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1 (* Author: Jaime Mendizabal Roche *) |
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2 |
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3 theory Quadratic_Reciprocity |
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4 imports Gauss |
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5 begin |
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6 |
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7 text {* The proof is based on Gauss's fifth proof, which can be found at http://www.lehigh.edu/~shw2/q-recip/gauss5.pdf *} |
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8 |
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9 locale QR = |
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10 fixes p :: "nat" |
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11 fixes q :: "nat" |
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12 |
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13 assumes p_prime: "prime p" |
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14 assumes p_ge_2: "2 < p" |
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15 assumes q_prime: "prime q" |
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16 assumes q_ge_2: "2 < q" |
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17 assumes pq_neq: "p \<noteq> q" |
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18 begin |
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19 |
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20 lemma odd_p: "odd p" using p_ge_2 p_prime prime_odd_nat by blast |
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21 |
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22 lemma p_ge_0: "0 < int p" |
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23 using p_prime not_prime_0[where 'a = nat] by fastforce+ |
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24 |
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25 lemma p_eq2: "int p = (2 * ((int p - 1) div 2)) + 1" using odd_p by simp |
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26 |
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27 lemma odd_q: "odd q" using q_ge_2 q_prime prime_odd_nat by blast |
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28 |
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29 lemma q_ge_0: "0 < int q" using q_prime not_prime_0[where 'a = nat] by fastforce+ |
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30 |
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31 lemma q_eq2: "int q = (2 * ((int q - 1) div 2)) + 1" using odd_q by simp |
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32 |
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33 lemma pq_eq2: "int p * int q = (2 * ((int p * int q - 1) div 2)) + 1" using odd_p odd_q by simp |
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34 |
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35 lemma pq_coprime: "coprime p q" |
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36 using pq_neq p_prime primes_coprime_nat q_prime by blast |
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37 |
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38 lemma pq_coprime_int: "coprime (int p) (int q)" |
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39 using pq_coprime transfer_int_nat_gcd(1) by presburger |
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40 |
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41 lemma qp_ineq: "(int p * k \<le> (int p * int q - 1) div 2) = (k \<le> (int q - 1) div 2)" |
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42 proof - |
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43 have "(2 * int p * k \<le> int p * int q - 1) = (2 * k \<le> int q - 1)" using p_ge_0 by auto |
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44 thus ?thesis by auto |
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45 qed |
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46 |
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47 lemma QRqp: "QR q p" using QR_def QR_axioms by simp |
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48 |
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49 lemma pq_commute: "int p * int q = int q * int p" by simp |
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50 |
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51 lemma pq_ge_0: "int p * int q > 0" using p_ge_0 q_ge_0 mult_pos_pos by blast |
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52 |
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53 definition "r = ((p - 1) div 2)*((q - 1) div 2)" |
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54 definition "m = card (GAUSS.E p q)" |
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55 definition "n = card (GAUSS.E q p)" |
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56 |
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57 abbreviation "Res (k::int) \<equiv> {0 .. k - 1}" |
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58 abbreviation "Res_ge_0 (k::int) \<equiv> {0 <.. k - 1}" |
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59 abbreviation "Res_0 (k::int) \<equiv> {0::int}" |
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60 abbreviation "Res_l (k::int) \<equiv> {0 <.. (k - 1) div 2}" |
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61 abbreviation "Res_h (k::int) \<equiv> {(k - 1) div 2 <.. k - 1}" |
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62 |
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63 abbreviation "Sets_pq r0 r1 r2 \<equiv> |
