src/HOL/Library/Pocklington.thy
 author nipkow Tue Mar 03 17:05:18 2009 +0100 (2009-03-03) changeset 30224 79136ce06bdb parent 30042 31039ee583fa child 30242 aea5d7fa7ef5 permissions -rw-r--r--
removed and renamed redundant lemmas
1 (*  Title:      HOL/Library/Pocklington.thy
2     ID:         \$Id\$
3     Author:     Amine Chaieb
4 *)
6 header {* Pocklington's Theorem for Primes *}
9 theory Pocklington
10 imports Plain "~~/src/HOL/List" "~~/src/HOL/Primes"
11 begin
13 definition modeq:: "nat => nat => nat => bool"    ("(1[_ = _] '(mod _'))")
14   where "[a = b] (mod p) == ((a mod p) = (b mod p))"
16 definition modneq:: "nat => nat => nat => bool"    ("(1[_ \<noteq> _] '(mod _'))")
17   where "[a \<noteq> b] (mod p) == ((a mod p) \<noteq> (b mod p))"
19 lemma modeq_trans:
20   "\<lbrakk> [a = b] (mod p); [b = c] (mod p) \<rbrakk> \<Longrightarrow> [a = c] (mod p)"
21   by (simp add:modeq_def)
24 lemma nat_mod_lemma: assumes xyn: "[x = y] (mod n)" and xy:"y \<le> x"
25   shows "\<exists>q. x = y + n * q"
26 using xyn xy unfolding modeq_def using nat_mod_eq_lemma by blast
28 lemma nat_mod[algebra]: "[x = y] (mod n) \<longleftrightarrow> (\<exists>q1 q2. x + n * q1 = y + n * q2)"
29 unfolding modeq_def nat_mod_eq_iff ..
31 (* Lemmas about previously defined terms.                                    *)
33 lemma prime: "prime p \<longleftrightarrow> p \<noteq> 0 \<and> p\<noteq>1 \<and> (\<forall>m. 0 < m \<and> m < p \<longrightarrow> coprime p m)"
34   (is "?lhs \<longleftrightarrow> ?rhs")
35 proof-
36   {assume "p=0 \<or> p=1" hence ?thesis using prime_0 prime_1 by (cases "p=0", simp_all)}
37   moreover
38   {assume p0: "p\<noteq>0" "p\<noteq>1"
39     {assume H: "?lhs"
40       {fix m assume m: "m > 0" "m < p"
41 	{assume "m=1" hence "coprime p m" by simp}
42 	moreover
43 	{assume "p dvd m" hence "p \<le> m" using dvd_imp_le m by blast with m(2)
44 	  have "coprime p m" by simp}
45 	ultimately have "coprime p m" using prime_coprime[OF H, of m] by blast}
46       hence ?rhs using p0 by auto}
47     moreover
48     { assume H: "\<forall>m. 0 < m \<and> m < p \<longrightarrow> coprime p m"
49       from prime_factor[OF p0(2)] obtain q where q: "prime q" "q dvd p" by blast
50       from prime_ge_2[OF q(1)] have q0: "q > 0" by arith
51       from dvd_imp_le[OF q(2)] p0 have qp: "q \<le> p" by arith
52       {assume "q = p" hence ?lhs using q(1) by blast}
53       moreover
54       {assume "q\<noteq>p" with qp have qplt: "q < p" by arith
55 	from H[rule_format, of q] qplt q0 have "coprime p q" by arith
56 	with coprime_prime[of p q q] q have False by simp hence ?lhs by blast}
57       ultimately have ?lhs by blast}
58     ultimately have ?thesis by blast}
59   ultimately show ?thesis  by (cases"p=0 \<or> p=1", auto)
60 qed
62 lemma finite_number_segment: "card { m. 0 < m \<and> m < n } = n - 1"
63 proof-
64   have "{ m. 0 < m \<and> m < n } = {1..<n}" by auto
65   thus ?thesis by simp
66 qed
68 lemma coprime_mod: assumes n: "n \<noteq> 0" shows "coprime (a mod n) n \<longleftrightarrow> coprime a n"
69   using n dvd_mod_iff[of _ n a] by (auto simp add: coprime)
71 (* Congruences.                                                              *)
73 lemma cong_mod_01[simp,presburger]:
74   "[x = y] (mod 0) \<longleftrightarrow> x = y" "[x = y] (mod 1)" "[x = 0] (mod n) \<longleftrightarrow> n dvd x"
75   by (simp_all add: modeq_def, presburger)
77 lemma cong_sub_cases:
78   "[x = y] (mod n) \<longleftrightarrow> (if x <= y then [y - x = 0] (mod n) else [x - y = 0] (mod n))"
79 apply (auto simp add: nat_mod)
80 apply (rule_tac x="q2" in exI)
81 apply (rule_tac x="q1" in exI, simp)
82 apply (rule_tac x="q2" in exI)
83 apply (rule_tac x="q1" in exI, simp)
84 apply (rule_tac x="q1" in exI)
85 apply (rule_tac x="q2" in exI, simp)
86 apply (rule_tac x="q1" in exI)
87 apply (rule_tac x="q2" in exI, simp)
88 done
90 lemma cong_mult_lcancel: assumes an: "coprime a n" and axy:"[a * x = a * y] (mod n)"
91   shows "[x = y] (mod n)"
92 proof-
93   {assume "a = 0" with an axy coprime_0'[of n] have ?thesis by (simp add: modeq_def) }
94   moreover
95   {assume az: "a\<noteq>0"
96     {assume xy: "x \<le> y" hence axy': "a*x \<le> a*y" by simp
97       with axy cong_sub_cases[of "a*x" "a*y" n]  have "[a*(y - x) = 0] (mod n)"
98 	by (simp only: if_True diff_mult_distrib2)
99       hence th: "n dvd a*(y -x)" by simp
100       from coprime_divprod[OF th] an have "n dvd y - x"
101 	by (simp add: coprime_commute)
102       hence ?thesis using xy cong_sub_cases[of x y n] by simp}
103     moreover
104     {assume H: "\<not>x \<le> y" hence xy: "y \<le> x"  by arith
105       from H az have axy': "\<not> a*x \<le> a*y" by auto
106       with axy H cong_sub_cases[of "a*x" "a*y" n]  have "[a*(x - y) = 0] (mod n)"
107 	by (simp only: if_False diff_mult_distrib2)
108       hence th: "n dvd a*(x - y)" by simp
109       from coprime_divprod[OF th] an have "n dvd x - y"
110 	by (simp add: coprime_commute)
111       hence ?thesis using xy cong_sub_cases[of x y n] by simp}
112     ultimately have ?thesis by blast}
113   ultimately show ?thesis by blast
114 qed
116 lemma cong_mult_rcancel: assumes an: "coprime a n" and axy:"[x*a = y*a] (mod n)"
117   shows "[x = y] (mod n)"
118   using cong_mult_lcancel[OF an axy[unfolded mult_commute[of _a]]] .
120 lemma cong_refl: "[x = x] (mod n)" by (simp add: modeq_def)
122 lemma eq_imp_cong: "a = b \<Longrightarrow> [a = b] (mod n)" by (simp add: cong_refl)
124 lemma cong_commute: "[x = y] (mod n) \<longleftrightarrow> [y = x] (mod n)"
125   by (auto simp add: modeq_def)
127 lemma cong_trans[trans]: "[x = y] (mod n) \<Longrightarrow> [y = z] (mod n) \<Longrightarrow> [x = z] (mod n)"
128   by (simp add: modeq_def)
130 lemma cong_add: assumes xx': "[x = x'] (mod n)" and yy':"[y = y'] (mod n)"
131   shows "[x + y = x' + y'] (mod n)"
132 proof-
133   have "(x + y) mod n = (x mod n + y mod n) mod n"
134     by (simp add: mod_add_left_eq[of x y n] mod_add_right_eq[of "x mod n" y n])
135   also have "\<dots> = (x' mod n + y' mod n) mod n" using xx' yy' modeq_def by simp
136   also have "\<dots> = (x' + y') mod n"
137     by (simp add: mod_add_left_eq[of x' y' n] mod_add_right_eq[of "x' mod n" y' n])
138   finally show ?thesis unfolding modeq_def .
139 qed
141 lemma cong_mult: assumes xx': "[x = x'] (mod n)" and yy':"[y = y'] (mod n)"
142   shows "[x * y = x' * y'] (mod n)"
143 proof-
144   have "(x * y) mod n = (x mod n) * (y mod n) mod n"
145     by (simp add: mod_mult_left_eq[of x y n] mod_mult_right_eq[of "x mod n" y n])
146   also have "\<dots> = (x' mod n) * (y' mod n) mod n" using xx'[unfolded modeq_def] yy'[unfolded modeq_def] by simp
147   also have "\<dots> = (x' * y') mod n"
148     by (simp add: mod_mult_left_eq[of x' y' n] mod_mult_right_eq[of "x' mod n" y' n])
149   finally show ?thesis unfolding modeq_def .
150 qed
152 lemma cong_exp: "[x = y] (mod n) \<Longrightarrow> [x^k = y^k] (mod n)"
153   by (induct k, auto simp add: cong_refl cong_mult)
154 lemma cong_sub: assumes xx': "[x = x'] (mod n)" and yy': "[y = y'] (mod n)"
155   and yx: "y <= x" and yx': "y' <= x'"
156   shows "[x - y = x' - y'] (mod n)"
157 proof-
158   { fix x a x' a' y b y' b'
159     have "(x::nat) + a = x' + a' \<Longrightarrow> y + b = y' + b' \<Longrightarrow> y <= x \<Longrightarrow> y' <= x'
160       \<Longrightarrow> (x - y) + (a + b') = (x' - y') + (a' + b)" by arith}
161   note th = this
162   from xx' yy' obtain q1 q2 q1' q2' where q12: "x + n*q1 = x'+n*q2"
163     and q12': "y + n*q1' = y'+n*q2'" unfolding nat_mod by blast+
164   from th[OF q12 q12' yx yx']
165   have "(x - y) + n*(q1 + q2') = (x' - y') + n*(q2 + q1')"
166     by (simp add: right_distrib)
167   thus ?thesis unfolding nat_mod by blast
168 qed
170 lemma cong_mult_lcancel_eq: assumes an: "coprime a n"
171   shows "[a * x = a * y] (mod n) \<longleftrightarrow> [x = y] (mod n)" (is "?lhs \<longleftrightarrow> ?rhs")
172 proof
173   assume H: "?rhs" from cong_mult[OF cong_refl[of a n] H] show ?lhs .
