src/HOL/GCD.thy
 author blanchet Wed Mar 04 11:05:29 2009 +0100 (2009-03-04) changeset 30242 aea5d7fa7ef5 parent 30240 5b25fee0362c parent 30082 43c5b7bfc791 child 30738 0842e906300c permissions -rw-r--r--
Merge.
1 (*  Title:      HOL/GCD.thy
2     Author:     Christophe Tabacznyj and Lawrence C Paulson
3     Copyright   1996  University of Cambridge
4 *)
6 header {* The Greatest Common Divisor *}
8 theory GCD
9 imports Plain Presburger Main
10 begin
12 text {*
13   See \cite{davenport92}. \bigskip
14 *}
16 subsection {* Specification of GCD on nats *}
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)"
23 text {* Uniqueness *}
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)
28 text {* Connection to divides relation *}
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)
33 text {* Commutativity *}
35 lemma is_gcd_commute: "is_gcd m n k = is_gcd n m k"
36   by (auto simp add: is_gcd_def)
39 subsection {* GCD on nat by Euclid's algorithm *}
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
54 lemma gcd_0 [simp, algebra]: "gcd m 0 = m"
55   by simp
57 lemma gcd_0_left [simp,algebra]: "gcd 0 m = m"
58   by simp
60 lemma gcd_non_0: "n > 0 \<Longrightarrow> gcd m n = gcd n (m mod n)"
61   by simp
63 lemma gcd_1 [simp, algebra]: "gcd m (Suc 0) = Suc 0"
64   by simp
66 lemma nat_gcd_1_right [simp, algebra]: "gcd m 1 = 1"
67   unfolding One_nat_def by (rule gcd_1)
69 declare gcd.simps [simp del]
71 text {*
72   \medskip @{term "gcd m n"} divides @{text m} and @{text n}.  The
73   conjunctions don't seem provable separately.
74 *}
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)
80   apply (blast dest: dvd_mod_imp_dvd)
81   done
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 *}
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)
92 text {*
93   \medskip Function gcd yields the Greatest Common Divisor.
94 *}
96 lemma is_gcd: "is_gcd m n (gcd m n) "
97   by (simp add: is_gcd_def gcd_greatest)
100 subsection {* Derived laws for GCD *}
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)
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)
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)
113   done
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)
119   apply (blast intro: dvd_trans)
120   done
122 lemma gcd_1_left [simp, algebra]: "gcd (Suc 0) m = Suc 0"
125 lemma nat_gcd_1_left [simp, algebra]: "gcd 1 m = 1"
126   unfolding One_nat_def by (rule gcd_1_left)
128 text {*
129   \medskip Multiplication laws
130 *}
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
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
144 lemma gcd_self [simp, algebra]: "gcd k k = k"
145   apply (rule gcd_mult [of k 1, simplified])
146   done
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
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)
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)
165   done
168 text {* \medskip Addition laws *}
170 lemma gcd_add1 [simp, algebra]: "gcd (m + n) n = gcd m n"
171   by (cases "n = 0") (auto simp add: gcd_non_0)
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
182 lemma gcd_add2' [simp, algebra]: "gcd m (n + m) = gcd m n"
185   done
187 lemma gcd_add_mult[algebra]: "gcd m (k * m + n) = gcd m n"
190 lemma gcd_dvd_prod: "gcd m n dvd m * n"
191   using mult_dvd_mono [of 1] by auto
193 text {*
194   \medskip Division by gcd yields rrelatively primes.
