src/HOL/Old_Number_Theory/Pocklington.thy
 author huffman Sun Apr 01 16:09:58 2012 +0200 (2012-04-01) changeset 47255 30a1692557b0 parent 41541 1fa4725c4656 child 49962 a8cc904a6820 permissions -rw-r--r--
removed Nat_Numeral.thy, moving all theorems elsewhere
1 (*  Title:      HOL/Old_Number_Theory/Pocklington.thy
2     Author:     Amine Chaieb
3 *)
5 header {* Pocklington's Theorem for Primes *}
7 theory Pocklington
8 imports Primes
9 begin
11 definition modeq:: "nat => nat => nat => bool"    ("(1[_ = _] '(mod _'))")
12   where "[a = b] (mod p) == ((a mod p) = (b mod p))"
14 definition modneq:: "nat => nat => nat => bool"    ("(1[_ \<noteq> _] '(mod _'))")
15   where "[a \<noteq> b] (mod p) == ((a mod p) \<noteq> (b mod p))"
17 lemma modeq_trans:
18   "\<lbrakk> [a = b] (mod p); [b = c] (mod p) \<rbrakk> \<Longrightarrow> [a = c] (mod p)"
19   by (simp add:modeq_def)
22 lemma nat_mod_lemma: assumes xyn: "[x = y] (mod n)" and xy:"y \<le> x"
23   shows "\<exists>q. x = y + n * q"
24 using xyn xy unfolding modeq_def using nat_mod_eq_lemma by blast
26 lemma nat_mod[algebra]: "[x = y] (mod n) \<longleftrightarrow> (\<exists>q1 q2. x + n * q1 = y + n * q2)"
27 unfolding modeq_def nat_mod_eq_iff ..
29 (* Lemmas about previously defined terms.                                    *)
31 lemma prime: "prime p \<longleftrightarrow> p \<noteq> 0 \<and> p\<noteq>1 \<and> (\<forall>m. 0 < m \<and> m < p \<longrightarrow> coprime p m)"
32   (is "?lhs \<longleftrightarrow> ?rhs")
33 proof-
34   {assume "p=0 \<or> p=1" hence ?thesis using prime_0 prime_1 by (cases "p=0", simp_all)}
35   moreover
36   {assume p0: "p\<noteq>0" "p\<noteq>1"
37     {assume H: "?lhs"
38       {fix m assume m: "m > 0" "m < p"
39         {assume "m=1" hence "coprime p m" by simp}
40         moreover
41         {assume "p dvd m" hence "p \<le> m" using dvd_imp_le m by blast with m(2)
42           have "coprime p m" by simp}
43         ultimately have "coprime p m" using prime_coprime[OF H, of m] by blast}
44       hence ?rhs using p0 by auto}
45     moreover
46     { assume H: "\<forall>m. 0 < m \<and> m < p \<longrightarrow> coprime p m"
47       from prime_factor[OF p0(2)] obtain q where q: "prime q" "q dvd p" by blast
48       from prime_ge_2[OF q(1)] have q0: "q > 0" by arith
49       from dvd_imp_le[OF q(2)] p0 have qp: "q \<le> p" by arith
50       {assume "q = p" hence ?lhs using q(1) by blast}
51       moreover
52       {assume "q\<noteq>p" with qp have qplt: "q < p" by arith
53         from H[rule_format, of q] qplt q0 have "coprime p q" by arith
54         with coprime_prime[of p q q] q have False by simp hence ?lhs by blast}
55       ultimately have ?lhs by blast}
56     ultimately have ?thesis by blast}
57   ultimately show ?thesis  by (cases"p=0 \<or> p=1", auto)
58 qed
60 lemma finite_number_segment: "card { m. 0 < m \<and> m < n } = n - 1"
61 proof-
62   have "{ m. 0 < m \<and> m < n } = {1..<n}" by auto
63   thus ?thesis by simp
64 qed
66 lemma coprime_mod: assumes n: "n \<noteq> 0" shows "coprime (a mod n) n \<longleftrightarrow> coprime a n"
67   using n dvd_mod_iff[of _ n a] by (auto simp add: coprime)
69 (* Congruences.                                                              *)
71 lemma cong_mod_01[simp,presburger]:
72   "[x = y] (mod 0) \<longleftrightarrow> x = y" "[x = y] (mod 1)" "[x = 0] (mod n) \<longleftrightarrow> n dvd x"
73   by (simp_all add: modeq_def, presburger)
75 lemma cong_sub_cases:
76   "[x = y] (mod n) \<longleftrightarrow> (if x <= y then [y - x = 0] (mod n) else [x - y = 0] (mod n))"
77 apply (auto simp add: nat_mod)
78 apply (rule_tac x="q2" in exI)
79 apply (rule_tac x="q1" in exI, simp)
80 apply (rule_tac x="q2" in exI)
81 apply (rule_tac x="q1" in exI, simp)
82 apply (rule_tac x="q1" in exI)
83 apply (rule_tac x="q2" in exI, simp)
84 apply (rule_tac x="q1" in exI)
85 apply (rule_tac x="q2" in exI, simp)
86 done
88 lemma cong_mult_lcancel: assumes an: "coprime a n" and axy:"[a * x = a * y] (mod n)"
89   shows "[x = y] (mod n)"
90 proof-
91   {assume "a = 0" with an axy coprime_0'[of n] have ?thesis by (simp add: modeq_def) }
92   moreover
93   {assume az: "a\<noteq>0"
94     {assume xy: "x \<le> y" hence axy': "a*x \<le> a*y" by simp
95       with axy cong_sub_cases[of "a*x" "a*y" n]  have "[a*(y - x) = 0] (mod n)"
96         by (simp only: if_True diff_mult_distrib2)
97       hence th: "n dvd a*(y -x)" by simp
98       from coprime_divprod[OF th] an have "n dvd y - x"
99         by (simp add: coprime_commute)
100       hence ?thesis using xy cong_sub_cases[of x y n] by simp}
101     moreover
102     {assume H: "\<not>x \<le> y" hence xy: "y \<le> x"  by arith
103       from H az have axy': "\<not> a*x \<le> a*y" by auto
104       with axy H cong_sub_cases[of "a*x" "a*y" n]  have "[a*(x - y) = 0] (mod n)"
105         by (simp only: if_False diff_mult_distrib2)
106       hence th: "n dvd a*(x - y)" by simp
107       from coprime_divprod[OF th] an have "n dvd x - y"
108         by (simp add: coprime_commute)
109       hence ?thesis using xy cong_sub_cases[of x y n] by simp}
110     ultimately have ?thesis by blast}
111   ultimately show ?thesis by blast
112 qed
114 lemma cong_mult_rcancel: assumes an: "coprime a n" and axy:"[x*a = y*a] (mod n)"
115   shows "[x = y] (mod n)"
116   using cong_mult_lcancel[OF an axy[unfolded mult_commute[of _a]]] .
118 lemma cong_refl: "[x = x] (mod n)" by (simp add: modeq_def)
120 lemma eq_imp_cong: "a = b \<Longrightarrow> [a = b] (mod n)" by (simp add: cong_refl)
122 lemma cong_commute: "[x = y] (mod n) \<longleftrightarrow> [y = x] (mod n)"
123   by (auto simp add: modeq_def)
125 lemma cong_trans[trans]: "[x = y] (mod n) \<Longrightarrow> [y = z] (mod n) \<Longrightarrow> [x = z] (mod n)"
126   by (simp add: modeq_def)
128 lemma cong_add: assumes xx': "[x = x'] (mod n)" and yy':"[y = y'] (mod n)"
129   shows "[x + y = x' + y'] (mod n)"
130 proof-
131   have "(x + y) mod n = (x mod n + y mod n) mod n"
132     by (simp add: mod_add_left_eq[of x y n] mod_add_right_eq[of "x mod n" y n])
133   also have "\<dots> = (x' mod n + y' mod n) mod n" using xx' yy' modeq_def by simp
134   also have "\<dots> = (x' + y') mod n"
135     by (simp add: mod_add_left_eq[of x' y' n] mod_add_right_eq[of "x' mod n" y' n])
136   finally show ?thesis unfolding modeq_def .
137 qed
139 lemma cong_mult: assumes xx': "[x = x'] (mod n)" and yy':"[y = y'] (mod n)"
140   shows "[x * y = x' * y'] (mod n)"
141 proof-
142   have "(x * y) mod n = (x mod n) * (y mod n) mod n"
143     by (simp add: mod_mult_left_eq[of x y n] mod_mult_right_eq[of "x mod n" y n])
144   also have "\<dots> = (x' mod n) * (y' mod n) mod n" using xx'[unfolded modeq_def] yy'[unfolded modeq_def] by simp
145   also have "\<dots> = (x' * y') mod n"
146     by (simp add: mod_mult_left_eq[of x' y' n] mod_mult_right_eq[of "x' mod n" y' n])
147   finally show ?thesis unfolding modeq_def .