174 next
175   assume H: "?lhs" hence H': "[x*a = y*a] (mod n)" by (simp add: mult_commute)
176   from cong_mult_rcancel[OF an H'] show ?rhs  .
177 qed
179 lemma cong_mult_rcancel_eq: assumes an: "coprime a n"
180   shows "[x * a = y * a] (mod n) \<longleftrightarrow> [x = y] (mod n)"
181 using cong_mult_lcancel_eq[OF an, of x y] by (simp add: mult_commute)
183 lemma cong_add_lcancel_eq: "[a + x = a + y] (mod n) \<longleftrightarrow> [x = y] (mod n)"
184   by (simp add: nat_mod)
186 lemma cong_add_rcancel_eq: "[x + a = y + a] (mod n) \<longleftrightarrow> [x = y] (mod n)"
187   by (simp add: nat_mod)
189 lemma cong_add_rcancel: "[x + a = y + a] (mod n) \<Longrightarrow> [x = y] (mod n)"
190   by (simp add: nat_mod)
192 lemma cong_add_lcancel: "[a + x = a + y] (mod n) \<Longrightarrow> [x = y] (mod n)"
193   by (simp add: nat_mod)
195 lemma cong_add_lcancel_eq_0: "[a + x = a] (mod n) \<longleftrightarrow> [x = 0] (mod n)"
196   by (simp add: nat_mod)
198 lemma cong_add_rcancel_eq_0: "[x + a = a] (mod n) \<longleftrightarrow> [x = 0] (mod n)"
199   by (simp add: nat_mod)
201 lemma cong_imp_eq: assumes xn: "x < n" and yn: "y < n" and xy: "[x = y] (mod n)"
202   shows "x = y"
203   using xy[unfolded modeq_def mod_less[OF xn] mod_less[OF yn]] .
205 lemma cong_divides_modulus: "[x = y] (mod m) \<Longrightarrow> n dvd m ==> [x = y] (mod n)"
206   apply (auto simp add: nat_mod dvd_def)
207   apply (rule_tac x="k*q1" in exI)
208   apply (rule_tac x="k*q2" in exI)
209   by simp
211 lemma cong_0_divides: "[x = 0] (mod n) \<longleftrightarrow> n dvd x" by simp
213 lemma cong_1_divides:"[x = 1] (mod n) ==> n dvd x - 1"
214   apply (cases "x\<le>1", simp_all)
215   using cong_sub_cases[of x 1 n] by auto
217 lemma cong_divides: "[x = y] (mod n) \<Longrightarrow> n dvd x \<longleftrightarrow> n dvd y"
218 apply (auto simp add: nat_mod dvd_def)
219 apply (rule_tac x="k + q1 - q2" in exI, simp add: add_mult_distrib2 diff_mult_distrib2)
220 apply (rule_tac x="k + q2 - q1" in exI, simp add: add_mult_distrib2 diff_mult_distrib2)
221 done
223 lemma cong_coprime: assumes xy: "[x = y] (mod n)"
224   shows "coprime n x \<longleftrightarrow> coprime n y"
225 proof-
226   {assume "n=0" hence ?thesis using xy by simp}
227   moreover
228   {assume nz: "n \<noteq> 0"
229   have "coprime n x \<longleftrightarrow> coprime (x mod n) n"
230     by (simp add: coprime_mod[OF nz, of x] coprime_commute[of n x])
231   also have "\<dots> \<longleftrightarrow> coprime (y mod n) n" using xy[unfolded modeq_def] by simp
232   also have "\<dots> \<longleftrightarrow> coprime y n" by (simp add: coprime_mod[OF nz, of y])
233   finally have ?thesis by (simp add: coprime_commute) }
234 ultimately show ?thesis by blast
235 qed
237 lemma cong_mod: "~(n = 0) \<Longrightarrow> [a mod n = a] (mod n)" by (simp add: modeq_def)
239 lemma mod_mult_cong: "~(a = 0) \<Longrightarrow> ~(b = 0)
240   \<Longrightarrow> [x mod (a * b) = y] (mod a) \<longleftrightarrow> [x = y] (mod a)"
241   by (simp add: modeq_def mod_mult2_eq mod_add_left_eq)
243 lemma cong_mod_mult: "[x = y] (mod n) \<Longrightarrow> m dvd n \<Longrightarrow> [x = y] (mod m)"
244   apply (auto simp add: nat_mod dvd_def)
245   apply (rule_tac x="k*q1" in exI)
246   apply (rule_tac x="k*q2" in exI, simp)
247   done
249 (* Some things when we know more about the order.                            *)
251 lemma cong_le: "y <= x \<Longrightarrow> [x = y] (mod n) \<longleftrightarrow> (\<exists>q. x = q * n + y)"
252   using nat_mod_lemma[of x y n]
253   apply auto
254   apply (simp add: nat_mod)
255   apply (rule_tac x="q" in exI)
256   apply (rule_tac x="q + q" in exI)
257   by (auto simp: algebra_simps)
259 lemma cong_to_1: "[a = 1] (mod n) \<longleftrightarrow> a = 0 \<and> n = 1 \<or> (\<exists>m. a = 1 + m * n)"
260 proof-
261   {assume "n = 0 \<or> n = 1\<or> a = 0 \<or> a = 1" hence ?thesis
262       apply (cases "n=0", simp_all add: cong_commute)
263       apply (cases "n=1", simp_all add: cong_commute modeq_def)
264       apply arith
265       by (cases "a=1", simp_all add: modeq_def cong_commute)}
266   moreover
267   {assume n: "n\<noteq>0" "n\<noteq>1" and a:"a\<noteq>0" "a \<noteq> 1" hence a': "a \<ge> 1" by simp
268     hence ?thesis using cong_le[OF a', of n] by auto }
269   ultimately show ?thesis by auto
270 qed
272 (* Some basic theorems about solving congruences.                            *)
275 lemma cong_solve: assumes an: "coprime a n" shows "\<exists>x. [a * x = b] (mod n)"
276 proof-
277   {assume "a=0" hence ?thesis using an by (simp add: modeq_def)}
278   moreover
279   {assume az: "a\<noteq>0"
280   from bezout_add_strong[OF az, of n]
281   obtain d x y where dxy: "d dvd a" "d dvd n" "a*x = n*y + d" by blast
282   from an[unfolded coprime, rule_format, of d] dxy(1,2) have d1: "d = 1" by blast
283   hence "a*x*b = (n*y + 1)*b" using dxy(3) by simp
284   hence "a*(x*b) = n*(y*b) + b" by algebra
285   hence "a*(x*b) mod n = (n*(y*b) + b) mod n" by simp
286   hence "a*(x*b) mod n = b mod n" by (simp add: mod_add_left_eq)
287   hence "[a*(x*b) = b] (mod n)" unfolding modeq_def .
288   hence ?thesis by blast}
289 ultimately  show ?thesis by blast
290 qed
292 lemma cong_solve_unique: assumes an: "coprime a n" and nz: "n \<noteq> 0"
293   shows "\<exists>!x. x < n \<and> [a * x = b] (mod n)"
294 proof-
295   let ?P = "\<lambda>x. x < n \<and> [a * x = b] (mod n)"
296   from cong_solve[OF an] obtain x where x: "[a*x = b] (mod n)" by blast
297   let ?x = "x mod n"
298   from x have th: "[a * ?x = b] (mod n)"
299     by (simp add: modeq_def mod_mult_right_eq[of a x n])
300   from mod_less_divisor[ of n x] nz th have Px: "?P ?x" by simp
301   {fix y assume Py: "y < n" "[a * y = b] (mod n)"
302     from Py(2) th have "[a * y = a*?x] (mod n)" by (simp add: modeq_def)
303     hence "[y = ?x] (mod n)" by (simp add: cong_mult_lcancel_eq[OF an])
304     with mod_less[OF Py(1)] mod_less_divisor[ of n x] nz
305     have "y = ?x" by (simp add: modeq_def)}
306   with Px show ?thesis by blast
307 qed
309 lemma cong_solve_unique_nontrivial:
310   assumes p: "prime p" and pa: "coprime p a" and x0: "0 < x" and xp: "x < p"
311   shows "\<exists>!y. 0 < y \<and> y < p \<and> [x * y = a] (mod p)"
312 proof-
313   from p have p1: "p > 1" using prime_ge_2[OF p] by arith
314   hence p01: "p \<noteq> 0" "p \<noteq> 1" by arith+
315   from pa have ap: "coprime a p" by (simp add: coprime_commute)
316   from prime_coprime[OF p, of x] dvd_imp_le[of p x] x0 xp have px:"coprime x p"
317     by (auto simp add: coprime_commute)
318   from cong_solve_unique[OF px p01(1)]
319   obtain y where y: "y < p" "[x * y = a] (mod p)" "\<forall>z. z < p \<and> [x * z = a] (mod p) \<longrightarrow> z = y" by blast
320   {assume y0: "y = 0"
321     with y(2) have th: "p dvd a" by (simp add: cong_commute[of 0 a p])
322     with p coprime_prime[OF pa, of p] have False by simp}
323   with y show ?thesis unfolding Ex1_def using neq0_conv by blast
324 qed
325 lemma cong_unique_inverse_prime:
326   assumes p: "prime p" and x0: "0 < x" and xp: "x < p"
327   shows "\<exists>!y. 0 < y \<and> y < p \<and> [x * y = 1] (mod p)"
328   using cong_solve_unique_nontrivial[OF p coprime_1[of p] x0 xp] .