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64 {(x::int). x \<in> r0 (int p * int q) \<and> x mod p \<in> r1 (int p) \<and> x mod q \<in> r2 (int q)}" |
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65 |
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66 definition "A = Sets_pq Res_l Res_l Res_h" |
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67 definition "B = Sets_pq Res_l Res_h Res_l" |
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68 definition "C = Sets_pq Res_h Res_h Res_l" |
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69 definition "D = Sets_pq Res_l Res_h Res_h" |
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70 definition "E = Sets_pq Res_l Res_0 Res_h" |
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71 definition "F = Sets_pq Res_l Res_h Res_0" |
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72 |
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73 definition "a = card A" |
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74 definition "b = card B" |
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75 definition "c = card C" |
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76 definition "d = card D" |
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77 definition "e = card E" |
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78 definition "f = card F" |
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79 |
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80 lemma Gpq: "GAUSS p q" unfolding GAUSS_def |
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81 using p_prime pq_neq p_ge_2 q_prime |
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82 by (auto simp: cong_altdef_int zdvd_int [symmetric] dest: primes_dvd_imp_eq) |
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83 |
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84 lemma Gqp: "GAUSS q p" using QRqp QR.Gpq by simp |
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85 |
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86 lemma QR_lemma_01: "(\<lambda>x. x mod q) ` E = GAUSS.E q p" |
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87 proof |
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88 { |
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89 fix x |
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90 assume a1: "x \<in> E" |
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91 then obtain k where k: "x = int p * k" unfolding E_def by blast |
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92 have "x \<in> Res_l (int p * int q)" using a1 E_def by blast |
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93 hence "k \<in> GAUSS.A q" using Gqp GAUSS.A_def k qp_ineq by (simp add: zero_less_mult_iff) |
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94 hence "x mod q \<in> GAUSS.E q p" |
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95 using GAUSS.C_def[of q p] Gqp k GAUSS.B_def[of q p] a1 GAUSS.E_def[of q p] |
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96 unfolding E_def by force |
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97 hence "x \<in> E \<longrightarrow> x mod int q \<in> GAUSS.E q p" by auto |
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98 } |
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99 thus "(\<lambda>x. x mod int q) ` E \<subseteq> GAUSS.E q p" by auto |
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100 next |
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101 show "GAUSS.E q p \<subseteq> (\<lambda>x. x mod q) ` E" |
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102 proof |
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103 fix x |
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104 assume a1: "x \<in> GAUSS.E q p" |
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105 then obtain ka where ka: "ka \<in> GAUSS.A q" "x = (ka * p) mod q" |
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106 using Gqp GAUSS.B_def GAUSS.C_def GAUSS.E_def by auto |
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107 hence "ka * p \<in> Res_l (int p * int q)" |
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108 using GAUSS.A_def Gqp p_ge_0 qp_ineq by (simp add: Groups.mult_ac(2)) |
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109 thus "x \<in> (\<lambda>x. x mod q) ` E" unfolding E_def using ka a1 Gqp GAUSS.E_def q_ge_0 by force |
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110 qed |
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111 qed |
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112 |
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113 lemma QR_lemma_02: "e= n" |
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114 proof - |
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115 { |
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116 fix x y |
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117 assume a: "x \<in> E" "y \<in> E" "x mod q = y mod q" |
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118 obtain p_inv where p_inv: "[int p * p_inv = 1] (mod int q)" |
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119 using pq_coprime_int cong_solve_coprime_int by blast |
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120 obtain kx ky where k: "x = int p * kx" "y = int p * ky" using a E_def dvd_def[of p x] by blast |
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121 hence "0 < x" "int p * kx \<le> (int p * int q - 1) div 2" |
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122 "0 < y" "int p * ky \<le> (int p * int q - 1) div 2" |
