
       
   234 qed
       
   235 
       
   236 lemma ind_euclid: 
       
   237   assumes c: " \<forall>a b. P (a::nat) b \<longleftrightarrow> P b a" and z: "\<forall>a. P a 0" 
       
   238   and add: "\<forall>a b. P a b \<longrightarrow> P a (a + b)" 
       
   239   shows "P a b"
       
   240 proof(induct n\<equiv>"a+b" arbitrary: a b rule: nat_less_induct)
       
   241   fix n a b
       
   242   assume H: "\<forall>m < n. \<forall>a b. m = a + b \<longrightarrow> P a b" "n = a + b"
       
   243   have "a = b \<or> a < b \<or> b < a" by arith
       
   244   moreover {assume eq: "a= b"
       
   245     from add[rule_format, OF z[rule_format, of a]] have "P a b" using eq by simp}
       
   246   moreover
       
   247   {assume lt: "a < b"
       
   248     hence "a + b - a < n \<or> a = 0"  using H(2) by arith
       
   249     moreover
       
   250     {assume "a =0" with z c have "P a b" by blast }
       
   251     moreover
       
   252     {assume ab: "a + b - a < n"
       
   253       have th0: "a + b - a = a + (b - a)" using lt by arith
       
   254       from add[rule_format, OF H(1)[rule_format, OF ab th0]]
       
   255       have "P a b" by (simp add: th0[symmetric])}
       
   256     ultimately have "P a b" by blast}
       
   257   moreover
       
   258   {assume lt: "a > b"
       
   259     hence "b + a - b < n \<or> b = 0"  using H(2) by arith
       
   260     moreover
       
   261     {assume "b =0" with z c have "P a b" by blast }
       
   262     moreover
       
   263     {assume ab: "b + a - b < n"
       
   264       have th0: "b + a - b = b + (a - b)" using lt by arith
       
   265       from add[rule_format, OF H(1)[rule_format, OF ab th0]]
       
   266       have "P b a" by (simp add: th0[symmetric])
       
   267       hence "P a b" using c by blast }
       
   268     ultimately have "P a b" by blast}
       
   269 ultimately  show "P a b" by blast
       
   270 qed
       
   271 
       
   272 lemma bezout_lemma: 
       
   273   assumes ex: "\<exists>(d::nat) x y. d dvd a \<and> d dvd b \<and> (a * x = b * y + d \<or> b * x = a * y + d)"
       
   274   shows "\<exists>d x y. d dvd a \<and> d dvd a + b \<and> (a * x = (a + b) * y + d \<or> (a + b) * x = a * y + d)"
       
   275 using ex
       
   276 apply clarsimp
       
   277 apply (rule_tac x="d" in exI, simp add: dvd_add)
       
   278 apply (case_tac "a * x = b * y + d" , simp_all)
       
   279 apply (rule_tac x="x + y" in exI)
       
   280 apply (rule_tac x="y" in exI)
       
   281 apply algebra
       
   282 apply (rule_tac x="x" in exI)
       
   283 apply (rule_tac x="x + y" in exI)
       
   284 apply algebra
       
   285 done
       
   286 
       
   287 lemma bezout_add: "\<exists>(d::nat) x y. d dvd a \<and> d dvd b \<and> (a * x = b * y + d \<or> b * x = a * y + d)"
       
   288 apply(induct a b rule: ind_euclid)
       
   289 apply blast
       
   290 apply clarify
       
   291 apply (rule_tac x="a" in exI, simp add: dvd_add)
       
   292 apply clarsimp
       
   293 apply (rule_tac x="d" in exI)
       
   294 apply (case_tac "a * x = b * y + d", simp_all add: dvd_add)
       
   295 apply (rule_tac x="x+y" in exI)
       
   296 apply (rule_tac x="y" in exI)
       
   297 apply algebra
       
   298 apply (rule_tac x="x" in exI)
       
   299 apply (rule_tac x="x+y" in exI)
       
   300 apply algebra
       
   301 done
       
   302 
       
   303 lemma bezout: "\<exists>(d::nat) x y. d dvd a \<and> d dvd b \<and> (a * x - b * y = d \<or> b * x - a * y = d)"
       
   304 using bezout_add[of a b]
       
   305 apply clarsimp
       
   306 apply (rule_tac x="d" in exI, simp)
       
   307 apply (rule_tac x="x" in exI)
       