148 qed
150 lemma cong_exp: "[x = y] (mod n) \<Longrightarrow> [x^k = y^k] (mod n)"
151   by (induct k, auto simp add: cong_refl cong_mult)
152 lemma cong_sub: assumes xx': "[x = x'] (mod n)" and yy': "[y = y'] (mod n)"
153   and yx: "y <= x" and yx': "y' <= x'"
154   shows "[x - y = x' - y'] (mod n)"
155 proof-
156   { fix x a x' a' y b y' b'
157     have "(x::nat) + a = x' + a' \<Longrightarrow> y + b = y' + b' \<Longrightarrow> y <= x \<Longrightarrow> y' <= x'
158       \<Longrightarrow> (x - y) + (a + b') = (x' - y') + (a' + b)" by arith}
159   note th = this
160   from xx' yy' obtain q1 q2 q1' q2' where q12: "x + n*q1 = x'+n*q2"
161     and q12': "y + n*q1' = y'+n*q2'" unfolding nat_mod by blast+
162   from th[OF q12 q12' yx yx']
163   have "(x - y) + n*(q1 + q2') = (x' - y') + n*(q2 + q1')"
164     by (simp add: right_distrib)
165   thus ?thesis unfolding nat_mod by blast
166 qed
168 lemma cong_mult_lcancel_eq: assumes an: "coprime a n"
169   shows "[a * x = a * y] (mod n) \<longleftrightarrow> [x = y] (mod n)" (is "?lhs \<longleftrightarrow> ?rhs")
170 proof
171   assume H: "?rhs" from cong_mult[OF cong_refl[of a n] H] show ?lhs .
172 next
173   assume H: "?lhs" hence H': "[x*a = y*a] (mod n)" by (simp add: mult_commute)
174   from cong_mult_rcancel[OF an H'] show ?rhs  .
175 qed
177 lemma cong_mult_rcancel_eq: assumes an: "coprime a n"
178   shows "[x * a = y * a] (mod n) \<longleftrightarrow> [x = y] (mod n)"
179 using cong_mult_lcancel_eq[OF an, of x y] by (simp add: mult_commute)
181 lemma cong_add_lcancel_eq: "[a + x = a + y] (mod n) \<longleftrightarrow> [x = y] (mod n)"
182   by (simp add: nat_mod)
184 lemma cong_add_rcancel_eq: "[x + a = y + a] (mod n) \<longleftrightarrow> [x = y] (mod n)"
185   by (simp add: nat_mod)
187 lemma cong_add_rcancel: "[x + a = y + a] (mod n) \<Longrightarrow> [x = y] (mod n)"
188   by (simp add: nat_mod)
190 lemma cong_add_lcancel: "[a + x = a + y] (mod n) \<Longrightarrow> [x = y] (mod n)"
191   by (simp add: nat_mod)
193 lemma cong_add_lcancel_eq_0: "[a + x = a] (mod n) \<longleftrightarrow> [x = 0] (mod n)"
194   by (simp add: nat_mod)
196 lemma cong_add_rcancel_eq_0: "[x + a = a] (mod n) \<longleftrightarrow> [x = 0] (mod n)"
197   by (simp add: nat_mod)
199 lemma cong_imp_eq: assumes xn: "x < n" and yn: "y < n" and xy: "[x = y] (mod n)"
200   shows "x = y"
201   using xy[unfolded modeq_def mod_less[OF xn] mod_less[OF yn]] .
203 lemma cong_divides_modulus: "[x = y] (mod m) \<Longrightarrow> n dvd m ==> [x = y] (mod n)"
204   apply (auto simp add: nat_mod dvd_def)
205   apply (rule_tac x="k*q1" in exI)
206   apply (rule_tac x="k*q2" in exI)
207   by simp
209 lemma cong_0_divides: "[x = 0] (mod n) \<longleftrightarrow> n dvd x" by simp
211 lemma cong_1_divides:"[x = 1] (mod n) ==> n dvd x - 1"
212   apply (cases "x\<le>1", simp_all)
213   using cong_sub_cases[of x 1 n] by auto
215 lemma cong_divides: "[x = y] (mod n) \<Longrightarrow> n dvd x \<longleftrightarrow> n dvd y"
216 apply (auto simp add: nat_mod dvd_def)
217 apply (rule_tac x="k + q1 - q2" in exI, simp add: add_mult_distrib2 diff_mult_distrib2)
218 apply (rule_tac x="k + q2 - q1" in exI, simp add: add_mult_distrib2 diff_mult_distrib2)
219 done
221 lemma cong_coprime: assumes xy: "[x = y] (mod n)"
222   shows "coprime n x \<longleftrightarrow> coprime n y"
223 proof-
224   {assume "n=0" hence ?thesis using xy by simp}
225   moreover
226   {assume nz: "n \<noteq> 0"
227   have "coprime n x \<longleftrightarrow> coprime (x mod n) n"
228     by (simp add: coprime_mod[OF nz, of x] coprime_commute[of n x])
229   also have "\<dots> \<longleftrightarrow> coprime (y mod n) n" using xy[unfolded modeq_def] by simp
230   also have "\<dots> \<longleftrightarrow> coprime y n" by (simp add: coprime_mod[OF nz, of y])
231   finally have ?thesis by (simp add: coprime_commute) }
232 ultimately show ?thesis by blast
233 qed
235 lemma cong_mod: "~(n = 0) \<Longrightarrow> [a mod n = a] (mod n)" by (simp add: modeq_def)
237 lemma mod_mult_cong: "~(a = 0) \<Longrightarrow> ~(b = 0)
238   \<Longrightarrow> [x mod (a * b) = y] (mod a) \<longleftrightarrow> [x = y] (mod a)"
239   by (simp add: modeq_def mod_mult2_eq mod_add_left_eq)
241 lemma cong_mod_mult: "[x = y] (mod n) \<Longrightarrow> m dvd n \<Longrightarrow> [x = y] (mod m)"
242   apply (auto simp add: nat_mod dvd_def)
243   apply (rule_tac x="k*q1" in exI)
244   apply (rule_tac x="k*q2" in exI, simp)
245   done
247 (* Some things when we know more about the order.                            *)
249 lemma cong_le: "y <= x \<Longrightarrow> [x = y] (mod n) \<longleftrightarrow> (\<exists>q. x = q * n + y)"
250   using nat_mod_lemma[of x y n]
251   apply auto
252   apply (simp add: nat_mod)
253   apply (rule_tac x="q" in exI)
254   apply (rule_tac x="q + q" in exI)
255   by (auto simp: algebra_simps)
257 lemma cong_to_1: "[a = 1] (mod n) \<longleftrightarrow> a = 0 \<and> n = 1 \<or> (\<exists>m. a = 1 + m * n)"
258 proof-
259   {assume "n = 0 \<or> n = 1\<or> a = 0 \<or> a = 1" hence ?thesis
260       apply (cases "n=0", simp_all add: cong_commute)
261       apply (cases "n=1", simp_all add: cong_commute modeq_def)
262       apply arith
263       apply (cases "a=1")
264       apply (simp_all add: modeq_def cong_commute)
265       done }
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 }"
388 lemma phi_0[simp]: "\<phi> 0 = 0"
389   unfolding phi_def by auto
391 lemma phi_finite[simp]: "finite ({ m. 0 < m \<and> m <= n \<and> coprime m n })"
392 proof-
393   have "{ m. 0 < m \<and> m <= n \<and> coprime m n } \<subseteq> {0..n}" by auto
394   thus ?thesis by (auto intro: finite_subset)
395 qed
397 declare coprime_1[presburger]
398 lemma phi_1[simp]: "\<phi> 1 = 1"
399 proof-
400   {fix m
401     have "0 < m \<and> m <= 1 \<and> coprime m 1 \<longleftrightarrow> m = 1" by presburger }
402   thus ?thesis by (simp add: phi_def)
403 qed
405 lemma [simp]: "\<phi> (Suc 0) = Suc 0" using phi_1 by simp
407 lemma phi_alt: "\<phi>(n) = card { m. coprime m n \<and> m < n}"
408 proof-
409   {assume "n=0 \<or> n=1" hence ?thesis by (cases "n=0", simp_all)}
410   moreover
411   {assume n: "n\<noteq>0" "n\<noteq>1"
412     {fix m
413       from n have "0 < m \<and> m <= n \<and> coprime m n \<longleftrightarrow> coprime m n \<and> m < n"
414         apply (cases "m = 0", simp_all)
415         apply (cases "m = 1", simp_all)
416         apply (cases "m = n", auto)
417         done }
418     hence ?thesis unfolding phi_def by simp}