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123 using E_def a greaterThanAtMost_iff mem_Collect_eq by blast+ |
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124 hence "0 \<le> kx" "kx < q" "0 \<le> ky" "ky < q" using qp_ineq k by (simp add: zero_less_mult_iff)+ |
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125 moreover have "(p_inv * (p * kx)) mod q = (p_inv * (p * ky)) mod q" |
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126 using a(3) mod_mult_cong k by blast |
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127 hence "(p * p_inv * kx) mod q = (p * p_inv * ky) mod q" by (simp add:algebra_simps) |
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128 hence "kx mod q = ky mod q" |
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129 using p_inv mod_mult_cong[of "p * p_inv" "q" "1"] cong_int_def by auto |
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130 hence "[kx = ky] (mod q)" using cong_int_def by blast |
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131 ultimately have "x = y" using cong_less_imp_eq_int k by blast |
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132 } |
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133 hence "inj_on (\<lambda>x. x mod q) E" unfolding inj_on_def by auto |
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134 thus ?thesis using QR_lemma_01 card_image e_def n_def by fastforce |
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135 qed |
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136 |
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137 lemma QR_lemma_03: "f = m" |
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138 proof - |
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139 have "F = QR.E q p" unfolding F_def pq_commute using QRqp QR.E_def[of q p] by fastforce |
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140 hence "f = QR.e q p" unfolding f_def using QRqp QR.e_def[of q p] by presburger |
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141 thus ?thesis using QRqp QR.QR_lemma_02 m_def QRqp QR.n_def by presburger |
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142 qed |
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143 |
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144 definition f_1 :: "int \<Rightarrow> int \<times> int" where |
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145 "f_1 x = ((x mod p), (x mod q))" |
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146 |
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147 definition P_1 :: "int \<times> int \<Rightarrow> int \<Rightarrow> bool" where |
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148 "P_1 res x \<longleftrightarrow> x mod p = fst res & x mod q = snd res & x \<in> Res (int p * int q)" |
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149 |
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150 definition g_1 :: "int \<times> int \<Rightarrow> int" where |
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151 "g_1 res = (THE x. P_1 res x)" |
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152 |
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153 lemma P_1_lemma: assumes "0 \<le> fst res" "fst res < p" "0 \<le> snd res" "snd res < q" |
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154 shows "\<exists>! x. P_1 res x" |
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155 proof - |
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156 obtain y k1 k2 where yk: "y = nat (fst res) + k1 * p" "y = nat (snd res) + k2 * q" |
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157 using chinese_remainder[of p q] pq_coprime p_ge_0 q_ge_0 by fastforce |
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158 have h1: "[y = fst res] (mod p)" "[y = snd res] (mod q)" |
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159 using yk(1) assms(1) cong_iff_lin_int[of "fst res"] cong_sym_int apply simp |
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160 using yk(2) assms(3) cong_iff_lin_int[of "snd res"] cong_sym_int by simp |
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161 have "(y mod (int p * int q)) mod int p = fst res" "(y mod (int p * int q)) mod int q = snd res" |
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162 using h1(1) mod_mod_cancel[of "int p"] assms(1) assms(2) cong_int_def apply simp |
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163 using h1(2) mod_mod_cancel[of "int q"] assms(3) assms(4) cong_int_def by simp |
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164 then obtain x where "P_1 res x" unfolding P_1_def |
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165 using Divides.pos_mod_bound Divides.pos_mod_sign pq_ge_0 by fastforce |
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166 moreover { |
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167 fix a b |
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168 assume a: "P_1 res a" "P_1 res b" |
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169 hence "int p * int q dvd a - b" |
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170 using divides_mult[of "int p" "a - b" "int q"] pq_coprime_int zmod_eq_dvd_iff[of a _ b] |
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171 unfolding P_1_def by force |
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172 hence "a = b" using dvd_imp_le_int[of "a - b"] a unfolding P_1_def by fastforce |
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173 } |
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174 ultimately show ?thesis by auto |