   308 apply (rule_tac x="y" in exI)
       
   309 apply auto
       
   310 done
       
   311 
       
   312 
       
   313 text {* We can get a stronger version with a nonzeroness assumption. *}
       
   314 lemma divides_le: "m dvd n ==> m <= n \<or> n = (0::nat)" by (auto simp add: dvd_def)
       
   315 
       
   316 lemma bezout_add_strong: assumes nz: "a \<noteq> (0::nat)"
       
   317   shows "\<exists>d x y. d dvd a \<and> d dvd b \<and> a * x = b * y + d"
       
   318 proof-
       
   319   from nz have ap: "a > 0" by simp
       
   320  from bezout_add[of a b] 
       
   321  have "(\<exists>d x y. d dvd a \<and> d dvd b \<and> a * x = b * y + d) \<or> (\<exists>d x y. d dvd a \<and> d dvd b \<and> b * x = a * y + d)" by blast
       
   322  moreover
       
   323  {fix d x y assume H: "d dvd a" "d dvd b" "a * x = b * y + d"
       
   324    from H have ?thesis by blast }
       
   325  moreover
       
   326  {fix d x y assume H: "d dvd a" "d dvd b" "b * x = a * y + d"
       
   327    {assume b0: "b = 0" with H  have ?thesis by simp}
       
   328    moreover 
       
   329    {assume b: "b \<noteq> 0" hence bp: "b > 0" by simp
       
   330      from divides_le[OF H(2)] b have "d < b \<or> d = b" using le_less by blast
       
   331      moreover
       
   332      {assume db: "d=b"
       
   333        from prems have ?thesis apply simp
       
   334 	 apply (rule exI[where x = b], simp)
       
   335 	 apply (rule exI[where x = b])
       
   336 	by (rule exI[where x = "a - 1"], simp add: diff_mult_distrib2)}
       
   337     moreover
       
   338     {assume db: "d < b" 
       
   339 	{assume "x=0" hence ?thesis  using prems by simp }
       
   340 	moreover
       
   341 	{assume x0: "x \<noteq> 0" hence xp: "x > 0" by simp
       
   342 	  
       
   343 	  from db have "d \<le> b - 1" by simp
       
   344 	  hence "d*b \<le> b*(b - 1)" by simp
       
   345 	  with xp mult_mono[of "1" "x" "d*b" "b*(b - 1)"]
       
   346 	  have dble: "d*b \<le> x*b*(b - 1)" using bp by simp
       
   347 	  from H (3) have "a * ((b - 1) * y) + d * (b - 1 + 1) = d + x*b*(b - 1)" by algebra
       
   348 	  hence "a * ((b - 1) * y) = d + x*b*(b - 1) - d*b" using bp by simp
       
   349 	  hence "a * ((b - 1) * y) = d + (x*b*(b - 1) - d*b)" 
       
   350 	    by (simp only: diff_add_assoc[OF dble, of d, symmetric])
       
   351 	  hence "a * ((b - 1) * y) = b*(x*(b - 1) - d) + d"
       
   352 	    by (simp only: diff_mult_distrib2 add_commute mult_ac)
       
   353 	  hence ?thesis using H(1,2)
       
   354 	    apply -
       
   355 	    apply (rule exI[where x=d], simp)
       
   356 	    apply (rule exI[where x="(b - 1) * y"])
       
   357 	    by (rule exI[where x="x*(b - 1) - d"], simp)}
       
   358 	ultimately have ?thesis by blast}
       
   359     ultimately have ?thesis by blast}
       
   360   ultimately have ?thesis by blast}
       
   361  ultimately show ?thesis by blast
       
   362 qed
       
   363 
       
   364 
       
   365 lemma bezout_gcd: "\<exists>x y. a * x - b * y = gcd a b \<or> b * x - a * y = gcd a b"
       
   366 proof-
       
   367   let ?g = "gcd a b"
       
   368   from bezout[of a b] obtain d x y where d: "d dvd a" "d dvd b" "a * x - b * y = d \<or> b * x - a * y = d" by blast
       
   369   from d(1,2) have "d dvd ?g" by simp
       
   370   then obtain k where k: "?g = d*k" unfolding dvd_def by blast
       
   371   from d(3) have "(a * x - b * y)*k = d*k \<or> (b * x - a * y)*k = d*k" by blast 
       
