src/HOL/Old_Number_Theory/Pocklington.thy
author bulwahn
Fri Oct 21 11:17:14 2011 +0200 (2011-10-21)
changeset 45231 d85a2fdc586c
parent 41541 1fa4725c4656
child 49962 a8cc904a6820
permissions -rw-r--r--
replacing code_inline by code_unfold, removing obsolete code_unfold, code_inline del now that the ancient code generator is removed
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(*  Title:      HOL/Old_Number_Theory/Pocklington.thy
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    Author:     Amine Chaieb
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*)
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header {* Pocklington's Theorem for Primes *}
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theory Pocklington
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imports Primes
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begin
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definition modeq:: "nat => nat => nat => bool"    ("(1[_ = _] '(mod _'))")
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  where "[a = b] (mod p) == ((a mod p) = (b mod p))"
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definition modneq:: "nat => nat => nat => bool"    ("(1[_ \<noteq> _] '(mod _'))")
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  where "[a \<noteq> b] (mod p) == ((a mod p) \<noteq> (b mod p))"
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lemma modeq_trans:
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  "\<lbrakk> [a = b] (mod p); [b = c] (mod p) \<rbrakk> \<Longrightarrow> [a = c] (mod p)"
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  by (simp add:modeq_def)
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lemma nat_mod_lemma: assumes xyn: "[x = y] (mod n)" and xy:"y \<le> x"
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  shows "\<exists>q. x = y + n * q"
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using xyn xy unfolding modeq_def using nat_mod_eq_lemma by blast
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lemma nat_mod[algebra]: "[x = y] (mod n) \<longleftrightarrow> (\<exists>q1 q2. x + n * q1 = y + n * q2)"
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unfolding modeq_def nat_mod_eq_iff ..
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(* Lemmas about previously defined terms.                                    *)
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lemma prime: "prime p \<longleftrightarrow> p \<noteq> 0 \<and> p\<noteq>1 \<and> (\<forall>m. 0 < m \<and> m < p \<longrightarrow> coprime p m)"
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  (is "?lhs \<longleftrightarrow> ?rhs")
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proof-
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  {assume "p=0 \<or> p=1" hence ?thesis using prime_0 prime_1 by (cases "p=0", simp_all)}
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  moreover
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  {assume p0: "p\<noteq>0" "p\<noteq>1"
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    {assume H: "?lhs"
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      {fix m assume m: "m > 0" "m < p"
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        {assume "m=1" hence "coprime p m" by simp}
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        moreover
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        {assume "p dvd m" hence "p \<le> m" using dvd_imp_le m by blast with m(2)
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          have "coprime p m" by simp}
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        ultimately have "coprime p m" using prime_coprime[OF H, of m] by blast}
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      hence ?rhs using p0 by auto}
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    moreover
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    { assume H: "\<forall>m. 0 < m \<and> m < p \<longrightarrow> coprime p m"
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      from prime_factor[OF p0(2)] obtain q where q: "prime q" "q dvd p" by blast
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      from prime_ge_2[OF q(1)] have q0: "q > 0" by arith
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      from dvd_imp_le[OF q(2)] p0 have qp: "q \<le> p" by arith
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      {assume "q = p" hence ?lhs using q(1) by blast}
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      moreover
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      {assume "q\<noteq>p" with qp have qplt: "q < p" by arith
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        from H[rule_format, of q] qplt q0 have "coprime p q" by arith
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        with coprime_prime[of p q q] q have False by simp hence ?lhs by blast}
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      ultimately have ?lhs by blast}
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    ultimately have ?thesis by blast}
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  ultimately show ?thesis  by (cases"p=0 \<or> p=1", auto)
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qed
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lemma finite_number_segment: "card { m. 0 < m \<and> m < n } = n - 1"
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proof-
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  have "{ m. 0 < m \<and> m < n } = {1..<n}" by auto
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  thus ?thesis by simp
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qed
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lemma coprime_mod: assumes n: "n \<noteq> 0" shows "coprime (a mod n) n \<longleftrightarrow> coprime a n"
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  using n dvd_mod_iff[of _ n a] by (auto simp add: coprime)