195 *}
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
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
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
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
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
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
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
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
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)"
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
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)
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
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
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
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"
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
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
392 lemma gcd_mult_distrib: "gcd(a * c) (b * c) = c * gcd a b"
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 nat_dvd_diff[OF dv(1,2)] nat_dvd_diff[OF dv(3,4)] H
413     have ?rhs by auto}
414   ultimately show ?thesis by blast
415 qed
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
423     show ?thesis by auto
424 qed
426 lemma gcd_mult': "gcd b (a * b) = b"
427 by (simp add: gcd_mult mult_commute[of a b])
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"
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"
447 qed
450 subsection {* LCM defined by GCD *}
453 definition
454   lcm :: "nat \<Rightarrow> nat \<Rightarrow> nat"
455 where
456   lcm_def: "lcm m n = m * n div gcd m n"
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])
462 lemma lcm_0 [simp]: "lcm m 0 = 0"
463   unfolding lcm_def by simp
465 lemma lcm_1 [simp]: "lcm m 1 = m"
466   unfolding lcm_def by simp
468 lemma lcm_0_left [simp]: "lcm 0 n = 0"
469   unfolding lcm_def by simp
471 lemma lcm_1_left [simp]: "lcm 1 m = m"
472   unfolding lcm_def by simp
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
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"
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"
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
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
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
554 lemma gcd_add1_eq: "gcd (m + k) k = gcd (m + k) m"
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
563 subsection {* GCD and LCM on integers *}
565 definition
566   zgcd :: "int \<Rightarrow> int \<Rightarrow> int" where
567   "zgcd i j = int (gcd (nat (abs i)) (nat (abs j)))"
569 lemma zgcd_zdvd1 [iff,simp, algebra]: "zgcd i j dvd i"
570 by (simp add: zgcd_def int_dvd_iff)
572 lemma zgcd_zdvd2 [iff,simp, algebra]: "zgcd i j dvd j"
573 by (simp add: zgcd_def int_dvd_iff)
575 lemma zgcd_pos: "zgcd i j \<ge> 0"
578 lemma zgcd0 [simp,algebra]: "(zgcd i j = 0) = (i = 0 \<and> j = 0)"
579 by (simp add: zgcd_def gcd_zero)
581 lemma zgcd_commute: "zgcd i j = zgcd j i"
582 unfolding zgcd_def by (simp add: gcd_commute)
584 lemma zgcd_zminus [simp, algebra]: "zgcd (- i) j = zgcd i j"
585 unfolding zgcd_def by simp
587 lemma zgcd_zminus2 [simp, algebra]: "zgcd i (- j) = zgcd i j"
588 unfolding zgcd_def by simp
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
612 lemma int_nat_abs: "int (nat (abs x)) = abs x" by arith
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
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
654 lemma zgcd_0 [simp, algebra]: "zgcd m 0 = abs m"
655   by (simp add: zgcd_def abs_if)
657 lemma zgcd_0_left [simp, algebra]: "zgcd 0 m = abs m"
658   by (simp add: zgcd_def abs_if)
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
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
674 lemma zgcd_1 [simp, algebra]: "zgcd m 1 = 1"
675   by (simp add: zgcd_def abs_if)
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)
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)
683 lemma zgcd_1_left [simp, algebra]: "zgcd 1 m = 1"
684   by (simp add: zgcd_def gcd_1_left)
686 lemma zgcd_assoc: "zgcd (zgcd k m) n = zgcd k (zgcd m n)"
687   by (simp add: zgcd_def gcd_assoc)
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
695 lemmas zgcd_ac = zgcd_assoc zgcd_commute zgcd_left_commute
696   -- {* addition is an AC-operator *}
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])
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)
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)
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)
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)
716 definition "zlcm i j = int (lcm(nat(abs i)) (nat(abs j)))"
718 lemma dvd_zlcm_self1[simp, algebra]: "i dvd zlcm i j"
721 lemma dvd_zlcm_self2[simp, algebra]: "j dvd zlcm i j"
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`
730   thus ?thesis by(simp add:zlcm_def dvd_int_iff)(blast intro: dvd_trans)
731 qed
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`
738   thus ?thesis by(simp add:zlcm_def dvd_int_iff)(blast intro: dvd_trans)
739 qed
742 lemma zdvd_self_abs1: "(d::int) dvd (abs d)"
743 by (case_tac "d <0", simp_all)
745 lemma zdvd_self_abs2: "(abs (d::int)) dvd d"
746 by (case_tac "d<0", simp_all)
748 (* lcm a b is positive for positive a and b *)
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
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
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)
787 end