   372   hence "a * x * k - b * y*k = d*k \<or> b * x * k - a * y*k = d*k" 
       
   373     by (algebra add: diff_mult_distrib)
       
   374   hence "a * (x * k) - b * (y*k) = ?g \<or> b * (x * k) - a * (y*k) = ?g" 
       
   375     by (simp add: k mult_assoc)
       
   376   thus ?thesis by blast
       
   377 qed
       
   378 
       
   379 lemma bezout_gcd_strong: assumes a: "a \<noteq> 0" 
       
   380   shows "\<exists>x y. a * x = b * y + gcd a b"
       
   381 proof-
       
   382   let ?g = "gcd a b"
       
   383   from bezout_add_strong[OF a, of b]
       
   384   obtain d x y where d: "d dvd a" "d dvd b" "a * x = b * y + d" by blast
       
   385   from d(1,2) have "d dvd ?g" by simp
       
   386   then obtain k where k: "?g = d*k" unfolding dvd_def by blast
       
   387   from d(3) have "a * x * k = (b * y + d) *k " by algebra
       
   388   hence "a * (x * k) = b * (y*k) + ?g" by (algebra add: k)
       
   389   thus ?thesis by blast
       
   390 qed
       
   391 
       
   392 lemma gcd_mult_distrib: "gcd(a * c) (b * c) = c * gcd a b"
       
   393 by(simp add: gcd_mult_distrib2 mult_commute)
       
   394 
       
   395 lemma gcd_bezout: "(\<exists>x y. a * x - b * y = d \<or> b * x - a * y = d) \<longleftrightarrow> gcd a b dvd d"
       
   396   (is "?lhs \<longleftrightarrow> ?rhs")
       
   397 proof-
       
   398   let ?g = "gcd a b"
       
   399   {assume H: ?rhs then obtain k where k: "d = ?g*k" unfolding dvd_def by blast
       
   400     from bezout_gcd[of a b] obtain x y where xy: "a * x - b * y = ?g \<or> b * x - a * y = ?g"
       
   401       by blast
       
   402     hence "(a * x - b * y)*k = ?g*k \<or> (b * x - a * y)*k = ?g*k" by auto
       
   403     hence "a * x*k - b * y*k = ?g*k \<or> b * x * k - a * y*k = ?g*k" 
       
   404       by (simp only: diff_mult_distrib)
       
   405     hence "a * (x*k) - b * (y*k) = d \<or> b * (x * k) - a * (y*k) = d"
       
   406       by (simp add: k[symmetric] mult_assoc)
       
   407     hence ?lhs by blast}
       
   408   moreover
       
   409   {fix x y assume H: "a * x - b * y = d \<or> b * x - a * y = d"
       
   410     have dv: "?g dvd a*x" "?g dvd b * y" "?g dvd b*x" "?g dvd a * y"
       
   411       using dvd_mult2[OF gcd_dvd1[of a b]] dvd_mult2[OF gcd_dvd2[of a b]] by simp_all
       
   412     from dvd_diff_nat[OF dv(1,2)] dvd_diff_nat[OF dv(3,4)] H
       
   413     have ?rhs by auto}
       
   414   ultimately show ?thesis by blast
       
   415 qed
       
   416 
       
   417 lemma gcd_bezout_sum: assumes H:"a * x + b * y = d" shows "gcd a b dvd d"
       
   418 proof-
       
   419   let ?g = "gcd a b"
       
   420     have dv: "?g dvd a*x" "?g dvd b * y" 
       
   421       using dvd_mult2[OF gcd_dvd1[of a b]] dvd_mult2[OF gcd_dvd2[of a b]] by simp_all
       
   422     from dvd_add[OF dv] H
       
   423     show ?thesis by auto
       
   424 qed
       
   425 
       
   426 lemma gcd_mult': "gcd b (a * b) = b"
       
   427 by (simp add: gcd_mult mult_commute[of a b]) 
       
   428 
       
   429 lemma gcd_add: "gcd(a + b) b = gcd a b" 
       
   430   "gcd(b + a) b = gcd a b" "gcd a (a + b) = gcd a b" "gcd a (b + a) = gcd a b"
       
   431 apply (simp_all add: gcd_add1)
       
   432 by (simp add: gcd_commute gcd_add1)
       