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175 qed |
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176 |
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177 lemma g_1_lemma: assumes "0 \<le> fst res" "fst res < p" "0 \<le> snd res" "snd res < q" |
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178 shows "P_1 res (g_1 res)" using assms P_1_lemma theI'[of "P_1 res"] g_1_def by presburger |
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179 |
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180 definition "BuC = Sets_pq Res_ge_0 Res_h Res_l" |
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181 |
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182 lemma QR_lemma_04: "card BuC = card ((Res_h p) \<times> (Res_l q))" |
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183 using card_bij_eq[of f_1 "BuC" "(Res_h p) \<times> (Res_l q)" g_1] |
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184 proof |
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185 { |
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186 fix x y |
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187 assume a: "x \<in> BuC" "y \<in> BuC" "f_1 x = f_1 y" |
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188 hence "int p * int q dvd x - y" |
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189 using f_1_def pq_coprime_int divides_mult[of "int p" "x - y" "int q"] |
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190 zmod_eq_dvd_iff[of x _ y] by auto |
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191 hence "x = y" |
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192 using dvd_imp_le_int[of "x - y" "int p * int q"] a unfolding BuC_def by force |
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193 } |
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194 thus "inj_on f_1 BuC" unfolding inj_on_def by auto |
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195 next |
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196 { |
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197 fix x y |
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198 assume a: "x \<in> (Res_h p) \<times> (Res_l q)" "y \<in> (Res_h p) \<times> (Res_l q)" "g_1 x = g_1 y" |
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199 hence "0 \<le> fst x" "fst x < p" "0 \<le> snd x" "snd x < q" |
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200 "0 \<le> fst y" "fst y < p" "0 \<le> snd y" "snd y < q" |
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201 using mem_Sigma_iff prod.collapse by fastforce+ |
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202 hence "x = y" using g_1_lemma[of x] g_1_lemma[of y] a P_1_def by fastforce |
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203 } |
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204 thus "inj_on g_1 ((Res_h p) \<times> (Res_l q))" unfolding inj_on_def by auto |
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205 next |
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206 show "g_1 ` ((Res_h p) \<times> (Res_l q)) \<subseteq> BuC" |
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207 proof |
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208 fix y |
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209 assume "y \<in> g_1 ` ((Res_h p) \<times> (Res_l q))" |
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210 then obtain x where x: "y = g_1 x" "x \<in> ((Res_h p) \<times> (Res_l q))" by blast |
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211 hence "P_1 x y" using g_1_lemma by fastforce |
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212 thus "y \<in> BuC" unfolding P_1_def BuC_def mem_Collect_eq using x SigmaE prod.sel by fastforce |
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213 qed |
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214 qed (auto simp: BuC_def finite_subset f_1_def) |
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215 |
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216 lemma QR_lemma_05: "card ((Res_h p) \<times> (Res_l q)) = r" |
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217 proof - |
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218 have "card (Res_l q) = (q - 1) div 2" "card (Res_h p) = (p - 1) div 2" using p_eq2 by force+ |
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219 thus ?thesis unfolding r_def using card_cartesian_product[of "Res_h p" "Res_l q"] by presburger |
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220 qed |
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221 |
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222 lemma QR_lemma_06: "b + c = r" |
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223 proof - |
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224 have "B \<inter> C = {}" "finite B" "finite C" "B \<union> C = BuC" unfolding B_def C_def BuC_def by fastforce+ |
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225 thus ?thesis |
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226 unfolding b_def c_def using card_empty card_Un_Int QR_lemma_04 QR_lemma_05 by fastforce |
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227 qed |
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228 |
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229 definition f_2:: "int \<Rightarrow> int" where |
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230 "f_2 x = (int p * int q) - x" |
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231 |