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(* Congruences.                                                              *)
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lemma cong_mod_01[simp,presburger]:
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  "[x = y] (mod 0) \<longleftrightarrow> x = y" "[x = y] (mod 1)" "[x = 0] (mod n) \<longleftrightarrow> n dvd x"
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  by (simp_all add: modeq_def, presburger)
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lemma cong_sub_cases:
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  "[x = y] (mod n) \<longleftrightarrow> (if x <= y then [y - x = 0] (mod n) else [x - y = 0] (mod n))"
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apply (auto simp add: nat_mod)
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apply (rule_tac x="q2" in exI)
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apply (rule_tac x="q1" in exI, simp)
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apply (rule_tac x="q2" in exI)
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apply (rule_tac x="q1" in exI, simp)
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apply (rule_tac x="q1" in exI)
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apply (rule_tac x="q2" in exI, simp)
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apply (rule_tac x="q1" in exI)
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apply (rule_tac x="q2" in exI, simp)
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done
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lemma cong_mult_lcancel: assumes an: "coprime a n" and axy:"[a * x = a * y] (mod n)"
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  shows "[x = y] (mod n)"
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proof-
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  {assume "a = 0" with an axy coprime_0'[of n] have ?thesis by (simp add: modeq_def) }
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  moreover
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  {assume az: "a\<noteq>0"
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    {assume xy: "x \<le> y" hence axy': "a*x \<le> a*y" by simp
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      with axy cong_sub_cases[of "a*x" "a*y" n]  have "[a*(y - x) = 0] (mod n)"
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        by (simp only: if_True diff_mult_distrib2)
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      hence th: "n dvd a*(y -x)" by simp
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      from coprime_divprod[OF th] an have "n dvd y - x"
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        by (simp add: coprime_commute)
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      hence ?thesis using xy cong_sub_cases[of x y n] by simp}
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    moreover
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    {assume H: "\<not>x \<le> y" hence xy: "y \<le> x"  by arith
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      from H az have axy': "\<not> a*x \<le> a*y" by auto
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      with axy H cong_sub_cases[of "a*x" "a*y" n]  have "[a*(x - y) = 0] (mod n)"
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        by (simp only: if_False diff_mult_distrib2)
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      hence th: "n dvd a*(x - y)" by simp
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      from coprime_divprod[OF th] an have "n dvd x - y"
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        by (simp add: coprime_commute)
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      hence ?thesis using xy cong_sub_cases[of x y n] by simp}
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    ultimately have ?thesis by blast}
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  ultimately show ?thesis by blast
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qed
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lemma cong_mult_rcancel: assumes an: "coprime a n" and axy:"[x*a = y*a] (mod n)"
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  shows "[x = y] (mod n)"
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  using cong_mult_lcancel[OF an axy[unfolded mult_commute[of _a]]] .
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lemma cong_refl: "[x = x] (mod n)" by (simp add: modeq_def)
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lemma eq_imp_cong: "a = b \<Longrightarrow> [a = b] (mod n)" by (simp add: cong_refl)
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lemma cong_commute: "[x = y] (mod n) \<longleftrightarrow> [y = x] (mod n)"
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  by (auto simp add: modeq_def)
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lemma cong_trans[trans]: "[x = y] (mod n) \<Longrightarrow> [y = z] (mod n) \<Longrightarrow> [x = z] (mod n)"
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  by (simp add: modeq_def)
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lemma cong_add: assumes xx': "[x = x'] (mod n)" and yy':"[y = y'] (mod n)"
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  shows "[x + y = x' + y'] (mod n)"
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proof-
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  have "(x + y) mod n = (x mod n + y mod n) mod n"
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    by (simp add: mod_add_left_eq[of x y n] mod_add_right_eq[of "x mod n" y n])
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  also have "\<dots> = (x' mod n + y' mod n) mod n" using xx' yy' modeq_def by simp
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  also have "\<dots> = (x' + y') mod n"
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    by (simp add: mod_add_left_eq[of x' y' n] mod_add_right_eq[of "x' mod n" y' n])
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  finally show ?thesis unfolding modeq_def .