330 (* Forms of the Chinese remainder theorem.                                   *)
332 lemma cong_chinese:
333   assumes ab: "coprime a b" and  xya: "[x = y] (mod a)"
334   and xyb: "[x = y] (mod b)"
335   shows "[x = y] (mod a*b)"
336   using ab xya xyb
337   by (simp add: cong_sub_cases[of x y a] cong_sub_cases[of x y b]
338     cong_sub_cases[of x y "a*b"])
339 (cases "x \<le> y", simp_all add: divides_mul[of a _ b])
341 lemma chinese_remainder_unique:
342   assumes ab: "coprime a b" and az: "a \<noteq> 0" and bz: "b\<noteq>0"
343   shows "\<exists>!x. x < a * b \<and> [x = m] (mod a) \<and> [x = n] (mod b)"
344 proof-
345   from az bz have abpos: "a*b > 0" by simp
346   from chinese_remainder[OF ab az bz] obtain x q1 q2 where
347     xq12: "x = m + q1 * a" "x = n + q2 * b" by blast
348   let ?w = "x mod (a*b)"
349   have wab: "?w < a*b" by (simp add: mod_less_divisor[OF abpos])
350   from xq12(1) have "?w mod a = ((m + q1 * a) mod (a*b)) mod a" by simp
351   also have "\<dots> = m mod a" apply (simp add: mod_mult2_eq)
352     apply (subst mod_add_left_eq)
353     by simp
354   finally have th1: "[?w = m] (mod a)" by (simp add: modeq_def)
355   from xq12(2) have "?w mod b = ((n + q2 * b) mod (a*b)) mod b" by simp
356   also have "\<dots> = ((n + q2 * b) mod (b*a)) mod b" by (simp add: mult_commute)
357   also have "\<dots> = n mod b" apply (simp add: mod_mult2_eq)
358     apply (subst mod_add_left_eq)
359     by simp
360   finally have th2: "[?w = n] (mod b)" by (simp add: modeq_def)
361   {fix y assume H: "y < a*b" "[y = m] (mod a)" "[y = n] (mod b)"
362     with th1 th2 have H': "[y = ?w] (mod a)" "[y = ?w] (mod b)"
363       by (simp_all add: modeq_def)
364     from cong_chinese[OF ab H'] mod_less[OF H(1)] mod_less[OF wab]
365     have "y = ?w" by (simp add: modeq_def)}
366   with th1 th2 wab show ?thesis by blast
367 qed
369 lemma chinese_remainder_coprime_unique:
370   assumes ab: "coprime a b" and az: "a \<noteq> 0" and bz: "b \<noteq> 0"
371   and ma: "coprime m a" and nb: "coprime n b"
372   shows "\<exists>!x. coprime x (a * b) \<and> x < a * b \<and> [x = m] (mod a) \<and> [x = n] (mod b)"
373 proof-
374   let ?P = "\<lambda>x. x < a * b \<and> [x = m] (mod a) \<and> [x = n] (mod b)"
375   from chinese_remainder_unique[OF ab az bz]
376   obtain x where x: "x < a * b" "[x = m] (mod a)" "[x = n] (mod b)"
377     "\<forall>y. ?P y \<longrightarrow> y = x" by blast
378   from ma nb cong_coprime[OF x(2)] cong_coprime[OF x(3)]
379   have "coprime x a" "coprime x b" by (simp_all add: coprime_commute)
380   with coprime_mul[of x a b] have "coprime x (a*b)" by simp
381   with x show ?thesis by blast
382 qed
384 (* Euler totient function.                                                   *)
386 definition phi_def: "\<phi> n = card { m. 0 < m \<and> m <= n \<and> coprime m n }"
387 lemma phi_0[simp]: "\<phi> 0 = 0"
388   unfolding phi_def by (auto simp add: card_eq_0_iff)
390 lemma phi_finite[simp]: "finite ({ m. 0 < m \<and> m <= n \<and> coprime m n })"
391 proof-
392   have "{ m. 0 < m \<and> m <= n \<and> coprime m n } \<subseteq> {0..n}" by auto
393   thus ?thesis by (auto intro: finite_subset)
394 qed
396 declare coprime_1[presburger]
397 lemma phi_1[simp]: "\<phi> 1 = 1"
398 proof-
399   {fix m
400     have "0 < m \<and> m <= 1 \<and> coprime m 1 \<longleftrightarrow> m = 1" by presburger }
401   thus ?thesis by (simp add: phi_def)
402 qed
404 lemma [simp]: "\<phi> (Suc 0) = Suc 0" using phi_1 by simp
406 lemma phi_alt: "\<phi>(n) = card { m. coprime m n \<and> m < n}"
407 proof-
408   {assume "n=0 \<or> n=1" hence ?thesis by (cases "n=0", simp_all)}
409   moreover
410   {assume n: "n\<noteq>0" "n\<noteq>1"
411     {fix m
412       from n have "0 < m \<and> m <= n \<and> coprime m n \<longleftrightarrow> coprime m n \<and> m < n"
413 	apply (cases "m = 0", simp_all)
414 	apply (cases "m = 1", simp_all)
415 	apply (cases "m = n", auto)
416 	done }
417     hence ?thesis unfolding phi_def by simp}
418   ultimately show ?thesis by auto
419 qed
421 lemma phi_finite_lemma[simp]: "finite {m. coprime m n \<and>  m < n}" (is "finite ?S")
422   by (rule finite_subset[of "?S" "{0..n}"], auto)
424 lemma phi_another: assumes n: "n\<noteq>1"
425   shows "\<phi> n = card {m. 0 < m \<and> m < n \<and> coprime m n }"
426 proof-
427   {fix m
428     from n have "0 < m \<and> m < n \<and> coprime m n \<longleftrightarrow> coprime m n \<and> m < n"
429       by (cases "m=0", auto)}
430   thus ?thesis unfolding phi_alt by auto
431 qed
433 lemma phi_limit: "\<phi> n \<le> n"
434 proof-
435   have "{ m. coprime m n \<and> m < n} \<subseteq> {0 ..<n}" by auto
436   with card_mono[of "{0 ..<n}" "{ m. coprime m n \<and> m < n}"]
437   show ?thesis unfolding phi_alt by auto
438 qed
440 lemma stupid[simp]: "{m. (0::nat) < m \<and> m < n} = {1..<n}"
441   by auto
443 lemma phi_limit_strong: assumes n: "n\<noteq>1"
444   shows "\<phi>(n) \<le> n - 1"
445 proof-
446   show ?thesis
447     unfolding phi_another[OF n] finite_number_segment[of n, symmetric]
448     by (rule card_mono[of "{m. 0 < m \<and> m < n}" "{m. 0 < m \<and> m < n \<and> coprime m n}"], auto)
449 qed
451 lemma phi_lowerbound_1_strong: assumes n: "n \<ge> 1"
452   shows "\<phi>(n) \<ge> 1"
453 proof-
454   let ?S = "{ m. 0 < m \<and> m <= n \<and> coprime m n }"
455   from card_0_eq[of ?S] n have "\<phi> n \<noteq> 0" unfolding phi_alt
456     apply auto
457     apply (cases "n=1", simp_all)
458     apply (rule exI[where x=1], simp)
459     done
460   thus ?thesis by arith
461 qed
463 lemma phi_lowerbound_1: "2 <= n ==> 1 <= \<phi>(n)"
464   using phi_lowerbound_1_strong[of n] by auto
466 lemma phi_lowerbound_2: assumes n: "3 <= n" shows "2 <= \<phi> (n)"
467 proof-
468   let ?S = "{ m. 0 < m \<and> m <= n \<and> coprime m n }"
469   have inS: "{1, n - 1} \<subseteq> ?S" using n coprime_plus1[of "n - 1"]
470     by (auto simp add: coprime_commute)
471   from n have c2: "card {1, n - 1} = 2" by (auto simp add: card_insert_if)
472   from card_mono[of ?S "{1, n - 1}", simplified inS c2] show ?thesis
473     unfolding phi_def by auto
474 qed
476 lemma phi_prime: "\<phi> n = n - 1 \<and> n\<noteq>0 \<and> n\<noteq>1 \<longleftrightarrow> prime n"
477 proof-
478   {assume "n=0 \<or> n=1" hence ?thesis by (cases "n=1", simp_all)}
479   moreover
480   {assume n: "n\<noteq>0" "n\<noteq>1"
481     let ?S = "{m. 0 < m \<and> m < n}"
482     have fS: "finite ?S" by simp
483     let ?S' = "{m. 0 < m \<and> m < n \<and> coprime m n}"
484     have fS':"finite ?S'" apply (rule finite_subset[of ?S' ?S]) by auto
485     {assume H: "\<phi> n = n - 1 \<and> n\<noteq>0 \<and> n\<noteq>1"
486       hence ceq: "card ?S' = card ?S"
487       using n finite_number_segment[of n] phi_another[OF n(2)] by simp
488       {fix m assume m: "0 < m" "m < n" "\<not> coprime m n"
489 	hence mS': "m \<notin> ?S'" by auto
490 	have "insert m ?S' \<le> ?S" using m by auto
491 	from m have "card (insert m ?S') \<le> card ?S"
492 	  by - (rule card_mono[of ?S "insert m ?S'"], auto)
493 	hence False
494 	  unfolding card_insert_disjoint[of "?S'" m, OF fS' mS'] ceq
495 	  by simp }
496       hence "\<forall>m. 0 <m \<and> m < n \<longrightarrow> coprime m n" by blast
497       hence "prime n" unfolding prime using n by (simp add: coprime_commute)}
498     moreover
499     {assume H: "prime n"
500       hence "?S = ?S'" unfolding prime using n
501 	by (auto simp add: coprime_commute)
502       hence "card ?S = card ?S'" by simp
503       hence "\<phi> n = n - 1" unfolding phi_another[OF n(2)] by simp}
504     ultimately have ?thesis using n by blast}
505   ultimately show ?thesis by (cases "n=0") blast+
506 qed