   433 
       
   434 lemma gcd_sub: "b <= a ==> gcd(a - b) b = gcd a b" "a <= b ==> gcd a (b - a) = gcd a b"
       
   435 proof-
       
   436   {fix a b assume H: "b \<le> (a::nat)"
       
   437     hence th: "a - b + b = a" by arith
       
   438     from gcd_add(1)[of "a - b" b] th  have "gcd(a - b) b = gcd a b" by simp}
       
   439   note th = this
       
   440 {
       
   441   assume ab: "b \<le> a"
       
   442   from th[OF ab] show "gcd (a - b)  b = gcd a b" by blast
       
   443 next
       
   444   assume ab: "a \<le> b"
       
   445   from th[OF ab] show "gcd a (b - a) = gcd a b" 
       
   446     by (simp add: gcd_commute)}
       
   447 qed
       
   448 
       
   449 
       
   450 subsection {* LCM defined by GCD *}
       
   451 
       
   452 
       
   453 definition
       
   454   lcm :: "nat \<Rightarrow> nat \<Rightarrow> nat"
       
   455 where
       
   456   lcm_def: "lcm m n = m * n div gcd m n"
       
   457 
       
   458 lemma prod_gcd_lcm:
       
   459   "m * n = gcd m n * lcm m n"
       
   460   unfolding lcm_def by (simp add: dvd_mult_div_cancel [OF gcd_dvd_prod])
       
   461 
       
   462 lemma lcm_0 [simp]: "lcm m 0 = 0"
       
   463   unfolding lcm_def by simp
       
   464 
       
   465 lemma lcm_1 [simp]: "lcm m 1 = m"
       
   466   unfolding lcm_def by simp
       
   467 
       
   468 lemma lcm_0_left [simp]: "lcm 0 n = 0"
       
   469   unfolding lcm_def by simp
       
   470 
       
   471 lemma lcm_1_left [simp]: "lcm 1 m = m"
       
   472   unfolding lcm_def by simp
       
   473 
       
   474 lemma dvd_pos:
       
   475   fixes n m :: nat
       
   476   assumes "n > 0" and "m dvd n"
       
   477   shows "m > 0"
       
   478 using assms by (cases m) auto
       
   479 
       
   480 lemma lcm_least:
       
   481   assumes "m dvd k" and "n dvd k"
       
   482   shows "lcm m n dvd k"
       
   483 proof (cases k)
       
   484   case 0 then show ?thesis by auto
       
   485 next
       
   486   case (Suc _) then have pos_k: "k > 0" by auto
       
   487   from assms dvd_pos [OF this] have pos_mn: "m > 0" "n > 0" by auto
       
   488   with gcd_zero [of m n] have pos_gcd: "gcd m n > 0" by simp
       
   489   from assms obtain p where k_m: "k = m * p" using dvd_def by blast
       
   490   from assms obtain q where k_n: "k = n * q" using dvd_def by blast
       
   491   from pos_k k_m have pos_p: "p > 0" by auto
       
   492   from pos_k k_n have pos_q: "q > 0" by auto
       
   493   have "k * k * gcd q p = k * gcd (k * q) (k * p)"
       
   494     by (simp add: mult_ac gcd_mult_distrib2)
       
   495   also have "\<dots> = k * gcd (m * p * q) (n * q * p)"
       
   496     by (simp add: k_m [symmetric] k_n [symmetric])
       
   497   also have "\<dots> = k * p * q * gcd m n"
       
   498     by (simp add: mult_ac gcd_mult_distrib2)
       
   499   finally have "(m * p) * (n * q) * gcd q p = k * p * q * gcd m n"
       
   500     by (simp only: k_m [symmetric] k_n [symmetric])
       
   501   then have "p * q * m * n * gcd q p = p * q * k * gcd m n"
       
   502     by (simp add: mult_ac)
       
   503   with pos_p pos_q have "m * n * gcd q p = k * gcd m n"
       
   504     by simp
       
   505   with prod_gcd_lcm [of m n]
       
   506   have "lcm m n * gcd q p * gcd m n = k * gcd m n"
       
   507     by (simp add: mult_ac)
       
   508   with pos_gcd have "lcm m n * gcd q p = k" by simp
       
   509   then show ?thesis using dvd_def by auto
       
   510 qed
       
   511 
       
   512 lemma lcm_dvd1 [iff]:
       
   513   "m dvd lcm m n"
       
   514 proof (cases m)
       