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qed
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lemma cong_mult: assumes xx': "[x = x'] (mod n)" and yy':"[y = y'] (mod n)"
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  shows "[x * y = x' * y'] (mod n)"
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proof-
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  have "(x * y) mod n = (x mod n) * (y mod n) mod n"
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    by (simp add: mod_mult_left_eq[of x y n] mod_mult_right_eq[of "x mod n" y n])
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  also have "\<dots> = (x' mod n) * (y' mod n) mod n" using xx'[unfolded modeq_def] yy'[unfolded modeq_def] by simp
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  also have "\<dots> = (x' * y') mod n"
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    by (simp add: mod_mult_left_eq[of x' y' n] mod_mult_right_eq[of "x' mod n" y' n])
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  finally show ?thesis unfolding modeq_def .
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qed
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lemma cong_exp: "[x = y] (mod n) \<Longrightarrow> [x^k = y^k] (mod n)"
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  by (induct k, auto simp add: cong_refl cong_mult)
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lemma cong_sub: assumes xx': "[x = x'] (mod n)" and yy': "[y = y'] (mod n)"
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  and yx: "y <= x" and yx': "y' <= x'"
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  shows "[x - y = x' - y'] (mod n)"
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proof-
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  { fix x a x' a' y b y' b'
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    have "(x::nat) + a = x' + a' \<Longrightarrow> y + b = y' + b' \<Longrightarrow> y <= x \<Longrightarrow> y' <= x'
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      \<Longrightarrow> (x - y) + (a + b') = (x' - y') + (a' + b)" by arith}
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  note th = this
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  from xx' yy' obtain q1 q2 q1' q2' where q12: "x + n*q1 = x'+n*q2"
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    and q12': "y + n*q1' = y'+n*q2'" unfolding nat_mod by blast+
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  from th[OF q12 q12' yx yx']
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  have "(x - y) + n*(q1 + q2') = (x' - y') + n*(q2 + q1')"
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    by (simp add: right_distrib)
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  thus ?thesis unfolding nat_mod by blast
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qed
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lemma cong_mult_lcancel_eq: assumes an: "coprime a n"
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  shows "[a * x = a * y] (mod n) \<longleftrightarrow> [x = y] (mod n)" (is "?lhs \<longleftrightarrow> ?rhs")
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proof
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  assume H: "?rhs" from cong_mult[OF cong_refl[of a n] H] show ?lhs .
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next
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  assume H: "?lhs" hence H': "[x*a = y*a] (mod n)" by (simp add: mult_commute)
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  from cong_mult_rcancel[OF an H'] show ?rhs  .
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qed
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lemma cong_mult_rcancel_eq: assumes an: "coprime a n"
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  shows "[x * a = y * a] (mod n) \<longleftrightarrow> [x = y] (mod n)"
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using cong_mult_lcancel_eq[OF an, of x y] by (simp add: mult_commute)
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lemma cong_add_lcancel_eq: "[a + x = a + y] (mod n) \<longleftrightarrow> [x = y] (mod n)"
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  by (simp add: nat_mod)
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lemma cong_add_rcancel_eq: "[x + a = y + a] (mod n) \<longleftrightarrow> [x = y] (mod n)"
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  by (simp add: nat_mod)
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lemma cong_add_rcancel: "[x + a = y + a] (mod n) \<Longrightarrow> [x = y] (mod n)"
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  by (simp add: nat_mod)
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lemma cong_add_lcancel: "[a + x = a + y] (mod n) \<Longrightarrow> [x = y] (mod n)"
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  by (simp add: nat_mod)
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lemma cong_add_lcancel_eq_0: "[a + x = a] (mod n) \<longleftrightarrow> [x = 0] (mod n)"
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  by (simp add: nat_mod)
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lemma cong_add_rcancel_eq_0: "[x + a = a] (mod n) \<longleftrightarrow> [x = 0] (mod n)"
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  by (simp add: nat_mod)
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lemma cong_imp_eq: assumes xn: "x < n" and yn: "y < n" and xy: "[x = y] (mod n)"
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  shows "x = y"
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  using xy[unfolded modeq_def mod_less[OF xn] mod_less[OF yn]] .