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232 lemma f_2_lemma_1: "\<And>x. f_2 (f_2 x) = x" unfolding f_2_def by simp |
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233 |
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234 lemma f_2_lemma_2: "[f_2 x = int p - x] (mod p)" unfolding f_2_def using cong_altdef_int by simp |
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235 |
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236 lemma f_2_lemma_3: "f_2 x \<in> S \<Longrightarrow> x \<in> f_2 ` S" |
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237 using f_2_lemma_1[of x] image_eqI[of x f_2 "f_2 x" S] by presburger |
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238 |
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239 lemma QR_lemma_07: "f_2 ` Res_l (int p * int q) = Res_h (int p * int q)" |
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240 "f_2 ` Res_h (int p * int q) = Res_l (int p * int q)" |
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241 proof - |
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242 have h1: "f_2 ` Res_l (int p * int q) \<subseteq> Res_h (int p * int q)" using f_2_def by force |
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243 have h2: "f_2 ` Res_h (int p * int q) \<subseteq> Res_l (int p * int q)" using f_2_def pq_eq2 by fastforce |
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244 have h3: "Res_h (int p * int q) \<subseteq> f_2 ` Res_l (int p * int q)" using h2 f_2_lemma_3 by blast |
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245 have h4: "Res_l (int p * int q) \<subseteq> f_2 ` Res_h (int p * int q)" using h1 f_2_lemma_3 by blast |
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246 show "f_2 ` Res_l (int p * int q) = Res_h (int p * int q)" using h1 h3 by blast |
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247 show "f_2 ` Res_h (int p * int q) = Res_l (int p * int q)" using h2 h4 by blast |
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248 qed |
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249 |
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250 lemma QR_lemma_08: "(f_2 x mod p \<in> Res_l p) = (x mod p \<in> Res_h p)" |
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251 "(f_2 x mod p \<in> Res_h p) = (x mod p \<in> Res_l p)" |
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252 using f_2_lemma_2[of x] cong_int_def[of "f_2 x" "p - x" p] minus_mod_self2[of x p] |
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253 zmod_zminus1_eq_if[of x p] p_eq2 by auto |
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254 |
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255 lemma QR_lemma_09: "(f_2 x mod q \<in> Res_l q) = (x mod q \<in> Res_h q)" |
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256 "(f_2 x mod q \<in> Res_h q) = (x mod q \<in> Res_l q)" |
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257 using QRqp QR.QR_lemma_08 f_2_def QR.f_2_def pq_commute by auto+ |
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258 |
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259 lemma QR_lemma_10: "a = c" unfolding a_def c_def apply (rule card_bij_eq[of f_2 A C f_2]) |
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260 unfolding A_def C_def |
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261 using QR_lemma_07 QR_lemma_08 QR_lemma_09 apply ((simp add: inj_on_def f_2_def),blast)+ |
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262 by fastforce+ |
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263 |
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264 definition "BuD = Sets_pq Res_l Res_h Res_ge_0" |
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265 definition "BuDuF = Sets_pq Res_l Res_h Res" |
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266 |
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267 definition f_3 :: "int \<Rightarrow> int \<times> int" where |
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268 "f_3 x = (x mod p, x div p + 1)" |
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269 |
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270 definition g_3 :: "int \<times> int \<Rightarrow> int" where |
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271 "g_3 x = fst x + (snd x - 1) * p" |
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272 |
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273 lemma QR_lemma_11: "card BuDuF = card ((Res_h p) \<times> (Res_l q))" |
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274 using card_bij_eq[of f_3 BuDuF "(Res_h p) \<times> (Res_l q)" g_3] |
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275 proof |
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276 show "f_3 ` BuDuF \<subseteq> (Res_h p) \<times> (Res_l q)" |
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277 proof |
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278 fix y |
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279 assume "y \<in> f_3 ` BuDuF" |
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280 then obtain x where x: "y = f_3 x" "x \<in> BuDuF" by blast |
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281 hence "x \<le> int p * (int q - 1) div 2 + (int p - 1) div 2" |
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282 unfolding BuDuF_def using p_eq2 int_distrib(4) by auto |
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283 moreover have "(int p - 1) div 2 \<le> - 1 + x mod p" using x BuDuF_def by auto |
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284 moreover have "int p * (int q - 1) div 2 = int p * ((int q - 1) div 2)" |