   515   case 0 then show ?thesis by simp
       
   516 next
       
   517   case (Suc _)
       
   518   then have mpos: "m > 0" by simp
       
   519   show ?thesis
       
   520   proof (cases n)
       
   521     case 0 then show ?thesis by simp
       
   522   next
       
   523     case (Suc _)
       
   524     then have npos: "n > 0" by simp
       
   525     have "gcd m n dvd n" by simp
       
   526     then obtain k where "n = gcd m n * k" using dvd_def by auto
       
   527     then have "m * n div gcd m n = m * (gcd m n * k) div gcd m n" by (simp add: mult_ac)
       
   528     also have "\<dots> = m * k" using mpos npos gcd_zero by simp
       
   529     finally show ?thesis by (simp add: lcm_def)
       
   530   qed
       
   531 qed
       
   532 
       
   533 lemma lcm_dvd2 [iff]: 
       
   534   "n dvd lcm m n"
       
   535 proof (cases n)
       
   536   case 0 then show ?thesis by simp
       
   537 next
       
   538   case (Suc _)
       
   539   then have npos: "n > 0" by simp
       
   540   show ?thesis
       
   541   proof (cases m)
       
   542     case 0 then show ?thesis by simp
       
   543   next
       
   544     case (Suc _)
       
   545     then have mpos: "m > 0" by simp
       
   546     have "gcd m n dvd m" by simp
       
   547     then obtain k where "m = gcd m n * k" using dvd_def by auto
       
   548     then have "m * n div gcd m n = (gcd m n * k) * n div gcd m n" by (simp add: mult_ac)
       
   549     also have "\<dots> = n * k" using mpos npos gcd_zero by simp
       
   550     finally show ?thesis by (simp add: lcm_def)
       
   551   qed
       
   552 qed
       
   553 
       
   554 lemma gcd_add1_eq: "gcd (m + k) k = gcd (m + k) m"
       
   555   by (simp add: gcd_commute)
       
   556 
       
   557 lemma gcd_diff2: "m \<le> n ==> gcd n (n - m) = gcd n m"
       
   558   apply (subgoal_tac "n = m + (n - m)")
       
   559   apply (erule ssubst, rule gcd_add1_eq, simp)  
       
   560   done
       
   561 
       
   562 
       
   563 subsection {* GCD and LCM on integers *}
       
   564 
       
   565 definition
       
   566   zgcd :: "int \<Rightarrow> int \<Rightarrow> int" where
       
   567   "zgcd i j = int (gcd (nat (abs i)) (nat (abs j)))"
       
   568 
       
   569 lemma zgcd_zdvd1 [iff,simp, algebra]: "zgcd i j dvd i"
       
   570 by (simp add: zgcd_def int_dvd_iff)
       
   571 
       
   572 lemma zgcd_zdvd2 [iff,simp, algebra]: "zgcd i j dvd j"
       
   573 by (simp add: zgcd_def int_dvd_iff)
       
   574 
       
   575 lemma zgcd_pos: "zgcd i j \<ge> 0"
       
   576 by (simp add: zgcd_def)
       
   577 
       
   578 lemma zgcd0 [simp,algebra]: "(zgcd i j = 0) = (i = 0 \<and> j = 0)"
       
   579 by (simp add: zgcd_def gcd_zero)
       
   580 
       
   581 lemma zgcd_commute: "zgcd i j = zgcd j i"
       
   582 unfolding zgcd_def by (simp add: gcd_commute)
       
   583 
       
   584 lemma zgcd_zminus [simp, algebra]: "zgcd (- i) j = zgcd i j"
       
   585 unfolding zgcd_def by simp
       
   586 
       
   587 lemma zgcd_zminus2 [simp, algebra]: "zgcd i (- j) = zgcd i j"
       
   588 unfolding zgcd_def by simp
       
   589 
       
   590   (* should be solved by algebra*)
       
   591 lemma zrelprime_dvd_mult: "zgcd i j = 1 \<Longrightarrow> i dvd k * j \<Longrightarrow> i dvd k"
       
   592   unfolding zgcd_def
       
   593 proof -
       
   594   assume "int (gcd (nat \<bar>i\<bar>) (nat \<bar>j\<bar>)) = 1" "i dvd k * j"
       