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lemma cong_divides_modulus: "[x = y] (mod m) \<Longrightarrow> n dvd m ==> [x = y] (mod n)"
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  apply (auto simp add: nat_mod dvd_def)
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  apply (rule_tac x="k*q1" in exI)
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  apply (rule_tac x="k*q2" in exI)
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  by simp
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lemma cong_0_divides: "[x = 0] (mod n) \<longleftrightarrow> n dvd x" by simp
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lemma cong_1_divides:"[x = 1] (mod n) ==> n dvd x - 1"
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  apply (cases "x\<le>1", simp_all)
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  using cong_sub_cases[of x 1 n] by auto
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lemma cong_divides: "[x = y] (mod n) \<Longrightarrow> n dvd x \<longleftrightarrow> n dvd y"
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apply (auto simp add: nat_mod dvd_def)
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apply (rule_tac x="k + q1 - q2" in exI, simp add: add_mult_distrib2 diff_mult_distrib2)
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apply (rule_tac x="k + q2 - q1" in exI, simp add: add_mult_distrib2 diff_mult_distrib2)
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done
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lemma cong_coprime: assumes xy: "[x = y] (mod n)"
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  shows "coprime n x \<longleftrightarrow> coprime n y"
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proof-
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  {assume "n=0" hence ?thesis using xy by simp}
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  moreover
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  {assume nz: "n \<noteq> 0"
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  have "coprime n x \<longleftrightarrow> coprime (x mod n) n"
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    by (simp add: coprime_mod[OF nz, of x] coprime_commute[of n x])
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  also have "\<dots> \<longleftrightarrow> coprime (y mod n) n" using xy[unfolded modeq_def] by simp
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  also have "\<dots> \<longleftrightarrow> coprime y n" by (simp add: coprime_mod[OF nz, of y])
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  finally have ?thesis by (simp add: coprime_commute) }
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ultimately show ?thesis by blast
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qed
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lemma cong_mod: "~(n = 0) \<Longrightarrow> [a mod n = a] (mod n)" by (simp add: modeq_def)
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lemma mod_mult_cong: "~(a = 0) \<Longrightarrow> ~(b = 0)
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  \<Longrightarrow> [x mod (a * b) = y] (mod a) \<longleftrightarrow> [x = y] (mod a)"
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  by (simp add: modeq_def mod_mult2_eq mod_add_left_eq)