419   ultimately show ?thesis by auto
420 qed
422 lemma phi_finite_lemma[simp]: "finite {m. coprime m n \<and>  m < n}" (is "finite ?S")
423   by (rule finite_subset[of "?S" "{0..n}"], auto)
425 lemma phi_another: assumes n: "n\<noteq>1"
426   shows "\<phi> n = card {m. 0 < m \<and> m < n \<and> coprime m n }"
427 proof-
428   {fix m
429     from n have "0 < m \<and> m < n \<and> coprime m n \<longleftrightarrow> coprime m n \<and> m < n"
430       by (cases "m=0", auto)}
431   thus ?thesis unfolding phi_alt by auto
432 qed
434 lemma phi_limit: "\<phi> n \<le> n"
435 proof-
436   have "{ m. coprime m n \<and> m < n} \<subseteq> {0 ..<n}" by auto
437   with card_mono[of "{0 ..<n}" "{ m. coprime m n \<and> m < n}"]
438   show ?thesis unfolding phi_alt by auto
439 qed
441 lemma stupid[simp]: "{m. (0::nat) < m \<and> m < n} = {1..<n}"
442   by auto
444 lemma phi_limit_strong: assumes n: "n\<noteq>1"
445   shows "\<phi>(n) \<le> n - 1"
446 proof-
447   show ?thesis
448     unfolding phi_another[OF n] finite_number_segment[of n, symmetric]
449     by (rule card_mono[of "{m. 0 < m \<and> m < n}" "{m. 0 < m \<and> m < n \<and> coprime m n}"], auto)
450 qed
452 lemma phi_lowerbound_1_strong: assumes n: "n \<ge> 1"
453   shows "\<phi>(n) \<ge> 1"
454 proof-
455   let ?S = "{ m. 0 < m \<and> m <= n \<and> coprime m n }"
456   from card_0_eq[of ?S] n have "\<phi> n \<noteq> 0" unfolding phi_alt
457     apply auto
458     apply (cases "n=1", simp_all)
459     apply (rule exI[where x=1], simp)
460     done
461   thus ?thesis by arith
462 qed
464 lemma phi_lowerbound_1: "2 <= n ==> 1 <= \<phi>(n)"
465   using phi_lowerbound_1_strong[of n] by auto
467 lemma phi_lowerbound_2: assumes n: "3 <= n" shows "2 <= \<phi> (n)"
468 proof-
469   let ?S = "{ m. 0 < m \<and> m <= n \<and> coprime m n }"
470   have inS: "{1, n - 1} \<subseteq> ?S" using n coprime_plus1[of "n - 1"]
471     by (auto simp add: coprime_commute)
472   from n have c2: "card {1, n - 1} = 2" by (auto simp add: card_insert_if)
473   from card_mono[of ?S "{1, n - 1}", simplified inS c2] show ?thesis
474     unfolding phi_def by auto
475 qed
477 lemma phi_prime: "\<phi> n = n - 1 \<and> n\<noteq>0 \<and> n\<noteq>1 \<longleftrightarrow> prime n"
478 proof-
479   {assume "n=0 \<or> n=1" hence ?thesis by (cases "n=1", simp_all)}
480   moreover
481   {assume n: "n\<noteq>0" "n\<noteq>1"
482     let ?S = "{m. 0 < m \<and> m < n}"
483     have fS: "finite ?S" by simp
484     let ?S' = "{m. 0 < m \<and> m < n \<and> coprime m n}"
485     have fS':"finite ?S'" apply (rule finite_subset[of ?S' ?S]) by auto
486     {assume H: "\<phi> n = n - 1 \<and> n\<noteq>0 \<and> n\<noteq>1"
487       hence ceq: "card ?S' = card ?S"
488       using n finite_number_segment[of n] phi_another[OF n(2)] by simp
489       {fix m assume m: "0 < m" "m < n" "\<not> coprime m n"
490         hence mS': "m \<notin> ?S'" by auto
491         have "insert m ?S' \<le> ?S" using m by auto
492         from m have "card (insert m ?S') \<le> card ?S"
493           by - (rule card_mono[of ?S "insert m ?S'"], auto)
494         hence False
495           unfolding card_insert_disjoint[of "?S'" m, OF fS' mS'] ceq
496           by simp }
497       hence "\<forall>m. 0 <m \<and> m < n \<longrightarrow> coprime m n" by blast
498       hence "prime n" unfolding prime using n by (simp add: coprime_commute)}
499     moreover
500     {assume H: "prime n"
501       hence "?S = ?S'" unfolding prime using n
502         by (auto simp add: coprime_commute)
503       hence "card ?S = card ?S'" by simp
504       hence "\<phi> n = n - 1" unfolding phi_another[OF n(2)] by simp}
505     ultimately have ?thesis using n by blast}
506   ultimately show ?thesis by (cases "n=0") blast+
507 qed
509 (* Multiplicativity property.                                                *)
511 lemma phi_multiplicative: assumes ab: "coprime a b"
512   shows "\<phi> (a * b) = \<phi> a * \<phi> b"
513 proof-
514   {assume "a = 0 \<or> b = 0 \<or> a = 1 \<or> b = 1"
515     hence ?thesis
516       by (cases "a=0", simp, cases "b=0", simp, cases"a=1", simp_all) }
517   moreover
518   {assume a: "a\<noteq>0" "a\<noteq>1" and b: "b\<noteq>0" "b\<noteq>1"
519     hence ab0: "a*b \<noteq> 0" by simp
520     let ?S = "\<lambda>k. {m. coprime m k \<and> m < k}"
521     let ?f = "\<lambda>x. (x mod a, x mod b)"
522     have eq: "?f ` (?S (a*b)) = (?S a \<times> ?S b)"
523     proof-
524       {fix x assume x:"x \<in> ?S (a*b)"
525         hence x': "coprime x (a*b)" "x < a*b" by simp_all
526         hence xab: "coprime x a" "coprime x b" by (simp_all add: coprime_mul_eq)
527         from mod_less_divisor a b have xab':"x mod a < a" "x mod b < b" by auto
528         from xab xab' have "?f x \<in> (?S a \<times> ?S b)"
529           by (simp add: coprime_mod[OF a(1)] coprime_mod[OF b(1)])}
530       moreover
531       {fix x y assume x: "x \<in> ?S a" and y: "y \<in> ?S b"
532         hence x': "coprime x a" "x < a" and y': "coprime y b" "y < b" by simp_all
533         from chinese_remainder_coprime_unique[OF ab a(1) b(1) x'(1) y'(1)]
534         obtain z where z: "coprime z (a * b)" "z < a * b" "[z = x] (mod a)"
535           "[z = y] (mod b)" by blast
536         hence "(x,y) \<in> ?f ` (?S (a*b))"
537           using y'(2) mod_less_divisor[of b y] x'(2) mod_less_divisor[of a x]
538           by (auto simp add: image_iff modeq_def)}
539       ultimately show ?thesis by auto
540     qed
541     have finj: "inj_on ?f (?S (a*b))"
542       unfolding inj_on_def
543     proof(clarify)
544       fix x y assume H: "coprime x (a * b)" "x < a * b" "coprime y (a * b)"
545         "y < a * b" "x mod a = y mod a" "x mod b = y mod b"
546       hence cp: "coprime x a" "coprime x b" "coprime y a" "coprime y b"
547         by (simp_all add: coprime_mul_eq)
548       from chinese_remainder_coprime_unique[OF ab a(1) b(1) cp(3,4)] H
549       show "x = y" unfolding modeq_def by blast
550     qed
551     from card_image[OF finj, unfolded eq] have ?thesis
552       unfolding phi_alt by simp }
553   ultimately show ?thesis by auto
554 qed
556 (* Fermat's Little theorem / Fermat-Euler theorem.                           *)
559 lemma nproduct_mod:
560   assumes fS: "finite S" and n0: "n \<noteq> 0"
561   shows "[setprod (\<lambda>m. a(m) mod n) S = setprod a S] (mod n)"
562 proof-
563   have th1:"[1 = 1] (mod n)" by (simp add: modeq_def)
564   from cong_mult
565   have th3:"\<forall>x1 y1 x2 y2.