508 (* Multiplicativity property.                                                *)
510 lemma phi_multiplicative: assumes ab: "coprime a b"
511   shows "\<phi> (a * b) = \<phi> a * \<phi> b"
512 proof-
513   {assume "a = 0 \<or> b = 0 \<or> a = 1 \<or> b = 1"
514     hence ?thesis
515       by (cases "a=0", simp, cases "b=0", simp, cases"a=1", simp_all) }
516   moreover
517   {assume a: "a\<noteq>0" "a\<noteq>1" and b: "b\<noteq>0" "b\<noteq>1"
518     hence ab0: "a*b \<noteq> 0" by simp
519     let ?S = "\<lambda>k. {m. coprime m k \<and> m < k}"
520     let ?f = "\<lambda>x. (x mod a, x mod b)"
521     have eq: "?f ` (?S (a*b)) = (?S a \<times> ?S b)"
522     proof-
523       {fix x assume x:"x \<in> ?S (a*b)"
524 	hence x': "coprime x (a*b)" "x < a*b" by simp_all
525 	hence xab: "coprime x a" "coprime x b" by (simp_all add: coprime_mul_eq)
526 	from mod_less_divisor a b have xab':"x mod a < a" "x mod b < b" by auto
527 	from xab xab' have "?f x \<in> (?S a \<times> ?S b)"
528 	  by (simp add: coprime_mod[OF a(1)] coprime_mod[OF b(1)])}
529       moreover
530       {fix x y assume x: "x \<in> ?S a" and y: "y \<in> ?S b"
531 	hence x': "coprime x a" "x < a" and y': "coprime y b" "y < b" by simp_all
532 	from chinese_remainder_coprime_unique[OF ab a(1) b(1) x'(1) y'(1)]
533 	obtain z where z: "coprime z (a * b)" "z < a * b" "[z = x] (mod a)"
534 	  "[z = y] (mod b)" by blast
535 	hence "(x,y) \<in> ?f ` (?S (a*b))"
536 	  using y'(2) mod_less_divisor[of b y] x'(2) mod_less_divisor[of a x]
537 	  by (auto simp add: image_iff modeq_def)}
538       ultimately show ?thesis by auto
539     qed
540     have finj: "inj_on ?f (?S (a*b))"
541       unfolding inj_on_def
542     proof(clarify)
543       fix x y assume H: "coprime x (a * b)" "x < a * b" "coprime y (a * b)"
544 	"y < a * b" "x mod a = y mod a" "x mod b = y mod b"
545       hence cp: "coprime x a" "coprime x b" "coprime y a" "coprime y b"
546 	by (simp_all add: coprime_mul_eq)
547       from chinese_remainder_coprime_unique[OF ab a(1) b(1) cp(3,4)] H
548       show "x = y" unfolding modeq_def by blast
549     qed
550     from card_image[OF finj, unfolded eq] have ?thesis
551       unfolding phi_alt by simp }
552   ultimately show ?thesis by auto
553 qed
555 (* Fermat's Little theorem / Fermat-Euler theorem.                           *)
557 lemma (in comm_monoid_mult) fold_image_related:
558   assumes Re: "R e e"
559   and Rop: "\<forall>x1 y1 x2 y2. R x1 x2 \<and> R y1 y2 \<longrightarrow> R (x1 * y1) (x2 * y2)"
560   and fS: "finite S" and Rfg: "\<forall>x\<in>S. R (h x) (g x)"
561   shows "R (fold_image (op *) h e S) (fold_image (op *) g e S)"
562   using fS by (rule finite_subset_induct) (insert assms, auto)
564 lemma nproduct_mod:
565   assumes fS: "finite S" and n0: "n \<noteq> 0"
566   shows "[setprod (\<lambda>m. a(m) mod n) S = setprod a S] (mod n)"
567 proof-
568   have th1:"[1 = 1] (mod n)" by (simp add: modeq_def)
569   from cong_mult
570   have th3:"\<forall>x1 y1 x2 y2.
571     [x1 = x2] (mod n) \<and> [y1 = y2] (mod n) \<longrightarrow> [x1 * y1 = x2 * y2] (mod n)"
572     by blast
573   have th4:"\<forall>x\<in>S. [a x mod n = a x] (mod n)" by (simp add: modeq_def)
574   from fold_image_related[where h="(\<lambda>m. a(m) mod n)" and g=a, OF th1 th3 fS, OF th4] show ?thesis unfolding setprod_def by (simp add: fS)
575 qed
577 lemma nproduct_cmul:
578   assumes fS:"finite S"
579   shows "setprod (\<lambda>m. (c::'a::{comm_monoid_mult,recpower})* a(m)) S = c ^ (card S) * setprod a S"
580 unfolding setprod_timesf setprod_constant[OF fS, of c] ..
582 lemma coprime_nproduct:
583   assumes fS: "finite S" and Sn: "\<forall>x\<in>S. coprime n (a x)"
584   shows "coprime n (setprod a S)"
585   using fS unfolding setprod_def by (rule finite_subset_induct)
586     (insert Sn, auto simp add: coprime_mul)
588 lemma (in comm_monoid_mult)
589   fold_image_eq_general:
590   assumes fS: "finite S"
591   and h: "\<forall>y\<in>S'. \<exists>!x. x\<in> S \<and> h(x) = y"
592   and f12:  "\<forall>x\<in>S. h x \<in> S' \<and> f2(h x) = f1 x"
593   shows "fold_image (op *) f1 e S = fold_image (op *) f2 e S'"
594 proof-
595   from h f12 have hS: "h ` S = S'" by auto
596   {fix x y assume H: "x \<in> S" "y \<in> S" "h x = h y"
597     from f12 h H  have "x = y" by auto }
598   hence hinj: "inj_on h S" unfolding inj_on_def Ex1_def by blast
599   from f12 have th: "\<And>x. x \<in> S \<Longrightarrow> (f2 \<circ> h) x = f1 x" by auto
600   from hS have "fold_image (op *) f2 e S' = fold_image (op *) f2 e (h ` S)" by simp
601   also have "\<dots> = fold_image (op *) (f2 o h) e S"
602     using fold_image_reindex[OF fS hinj, of f2 e] .
603   also have "\<dots> = fold_image (op *) f1 e S " using th fold_image_cong[OF fS, of "f2 o h" f1 e]
604     by blast
605   finally show ?thesis ..
606 qed
608 lemma fermat_little: assumes an: "coprime a n"
609   shows "[a ^ (\<phi> n) = 1] (mod n)"
610 proof-
611   {assume "n=0" hence ?thesis by simp}
612   moreover
613   {assume "n=1" hence ?thesis by (simp add: modeq_def)}
614   moreover
615   {assume nz: "n \<noteq> 0" and n1: "n \<noteq> 1"
616     let ?S = "{m. coprime m n \<and> m < n}"
617     let ?P = "\<Prod> ?S"
618     have fS: "finite ?S" by simp
619     have cardfS: "\<phi> n = card ?S" unfolding phi_alt ..
620     {fix m assume m: "m \<in> ?S"
621       hence "coprime m n" by simp
622       with coprime_mul[of n a m] an have "coprime (a*m) n"
623 	by (simp add: coprime_commute)}
624     hence Sn: "\<forall>m\<in> ?S. coprime (a*m) n " by blast
625     from coprime_nproduct[OF fS, of n "\<lambda>m. m"] have nP:"coprime ?P n"
626       by (simp add: coprime_commute)
627     have Paphi: "[?P*a^ (\<phi> n) = ?P*1] (mod n)"
628     proof-
629       let ?h = "\<lambda>m. m mod n"
630       {fix m assume mS: "m\<in> ?S"
631 	hence "?h m \<in> ?S" by simp}
632       hence hS: "?h ` ?S = ?S"by (auto simp add: image_iff)
633       have "a\<noteq>0" using an n1 nz apply- apply (rule ccontr) by simp
634       hence inj: "inj_on (op * a) ?S" unfolding inj_on_def by simp
636       have eq0: "fold_image op * (?h \<circ> op * a) 1 {m. coprime m n \<and> m < n} =
637      fold_image op * (\<lambda>m. m) 1 {m. coprime m n \<and> m < n}"
638       proof (rule fold_image_eq_general[where h="?h o (op * a)"])
639 	show "finite ?S" using fS .
640       next
641 	{fix y assume yS: "y \<in> ?S" hence y: "coprime y n" "y < n" by simp_all
642 	  from cong_solve_unique[OF an nz, of y]
643 	  obtain x where x:"x < n" "[a * x = y] (mod n)" "\<forall>z. z < n \<and> [a * z = y] (mod n) \<longrightarrow> z=x" by blast
644 	  from cong_coprime[OF x(2)] y(1)
645 	  have xm: "coprime x n" by (simp add: coprime_mul_eq coprime_commute)
646 	  {fix z assume "z \<in> ?S" "(?h \<circ> op * a) z = y"
647 	    hence z: "coprime z n" "z < n" "(?h \<circ> op * a) z = y" by simp_all
648 	    from x(3)[rule_format, of z] z(2,3) have "z=x"
649 	      unfolding modeq_def mod_less[OF y(2)] by simp}
650 	  with xm x(1,2) have "\<exists>!x. x \<in> ?S \<and> (?h \<circ> op * a) x = y"
651 	    unfolding modeq_def mod_less[OF y(2)] by auto }
652 	thus "\<forall>y\<in>{m. coprime m n \<and> m < n}.
653        \<exists>!x. x \<in> {m. coprime m n \<and> m < n} \<and> ((\<lambda>m. m mod n) \<circ> op * a) x = y" by blast
654       next
655 	{fix x assume xS: "x\<in> ?S"
656 	  hence x: "coprime x n" "x < n" by simp_all
657 	  with an have "coprime (a*x) n"
658 	    by (simp add: coprime_mul_eq[of n a x] coprime_commute)
659 	  hence "?h (a*x) \<in> ?S" using nz
660 	    by (simp add: coprime_mod[OF nz] mod_less_divisor)}
661 	thus " \<forall>x\<in>{m. coprime m n \<and> m < n}.
