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 *)
     5 
     6 header {* Pocklington's Theorem for Primes *}
     7 
     8 
     9 theory Pocklington
    10 imports Plain "~~/src/HOL/List" "~~/src/HOL/Primes"
    11 begin
    12 
    13 definition modeq:: "nat => nat => nat => bool"    ("(1[_ = _] '(mod _'))")
    14   where "[a = b] (mod p) == ((a mod p) = (b mod p))"
    15 
    16 definition modneq:: "nat => nat => nat => bool"    ("(1[_ \<noteq> _] '(mod _'))")
    17   where "[a \<noteq> b] (mod p) == ((a mod p) \<noteq> (b mod p))"
    18 
    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)
    22 
    23 
    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
    27 
    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 ..
    30 
    31 (* Lemmas about previously defined terms.                                    *)
    32 
    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
    61 
    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
    67 
    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)
    70 
    71 (* Congruences.                                                              *)
    72 
    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)
    76 
    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
    89 
    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
   115 
   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]]] .
   119 
   120 lemma cong_refl: "[x = x] (mod n)" by (simp add: modeq_def)
   121 
   122 lemma eq_imp_cong: "a = b \<Longrightarrow> [a = b] (mod n)" by (simp add: cong_refl)
   123 
   124 lemma cong_commute: "[x = y] (mod n) \<longleftrightarrow> [y = x] (mod n)" 
   125   by (auto simp add: modeq_def)
   126 
   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)
   129 
   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
   140 
   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
   151 
   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
   169 
   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
   178 
   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)
   182 
   183 lemma cong_add_lcancel_eq: "[a + x = a + y] (mod n) \<longleftrightarrow> [x = y] (mod n)" 
   184   by (simp add: nat_mod)
   185 
   186 lemma cong_add_rcancel_eq: "[x + a = y + a] (mod n) \<longleftrightarrow> [x = y] (mod n)"
   187   by (simp add: nat_mod)
   188 
   189 lemma cong_add_rcancel: "[x + a = y + a] (mod n) \<Longrightarrow> [x = y] (mod n)" 
   190   by (simp add: nat_mod)
   191 
   192 lemma cong_add_lcancel: "[a + x = a + y] (mod n) \<Longrightarrow> [x = y] (mod n)"
   193   by (simp add: nat_mod)
   194 
   195 lemma cong_add_lcancel_eq_0: "[a + x = a] (mod n) \<longleftrightarrow> [x = 0] (mod n)" 
   196   by (simp add: nat_mod)
   197 
   198 lemma cong_add_rcancel_eq_0: "[x + a = a] (mod n) \<longleftrightarrow> [x = 0] (mod n)"
   199   by (simp add: nat_mod)
   200 
   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]] . 
   204 
   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
   210   
   211 lemma cong_0_divides: "[x = 0] (mod n) \<longleftrightarrow> n dvd x" by simp
   212 
   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
   216 
   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
   222 
   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
   236 
   237 lemma cong_mod: "~(n = 0) \<Longrightarrow> [a mod n = a] (mod n)" by (simp add: modeq_def)
   238 
   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)
   242 
   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
   248 
   249 (* Some things when we know more about the order.                            *)
   250 
   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)
   258 
   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
   271 
   272 (* Some basic theorems about solving congruences.                            *)
   273 
   274 
   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
   291 
   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
   308 
   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] .
   329 
   330 (* Forms of the Chinese remainder theorem.                                   *)
   331 
   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])
   340 
   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
   368 
   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
   383 
   384 (* Euler totient function.                                                   *)
   385 
   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)
   389 
   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
   395 
   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
   403 
   404 lemma [simp]: "\<phi> (Suc 0) = Suc 0" using phi_1 by simp
   405 
   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
   420 
   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)
   423 
   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
   432 
   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
   439 
   440 lemma stupid[simp]: "{m. (0::nat) < m \<and> m < n} = {1..<n}"
   441   by auto
   442 
   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
   450 
   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
   462 
   463 lemma phi_lowerbound_1: "2 <= n ==> 1 <= \<phi>(n)"
   464   using phi_lowerbound_1_strong[of n] by auto
   465 
   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
   475 
   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
   507 
   508 (* Multiplicativity property.                                                *)
   509 
   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
   554 
   555 (* Fermat's Little theorem / Fermat-Euler theorem.                           *)
   556 
   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)
   563 
   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
   576 
   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] ..
   581 
   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)
   587 
   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
   607 
   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
   635       
   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
   678 
   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
   683 
   684 
   685 (* Lucas's theorem.                                                          *)
   686 
   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
   708 
   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
   724 
   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
   736 
   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
   749  
   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
   762 
   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
   816 
   817 (* Definition of the order of a number mod n (0 in non-coprime case).        *)
   818 
   819 definition "ord n a = (if coprime n a then Least (\<lambda>d. d > 0 \<and> [a ^d = 1] (mod n)) else 0)"
   820 
   821 (* This has the expected properties.                                         *)
   822 
   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)
   846 
   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)
   852 
   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
   906 
   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
   942 
   943 (* Another trivial primality characterization.                               *)
   944 
   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"
   952       
   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
   970 
   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. *)
  1033 
  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
  1100 
  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
  1119 
  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
  1155 
  1156 (* Prime factorizations.                                                     *)
  1157 
  1158 definition "primefact ps n = (foldr op * ps  1 = n \<and> (\<forall>p\<in> set ps. prime p))"
  1159 
  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
  1182 
  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
  1203 
  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)
  1205 
  1206 (* Variant of Lucas theorem.                                                 *)
  1207 
  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
  1221 
  1222 (* Variant of Pocklington theorem.                                           *)
  1223 
  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
  1230   
  1231 
  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
  1289 
  1290 
  1291 end