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285 using zdiv_zmult1_eq odd_q by auto |
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286 hence "p * (int q - 1) div 2 = p * ((int q + 1) div 2 - 1)" by fastforce |
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287 ultimately have "x \<le> p * ((int q + 1) div 2 - 1) - 1 + x mod p" by linarith |
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288 hence "x div p < (int q + 1) div 2 - 1" |
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289 using mult.commute[of "int p" "x div p"] p_ge_0 div_mult_mod_eq[of x p] |
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290 mult_less_cancel_left_pos[of p "x div p" "(int q + 1) div 2 - 1"] by linarith |
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291 moreover have "0 < x div p + 1" |
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292 using pos_imp_zdiv_neg_iff[of p x] p_ge_0 x mem_Collect_eq BuDuF_def by auto |
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293 ultimately show "y \<in> (Res_h p) \<times> (Res_l q)" using x BuDuF_def f_3_def by auto |
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294 qed |
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295 next |
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296 have h1: "\<And>x. x \<in> ((Res_h p) \<times> (Res_l q)) \<Longrightarrow> f_3 (g_3 x) = x" |
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297 proof - |
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298 fix x |
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299 assume a: "x \<in> ((Res_h p) \<times> (Res_l q))" |
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300 moreover have h: "(fst x + (snd x - 1) * int p) mod int p = fst x" using a by force |
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301 ultimately have "(fst x + (snd x - 1) * int p) div int p + 1 = snd x" |
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302 by (auto simp: semiring_numeral_div_class.div_less) |
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303 with h show "f_3 (g_3 x) = x" unfolding f_3_def g_3_def by simp |
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304 qed |
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305 show "inj_on g_3 ((Res_h p) \<times> (Res_l q))" apply (rule inj_onI[of "(Res_h p) \<times> (Res_l q)" g_3]) |
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306 proof - |
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307 fix x y |
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308 assume "x \<in> ((Res_h p) \<times> (Res_l q))" "y \<in> ((Res_h p) \<times> (Res_l q))" "g_3 x = g_3 y" |
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309 thus "x = y" using h1[of x] h1[of y] by presburger |
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310 qed |
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311 next |
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312 show "g_3 ` ((Res_h p) \<times> (Res_l q)) \<subseteq> BuDuF" |
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313 proof |
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314 fix y |
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315 assume "y \<in> g_3 ` ((Res_h p) \<times> (Res_l q))" |
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316 then obtain x where x: "y = g_3 x" "x \<in> (Res_h p) \<times> (Res_l q)" by blast |
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317 hence "snd x \<le> (int q - 1) div 2" by force |
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318 moreover have "int p * ((int q - 1) div 2) = (int p * int q - int p) div 2" |
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319 using int_distrib(4) zdiv_zmult1_eq[of "int p" "int q - 1" 2] odd_q by fastforce |
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320 ultimately have "(snd x) * int p \<le> (int q * int p - int p) div 2" |
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321 using mult_right_mono[of "snd x" "(int q - 1) div 2" p] mult.commute[of "(int q - 1) div 2" p] |
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322 pq_commute by presburger |
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323 hence "(snd x - 1) * int p \<le> (int q * int p - 1) div 2 - int p" |
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324 using p_ge_0 int_distrib(3) by auto |
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325 moreover have "fst x \<le> int p - 1" using x by force |
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326 ultimately have "fst x + (snd x - 1) * int p \<le> (int p * int q - 1) div 2" |
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327 using pq_commute by linarith |
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328 moreover have "0 < fst x" "0 \<le> (snd x - 1) * p" using x(2) by fastforce+ |
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329 ultimately show "y \<in> BuDuF" unfolding BuDuF_def using q_ge_0 x g_3_def x(1) by auto |
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330 qed |
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331 next |
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332 show "finite BuDuF" unfolding BuDuF_def by fastforce |
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333 qed (simp add: inj_on_inverseI[of BuDuF g_3] f_3_def g_3_def QR_lemma_05)+ |
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334 |
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335 lemma QR_lemma_12: "b + d + m = r" |