   595   then have g: "gcd (nat \<bar>i\<bar>) (nat \<bar>j\<bar>) = 1" by simp
       
   596   from `i dvd k * j` obtain h where h: "k*j = i*h" unfolding dvd_def by blast
       
   597   have th: "nat \<bar>i\<bar> dvd nat \<bar>k\<bar> * nat \<bar>j\<bar>"
       
   598     unfolding dvd_def
       
   599     by (rule_tac x= "nat \<bar>h\<bar>" in exI, simp add: h nat_abs_mult_distrib [symmetric])
       
   600   from relprime_dvd_mult [OF g th] obtain h' where h': "nat \<bar>k\<bar> = nat \<bar>i\<bar> * h'"
       
   601     unfolding dvd_def by blast
       
   602   from h' have "int (nat \<bar>k\<bar>) = int (nat \<bar>i\<bar> * h')" by simp
       
   603   then have "\<bar>k\<bar> = \<bar>i\<bar> * int h'" by (simp add: int_mult)
       
   604   then show ?thesis
       
   605     apply (subst abs_dvd_iff [symmetric])
       
   606     apply (subst dvd_abs_iff [symmetric])
       
   607     apply (unfold dvd_def)
       
   608     apply (rule_tac x = "int h'" in exI, simp)
       
   609     done
       
   610 qed
       
   611 
       
   612 lemma int_nat_abs: "int (nat (abs x)) = abs x" by arith
       
   613 
       
   614 lemma zgcd_greatest:
       
   615   assumes "k dvd m" and "k dvd n"
       
   616   shows "k dvd zgcd m n"
       
   617 proof -
       
   618   let ?k' = "nat \<bar>k\<bar>"
       
   619   let ?m' = "nat \<bar>m\<bar>"
       
   620   let ?n' = "nat \<bar>n\<bar>"
       
   621   from `k dvd m` and `k dvd n` have dvd': "?k' dvd ?m'" "?k' dvd ?n'"
       
   622     unfolding zdvd_int by (simp_all only: int_nat_abs abs_dvd_iff dvd_abs_iff)
       
   623   from gcd_greatest [OF dvd'] have "int (nat \<bar>k\<bar>) dvd zgcd m n"
       
   624     unfolding zgcd_def by (simp only: zdvd_int)
       
   625   then have "\<bar>k\<bar> dvd zgcd m n" by (simp only: int_nat_abs)
       
   626   then show "k dvd zgcd m n" by simp
       
   627 qed
       
   628 
       
   629 lemma div_zgcd_relprime:
       
   630   assumes nz: "a \<noteq> 0 \<or> b \<noteq> 0"
       
   631   shows "zgcd (a div (zgcd a b)) (b div (zgcd a b)) = 1"
       
   632 proof -
       
   633   from nz have nz': "nat \<bar>a\<bar> \<noteq> 0 \<or> nat \<bar>b\<bar> \<noteq> 0" by arith 
       
   634   let ?g = "zgcd a b"
       
   635   let ?a' = "a div ?g"
       
   636   let ?b' = "b div ?g"
       
   637   let ?g' = "zgcd ?a' ?b'"
       
   638   have dvdg: "?g dvd a" "?g dvd b" by (simp_all add: zgcd_zdvd1 zgcd_zdvd2)
       
   639   have dvdg': "?g' dvd ?a'" "?g' dvd ?b'" by (simp_all add: zgcd_zdvd1 zgcd_zdvd2)
       
   640   from dvdg dvdg' obtain ka kb ka' kb' where
       
   641    kab: "a = ?g*ka" "b = ?g*kb" "?a' = ?g'*ka'" "?b' = ?g' * kb'"
       
   642     unfolding dvd_def by blast
       
   643   then have "?g* ?a' = (?g * ?g') * ka'" "?g* ?b' = (?g * ?g') * kb'" by simp_all
       
   644   then have dvdgg':"?g * ?g' dvd a" "?g* ?g' dvd b"
       
   645     by (auto simp add: zdvd_mult_div_cancel [OF dvdg(1)]
       
   646       zdvd_mult_div_cancel [OF dvdg(2)] dvd_def)
       
   647   have "?g \<noteq> 0" using nz by simp
       
   648   then have gp: "?g \<noteq> 0" using zgcd_pos[where i="a" and j="b"] by arith
       
   649   from zgcd_greatest [OF dvdgg'] have "?g * ?g' dvd ?g" .
       