566     [x1 = x2] (mod n) \<and> [y1 = y2] (mod n) \<longrightarrow> [x1 * y1 = x2 * y2] (mod n)"
567     by blast
568   have th4:"\<forall>x\<in>S. [a x mod n = a x] (mod n)" by (simp add: modeq_def)
569   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)
570 qed
572 lemma nproduct_cmul:
573   assumes fS:"finite S"
574   shows "setprod (\<lambda>m. (c::'a::{comm_monoid_mult})* a(m)) S = c ^ (card S) * setprod a S"
575 unfolding setprod_timesf setprod_constant[OF fS, of c] ..
577 lemma coprime_nproduct:
578   assumes fS: "finite S" and Sn: "\<forall>x\<in>S. coprime n (a x)"
579   shows "coprime n (setprod a S)"
580   using fS unfolding setprod_def by (rule finite_subset_induct)
581     (insert Sn, auto simp add: coprime_mul)
583 lemma fermat_little: assumes an: "coprime a n"
584   shows "[a ^ (\<phi> n) = 1] (mod n)"
585 proof-
586   {assume "n=0" hence ?thesis by simp}
587   moreover
588   {assume "n=1" hence ?thesis by (simp add: modeq_def)}
589   moreover
590   {assume nz: "n \<noteq> 0" and n1: "n \<noteq> 1"
591     let ?S = "{m. coprime m n \<and> m < n}"
592     let ?P = "\<Prod> ?S"
593     have fS: "finite ?S" by simp
594     have cardfS: "\<phi> n = card ?S" unfolding phi_alt ..
595     {fix m assume m: "m \<in> ?S"
596       hence "coprime m n" by simp
597       with coprime_mul[of n a m] an have "coprime (a*m) n"
598         by (simp add: coprime_commute)}
599     hence Sn: "\<forall>m\<in> ?S. coprime (a*m) n " by blast
600     from coprime_nproduct[OF fS, of n "\<lambda>m. m"] have nP:"coprime ?P n"
601       by (simp add: coprime_commute)
602     have Paphi: "[?P*a^ (\<phi> n) = ?P*1] (mod n)"
603     proof-
604       let ?h = "\<lambda>m. m mod n"
605       {fix m assume mS: "m\<in> ?S"
606         hence "?h m \<in> ?S" by simp}
607       hence hS: "?h ` ?S = ?S"by (auto simp add: image_iff)
608       have "a\<noteq>0" using an n1 nz apply- apply (rule ccontr) by simp
609       hence inj: "inj_on (op * a) ?S" unfolding inj_on_def by simp
611       have eq0: "fold_image op * (?h \<circ> op * a) 1 {m. coprime m n \<and> m < n} =
612      fold_image op * (\<lambda>m. m) 1 {m. coprime m n \<and> m < n}"
613       proof (rule fold_image_eq_general[where h="?h o (op * a)"])
614         show "finite ?S" using fS .
615       next
616         {fix y assume yS: "y \<in> ?S" hence y: "coprime y n" "y < n" by simp_all
617           from cong_solve_unique[OF an nz, of y]
618           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
619           from cong_coprime[OF x(2)] y(1)
620           have xm: "coprime x n" by (simp add: coprime_mul_eq coprime_commute)
621           {fix z assume "z \<in> ?S" "(?h \<circ> op * a) z = y"
622             hence z: "coprime z n" "z < n" "(?h \<circ> op * a) z = y" by simp_all
623             from x(3)[rule_format, of z] z(2,3) have "z=x"
624               unfolding modeq_def mod_less[OF y(2)] by simp}
625           with xm x(1,2) have "\<exists>!x. x \<in> ?S \<and> (?h \<circ> op * a) x = y"
626             unfolding modeq_def mod_less[OF y(2)] by auto }
627         thus "\<forall>y\<in>{m. coprime m n \<and> m < n}.
628        \<exists>!x. x \<in> {m. coprime m n \<and> m < n} \<and> ((\<lambda>m. m mod n) \<circ> op * a) x = y" by blast
629       next
630         {fix x assume xS: "x\<in> ?S"
631           hence x: "coprime x n" "x < n" by simp_all
632           with an have "coprime (a*x) n"
633             by (simp add: coprime_mul_eq[of n a x] coprime_commute)
634           hence "?h (a*x) \<in> ?S" using nz
635             by (simp add: coprime_mod[OF nz])}
636         thus " \<forall>x\<in>{m. coprime m n \<and> m < n}.
637        ((\<lambda>m. m mod n) \<circ> op * a) x \<in> {m. coprime m n \<and> m < n} \<and>
638        ((\<lambda>m. m mod n) \<circ> op * a) x = ((\<lambda>m. m mod n) \<circ> op * a) x" by simp
639       qed
640       from nproduct_mod[OF fS nz, of "op * a"]
641       have "[(setprod (op *a) ?S) = (setprod (?h o (op * a)) ?S)] (mod n)"
642         unfolding o_def
643         by (simp add: cong_commute)
644       also have "[setprod (?h o (op * a)) ?S = ?P ] (mod n)"
645         using eq0 fS an by (simp add: setprod_def modeq_def o_def)
646       finally show "[?P*a^ (\<phi> n) = ?P*1] (mod n)"
647         unfolding cardfS mult_commute[of ?P "a^ (card ?S)"]
648           nproduct_cmul[OF fS, symmetric] mult_1_right by simp
649     qed
650     from cong_mult_lcancel[OF nP Paphi] have ?thesis . }
651   ultimately show ?thesis by blast
652 qed
654 lemma fermat_little_prime: assumes p: "prime p" and ap: "coprime a p"
655   shows "[a^ (p - 1) = 1] (mod p)"
656   using fermat_little[OF ap] p[unfolded phi_prime[symmetric]]
657 by simp
660 (* Lucas's theorem.                                                          *)
662 lemma lucas_coprime_lemma:
663   assumes m: "m\<noteq>0" and am: "[a^m = 1] (mod n)"
664   shows "coprime a n"
665 proof-
666   {assume "n=1" hence ?thesis by simp}
667   moreover
668   {assume "n = 0" hence ?thesis using am m exp_eq_1[of a m] by simp}
669   moreover
670   {assume n: "n\<noteq>0" "n\<noteq>1"
671     from m obtain m' where m': "m = Suc m'" by (cases m, blast+)
672     {fix d
673       assume d: "d dvd a" "d dvd n"
674       from n have n1: "1 < n" by arith
675       from am mod_less[OF n1] have am1: "a^m mod n = 1" unfolding modeq_def by simp
676       from dvd_mult2[OF d(1), of "a^m'"] have dam:"d dvd a^m" by (simp add: m')
677       from dvd_mod_iff[OF d(2), of "a^m"] dam am1
678       have "d = 1" by simp }
679     hence ?thesis unfolding coprime by auto
680   }
681   ultimately show ?thesis by blast
682 qed
684 lemma lucas_weak:
685   assumes n: "n \<ge> 2" and an:"[a^(n - 1) = 1] (mod n)"
686   and nm: "\<forall>m. 0 <m \<and> m < n - 1 \<longrightarrow> \<not> [a^m = 1] (mod n)"
687   shows "prime n"
688 proof-
689   from n have n1: "n \<noteq> 1" "n\<noteq>0" "n - 1 \<noteq> 0" "n - 1 > 0" "n - 1 < n" by arith+
690   from lucas_coprime_lemma[OF n1(3) an] have can: "coprime a n" .
691   from fermat_little[OF can] have afn: "[a ^ \<phi> n = 1] (mod n)" .
692   {assume "\<phi> n \<noteq> n - 1"
693     with phi_limit_strong[OF n1(1)] phi_lowerbound_1[OF n]
694     have c:"\<phi> n > 0 \<and> \<phi> n < n - 1" by arith
695     from nm[rule_format, OF c] afn have False ..}
696   hence "\<phi> n = n - 1" by blast
697   with phi_prime[of n] n1(1,2) show ?thesis by simp
698 qed
700 lemma nat_exists_least_iff: "(\<exists>(n::nat). P n) \<longleftrightarrow> (\<exists>n. P n \<and> (\<forall>m < n. \<not> P m))"
701   (is "?lhs \<longleftrightarrow> ?rhs")
702 proof
703   assume ?rhs thus ?lhs by blast
704 next
705   assume H: ?lhs then obtain n where n: "P n" by blast
706   let ?x = "Least P"
707   {fix m assume m: "m < ?x"
708     from not_less_Least[OF m] have "\<not> P m" .}
709   with LeastI_ex[OF H] show ?rhs by blast
710 qed
712 lemma nat_exists_least_iff': "(\<exists>(n::nat). P n) \<longleftrightarrow> (P (Least P) \<and> (\<forall>m < (Least P). \<not> P m))"
713   (is "?lhs \<longleftrightarrow> ?rhs")
714 proof-
715   {assume ?rhs hence ?lhs by blast}
716   moreover
717   { assume H: ?lhs then obtain n where n: "P n" by blast
718     let ?x = "Least P"
719     {fix m assume m: "m < ?x"
720       from not_less_Least[OF m] have "\<not> P m" .}
721     with LeastI_ex[OF H] have ?rhs by blast}
722   ultimately show ?thesis by blast
723 qed
725 lemma power_mod: "((x::nat) mod m)^n mod m = x^n mod m"
726 proof(induct n)
727   case 0 thus ?case by simp
728 next
729   case (Suc n)
730   have "(x mod m)^(Suc n) mod m = ((x mod m) * (((x mod m) ^ n) mod m)) mod m"
731     by (simp add: mod_mult_right_eq[symmetric])
732   also have "\<dots> = ((x mod m) * (x^n mod m)) mod m" using Suc.hyps by simp
733   also have "\<dots> = x^(Suc n) mod m"
734     by (simp add: mod_mult_left_eq[symmetric] mod_mult_right_eq[symmetric])
735   finally show ?case .