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lemma cong_mod_mult: "[x = y] (mod n) \<Longrightarrow> m dvd n \<Longrightarrow> [x = y] (mod m)"
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  apply (auto simp add: nat_mod dvd_def)
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  apply (rule_tac x="k*q1" in exI)
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  apply (rule_tac x="k*q2" in exI, simp)
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  done
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(* Some things when we know more about the order.                            *)
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lemma cong_le: "y <= x \<Longrightarrow> [x = y] (mod n) \<longleftrightarrow> (\<exists>q. x = q * n + y)"
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  using nat_mod_lemma[of x y n]
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  apply auto
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  apply (simp add: nat_mod)
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  apply (rule_tac x="q" in exI)
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  apply (rule_tac x="q + q" in exI)
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  by (auto simp: algebra_simps)
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lemma cong_to_1: "[a = 1] (mod n) \<longleftrightarrow> a = 0 \<and> n = 1 \<or> (\<exists>m. a = 1 + m * n)"
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proof-
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  {assume "n = 0 \<or> n = 1\<or> a = 0 \<or> a = 1" hence ?thesis
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      apply (cases "n=0", simp_all add: cong_commute)
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      apply (cases "n=1", simp_all add: cong_commute modeq_def)
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      apply arith
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      apply (cases "a=1")
wenzelm@41541
   264
      apply (simp_all add: modeq_def cong_commute)
wenzelm@41541
   265
      done }
chaieb@26126
   266
  moreover
chaieb@26126
   267
  {assume n: "n\<noteq>0" "n\<noteq>1" and a:"a\<noteq>0" "a \<noteq> 1" hence a': "a \<ge> 1" by simp
chaieb@26126
   268
    hence ?thesis using cong_le[OF a', of n] by auto }
chaieb@26126
   269
  ultimately show ?thesis by auto
chaieb@26126
   270
qed
chaieb@26126
   271
chaieb@26126
   272
(* Some basic theorems about solving congruences.                            *)
chaieb@26126
   273
chaieb@26126
   274
chaieb@26126
   275
lemma cong_solve: assumes an: "coprime a n" shows "\<exists>x. [a * x = b] (mod n)"
chaieb@26126
   276
proof-
chaieb@26126
   277
  {assume "a=0" hence ?thesis using an by (simp add: modeq_def)}
chaieb@26126
   278
  moreover
chaieb@26126
   279
  {assume az: "a\<noteq>0"
huffman@30488
   280
  from bezout_add_strong[OF az, of n]
chaieb@26126
   281
  obtain d x y where dxy: "d dvd a" "d dvd n" "a*x = n*y + d" by blast
chaieb@26126
   282
  from an[unfolded coprime, rule_format, of d] dxy(1,2) have d1: "d = 1" by blast
chaieb@26126
   283
  hence "a*x*b = (n*y + 1)*b" using dxy(3) by simp
chaieb@26126
   284
  hence "a*(x*b) = n*(y*b) + b" by algebra
chaieb@26126
   285
  hence "a*(x*b) mod n = (n*(y*b) + b) mod n" by simp