662        ((\<lambda>m. m mod n) \<circ> op * a) x \<in> {m. coprime m n \<and> m < n} \<and>
663        ((\<lambda>m. m mod n) \<circ> op * a) x = ((\<lambda>m. m mod n) \<circ> op * a) x" by simp
664       qed
665       from nproduct_mod[OF fS nz, of "op * a"]
666       have "[(setprod (op *a) ?S) = (setprod (?h o (op * a)) ?S)] (mod n)"
667 	unfolding o_def
668 	by (simp add: cong_commute)
669       also have "[setprod (?h o (op * a)) ?S = ?P ] (mod n)"
670 	using eq0 fS an by (simp add: setprod_def modeq_def o_def)
671       finally show "[?P*a^ (\<phi> n) = ?P*1] (mod n)"
672 	unfolding cardfS mult_commute[of ?P "a^ (card ?S)"]
673 	  nproduct_cmul[OF fS, symmetric] mult_1_right by simp
674     qed
675     from cong_mult_lcancel[OF nP Paphi] have ?thesis . }
676   ultimately show ?thesis by blast
677 qed
679 lemma fermat_little_prime: assumes p: "prime p" and ap: "coprime a p"
680   shows "[a^ (p - 1) = 1] (mod p)"
681   using fermat_little[OF ap] p[unfolded phi_prime[symmetric]]
682 by simp
685 (* Lucas's theorem.                                                          *)
687 lemma lucas_coprime_lemma:
688   assumes m: "m\<noteq>0" and am: "[a^m = 1] (mod n)"
689   shows "coprime a n"
690 proof-
691   {assume "n=1" hence ?thesis by simp}
692   moreover
693   {assume "n = 0" hence ?thesis using am m exp_eq_1[of a m] by simp}
694   moreover
695   {assume n: "n\<noteq>0" "n\<noteq>1"
696     from m obtain m' where m': "m = Suc m'" by (cases m, blast+)
697     {fix d
698       assume d: "d dvd a" "d dvd n"
699       from n have n1: "1 < n" by arith
700       from am mod_less[OF n1] have am1: "a^m mod n = 1" unfolding modeq_def by simp
701       from dvd_mult2[OF d(1), of "a^m'"] have dam:"d dvd a^m" by (simp add: m')
702       from dvd_mod_iff[OF d(2), of "a^m"] dam am1
703       have "d = 1" by simp }
704     hence ?thesis unfolding coprime by auto
705   }
706   ultimately show ?thesis by blast
707 qed
709 lemma lucas_weak:
710   assumes n: "n \<ge> 2" and an:"[a^(n - 1) = 1] (mod n)"
711   and nm: "\<forall>m. 0 <m \<and> m < n - 1 \<longrightarrow> \<not> [a^m = 1] (mod n)"
712   shows "prime n"
713 proof-
714   from n have n1: "n \<noteq> 1" "n\<noteq>0" "n - 1 \<noteq> 0" "n - 1 > 0" "n - 1 < n" by arith+
715   from lucas_coprime_lemma[OF n1(3) an] have can: "coprime a n" .
716   from fermat_little[OF can] have afn: "[a ^ \<phi> n = 1] (mod n)" .
717   {assume "\<phi> n \<noteq> n - 1"
718     with phi_limit_strong[OF n1(1)] phi_lowerbound_1[OF n]
719     have c:"\<phi> n > 0 \<and> \<phi> n < n - 1" by arith
720     from nm[rule_format, OF c] afn have False ..}
721   hence "\<phi> n = n - 1" by blast
722   with phi_prime[of n] n1(1,2) show ?thesis by simp
723 qed
725 lemma nat_exists_least_iff: "(\<exists>(n::nat). P n) \<longleftrightarrow> (\<exists>n. P n \<and> (\<forall>m < n. \<not> P m))"
726   (is "?lhs \<longleftrightarrow> ?rhs")
727 proof
728   assume ?rhs thus ?lhs by blast
729 next
730   assume H: ?lhs then obtain n where n: "P n" by blast
731   let ?x = "Least P"
732   {fix m assume m: "m < ?x"
733     from not_less_Least[OF m] have "\<not> P m" .}
734   with LeastI_ex[OF H] show ?rhs by blast
735 qed
737 lemma nat_exists_least_iff': "(\<exists>(n::nat). P n) \<longleftrightarrow> (P (Least P) \<and> (\<forall>m < (Least P). \<not> P m))"
738   (is "?lhs \<longleftrightarrow> ?rhs")
739 proof-
740   {assume ?rhs hence ?lhs by blast}
741   moreover
742   { assume H: ?lhs then obtain n where n: "P n" by blast
743     let ?x = "Least P"
744     {fix m assume m: "m < ?x"
745       from not_less_Least[OF m] have "\<not> P m" .}
746     with LeastI_ex[OF H] have ?rhs by blast}
747   ultimately show ?thesis by blast
748 qed
750 lemma power_mod: "((x::nat) mod m)^n mod m = x^n mod m"
751 proof(induct n)
752   case 0 thus ?case by simp
753 next
754   case (Suc n)
755   have "(x mod m)^(Suc n) mod m = ((x mod m) * (((x mod m) ^ n) mod m)) mod m"
756     by (simp add: mod_mult_right_eq[symmetric])
757   also have "\<dots> = ((x mod m) * (x^n mod m)) mod m" using Suc.hyps by simp
758   also have "\<dots> = x^(Suc n) mod m"
759     by (simp add: mod_mult_left_eq[symmetric] mod_mult_right_eq[symmetric])
760   finally show ?case .
761 qed
763 lemma lucas:
764   assumes n2: "n \<ge> 2" and an1: "[a^(n - 1) = 1] (mod n)"
765   and pn: "\<forall>p. prime p \<and> p dvd n - 1 \<longrightarrow> \<not> [a^((n - 1) div p) = 1] (mod n)"
766   shows "prime n"
767 proof-
768   from n2 have n01: "n\<noteq>0" "n\<noteq>1" "n - 1 \<noteq> 0" by arith+
769   from mod_less_divisor[of n 1] n01 have onen: "1 mod n = 1" by simp
770   from lucas_coprime_lemma[OF n01(3) an1] cong_coprime[OF an1]
771   have an: "coprime a n" "coprime (a^(n - 1)) n" by (simp_all add: coprime_commute)
772   {assume H0: "\<exists>m. 0 < m \<and> m < n - 1 \<and> [a ^ m = 1] (mod n)" (is "EX m. ?P m")
773     from H0[unfolded nat_exists_least_iff[of ?P]] obtain m where
774       m: "0 < m" "m < n - 1" "[a ^ m = 1] (mod n)" "\<forall>k <m. \<not>?P k" by blast
775     {assume nm1: "(n - 1) mod m > 0"
776       from mod_less_divisor[OF m(1)] have th0:"(n - 1) mod m < m" by blast
777       let ?y = "a^ ((n - 1) div m * m)"
778       note mdeq = mod_div_equality[of "(n - 1)" m]
779       from coprime_exp[OF an(1)[unfolded coprime_commute[of a n]],
780 	of "(n - 1) div m * m"]
781       have yn: "coprime ?y n" by (simp add: coprime_commute)
782       have "?y mod n = (a^m)^((n - 1) div m) mod n"
783 	by (simp add: algebra_simps power_mult)
784       also have "\<dots> = (a^m mod n)^((n - 1) div m) mod n"
785 	using power_mod[of "a^m" n "(n - 1) div m"] by simp
786       also have "\<dots> = 1" using m(3)[unfolded modeq_def onen] onen
787 	by (simp add: power_Suc0)
788       finally have th3: "?y mod n = 1"  .
789       have th2: "[?y * a ^ ((n - 1) mod m) = ?y* 1] (mod n)"
790 	using an1[unfolded modeq_def onen] onen
791 	  mod_div_equality[of "(n - 1)" m, symmetric]
792 	by (simp add:power_add[symmetric] modeq_def th3 del: One_nat_def)
793       from cong_mult_lcancel[of ?y n "a^((n - 1) mod m)" 1, OF yn th2]
794       have th1: "[a ^ ((n - 1) mod m) = 1] (mod n)"  .
795       from m(4)[rule_format, OF th0] nm1
796 	less_trans[OF mod_less_divisor[OF m(1), of "n - 1"] m(2)] th1
797       have False by blast }
798     hence "(n - 1) mod m = 0" by auto
799     then have mn: "m dvd n - 1" by presburger
800     then obtain r where r: "n - 1 = m*r" unfolding dvd_def by blast
801     from n01 r m(2) have r01: "r\<noteq>0" "r\<noteq>1" by - (rule ccontr, simp)+
802     from prime_factor[OF r01(2)] obtain p where p: "prime p" "p dvd r" by blast
803     hence th: "prime p \<and> p dvd n - 1" unfolding r by (auto intro: dvd_mult)
804     have "(a ^ ((n - 1) div p)) mod n = (a^(m*r div p)) mod n" using r
805       by (simp add: power_mult)
806     also have "\<dots> = (a^(m*(r div p))) mod n" using div_mult1_eq[of m r p] p(2)[unfolded dvd_eq_mod_eq_0] by simp
807     also have "\<dots> = ((a^m)^(r div p)) mod n" by (simp add: power_mult)
808     also have "\<dots> = ((a^m mod n)^(r div p)) mod n" using power_mod[of "a^m" "n" "r div p" ] ..
809     also have "\<dots> = 1" using m(3) onen by (simp add: modeq_def power_Suc0)
810     finally have "[(a ^ ((n - 1) div p))= 1] (mod n)"
811       using onen by (simp add: modeq_def)
812     with pn[rule_format, OF th] have False by blast}
813   hence th: "\<forall>m. 0 < m \<and> m < n - 1 \<longrightarrow> \<not> [a ^ m = 1] (mod n)" by blast
814   from lucas_weak[OF n2 an1 th] show ?thesis .