736 qed
738 lemma lucas:
739   assumes n2: "n \<ge> 2" and an1: "[a^(n - 1) = 1] (mod n)"
740   and pn: "\<forall>p. prime p \<and> p dvd n - 1 \<longrightarrow> \<not> [a^((n - 1) div p) = 1] (mod n)"
741   shows "prime n"
742 proof-
743   from n2 have n01: "n\<noteq>0" "n\<noteq>1" "n - 1 \<noteq> 0" by arith+
744   from mod_less_divisor[of n 1] n01 have onen: "1 mod n = 1" by simp
745   from lucas_coprime_lemma[OF n01(3) an1] cong_coprime[OF an1]
746   have an: "coprime a n" "coprime (a^(n - 1)) n" by (simp_all add: coprime_commute)
747   {assume H0: "\<exists>m. 0 < m \<and> m < n - 1 \<and> [a ^ m = 1] (mod n)" (is "EX m. ?P m")
748     from H0[unfolded nat_exists_least_iff[of ?P]] obtain m where
749       m: "0 < m" "m < n - 1" "[a ^ m = 1] (mod n)" "\<forall>k <m. \<not>?P k" by blast
750     {assume nm1: "(n - 1) mod m > 0"
751       from mod_less_divisor[OF m(1)] have th0:"(n - 1) mod m < m" by blast
752       let ?y = "a^ ((n - 1) div m * m)"
753       note mdeq = mod_div_equality[of "(n - 1)" m]
754       from coprime_exp[OF an(1)[unfolded coprime_commute[of a n]],
755         of "(n - 1) div m * m"]
756       have yn: "coprime ?y n" by (simp add: coprime_commute)
757       have "?y mod n = (a^m)^((n - 1) div m) mod n"
758         by (simp add: algebra_simps power_mult)
759       also have "\<dots> = (a^m mod n)^((n - 1) div m) mod n"
760         using power_mod[of "a^m" n "(n - 1) div m"] by simp
761       also have "\<dots> = 1" using m(3)[unfolded modeq_def onen] onen
762         by (simp add: power_Suc0)
763       finally have th3: "?y mod n = 1"  .
764       have th2: "[?y * a ^ ((n - 1) mod m) = ?y* 1] (mod n)"
765         using an1[unfolded modeq_def onen] onen
766           mod_div_equality[of "(n - 1)" m, symmetric]
767         by (simp add:power_add[symmetric] modeq_def th3 del: One_nat_def)
768       from cong_mult_lcancel[of ?y n "a^((n - 1) mod m)" 1, OF yn th2]
769       have th1: "[a ^ ((n - 1) mod m) = 1] (mod n)"  .
770       from m(4)[rule_format, OF th0] nm1
771         less_trans[OF mod_less_divisor[OF m(1), of "n - 1"] m(2)] th1
772       have False by blast }
773     hence "(n - 1) mod m = 0" by auto
774     then have mn: "m dvd n - 1" by presburger
775     then obtain r where r: "n - 1 = m*r" unfolding dvd_def by blast
776     from n01 r m(2) have r01: "r\<noteq>0" "r\<noteq>1" by - (rule ccontr, simp)+
777     from prime_factor[OF r01(2)] obtain p where p: "prime p" "p dvd r" by blast
778     hence th: "prime p \<and> p dvd n - 1" unfolding r by (auto intro: dvd_mult)
779     have "(a ^ ((n - 1) div p)) mod n = (a^(m*r div p)) mod n" using r
780       by (simp add: power_mult)
781     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
782     also have "\<dots> = ((a^m)^(r div p)) mod n" by (simp add: power_mult)
783     also have "\<dots> = ((a^m mod n)^(r div p)) mod n" using power_mod[of "a^m" "n" "r div p" ] ..
784     also have "\<dots> = 1" using m(3) onen by (simp add: modeq_def power_Suc0)
785     finally have "[(a ^ ((n - 1) div p))= 1] (mod n)"
786       using onen by (simp add: modeq_def)
787     with pn[rule_format, OF th] have False by blast}
788   hence th: "\<forall>m. 0 < m \<and> m < n - 1 \<longrightarrow> \<not> [a ^ m = 1] (mod n)" by blast
789   from lucas_weak[OF n2 an1 th] show ?thesis .
790 qed
792 (* Definition of the order of a number mod n (0 in non-coprime case).        *)
794 definition "ord n a = (if coprime n a then Least (\<lambda>d. d > 0 \<and> [a ^d = 1] (mod n)) else 0)"
796 (* This has the expected properties.                                         *)
798 lemma coprime_ord:
799   assumes na: "coprime n a"
800   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))"
801 proof-
802   let ?P = "\<lambda>d. 0 < d \<and> [a ^ d = 1] (mod n)"
803   from euclid[of a] obtain p where p: "prime p" "a < p" by blast
804   from na have o: "ord n a = Least ?P" by (simp add: ord_def)
805   {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)}
806   moreover
807   {assume "n\<noteq>0 \<and> n\<noteq>1" hence n2:"n \<ge> 2" by arith
808     from na have na': "coprime a n" by (simp add: coprime_commute)
809     from phi_lowerbound_1[OF n2] fermat_little[OF na']
810     have ex: "\<exists>m>0. ?P m" by - (rule exI[where x="\<phi> n"], auto) }
811   ultimately have ex: "\<exists>m>0. ?P m" by blast
812   from nat_exists_least_iff'[of ?P] ex na show ?thesis
813     unfolding o[symmetric] by auto
814 qed
815 (* With the special value 0 for non-coprime case, it's more convenient.      *)
816 lemma ord_works:
817  "[a ^ (ord n a) = 1] (mod n) \<and> (\<forall>m. 0 < m \<and> m < ord n a \<longrightarrow> ~[a^ m = 1] (mod n))"
818 apply (cases "coprime n a")
819 using coprime_ord[of n a]
820 by (blast, simp add: ord_def modeq_def)
822 lemma ord: "[a^(ord n a) = 1] (mod n)" using ord_works by blast
823 lemma ord_minimal: "0 < m \<Longrightarrow> m < ord n a \<Longrightarrow> ~[a^m = 1] (mod n)"
824   using ord_works by blast
825 lemma ord_eq_0: "ord n a = 0 \<longleftrightarrow> ~coprime n a"
826 by (cases "coprime n a", simp add: coprime_ord, simp add: ord_def)
828 lemma ord_divides:
829  "[a ^ d = 1] (mod n) \<longleftrightarrow> ord n a dvd d" (is "?lhs \<longleftrightarrow> ?rhs")
830 proof
831   assume rh: ?rhs
832   then obtain k where "d = ord n a * k" unfolding dvd_def by blast
833   hence "[a ^ d = (a ^ (ord n a) mod n)^k] (mod n)"
834     by (simp add : modeq_def power_mult power_mod)
835   also have "[(a ^ (ord n a) mod n)^k = 1] (mod n)"
836     using ord[of a n, unfolded modeq_def]
837     by (simp add: modeq_def power_mod power_Suc0)
838   finally  show ?lhs .