chaieb@26126
   286
  hence "a*(x*b) mod n = b mod n" by (simp add: mod_add_left_eq)
chaieb@26126
   287
  hence "[a*(x*b) = b] (mod n)" unfolding modeq_def .
chaieb@26126
   288
  hence ?thesis by blast}
chaieb@26126
   289
ultimately  show ?thesis by blast
chaieb@26126
   290
qed
chaieb@26126
   291
chaieb@26126
   292
lemma cong_solve_unique: assumes an: "coprime a n" and nz: "n \<noteq> 0"
chaieb@26126
   293
  shows "\<exists>!x. x < n \<and> [a * x = b] (mod n)"
chaieb@26126
   294
proof-
chaieb@26126
   295
  let ?P = "\<lambda>x. x < n \<and> [a * x = b] (mod n)"
chaieb@26126
   296
  from cong_solve[OF an] obtain x where x: "[a*x = b] (mod n)" by blast
chaieb@26126
   297
  let ?x = "x mod n"
chaieb@26126
   298
  from x have th: "[a * ?x = b] (mod n)"
nipkow@30224
   299
    by (simp add: modeq_def mod_mult_right_eq[of a x n])
chaieb@26126
   300
  from mod_less_divisor[ of n x] nz th have Px: "?P ?x" by simp
chaieb@26126
   301
  {fix y assume Py: "y < n" "[a * y = b] (mod n)"
chaieb@26126
   302
    from Py(2) th have "[a * y = a*?x] (mod n)" by (simp add: modeq_def)
chaieb@26126
   303
    hence "[y = ?x] (mod n)" by (simp add: cong_mult_lcancel_eq[OF an])
chaieb@26126
   304
    with mod_less[OF Py(1)] mod_less_divisor[ of n x] nz
chaieb@26126
   305
    have "y = ?x" by (simp add: modeq_def)}
chaieb@26126
   306
  with Px show ?thesis by blast
chaieb@26126
   307
qed
chaieb@26126
   308
chaieb@26126
   309
lemma cong_solve_unique_nontrivial:
chaieb@26126
   310
  assumes p: "prime p" and pa: "coprime p a" and x0: "0 < x" and xp: "x < p"
chaieb@26126
   311
  shows "\<exists>!y. 0 < y \<and> y < p \<and> [x * y = a] (mod p)"
chaieb@26126
   312
proof-
chaieb@26126
   313
  from p have p1: "p > 1" using prime_ge_2[OF p] by arith
chaieb@26126
   314
  hence p01: "p \<noteq> 0" "p \<noteq> 1" by arith+
chaieb@26126
   315
  from pa have ap: "coprime a p" by (simp add: coprime_commute)
chaieb@26126
   316
  from prime_coprime[OF p, of x] dvd_imp_le[of p x] x0 xp have px:"coprime x p"
chaieb@26126
   317
    by (auto simp add: coprime_commute)
huffman@30488
   318
  from cong_solve_unique[OF px p01(1)]
chaieb@26126
   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
chaieb@26126
   320
  {assume y0: "y = 0"
chaieb@26126
   321
    with y(2) have th: "p dvd a" by (simp add: cong_commute[of 0 a p])
chaieb@26126
   322
    with p coprime_prime[OF pa, of p] have False by simp}
huffman@30488
   323
  with y show ?thesis unfolding Ex1_def using neq0_conv by blast
chaieb@26126
   324
qed
chaieb@26126
   325
lemma cong_unique_inverse_prime:
chaieb@26126
   326
  assumes p: "prime p" and x0: "0 < x" and xp: "x < p"
chaieb@26126
   327
  shows "\<exists>!y. 0 < y \<and> y < p \<and> [x * y = 1] (mod p)"
chaieb@26126
   328
  using cong_solve_unique_nontrivial[OF p coprime_1[of p] x0 xp] .
chaieb@26126
   329
chaieb@26126
   330
(* Forms of the Chinese remainder theorem.                                   *)
chaieb@26126
   331
huffman@30488
   332
lemma cong_chinese:
huffman@30488
   333
  assumes ab: "coprime a b" and  xya: "[x = y] (mod a)"
chaieb@26126
   334
  and xyb: "[x = y] (mod b)"
chaieb@26126
   335
  shows "[x = y] (mod a*b)"
chaieb@26126
   336
  using ab xya xyb
huffman@30488
   337