   650   with zdvd_mult_cancel1 [OF gp] have "\<bar>?g'\<bar> = 1" by simp
       
   651   with zgcd_pos show "?g' = 1" by simp
       
   652 qed
       
   653 
       
   654 lemma zgcd_0 [simp, algebra]: "zgcd m 0 = abs m"
       
   655   by (simp add: zgcd_def abs_if)
       
   656 
       
   657 lemma zgcd_0_left [simp, algebra]: "zgcd 0 m = abs m"
       
   658   by (simp add: zgcd_def abs_if)
       
   659 
       
   660 lemma zgcd_non_0: "0 < n ==> zgcd m n = zgcd n (m mod n)"
       
   661   apply (frule_tac b = n and a = m in pos_mod_sign)
       
   662   apply (simp del: pos_mod_sign add: zgcd_def abs_if nat_mod_distrib)
       
   663   apply (auto simp add: gcd_non_0 nat_mod_distrib [symmetric] zmod_zminus1_eq_if)
       
   664   apply (frule_tac a = m in pos_mod_bound)
       
   665   apply (simp del: pos_mod_bound add: nat_diff_distrib gcd_diff2 nat_le_eq_zle)
       
   666   done
       
   667 
       
   668 lemma zgcd_eq: "zgcd m n = zgcd n (m mod n)"
       
   669   apply (case_tac "n = 0", simp add: DIVISION_BY_ZERO)
       
   670   apply (auto simp add: linorder_neq_iff zgcd_non_0)
       
   671   apply (cut_tac m = "-m" and n = "-n" in zgcd_non_0, auto)
       
   672   done
       
   673 
       
   674 lemma zgcd_1 [simp, algebra]: "zgcd m 1 = 1"
       
   675   by (simp add: zgcd_def abs_if)
       
   676 
       
   677 lemma zgcd_0_1_iff [simp, algebra]: "zgcd 0 m = 1 \<longleftrightarrow> \<bar>m\<bar> = 1"
       
   678   by (simp add: zgcd_def abs_if)
       
   679 
       
   680 lemma zgcd_greatest_iff[algebra]: "k dvd zgcd m n = (k dvd m \<and> k dvd n)"
       
   681   by (simp add: zgcd_def abs_if int_dvd_iff dvd_int_iff nat_dvd_iff)
       
   682 
       
   683 lemma zgcd_1_left [simp, algebra]: "zgcd 1 m = 1"
       
   684   by (simp add: zgcd_def gcd_1_left)
       
   685 
       
   686 lemma zgcd_assoc: "zgcd (zgcd k m) n = zgcd k (zgcd m n)"
       
   687   by (simp add: zgcd_def gcd_assoc)
       
   688 
       
   689 lemma zgcd_left_commute: "zgcd k (zgcd m n) = zgcd m (zgcd k n)"
       
   690   apply (rule zgcd_commute [THEN trans])
       
   691   apply (rule zgcd_assoc [THEN trans])
       
   692   apply (rule zgcd_commute [THEN arg_cong])
       
   693   done
       
   694 
       
   695 lemmas zgcd_ac = zgcd_assoc zgcd_commute zgcd_left_commute
       
   696   -- {* addition is an AC-operator *}
       
   697 
       
   698 lemma zgcd_zmult_distrib2: "0 \<le> k ==> k * zgcd m n = zgcd (k * m) (k * n)"
       
   699   by (simp del: minus_mult_right [symmetric]
       
   700       add: minus_mult_right nat_mult_distrib zgcd_def abs_if
       
   701           mult_less_0_iff gcd_mult_distrib2 [symmetric] zmult_int [symmetric])
       
   702 
       
   703 lemma zgcd_zmult_distrib2_abs: "zgcd (k * m) (k * n) = abs k * zgcd m n"
       
   704   by (simp add: abs_if zgcd_zmult_distrib2)
       
   705 
       
   706 lemma zgcd_self [simp]: "0 \<le> m ==> zgcd m m = m"
       
   707   by (cut_tac k = m and m = 1 and n = 1 in zgcd_zmult_distrib2, simp_all)
       
   708 
       
   709 lemma zgcd_zmult_eq_self [simp]: "0 \<le> k ==> zgcd k (k * n) = k"
       
   710   by (cut_tac k = k and m = 1 and n = n in zgcd_zmult_distrib2, simp_all)
       