839 next
840   assume lh: ?lhs
841   { assume H: "\<not> coprime n a"
842     hence o: "ord n a = 0" by (simp add: ord_def)
843     {assume d: "d=0" with o H have ?rhs by (simp add: modeq_def)}
844     moreover
845     {assume d0: "d\<noteq>0" then obtain d' where d': "d = Suc d'" by (cases d, auto)
846       from H[unfolded coprime]
847       obtain p where p: "p dvd n" "p dvd a" "p \<noteq> 1" by auto
848       from lh[unfolded nat_mod]
849       obtain q1 q2 where q12:"a ^ d + n * q1 = 1 + n * q2" by blast
850       hence "a ^ d + n * q1 - n * q2 = 1" by simp
851       with dvd_diff_nat [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
852       with p(3) have False by simp
853       hence ?rhs ..}
854     ultimately have ?rhs by blast}
855   moreover
856   {assume H: "coprime n a"
857     let ?o = "ord n a"
858     let ?q = "d div ord n a"
859     let ?r = "d mod ord n a"
860     from cong_exp[OF ord[of a n], of ?q]
861     have eqo: "[(a^?o)^?q = 1] (mod n)"  by (simp add: modeq_def power_Suc0)
862     from H have onz: "?o \<noteq> 0" by (simp add: ord_eq_0)
863     hence op: "?o > 0" by simp
864     from mod_div_equality[of d "ord n a"] lh
865     have "[a^(?o*?q + ?r) = 1] (mod n)" by (simp add: modeq_def mult_commute)
866     hence "[(a^?o)^?q * (a^?r) = 1] (mod n)"
867       by (simp add: modeq_def power_mult[symmetric] power_add[symmetric])
868     hence th: "[a^?r = 1] (mod n)"
869       using eqo mod_mult_left_eq[of "(a^?o)^?q" "a^?r" n]
870       apply (simp add: modeq_def del: One_nat_def)
871       by (simp add: mod_mult_left_eq[symmetric])
872     {assume r: "?r = 0" hence ?rhs by (simp add: dvd_eq_mod_eq_0)}
873     moreover
874     {assume r: "?r \<noteq> 0"
875       with mod_less_divisor[OF op, of d] have r0o:"?r >0 \<and> ?r < ?o" by simp
876       from conjunct2[OF ord_works[of a n], rule_format, OF r0o] th
877       have ?rhs by blast}
878     ultimately have ?rhs by blast}
879   ultimately  show ?rhs by blast
880 qed
882 lemma order_divides_phi: "coprime n a \<Longrightarrow> ord n a dvd \<phi> n"
883 using ord_divides fermat_little coprime_commute by simp
884 lemma order_divides_expdiff:
885   assumes na: "coprime n a"
886   shows "[a^d = a^e] (mod n) \<longleftrightarrow> [d = e] (mod (ord n a))"
887 proof-
888   {fix n a d e
889     assume na: "coprime n a" and ed: "(e::nat) \<le> d"
890     hence "\<exists>c. d = e + c" by arith
891     then obtain c where c: "d = e + c" by arith
892     from na have an: "coprime a n" by (simp add: coprime_commute)
893     from coprime_exp[OF na, of e]
894     have aen: "coprime (a^e) n" by (simp add: coprime_commute)
895     from coprime_exp[OF na, of c]
896     have acn: "coprime (a^c) n" by (simp add: coprime_commute)
897     have "[a^d = a^e] (mod n) \<longleftrightarrow> [a^(e + c) = a^(e + 0)] (mod n)"
898       using c by simp
899     also have "\<dots> \<longleftrightarrow> [a^e* a^c = a^e *a^0] (mod n)" by (simp add: power_add)
900     also have  "\<dots> \<longleftrightarrow> [a ^ c = 1] (mod n)"
901       using cong_mult_lcancel_eq[OF aen, of "a^c" "a^0"] by simp
902     also  have "\<dots> \<longleftrightarrow> ord n a dvd c" by (simp only: ord_divides)
903     also have "\<dots> \<longleftrightarrow> [e + c = e + 0] (mod ord n a)"
904       using cong_add_lcancel_eq[of e c 0 "ord n a", simplified cong_0_divides]
905       by simp
906     finally have "[a^d = a^e] (mod n) \<longleftrightarrow> [d = e] (mod (ord n a))"
907       using c by simp }
908   note th = this
909   have "e \<le> d \<or> d \<le> e" by arith
910   moreover
911   {assume ed: "e \<le> d" from th[OF na ed] have ?thesis .}
912   moreover
913   {assume de: "d \<le> e"
914     from th[OF na de] have ?thesis by (simp add: cong_commute) }
915   ultimately show ?thesis by blast
916 qed
918 (* Another trivial primality characterization.                               *)
920 lemma prime_prime_factor:
921   "prime n \<longleftrightarrow> n \<noteq> 1\<and> (\<forall>p. prime p \<and> p dvd n \<longrightarrow> p = n)"
922 proof-
923   {assume n: "n=0 \<or> n=1" hence ?thesis using prime_0 two_is_prime by auto}
924   moreover
925   {assume n: "n\<noteq>0" "n\<noteq>1"
926     {assume pn: "prime n"
928       from pn[unfolded prime_def] have "\<forall>p. prime p \<and> p dvd n \<longrightarrow> p = n"
929         using n
930         apply (cases "n = 0 \<or> n=1",simp)
931         by (clarsimp, erule_tac x="p" in allE, auto)}
932     moreover
933     {assume H: "\<forall>p. prime p \<and> p dvd n \<longrightarrow> p = n"
934       from n have n1: "n > 1" by arith
935       {fix m assume m: "m dvd n" "m\<noteq>1"
936         from prime_factor[OF m(2)] obtain p where
937           p: "prime p" "p dvd m" by blast
938         from dvd_trans[OF p(2) m(1)] p(1) H have "p = n" by blast
939         with p(2) have "n dvd m"  by simp
940         hence "m=n"  using dvd_antisym[OF m(1)] by simp }
941       with n1 have "prime n"  unfolding prime_def by auto }
942     ultimately have ?thesis using n by blast}
943   ultimately       show ?thesis by auto
944 qed
946 lemma prime_divisor_sqrt:
947   "prime n \<longleftrightarrow> n \<noteq> 1 \<and> (\<forall>d. d dvd n \<and> d^2 \<le> n \<longrightarrow> d = 1)"
948 proof-
949   {assume "n=0 \<or> n=1" hence ?thesis using prime_0 prime_1
950     by (auto simp add: nat_power_eq_0_iff)}
951   moreover
952   {assume n: "n\<noteq>0" "n\<noteq>1"
953     hence np: "n > 1" by arith
954     {fix d assume d: "d dvd n" "d^2 \<le> n" and H: "\<forall>m. m dvd n \<longrightarrow> m=1 \<or> m=n"
955       from H d have d1n: "d = 1 \<or> d=n" by blast
956       {assume dn: "d=n"
957         have "n^2 > n*1" using n by (simp add: power2_eq_square)
958         with dn d(2) have "d=1" by simp}
959       with d1n have "d = 1" by blast  }
960     moreover
961     {fix d assume d: "d dvd n" and H: "\<forall>d'. d' dvd n \<and> d'^2 \<le> n \<longrightarrow> d' = 1"
962       from d n have "d \<noteq> 0" apply - apply (rule ccontr) by simp
963       hence dp: "d > 0" by simp
964       from d[unfolded dvd_def] obtain e where e: "n= d*e" by blast
965       from n dp e have ep:"e > 0" by simp
966       have "d^2 \<le> n \<or> e^2 \<le> n" using dp ep
967         by (auto simp add: e power2_eq_square mult_le_cancel_left)
968       moreover
969       {assume h: "d^2 \<le> n"
970         from H[rule_format, of d] h d have "d = 1" by blast}
971       moreover
972       {assume h: "e^2 \<le> n"
973         from e have "e dvd n" unfolding dvd_def by (simp add: mult_commute)
974         with H[rule_format, of e] h have "e=1" by simp
975         with e have "d = n" by simp}
976       ultimately have "d=1 \<or> d=n"  by blast}
977     ultimately have ?thesis unfolding prime_def using np n(2) by blast}
978   ultimately show ?thesis by auto
979 qed
980 lemma prime_prime_factor_sqrt:
981   "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)"
982   (is "?lhs \<longleftrightarrow>?rhs")
983 proof-
984   {assume "n=0 \<or> n=1" hence ?thesis using prime_0 prime_1 by auto}
985   moreover
986   {assume n: "n\<noteq>0" "n\<noteq>1"
987     {assume H: ?lhs
988       from H[unfolded prime_divisor_sqrt] n
989       have ?rhs
990         apply clarsimp
991         apply (erule_tac x="p" in allE)
992         apply simp
993         done
994     }
995     moreover
996     {assume H: ?rhs
997       {fix d assume d: "d dvd n" "d^2 \<le> n" "d\<noteq>1"
998         from prime_factor[OF d(3)]
999         obtain p where p: "prime p" "p dvd d" by blast
1000         from n have np: "n > 0" by arith
1001         from d(1) n have "d \<noteq> 0" by - (rule ccontr, auto)
1002         hence dp: "d > 0" by arith
1003         from mult_mono[OF dvd_imp_le[OF p(2) dp] dvd_imp_le[OF p(2) dp]] d(2)
1004         have "p^2 \<le> n" unfolding power2_eq_square by arith
1005         with H n p(1) dvd_trans[OF p(2) d(1)] have False  by blast}
1006       with n prime_divisor_sqrt  have ?lhs by auto}
1007     ultimately have ?thesis by blast }
1008   ultimately show ?thesis by (cases "n=0 \<or> n=1", auto)
1009 qed
1010 (* Pocklington theorem. *)
1012 lemma pocklington_lemma:
1013   assumes n: "n \<ge> 2" and nqr: "n - 1 = q*r" and an: "[a^ (n - 1) = 1] (mod n)"
1014   and aq:"\<forall>p. prime p \<and> p dvd q \<longrightarrow> coprime (a^ ((n - 1) div p) - 1) n"
1015   and pp: "prime p" and pn: "p dvd n"
1016   shows "[p = 1] (mod q)"
1017 proof-
1018   from pp prime_0 prime_1 have p01: "p \<noteq> 0" "p \<noteq> 1" by - (rule ccontr, simp)+
1019   from cong_1_divides[OF an, unfolded nqr, unfolded dvd_def]
1020   obtain k where k: "a ^ (q * r) - 1 = n*k" by blast
1021   from pn[unfolded dvd_def] obtain l where l: "n = p*l" by blast
1022   {assume a0: "a = 0"
1023     hence "a^ (n - 1) = 0" using n by (simp add: power_0_left)
1024     with n an mod_less[of 1 n]  have False by (simp add: power_0_left modeq_def)}
1025   hence a0: "a\<noteq>0" ..