  by (simp add: cong_sub_cases[of x y a] cong_sub_cases[of x y b]
huffman@30488
   338
    cong_sub_cases[of x y "a*b"])
chaieb@26126
   339
(cases "x \<le> y", simp_all add: divides_mul[of a _ b])
chaieb@26126
   340
chaieb@26126
   341
lemma chinese_remainder_unique:
chaieb@26126
   342
  assumes ab: "coprime a b" and az: "a \<noteq> 0" and bz: "b\<noteq>0"
chaieb@26126
   343
  shows "\<exists>!x. x < a * b \<and> [x = m] (mod a) \<and> [x = n] (mod b)"
chaieb@26126
   344
proof-
chaieb@26126
   345
  from az bz have abpos: "a*b > 0" by simp
huffman@30488
   346
  from chinese_remainder[OF ab az bz] obtain x q1 q2 where
chaieb@26126
   347
    xq12: "x = m + q1 * a" "x = n + q2 * b" by blast
huffman@30488
   348
  let ?w = "x mod (a*b)"
chaieb@26126
   349
  have wab: "?w < a*b" by (simp add: mod_less_divisor[OF abpos])
chaieb@26126
   350
  from xq12(1) have "?w mod a = ((m + q1 * a) mod (a*b)) mod a" by simp
chaieb@26126
   351
  also have "\<dots> = m mod a" apply (simp add: mod_mult2_eq)
chaieb@26126
   352
    apply (subst mod_add_left_eq)
chaieb@26126
   353
    by simp
chaieb@26126
   354
  finally have th1: "[?w = m] (mod a)" by (simp add: modeq_def)
chaieb@26126
   355
  from xq12(2) have "?w mod b = ((n + q2 * b) mod (a*b)) mod b" by simp
chaieb@26126
   356
  also have "\<dots> = ((n + q2 * b) mod (b*a)) mod b" by (simp add: mult_commute)
chaieb@26126
   357
  also have "\<dots> = n mod b" apply (simp add: mod_mult2_eq)
chaieb@26126
   358
    apply (subst mod_add_left_eq)
chaieb@26126
   359
    by simp
chaieb@26126
   360
  finally have th2: "[?w = n] (mod b)" by (simp add: modeq_def)
chaieb@26126
   361
  {fix y assume H: "y < a*b" "[y = m] (mod a)" "[y = n] (mod b)"
chaieb@26126
   362
    with th1 th2 have H': "[y = ?w] (mod a)" "[y = ?w] (mod b)"
chaieb@26126
   363
      by (simp_all add: modeq_def)
huffman@30488
   364
    from cong_chinese[OF ab H'] mod_less[OF H(1)] mod_less[OF wab]
chaieb@26126
   365
    have "y = ?w" by (simp add: modeq_def)}
chaieb@26126
   366
  with th1 th2 wab show ?thesis by blast
chaieb@26126
   367
qed
chaieb@26126
   368
chaieb@26126
   369
lemma chinese_remainder_coprime_unique:
huffman@30488
   370
  assumes ab: "coprime a b" and az: "a \<noteq> 0" and bz: "b \<noteq> 0"
chaieb@26126
   371
  and ma: "coprime m a" and nb: "coprime n b"
chaieb@26126
   372
  shows "\<exists>!x. coprime x (a * b) \<and> x < a * b \<and> [x = m] (mod a) \<and> [x = n] (mod b)"
chaieb@26126
   373
proof-
chaieb@26126
   374
  let ?P = "\<lambda>x. x < a * b \<and> [x = m] (mod a) \<and> [x = n] (mod b)"
chaieb@26126
   375
  from chinese_remainder_unique[OF ab az bz]
huffman@30488
   376
  obtain x where x: "x < a * b" "[x = m] (mod a)" "[x = n] (mod b)"
chaieb@26126
   377
    "\<forall>y. ?P y \<longrightarrow> y = x" by blast
chaieb@26126
   378
  from ma nb cong_coprime[OF x(2)] cong_coprime[OF x(3)]
chaieb@26126
   379
  have "coprime x a" "coprime x b" by (simp_all add: coprime_commute)
chaieb@26126
   380
  with coprime_mul[of x a b] have "coprime x (a*b)" by simp
chaieb@26126
   381
  with x show ?thesis by blast
chaieb@26126
   382
qed
chaieb@26126
   383
chaieb@26126
   384
(* Euler totient function.                                                   *)
chaieb@26126
   385
chaieb@26126
   386