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336 proof - |
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337 have "B \<inter> D = {}" "finite B" "finite D" "B \<union> D = BuD" unfolding B_def D_def BuD_def by fastforce+ |
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338 hence "b + d = card BuD" unfolding b_def d_def using card_Un_Int by fastforce |
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339 moreover have "BuD \<inter> F = {}" "finite BuD" "finite F" unfolding BuD_def F_def by fastforce+ |
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340 moreover have "BuD \<union> F = BuDuF" unfolding BuD_def F_def BuDuF_def |
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341 using q_ge_0 ivl_disj_un_singleton(5)[of 0 "int q - 1"] by auto |
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342 ultimately show ?thesis using QR_lemma_03 QR_lemma_05 QR_lemma_11 card_Un_disjoint[of BuD F] |
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343 unfolding b_def d_def f_def by presburger |
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344 qed |
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345 |
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346 lemma QR_lemma_13: "a + d + n = r" |
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347 proof - |
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348 have "A = QR.B q p" unfolding A_def pq_commute using QRqp QR.B_def[of q p] by blast |
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349 hence "a = QR.b q p" using a_def QRqp QR.b_def[of q p] by presburger |
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350 moreover have "D = QR.D q p" unfolding D_def pq_commute using QRqp QR.D_def[of q p] by blast |
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351 hence "d = QR.d q p" using d_def QRqp QR.d_def[of q p] by presburger |
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352 moreover have "n = QR.m q p" using n_def QRqp QR.m_def[of q p] by presburger |
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353 moreover have "r = QR.r q p" unfolding r_def using QRqp QR.r_def[of q p] by auto |
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354 ultimately show ?thesis using QRqp QR.QR_lemma_12 by presburger |
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355 qed |
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356 |
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357 lemma QR_lemma_14: "(-1::int) ^ (m + n) = (-1) ^ r" |
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358 proof - |
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359 have "m + n + 2 * d = r" using QR_lemma_06 QR_lemma_10 QR_lemma_12 QR_lemma_13 by auto |
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360 thus ?thesis using power_add[of "-1::int" "m + n" "2 * d"] by fastforce |
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361 qed |
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362 |
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363 lemma Quadratic_Reciprocity: |
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364 "(Legendre p q) * (Legendre q p) = (-1::int) ^ ((p - 1) div 2 * ((q - 1) div 2))" |
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365 using Gpq Gqp GAUSS.gauss_lemma power_add[of "-1::int" m n] QR_lemma_14 |
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366 unfolding r_def m_def n_def by auto |
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367 |
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368 end |
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369 |
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370 theorem Quadratic_Reciprocity: assumes "prime p" "2 < p" "prime q" "2 < q" "p \<noteq> q" |
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371 shows "(Legendre p q) * (Legendre q p) = (-1::int) ^ ((p - 1) div 2 * ((q - 1) div 2))" |
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372 using QR.Quadratic_Reciprocity QR_def assms by blast |
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373 |
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374 theorem Quadratic_Reciprocity_int: assumes "prime (nat p)" "2 < p" "prime (nat q)" "2 < q" "p \<noteq> q" |
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375 shows "(Legendre p q) * (Legendre q p) = (-1::int) ^ (nat ((p - 1) div 2 * ((q - 1) div 2)))" |
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376 proof - |
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377 have "0 \<le> (p - 1) div 2" using assms by simp |
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378 moreover have "(nat p - 1) div 2 = nat ((p - 1) div 2)" "(nat q - 1) div 2 = nat ((q - 1) div 2)" |
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379 by fastforce+ |
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380 ultimately have "(nat p - 1) div 2 * ((nat q - 1) div 2) = nat ((p - 1) div 2 * ((q - 1) div 2))" |
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381 using nat_mult_distrib by presburger |
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382 moreover have "2 < nat p" "2 < nat q" "nat p \<noteq> nat q" "int (nat p) = p" "int (nat q) = q" |
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383 using assms by linarith+ |
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384 ultimately show ?thesis using Quadratic_Reciprocity[of "nat p" "nat q"] assms by presburger |
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385 qed |
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386 |
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387 end |