1026   from n nqr have aqr0: "a ^ (q * r) \<noteq> 0" using a0 by simp
1027   hence "(a ^ (q * r) - 1) + 1  = a ^ (q * r)" by simp
1028   with k l have "a ^ (q * r) = p*l*k + 1" by simp
1029   hence "a ^ (r * q) + p * 0 = 1 + p * (l*k)" by (simp add: mult_ac)
1030   hence odq: "ord p (a^r) dvd q"
1031     unfolding ord_divides[symmetric] power_mult[symmetric] nat_mod  by blast
1032   from odq[unfolded dvd_def] obtain d where d: "q = ord p (a^r) * d" by blast
1033   {assume d1: "d \<noteq> 1"
1034     from prime_factor[OF d1] obtain P where P: "prime P" "P dvd d" by blast
1035     from d dvd_mult[OF P(2), of "ord p (a^r)"] have Pq: "P dvd q" by simp
1036     from aq P(1) Pq have caP:"coprime (a^ ((n - 1) div P) - 1) n" by blast
1037     from Pq obtain s where s: "q = P*s" unfolding dvd_def by blast
1038     have P0: "P \<noteq> 0" using P(1) prime_0 by - (rule ccontr, simp)
1039     from P(2) obtain t where t: "d = P*t" unfolding dvd_def by blast
1040     from d s t P0  have s': "ord p (a^r) * t = s" by algebra
1041     have "ord p (a^r) * t*r = r * ord p (a^r) * t" by algebra
1042     hence exps: "a^(ord p (a^r) * t*r) = ((a ^ r) ^ ord p (a^r)) ^ t"
1043       by (simp only: power_mult)
1044     have "[((a ^ r) ^ ord p (a^r)) ^ t= 1^t] (mod p)"
1045       by (rule cong_exp, rule ord)
1046     then have th: "[((a ^ r) ^ ord p (a^r)) ^ t= 1] (mod p)"
1047       by (simp add: power_Suc0)
1048     from cong_1_divides[OF th] exps have pd0: "p dvd a^(ord p (a^r) * t*r) - 1" by simp
1049     from nqr s s' have "(n - 1) div P = ord p (a^r) * t*r" using P0 by simp
1050     with caP have "coprime (a^(ord p (a^r) * t*r) - 1) n" by simp
1051     with p01 pn pd0 have False unfolding coprime by auto}
1052   hence d1: "d = 1" by blast
1053   hence o: "ord p (a^r) = q" using d by simp
1054   from pp phi_prime[of p] have phip: " \<phi> p = p - 1" by simp
1055   {fix d assume d: "d dvd p" "d dvd a" "d \<noteq> 1"
1056     from pp[unfolded prime_def] d have dp: "d = p" by blast
1057     from n have n12:"Suc (n - 2) = n - 1" by arith
1058     with divides_rexp[OF d(2)[unfolded dp], of "n - 2"]
1059     have th0: "p dvd a ^ (n - 1)" by simp
1060     from n have n0: "n \<noteq> 0" by simp
1061     from d(2) an n12[symmetric] have a0: "a \<noteq> 0"
1062       by - (rule ccontr, simp add: modeq_def)
1063     have th1: "a^ (n - 1) \<noteq> 0" using n d(2) dp a0 by auto
1064     from coprime_minus1[OF th1, unfolded coprime]
1065       dvd_trans[OF pn cong_1_divides[OF an]] th0 d(3) dp
1066     have False by auto}
1067   hence cpa: "coprime p a" using coprime by auto
1068   from coprime_exp[OF cpa, of r] coprime_commute
1069   have arp: "coprime (a^r) p" by blast
1070   from fermat_little[OF arp, simplified ord_divides] o phip
1071   have "q dvd (p - 1)" by simp
1072   then obtain d where d:"p - 1 = q * d" unfolding dvd_def by blast
1073   from prime_0 pp have p0:"p \<noteq> 0" by -  (rule ccontr, auto)
1074   from p0 d have "p + q * 0 = 1 + q * d" by simp
1075   with nat_mod[of p 1 q, symmetric]
1076   show ?thesis by blast
1077 qed
1079 lemma pocklington:
1080   assumes n: "n \<ge> 2" and nqr: "n - 1 = q*r" and sqr: "n \<le> q^2"
1081   and an: "[a^ (n - 1) = 1] (mod n)"
1082   and aq:"\<forall>p. prime p \<and> p dvd q \<longrightarrow> coprime (a^ ((n - 1) div p) - 1) n"
1083   shows "prime n"
1084 unfolding prime_prime_factor_sqrt[of n]
1085 proof-
1086   let ?ths = "n \<noteq> 0 \<and> n \<noteq> 1 \<and> \<not> (\<exists>p. prime p \<and> p dvd n \<and> p\<twosuperior> \<le> n)"
1087   from n have n01: "n\<noteq>0" "n\<noteq>1" by arith+
1088   {fix p assume p: "prime p" "p dvd n" "p^2 \<le> n"
1089     from p(3) sqr have "p^(Suc 1) \<le> q^(Suc 1)" by (simp add: power2_eq_square)
1090     hence pq: "p \<le> q" unfolding exp_mono_le .
1091     from pocklington_lemma[OF n nqr an aq p(1,2)]  cong_1_divides
1092     have th: "q dvd p - 1" by blast
1093     have "p - 1 \<noteq> 0"using prime_ge_2[OF p(1)] by arith
1094     with divides_ge[OF th] pq have False by arith }
1095   with n01 show ?ths by blast
1096 qed
1098 (* Variant for application, to separate the exponentiation.                  *)
1099 lemma pocklington_alt:
1100   assumes n: "n \<ge> 2" and nqr: "n - 1 = q*r" and sqr: "n \<le> q^2"
1101   and an: "[a^ (n - 1) = 1] (mod n)"
1102   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)"
1103   shows "prime n"
1104 proof-
1105   {fix p assume p: "prime p" "p dvd q"
1106     from aq[rule_format] p obtain b where
1107       b: "[a^((n - 1) div p) = b] (mod n)" "coprime (b - 1) n" by blast
1108     {assume a0: "a=0"
1109       from n an have "[0 = 1] (mod n)" unfolding a0 power_0_left by auto
1110       hence False using n by (simp add: modeq_def dvd_eq_mod_eq_0[symmetric])}
1111     hence a0: "a\<noteq> 0" ..
1112     hence a1: "a \<ge> 1" by arith
1113     from one_le_power[OF a1] have ath: "1 \<le> a ^ ((n - 1) div p)" .
1114     {assume b0: "b = 0"
1115       from p(2) nqr have "(n - 1) mod p = 0"
1116         apply (simp only: dvd_eq_mod_eq_0[symmetric]) by (rule dvd_mult2, simp)
1117       with mod_div_equality[of "n - 1" p]
1118       have "(n - 1) div p * p= n - 1" by auto
1119       hence eq: "(a^((n - 1) div p))^p = a^(n - 1)"
1120         by (simp only: power_mult[symmetric])
1121       from prime_ge_2[OF p(1)] have pS: "Suc (p - 1) = p" by arith
1122       from b(1) have d: "n dvd a^((n - 1) div p)" unfolding b0 cong_0_divides .