815 qed
817 (* Definition of the order of a number mod n (0 in non-coprime case).        *)
819 definition "ord n a = (if coprime n a then Least (\<lambda>d. d > 0 \<and> [a ^d = 1] (mod n)) else 0)"
821 (* This has the expected properties.                                         *)
823 lemma coprime_ord:
824   assumes na: "coprime n a"
825   shows "ord n a > 0 \<and> [a ^(ord n a) = 1] (mod n) \<and> (\<forall>m. 0 < m \<and> m < ord n a \<longrightarrow> \<not> [a^ m = 1] (mod n))"
826 proof-
827   let ?P = "\<lambda>d. 0 < d \<and> [a ^ d = 1] (mod n)"
828   from euclid[of a] obtain p where p: "prime p" "a < p" by blast
829   from na have o: "ord n a = Least ?P" by (simp add: ord_def)
830   {assume "n=0 \<or> n=1" with na have "\<exists>m>0. ?P m" apply auto apply (rule exI[where x=1]) by (simp  add: modeq_def)}
831   moreover
832   {assume "n\<noteq>0 \<and> n\<noteq>1" hence n2:"n \<ge> 2" by arith
833     from na have na': "coprime a n" by (simp add: coprime_commute)
834     from phi_lowerbound_1[OF n2] fermat_little[OF na']
835     have ex: "\<exists>m>0. ?P m" by - (rule exI[where x="\<phi> n"], auto) }
836   ultimately have ex: "\<exists>m>0. ?P m" by blast
837   from nat_exists_least_iff'[of ?P] ex na show ?thesis
838     unfolding o[symmetric] by auto
839 qed
840 (* With the special value 0 for non-coprime case, it's more convenient.      *)
841 lemma ord_works:
842  "[a ^ (ord n a) = 1] (mod n) \<and> (\<forall>m. 0 < m \<and> m < ord n a \<longrightarrow> ~[a^ m = 1] (mod n))"
843 apply (cases "coprime n a")
844 using coprime_ord[of n a]
845 by (blast, simp add: ord_def modeq_def)
847 lemma ord: "[a^(ord n a) = 1] (mod n)" using ord_works by blast
848 lemma ord_minimal: "0 < m \<Longrightarrow> m < ord n a \<Longrightarrow> ~[a^m = 1] (mod n)"
849   using ord_works by blast
850 lemma ord_eq_0: "ord n a = 0 \<longleftrightarrow> ~coprime n a"
851 by (cases "coprime n a", simp add: neq0_conv coprime_ord, simp add: neq0_conv ord_def)
853 lemma ord_divides:
854  "[a ^ d = 1] (mod n) \<longleftrightarrow> ord n a dvd d" (is "?lhs \<longleftrightarrow> ?rhs")
855 proof
856   assume rh: ?rhs
857   then obtain k where "d = ord n a * k" unfolding dvd_def by blast
858   hence "[a ^ d = (a ^ (ord n a) mod n)^k] (mod n)"
859     by (simp add : modeq_def power_mult power_mod)
860   also have "[(a ^ (ord n a) mod n)^k = 1] (mod n)"
861     using ord[of a n, unfolded modeq_def]
862     by (simp add: modeq_def power_mod power_Suc0)
863   finally  show ?lhs .
864 next
865   assume lh: ?lhs
866   { assume H: "\<not> coprime n a"
867     hence o: "ord n a = 0" by (simp add: ord_def)
868     {assume d: "d=0" with o H have ?rhs by (simp add: modeq_def)}
869     moreover
870     {assume d0: "d\<noteq>0" then obtain d' where d': "d = Suc d'" by (cases d, auto)
871       from H[unfolded coprime]
872       obtain p where p: "p dvd n" "p dvd a" "p \<noteq> 1" by auto
873       from lh[unfolded nat_mod]
874       obtain q1 q2 where q12:"a ^ d + n * q1 = 1 + n * q2" by blast
875       hence "a ^ d + n * q1 - n * q2 = 1" by simp
876       with nat_dvd_diff [OF dvd_add [OF divides_rexp[OF p(2), of d'] dvd_mult2[OF p(1), of q1]] dvd_mult2[OF p(1), of q2]] d' have "p dvd 1" by simp
877       with p(3) have False by simp
878       hence ?rhs ..}
879     ultimately have ?rhs by blast}
880   moreover
881   {assume H: "coprime n a"
882     let ?o = "ord n a"
883     let ?q = "d div ord n a"
884     let ?r = "d mod ord n a"
885     from cong_exp[OF ord[of a n], of ?q]
886     have eqo: "[(a^?o)^?q = 1] (mod n)"  by (simp add: modeq_def power_Suc0)
887     from H have onz: "?o \<noteq> 0" by (simp add: ord_eq_0)
888     hence op: "?o > 0" by simp
889     from mod_div_equality[of d "ord n a"] lh
890     have "[a^(?o*?q + ?r) = 1] (mod n)" by (simp add: modeq_def mult_commute)
891     hence "[(a^?o)^?q * (a^?r) = 1] (mod n)"
892       by (simp add: modeq_def power_mult[symmetric] power_add[symmetric])
893     hence th: "[a^?r = 1] (mod n)"
894       using eqo mod_mult_left_eq[of "(a^?o)^?q" "a^?r" n]
895       apply (simp add: modeq_def del: One_nat_def)
896       by (simp add: mod_mult_left_eq[symmetric])
897     {assume r: "?r = 0" hence ?rhs by (simp add: dvd_eq_mod_eq_0)}
898     moreover
899     {assume r: "?r \<noteq> 0"
900       with mod_less_divisor[OF op, of d] have r0o:"?r >0 \<and> ?r < ?o" by simp
901       from conjunct2[OF ord_works[of a n], rule_format, OF r0o] th
902       have ?rhs by blast}
903     ultimately have ?rhs by blast}
904   ultimately  show ?rhs by blast
905 qed
907 lemma order_divides_phi: "coprime n a \<Longrightarrow> ord n a dvd \<phi> n"
908 using ord_divides fermat_little coprime_commute by simp
909 lemma order_divides_expdiff:
910   assumes na: "coprime n a"
911   shows "[a^d = a^e] (mod n) \<longleftrightarrow> [d = e] (mod (ord n a))"
912 proof-
913   {fix n a d e
914     assume na: "coprime n a" and ed: "(e::nat) \<le> d"
915     hence "\<exists>c. d = e + c" by arith
916     then obtain c where c: "d = e + c" by arith
917     from na have an: "coprime a n" by (simp add: coprime_commute)
918     from coprime_exp[OF na, of e]
919     have aen: "coprime (a^e) n" by (simp add: coprime_commute)
920     from coprime_exp[OF na, of c]
921     have acn: "coprime (a^c) n" by (simp add: coprime_commute)
922     have "[a^d = a^e] (mod n) \<longleftrightarrow> [a^(e + c) = a^(e + 0)] (mod n)"
923       using c by simp
924     also have "\<dots> \<longleftrightarrow> [a^e* a^c = a^e *a^0] (mod n)" by (simp add: power_add)
925     also have  "\<dots> \<longleftrightarrow> [a ^ c = 1] (mod n)"
926       using cong_mult_lcancel_eq[OF aen, of "a^c" "a^0"] by simp
927     also  have "\<dots> \<longleftrightarrow> ord n a dvd c" by (simp only: ord_divides)
928     also have "\<dots> \<longleftrightarrow> [e + c = e + 0] (mod ord n a)"
929       using cong_add_lcancel_eq[of e c 0 "ord n a", simplified cong_0_divides]
930       by simp
931     finally have "[a^d = a^e] (mod n) \<longleftrightarrow> [d = e] (mod (ord n a))"
932       using c by simp }
933   note th = this
934   have "e \<le> d \<or> d \<le> e" by arith
935   moreover
936   {assume ed: "e \<le> d" from th[OF na ed] have ?thesis .}
937   moreover
938   {assume de: "d \<le> e"
939     from th[OF na de] have ?thesis by (simp add: cong_commute) }
940   ultimately show ?thesis by blast
941 qed
943 (* Another trivial primality characterization.                               *)
945 lemma prime_prime_factor:
946   "prime n \<longleftrightarrow> n \<noteq> 1\<and> (\<forall>p. prime p \<and> p dvd n \<longrightarrow> p = n)"
947 proof-
948   {assume n: "n=0 \<or> n=1" hence ?thesis using prime_0 two_is_prime by auto}
949   moreover
950   {assume n: "n\<noteq>0" "n\<noteq>1"
951     {assume pn: "prime n"
953       from pn[unfolded prime_def] have "\<forall>p. prime p \<and> p dvd n \<longrightarrow> p = n"
954 	using n
955 	apply (cases "n = 0 \<or> n=1",simp)
956 	by (clarsimp, erule_tac x="p" in allE, auto)}
957     moreover
958     {assume H: "\<forall>p. prime p \<and> p dvd n \<longrightarrow> p = n"
959       from n have n1: "n > 1" by arith
960       {fix m assume m: "m dvd n" "m\<noteq>1"
961 	from prime_factor[OF m(2)] obtain p where
962 	  p: "prime p" "p dvd m" by blast
963 	from dvd_trans[OF p(2) m(1)] p(1) H have "p = n" by blast
964 	with p(2) have "n dvd m"  by simp
965 	hence "m=n"  using dvd_anti_sym[OF m(1)] by simp }
966       with n1 have "prime n"  unfolding prime_def by auto }
967     ultimately have ?thesis using n by blast}
968   ultimately       show ?thesis by auto
969 qed
971 lemma prime_divisor_sqrt:
972   "prime n \<longleftrightarrow> n \<noteq> 1 \<and> (\<forall>d. d dvd n \<and> d^2 \<le> n \<longrightarrow> d = 1)"
973 proof-
974   {assume "n=0 \<or> n=1" hence ?thesis using prime_0 prime_1
975     by (auto simp add: nat_power_eq_0_iff)}
976   moreover
977   {assume n: "n\<noteq>0" "n\<noteq>1"
978     hence np: "n > 1" by arith
979     {fix d assume d: "d dvd n" "d^2 \<le> n" and H: "\<forall>m. m dvd n \<longrightarrow> m=1 \<or> m=n"
980       from H d have d1n: "d = 1 \<or> d=n" by blast
981       {assume dn: "d=n"