   711 
       
   712 lemma zgcd_zmult_eq_self2 [simp]: "0 \<le> k ==> zgcd (k * n) k = k"
       
   713   by (cut_tac k = k and m = n and n = 1 in zgcd_zmult_distrib2, simp_all)
       
   714 
       
   715 
       
   716 definition "zlcm i j = int (lcm(nat(abs i)) (nat(abs j)))"
       
   717 
       
   718 lemma dvd_zlcm_self1[simp, algebra]: "i dvd zlcm i j"
       
   719 by(simp add:zlcm_def dvd_int_iff)
       
   720 
       
   721 lemma dvd_zlcm_self2[simp, algebra]: "j dvd zlcm i j"
       
   722 by(simp add:zlcm_def dvd_int_iff)
       
   723 
       
   724 
       
   725 lemma dvd_imp_dvd_zlcm1:
       
   726   assumes "k dvd i" shows "k dvd (zlcm i j)"
       
   727 proof -
       
   728   have "nat(abs k) dvd nat(abs i)" using `k dvd i`
       
   729     by(simp add:int_dvd_iff[symmetric] dvd_int_iff[symmetric])
       
   730   thus ?thesis by(simp add:zlcm_def dvd_int_iff)(blast intro: dvd_trans)
       
   731 qed
       
   732 
       
   733 lemma dvd_imp_dvd_zlcm2:
       
   734   assumes "k dvd j" shows "k dvd (zlcm i j)"
       
   735 proof -
       
   736   have "nat(abs k) dvd nat(abs j)" using `k dvd j`
       
   737     by(simp add:int_dvd_iff[symmetric] dvd_int_iff[symmetric])
       
   738   thus ?thesis by(simp add:zlcm_def dvd_int_iff)(blast intro: dvd_trans)
       
   739 qed
       
   740 
       
   741 
       
   742 lemma zdvd_self_abs1: "(d::int) dvd (abs d)"
       
   743 by (case_tac "d <0", simp_all)
       
   744 
       
   745 lemma zdvd_self_abs2: "(abs (d::int)) dvd d"
       
   746 by (case_tac "d<0", simp_all)
       
   747 
       
   748 (* lcm a b is positive for positive a and b *)
       
   749 
       
   750 lemma lcm_pos: 
       
   751   assumes mpos: "m > 0"
       
   752   and npos: "n>0"
       
   753   shows "lcm m n > 0"
       
   754 proof(rule ccontr, simp add: lcm_def gcd_zero)
       
   755 assume h:"m*n div gcd m n = 0"
       
   756 from mpos npos have "gcd m n \<noteq> 0" using gcd_zero by simp
       
   757 hence gcdp: "gcd m n > 0" by simp
       
   758 with h
       
   759 have "m*n < gcd m n"
       
   760   by (cases "m * n < gcd m n") (auto simp add: div_if[OF gcdp, where m="m*n"])
       
   761 moreover 
       
   762 have "gcd m n dvd m" by simp
       
   763  with mpos dvd_imp_le have t1:"gcd m n \<le> m" by simp
       
   764  with npos have t1:"gcd m n *n \<le> m*n" by simp
       
   765  have "gcd m n \<le> gcd m n*n" using npos by simp
       
   766  with t1 have "gcd m n \<le> m*n" by arith
       
   767 ultimately show "False" by simp
       
   768 qed
       
   769 
       
   770 lemma zlcm_pos: 
       
   771   assumes anz: "a \<noteq> 0"
       
   772   and bnz: "b \<noteq> 0" 
       
   773   shows "0 < zlcm a b"
       
   774 proof-
       
   775   let ?na = "nat (abs a)"
       
   776   let ?nb = "nat (abs b)"
       
   777   have nap: "?na >0" using anz by simp
       
   778   have nbp: "?nb >0" using bnz by simp
       
   779   have "0 < lcm ?na ?nb" by (rule lcm_pos[OF nap nbp])
       
   780   thus ?thesis by (simp add: zlcm_def)
       
   781 qed
       
   782 
       
   783 lemma zgcd_code [code]:
       
   784   "zgcd k l = \<bar>if l = 0 then k else zgcd l (\<bar>k\<bar> mod \<bar>l\<bar>)\<bar>"
       
   785   by (simp add: zgcd_def gcd.simps [of "nat \<bar>k\<bar>"] nat_mod_distrib)
       
   786 
       
   787 end