1123       from divides_rexp[OF d, of "p - 1"] pS eq cong_divides[OF an] n
1124       have False by simp}
1125     then have b0: "b \<noteq> 0" ..
1126     hence b1: "b \<ge> 1" by arith
1127     from cong_coprime[OF cong_sub[OF b(1) cong_refl[of 1] ath b1]] b(2) nqr
1128     have "coprime (a ^ ((n - 1) div p) - 1) n" by (simp add: coprime_commute)}
1129   hence th: "\<forall>p. prime p \<and> p dvd q \<longrightarrow> coprime (a ^ ((n - 1) div p) - 1) n "
1130     by blast
1131   from pocklington[OF n nqr sqr an th] show ?thesis .
1132 qed
1134 (* Prime factorizations.                                                     *)
1136 definition "primefact ps n = (foldr op * ps  1 = n \<and> (\<forall>p\<in> set ps. prime p))"
1138 lemma primefact: assumes n: "n \<noteq> 0"
1139   shows "\<exists>ps. primefact ps n"
1140 using n
1141 proof(induct n rule: nat_less_induct)
1142   fix n assume H: "\<forall>m<n. m \<noteq> 0 \<longrightarrow> (\<exists>ps. primefact ps m)" and n: "n\<noteq>0"
1143   let ?ths = "\<exists>ps. primefact ps n"
1144   {assume "n = 1"
1145     hence "primefact [] n" by (simp add: primefact_def)
1146     hence ?ths by blast }
1147   moreover
1148   {assume n1: "n \<noteq> 1"
1149     with n have n2: "n \<ge> 2" by arith
1150     from prime_factor[OF n1] obtain p where p: "prime p" "p dvd n" by blast
1151     from p(2) obtain m where m: "n = p*m" unfolding dvd_def by blast
1152     from n m have m0: "m > 0" "m\<noteq>0" by auto
1153     from prime_ge_2[OF p(1)] have "1 < p" by arith
1154     with m0 m have mn: "m < n" by auto
1155     from H[rule_format, OF mn m0(2)] obtain ps where ps: "primefact ps m" ..
1156     from ps m p(1) have "primefact (p#ps) n" by (simp add: primefact_def)
1157     hence ?ths by blast}
1158   ultimately show ?ths by blast
1159 qed
1161 lemma primefact_contains:
1162   assumes pf: "primefact ps n" and p: "prime p" and pn: "p dvd n"
1163   shows "p \<in> set ps"
1164   using pf p pn
1165 proof(induct ps arbitrary: p n)
1166   case Nil thus ?case by (auto simp add: primefact_def)
1167 next
1168   case (Cons q qs p n)
1169   from Cons.prems[unfolded primefact_def]
1170   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
1171   {assume "p dvd q"
1172     with p(1) q(1) have "p = q" unfolding prime_def by auto
1173     hence ?case by simp}
1174   moreover
1175   { assume h: "p dvd foldr op * qs 1"
1176     from q(3) have pqs: "primefact qs (foldr op * qs 1)"
1177       by (simp add: primefact_def)
1178     from Cons.hyps[OF pqs p(1) h] have ?case by simp}
1179   ultimately show ?case using prime_divprod[OF p] by blast
1180 qed
1182 lemma primefact_variant: "primefact ps n \<longleftrightarrow> foldr op * ps 1 = n \<and> list_all prime ps"
1183   by (auto simp add: primefact_def list_all_iff)
1185 (* Variant of Lucas theorem.                                                 *)
1187 lemma lucas_primefact:
1188   assumes n: "n \<ge> 2" and an: "[a^(n - 1) = 1] (mod n)"
1189   and psn: "foldr op * ps 1 = n - 1"
1190   and psp: "list_all (\<lambda>p. prime p \<and> \<not> [a^((n - 1) div p) = 1] (mod n)) ps"
1191   shows "prime n"
1192 proof-
1193   {fix p assume p: "prime p" "p dvd n - 1" "[a ^ ((n - 1) div p) = 1] (mod n)"
1194     from psn psp have psn1: "primefact ps (n - 1)"
1195       by (auto simp add: list_all_iff primefact_variant)
1196     from p(3) primefact_contains[OF psn1 p(1,2)] psp
1197     have False by (induct ps, auto)}
1198   with lucas[OF n an] show ?thesis by blast
1199 qed
1201 (* Variant of Pocklington theorem.                                           *)
1203 lemma mod_le: assumes n: "n \<noteq> (0::nat)" shows "m mod n \<le> m"
1204 proof-
1205     from mod_div_equality[of m n]
1206     have "\<exists>x. x + m mod n = m" by blast
1207     then show ?thesis by auto
1208 qed
1211 lemma pocklington_primefact:
1212   assumes n: "n \<ge> 2" and qrn: "q*r = n - 1" and nq2: "n \<le> q^2"
1213   and arnb: "(a^r) mod n = b" and psq: "foldr op * ps 1 = q"
1214   and bqn: "(b^q) mod n = 1"
1215   and psp: "list_all (\<lambda>p. prime p \<and> coprime ((b^(q div p)) mod n - 1) n) ps"
1216   shows "prime n"
1217 proof-
1218   from bqn psp qrn
1219   have bqn: "a ^ (n - 1) mod n = 1"
1220     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
1221     by (simp_all add: power_mult[symmetric] algebra_simps)
1222   from n  have n0: "n > 0" by arith
1223   from mod_div_equality[of "a^(n - 1)" n]
1224     mod_less_divisor[OF n0, of "a^(n - 1)"]
1225   have an1: "[a ^ (n - 1) = 1] (mod n)"
1226     unfolding nat_mod bqn
1227     apply -
1228     apply (rule exI[where x="0"])
1229     apply (rule exI[where x="a^(n - 1) div n"])
1230     by (simp add: algebra_simps)
1231   {fix p assume p: "prime p" "p dvd q"
1232     from psp psq have pfpsq: "primefact ps q"
1233       by (auto simp add: primefact_variant list_all_iff)
1234     from psp primefact_contains[OF pfpsq p]
1235     have p': "coprime (a ^ (r * (q div p)) mod n - 1) n"
1236       by (simp add: list_all_iff)
1237     from prime_ge_2[OF p(1)] have p01: "p \<noteq> 0" "p \<noteq> 1" "p =Suc(p - 1)" by arith+
1238     from div_mult1_eq[of r q p] p(2)
1239     have eq1: "r* (q div p) = (n - 1) div p"
1240       unfolding qrn[symmetric] dvd_eq_mod_eq_0 by (simp add: mult_commute)
1241     have ath: "\<And>a (b::nat). a <= b \<Longrightarrow> a \<noteq> 0 ==> 1 <= a \<and> 1 <= b" by arith
1242     from n0 have n00: "n \<noteq> 0" by arith
1243     from mod_le[OF n00]
1244     have th10: "a ^ ((n - 1) div p) mod n \<le> a ^ ((n - 1) div p)" .
1245     {assume "a ^ ((n - 1) div p) mod n = 0"
1246       then obtain s where s: "a ^ ((n - 1) div p) = n*s"
1247         unfolding mod_eq_0_iff by blast
1248       hence eq0: "(a^((n - 1) div p))^p = (n*s)^p" by simp
1249       from qrn[symmetric] have qn1: "q dvd n - 1" unfolding dvd_def by auto
1250       from dvd_trans[OF p(2) qn1] div_mod_equality'[of "n - 1" p]
1251       have npp: "(n - 1) div p * p = n - 1" by (simp add: dvd_eq_mod_eq_0)
1252       with eq0 have "a^ (n - 1) = (n*s)^p"
1253         by (simp add: power_mult[symmetric])
1254       hence "1 = (n*s)^(Suc (p - 1)) mod n" using bqn p01 by simp
1255       also have "\<dots> = 0" by (simp add: mult_assoc)
1256       finally have False by simp }
1257       then have th11: "a ^ ((n - 1) div p) mod n \<noteq> 0" by auto
1258     have th1: "[a ^ ((n - 1) div p) mod n = a ^ ((n - 1) div p)] (mod n)"
1259       unfolding modeq_def by simp
1260     from cong_sub[OF th1 cong_refl[of 1]]  ath[OF th10 th11]
1261     have th: "[a ^ ((n - 1) div p) mod n - 1 = a ^ ((n - 1) div p) - 1] (mod n)"
1262       by blast
1263     from cong_coprime[OF th] p'[unfolded eq1]
1264     have "coprime (a ^ ((n - 1) div p) - 1) n" by (simp add: coprime_commute) }
1265   with pocklington[OF n qrn[symmetric] nq2 an1]
1266   show ?thesis by blast
1267 qed
1269 end