982 	have "n^2 > n*1" using n
983 	  by (simp add: power2_eq_square mult_less_cancel1)
984 	with dn d(2) have "d=1" by simp}
985       with d1n have "d = 1" by blast  }
986     moreover
987     {fix d assume d: "d dvd n" and H: "\<forall>d'. d' dvd n \<and> d'^2 \<le> n \<longrightarrow> d' = 1"
988       from d n have "d \<noteq> 0" apply - apply (rule ccontr) by simp
989       hence dp: "d > 0" by simp
990       from d[unfolded dvd_def] obtain e where e: "n= d*e" by blast
991       from n dp e have ep:"e > 0" by simp
992       have "d^2 \<le> n \<or> e^2 \<le> n" using dp ep
993 	by (auto simp add: e power2_eq_square mult_le_cancel_left)
994       moreover
995       {assume h: "d^2 \<le> n"
996 	from H[rule_format, of d] h d have "d = 1" by blast}
997       moreover
998       {assume h: "e^2 \<le> n"
999 	from e have "e dvd n" unfolding dvd_def by (simp add: mult_commute)
1000 	with H[rule_format, of e] h have "e=1" by simp
1001 	with e have "d = n" by simp}
1002       ultimately have "d=1 \<or> d=n"  by blast}
1003     ultimately have ?thesis unfolding prime_def using np n(2) by blast}
1004   ultimately show ?thesis by auto
1005 qed
1006 lemma prime_prime_factor_sqrt:
1007   "prime n \<longleftrightarrow> n \<noteq> 0 \<and> n \<noteq> 1 \<and> \<not> (\<exists>p. prime p \<and> p dvd n \<and> p^2 \<le> n)"
1008   (is "?lhs \<longleftrightarrow>?rhs")
1009 proof-
1010   {assume "n=0 \<or> n=1" hence ?thesis using prime_0 prime_1 by auto}
1011   moreover
1012   {assume n: "n\<noteq>0" "n\<noteq>1"
1013     {assume H: ?lhs
1014       from H[unfolded prime_divisor_sqrt] n
1015       have ?rhs  apply clarsimp by (erule_tac x="p" in allE, simp add: prime_1)
1016     }
1017     moreover
1018     {assume H: ?rhs
1019       {fix d assume d: "d dvd n" "d^2 \<le> n" "d\<noteq>1"
1020 	from prime_factor[OF d(3)]
1021 	obtain p where p: "prime p" "p dvd d" by blast
1022 	from n have np: "n > 0" by arith
1023 	from d(1) n have "d \<noteq> 0" by - (rule ccontr, auto)
1024 	hence dp: "d > 0" by arith
1025 	from mult_mono[OF dvd_imp_le[OF p(2) dp] dvd_imp_le[OF p(2) dp]] d(2)
1026 	have "p^2 \<le> n" unfolding power2_eq_square by arith
1027 	with H n p(1) dvd_trans[OF p(2) d(1)] have False  by blast}
1028       with n prime_divisor_sqrt  have ?lhs by auto}
1029     ultimately have ?thesis by blast }
1030   ultimately show ?thesis by (cases "n=0 \<or> n=1", auto)
1031 qed
1032 (* Pocklington theorem. *)
1034 lemma pocklington_lemma:
1035   assumes n: "n \<ge> 2" and nqr: "n - 1 = q*r" and an: "[a^ (n - 1) = 1] (mod n)"
1036   and aq:"\<forall>p. prime p \<and> p dvd q \<longrightarrow> coprime (a^ ((n - 1) div p) - 1) n"
1037   and pp: "prime p" and pn: "p dvd n"
1038   shows "[p = 1] (mod q)"
1039 proof-
1040   from pp prime_0 prime_1 have p01: "p \<noteq> 0" "p \<noteq> 1" by - (rule ccontr, simp)+
1041   from cong_1_divides[OF an, unfolded nqr, unfolded dvd_def]
1042   obtain k where k: "a ^ (q * r) - 1 = n*k" by blast
1043   from pn[unfolded dvd_def] obtain l where l: "n = p*l" by blast
1044   {assume a0: "a = 0"
1045     hence "a^ (n - 1) = 0" using n by (simp add: power_0_left)
1046     with n an mod_less[of 1 n]  have False by (simp add: power_0_left modeq_def)}
1047   hence a0: "a\<noteq>0" ..
1048   from n nqr have aqr0: "a ^ (q * r) \<noteq> 0" using a0 by (simp add: neq0_conv)
1049   hence "(a ^ (q * r) - 1) + 1  = a ^ (q * r)" by simp
1050   with k l have "a ^ (q * r) = p*l*k + 1" by simp
1051   hence "a ^ (r * q) + p * 0 = 1 + p * (l*k)" by (simp add: mult_ac)
1052   hence odq: "ord p (a^r) dvd q"
1053     unfolding ord_divides[symmetric] power_mult[symmetric] nat_mod  by blast
1054   from odq[unfolded dvd_def] obtain d where d: "q = ord p (a^r) * d" by blast
1055   {assume d1: "d \<noteq> 1"
1056     from prime_factor[OF d1] obtain P where P: "prime P" "P dvd d" by blast
1057     from d dvd_mult[OF P(2), of "ord p (a^r)"] have Pq: "P dvd q" by simp
1058     from aq P(1) Pq have caP:"coprime (a^ ((n - 1) div P) - 1) n" by blast
1059     from Pq obtain s where s: "q = P*s" unfolding dvd_def by blast
1060     have P0: "P \<noteq> 0" using P(1) prime_0 by - (rule ccontr, simp)
1061     from P(2) obtain t where t: "d = P*t" unfolding dvd_def by blast
1062     from d s t P0  have s': "ord p (a^r) * t = s" by algebra
1063     have "ord p (a^r) * t*r = r * ord p (a^r) * t" by algebra
1064     hence exps: "a^(ord p (a^r) * t*r) = ((a ^ r) ^ ord p (a^r)) ^ t"
1065       by (simp only: power_mult)
1066     have "[((a ^ r) ^ ord p (a^r)) ^ t= 1^t] (mod p)"
1067       by (rule cong_exp, rule ord)
1068     then have th: "[((a ^ r) ^ ord p (a^r)) ^ t= 1] (mod p)"
1069       by (simp add: power_Suc0)
1070     from cong_1_divides[OF th] exps have pd0: "p dvd a^(ord p (a^r) * t*r) - 1" by simp
1071     from nqr s s' have "(n - 1) div P = ord p (a^r) * t*r" using P0 by simp
1072     with caP have "coprime (a^(ord p (a^r) * t*r) - 1) n" by simp
1073     with p01 pn pd0 have False unfolding coprime by auto}
1074   hence d1: "d = 1" by blast
1075   hence o: "ord p (a^r) = q" using d by simp
1076   from pp phi_prime[of p] have phip: " \<phi> p = p - 1" by simp
1077   {fix d assume d: "d dvd p" "d dvd a" "d \<noteq> 1"
1078     from pp[unfolded prime_def] d have dp: "d = p" by blast
1079     from n have n12:"Suc (n - 2) = n - 1" by arith
1080     with divides_rexp[OF d(2)[unfolded dp], of "n - 2"]
1081     have th0: "p dvd a ^ (n - 1)" by simp
1082     from n have n0: "n \<noteq> 0" by simp
1083     from d(2) an n12[symmetric] have a0: "a \<noteq> 0"
1084       by - (rule ccontr, simp add: modeq_def)
1085     have th1: "a^ (n - 1) \<noteq> 0" using n d(2) dp a0 by (auto simp add: neq0_conv)
1086     from coprime_minus1[OF th1, unfolded coprime]
1087       dvd_trans[OF pn cong_1_divides[OF an]] th0 d(3) dp
1088     have False by auto}
1089   hence cpa: "coprime p a" using coprime by auto
1090   from coprime_exp[OF cpa, of r] coprime_commute
1091   have arp: "coprime (a^r) p" by blast
1092   from fermat_little[OF arp, simplified ord_divides] o phip
1093   have "q dvd (p - 1)" by simp
1094   then obtain d where d:"p - 1 = q * d" unfolding dvd_def by blast
1095   from prime_0 pp have p0:"p \<noteq> 0" by -  (rule ccontr, auto)
1096   from p0 d have "p + q * 0 = 1 + q * d" by simp
1097   with nat_mod[of p 1 q, symmetric]
1098   show ?thesis by blast
1099 qed
1101 lemma pocklington:
1102   assumes n: "n \<ge> 2" and nqr: "n - 1 = q*r" and sqr: "n \<le> q^2"
1103   and an: "[a^ (n - 1) = 1] (mod n)"
1104   and aq:"\<forall>p. prime p \<and> p dvd q \<longrightarrow> coprime (a^ ((n - 1) div p) - 1) n"
1105   shows "prime n"
1106 unfolding prime_prime_factor_sqrt[of n]
1107 proof-
1108   let ?ths = "n \<noteq> 0 \<and> n \<noteq> 1 \<and> \<not> (\<exists>p. prime p \<and> p dvd n \<and> p\<twosuperior> \<le> n)"
1109   from n have n01: "n\<noteq>0" "n\<noteq>1" by arith+
1110   {fix p assume p: "prime p" "p dvd n" "p^2 \<le> n"
1111     from p(3) sqr have "p^(Suc 1) \<le> q^(Suc 1)" by (simp add: power2_eq_square)
1112     hence pq: "p \<le> q" unfolding exp_mono_le .
1113     from pocklington_lemma[OF n nqr an aq p(1,2)]  cong_1_divides
1114     have th: "q dvd p - 1" by blast
1115     have "p - 1 \<noteq> 0"using prime_ge_2[OF p(1)] by arith
1116     with divides_ge[OF th] pq have False by arith }
1117   with n01 show ?ths by blast
1118 qed
1120 (* Variant for application, to separate the exponentiation.                  *)
1121 lemma pocklington_alt:
1122   assumes n: "n \<ge> 2" and nqr: "n - 1 = q*r" and sqr: "n \<le> q^2"
1123   and an: "[a^ (n - 1) = 1] (mod n)"
1124   and aq:"\<forall>p. prime p \<and> p dvd q \<longrightarrow> (\<exists>b. [a^((n - 1) div p) = b] (mod n) \<and> coprime (b - 1) n)"
1125   shows "prime n"
1126 proof-
1127   {fix p assume p: "prime p" "p dvd q"
1128     from aq[rule_format] p obtain b where
1129       b: "[a^((n - 1) div p) = b] (mod n)" "coprime (b - 1) n" by blast
1130     {assume a0: "a=0"
1131       from n an have "[0 = 1] (mod n)" unfolding a0 power_0_left by auto
1132       hence False using n by (simp add: modeq_def dvd_eq_mod_eq_0[symmetric])}
1133     hence a0: "a\<noteq> 0" ..
1134     hence a1: "a \<ge> 1" by arith
1135     from one_le_power[OF a1] have ath: "1 \<le> a ^ ((n - 1) div p)" .
1136     {assume b0: "b = 0"
1137       from p(2) nqr have "(n - 1) mod p = 0"
1138 	apply (simp only: dvd_eq_mod_eq_0[symmetric]) by (rule dvd_mult2, simp)
1139       with mod_div_equality[of "n - 1" p]
1140       have "(n - 1) div p * p= n - 1" by auto
1141       hence eq: "(a^((n - 1) div p))^p = a^(n - 1)"
1142 	by (simp only: power_mult[symmetric])
1143       from prime_ge_2[OF p(1)] have pS: "Suc (p - 1) = p" by arith
1144       from b(1) have d: "n dvd a^((n - 1) div p)" unfolding b0 cong_0_divides .
1145       from divides_rexp[OF d, of "p - 1"] pS eq cong_divides[OF an] n
1146       have False by simp}
1147     then have b0: "b \<noteq> 0" ..
1148     hence b1: "b \<ge> 1" by arith
1149     from cong_coprime[OF cong_sub[OF b(1) cong_refl[of 1] ath b1]] b(2) nqr
1150     have "coprime (a ^ ((n - 1) div p) - 1) n" by (simp add: coprime_commute)}
1151   hence th: "\<forall>p. prime p \<and> p dvd q \<longrightarrow> coprime (a ^ ((n - 1) div p) - 1) n "
1152     by blast
1153   from pocklington[OF n nqr sqr an th] show ?thesis .
1154 qed
1156 (* Prime factorizations.                                                     *)
1158 definition "primefact ps n = (foldr op * ps  1 = n \<and> (\<forall>p\<in> set ps. prime p))"
1160 lemma primefact: assumes n: "n \<noteq> 0"
1161   shows "\<exists>ps. primefact ps n"
1162 using n
1163 proof(induct n rule: nat_less_induct)
1164   fix n assume H: "\<forall>m<n. m \<noteq> 0 \<longrightarrow> (\<exists>ps. primefact ps m)" and n: "n\<noteq>0"
1165   let ?ths = "\<exists>ps. primefact ps n"
1166   {assume "n = 1"
1167     hence "primefact [] n" by (simp add: primefact_def)
1168     hence ?ths by blast }
1169   moreover
1170   {assume n1: "n \<noteq> 1"
1171     with n have n2: "n \<ge> 2" by arith
1172     from prime_factor[OF n1] obtain p where p: "prime p" "p dvd n" by blast
1173     from p(2) obtain m where m: "n = p*m" unfolding dvd_def by blast
1174     from n m have m0: "m > 0" "m\<noteq>0" by auto
1175     from prime_ge_2[OF p(1)] have "1 < p" by arith
1176     with m0 m have mn: "m < n" by auto
1177     from H[rule_format, OF mn m0(2)] obtain ps where ps: "primefact ps m" ..
1178     from ps m p(1) have "primefact (p#ps) n" by (simp add: primefact_def)
1179     hence ?ths by blast}
1180   ultimately show ?ths by blast
1181 qed
1183 lemma primefact_contains:
1184   assumes pf: "primefact ps n" and p: "prime p" and pn: "p dvd n"
1185   shows "p \<in> set ps"
1186   using pf p pn
1187 proof(induct ps arbitrary: p n)
1188   case Nil thus ?case by (auto simp add: primefact_def)
1189 next
1190   case (Cons q qs p n)
1191   from Cons.prems[unfolded primefact_def]
1192   have q: "prime q" "q * foldr op * qs 1 = n" "\<forall>p \<in>set qs. prime p"  and p: "prime p" "p dvd q * foldr op * qs 1" by simp_all
1193   {assume "p dvd q"
1194     with p(1) q(1) have "p = q" unfolding prime_def by auto
1195     hence ?case by simp}
1196   moreover
1197   { assume h: "p dvd foldr op * qs 1"
1198     from q(3) have pqs: "primefact qs (foldr op * qs 1)"
1199       by (simp add: primefact_def)
1200     from Cons.hyps[OF pqs p(1) h] have ?case by simp}
1201   ultimately show ?case using prime_divprod[OF p] by blast
1202 qed
1204 lemma primefact_variant: "primefact ps n \<longleftrightarrow> foldr op * ps 1 = n \<and> list_all prime ps" by (auto simp add: primefact_def list_all_iff)
1206 (* Variant of Lucas theorem.                                                 *)
1208 lemma lucas_primefact:
1209   assumes n: "n \<ge> 2" and an: "[a^(n - 1) = 1] (mod n)"
1210   and psn: "foldr op * ps 1 = n - 1"
1211   and psp: "list_all (\<lambda>p. prime p \<and> \<not> [a^((n - 1) div p) = 1] (mod n)) ps"
1212   shows "prime n"
1213 proof-
1214   {fix p assume p: "prime p" "p dvd n - 1" "[a ^ ((n - 1) div p) = 1] (mod n)"
1215     from psn psp have psn1: "primefact ps (n - 1)"
1216       by (auto simp add: list_all_iff primefact_variant)
1217     from p(3) primefact_contains[OF psn1 p(1,2)] psp
1218     have False by (induct ps, auto)}
1219   with lucas[OF n an] show ?thesis by blast
1220 qed
1222 (* Variant of Pocklington theorem.                                           *)
1224 lemma mod_le: assumes n: "n \<noteq> (0::nat)" shows "m mod n \<le> m"
1225 proof-
1226     from mod_div_equality[of m n]
1227     have "\<exists>x. x + m mod n = m" by blast
1228     then show ?thesis by auto
1229 qed
1232 lemma pocklington_primefact:
1233   assumes n: "n \<ge> 2" and qrn: "q*r = n - 1" and nq2: "n \<le> q^2"
1234   and arnb: "(a^r) mod n = b" and psq: "foldr op * ps 1 = q"
1235   and bqn: "(b^q) mod n = 1"
1236   and psp: "list_all (\<lambda>p. prime p \<and> coprime ((b^(q div p)) mod n - 1) n) ps"
1237   shows "prime n"
1238 proof-
1239   from bqn psp qrn
1240   have bqn: "a ^ (n - 1) mod n = 1"
1241     and psp: "list_all (\<lambda>p. prime p \<and> coprime (a^(r *(q div p)) mod n - 1) n) ps"  unfolding arnb[symmetric] power_mod
1242     by (simp_all add: power_mult[symmetric] algebra_simps)
1243   from n  have n0: "n > 0" by arith
1244   from mod_div_equality[of "a^(n - 1)" n]
1245     mod_less_divisor[OF n0, of "a^(n - 1)"]
1246   have an1: "[a ^ (n - 1) = 1] (mod n)"
1247     unfolding nat_mod bqn
1248     apply -
1249     apply (rule exI[where x="0"])
1250     apply (rule exI[where x="a^(n - 1) div n"])
1251     by (simp add: algebra_simps)
1252   {fix p assume p: "prime p" "p dvd q"
1253     from psp psq have pfpsq: "primefact ps q"
1254       by (auto simp add: primefact_variant list_all_iff)
1255     from psp primefact_contains[OF pfpsq p]
1256     have p': "coprime (a ^ (r * (q div p)) mod n - 1) n"
1257       by (simp add: list_all_iff)
1258     from prime_ge_2[OF p(1)] have p01: "p \<noteq> 0" "p \<noteq> 1" "p =Suc(p - 1)" by arith+
1259     from div_mult1_eq[of r q p] p(2)
1260     have eq1: "r* (q div p) = (n - 1) div p"
1261       unfolding qrn[symmetric] dvd_eq_mod_eq_0 by (simp add: mult_commute)
1262     have ath: "\<And>a (b::nat). a <= b \<Longrightarrow> a \<noteq> 0 ==> 1 <= a \<and> 1 <= b" by arith
1263     from n0 have n00: "n \<noteq> 0" by arith
1264     from mod_le[OF n00]
1265     have th10: "a ^ ((n - 1) div p) mod n \<le> a ^ ((n - 1) div p)" .
1266     {assume "a ^ ((n - 1) div p) mod n = 0"
1267       then obtain s where s: "a ^ ((n - 1) div p) = n*s"
1268 	unfolding mod_eq_0_iff by blast
1269       hence eq0: "(a^((n - 1) div p))^p = (n*s)^p" by simp
1270       from qrn[symmetric] have qn1: "q dvd n - 1" unfolding dvd_def by auto
1271       from dvd_trans[OF p(2) qn1] div_mod_equality'[of "n - 1" p]
1272       have npp: "(n - 1) div p * p = n - 1" by (simp add: dvd_eq_mod_eq_0)
1273       with eq0 have "a^ (n - 1) = (n*s)^p"
1274 	by (simp add: power_mult[symmetric])
1275       hence "1 = (n*s)^(Suc (p - 1)) mod n" using bqn p01 by simp
1276       also have "\<dots> = 0" by (simp add: mult_assoc)
1277       finally have False by simp }
1278       then have th11: "a ^ ((n - 1) div p) mod n \<noteq> 0" by auto
1279     have th1: "[a ^ ((n - 1) div p) mod n = a ^ ((n - 1) div p)] (mod n)"
1280       unfolding modeq_def by simp
1281     from cong_sub[OF th1 cong_refl[of 1]]  ath[OF th10 th11]
1282     have th: "[a ^ ((n - 1) div p) mod n - 1 = a ^ ((n - 1) div p) - 1] (mod n)"
1283       by blast
1284     from cong_coprime[OF th] p'[unfolded eq1]
1285     have "coprime (a ^ ((n - 1) div p) - 1) n" by (simp add: coprime_commute) }
1286   with pocklington[OF n qrn[symmetric] nq2 an1]
1287   show ?thesis by blast
1288 qed
1291 end