definition phi_def: "\<phi> n = card { m. 0 < m \<and> m <= n \<and> coprime m n }"
nipkow@31197
   387
chaieb@26126
   388
lemma phi_0[simp]: "\<phi> 0 = 0"
nipkow@31197
   389
  unfolding phi_def by auto
chaieb@26126
   390
chaieb@26126
   391
lemma phi_finite[simp]: "finite ({ m. 0 < m \<and> m <= n \<and> coprime m n })"
chaieb@26126
   392
proof-
chaieb@26126
   393
  have "{ m. 0 < m \<and> m <= n \<and> coprime m n } \<subseteq> {0..n}" by auto
chaieb@26126
   394
  thus ?thesis by (auto intro: finite_subset)
chaieb@26126
   395
qed
chaieb@26126
   396
chaieb@26126
   397
declare coprime_1[presburger]
chaieb@26126
   398
lemma phi_1[simp]: "\<phi> 1 = 1"
chaieb@26126
   399
proof-
huffman@30488
   400
  {fix m
chaieb@26126
   401
    have "0 < m \<and> m <= 1 \<and> coprime m 1 \<longleftrightarrow> m = 1" by presburger }
chaieb@26126
   402
  thus ?thesis by (simp add: phi_def)
chaieb@26126
   403
qed
chaieb@26126
   404
chaieb@26126
   405
lemma [simp]: "\<phi> (Suc 0) = Suc 0" using phi_1 by simp
chaieb@26126
   406
chaieb@26126
   407
lemma phi_alt: "\<phi>(n) = card { m. coprime m n \<and> m < n}"
chaieb@26126
   408
proof-
chaieb@26126
   409
  {assume "n=0 \<or> n=1" hence ?thesis by (cases "n=0", simp_all)}
chaieb@26126
   410
  moreover
chaieb@26126
   411
  {assume n: "n\<noteq>0" "n\<noteq>1"
chaieb@26126
   412
    {fix m
chaieb@26126
   413
      from n have "0 < m \<and> m <= n \<and> coprime m n \<longleftrightarrow> coprime m n \<and> m < n"
wenzelm@32960
   414
        apply (cases "m = 0", simp_all)
wenzelm@32960
   415
        apply (cases "m = 1", simp_all)
wenzelm@32960
   416
        apply (cases "m = n", auto)
wenzelm@32960
   417
        done }
chaieb@26126
   418
    hence ?thesis unfolding phi_def by simp}
chaieb@26126
   419
  ultimately show ?thesis by auto
chaieb@26126
   420
qed
chaieb@26126
   421
chaieb@26126
   422
lemma phi_finite_lemma[simp]: "finite {m. coprime m n \<and>  m < n}" (is "finite ?S")
chaieb@26126
   423
  by (rule finite_subset[of "?S" "{0..n}"], auto)
chaieb@26126
   424
chaieb@26126
   425
lemma phi_another: assumes n: "n\<noteq>1"
chaieb@26126
   426
  shows "\<phi> n = card {m. 0 < m \<and> m < n \<and> coprime m n }"
chaieb@26126
   427
proof-
huffman@30488
   428
  {fix m
chaieb@26126
   429
    from n have "0 < m \<and> m < n \<and> coprime m n \<longleftrightarrow> coprime m n \<and> m < n"
chaieb@26126
   430
      by (cases "m=0", auto)}
chaieb@26126
   431
  thus ?thesis unfolding phi_alt by auto
chaieb@26126
   432
qed
chaieb@26126
   433
chaieb@26126
   434
lemma phi_limit: "\<phi> n \<le> n"
chaieb@26126
   435
proof-
chaieb@26126
   436
  have "{ m. coprime m n \<and> m < n} \<subseteq> {0 ..<n}" by auto
chaieb@26126
   437
  with card_mono[of "{0 ..<n}" "{ m. coprime m n \<and> m < n}"]
chaieb@26126
   438
  show ?thesis unfolding phi_alt by auto
chaieb@26126
   439
qed
chaieb@26126
   440
chaieb@26126
   441
lemma stupid[simp]: "{m. (0::nat) < m \<and> m < n} = {1..<n}"
chaieb@26126
   442
  by auto
chaieb@26126
   443
huffman@30488
   444
lemma phi_limit_strong: assumes n: "n\<noteq>1"
chaieb@26126
   445
  shows "\<phi>(n) \<le> n - 1"
chaieb@26126
   446
proof-
chaieb@26126
   447
  show ?thesis
huffman@30488
   448
    unfolding phi_another[OF n] finite_number_segment[of n, symmetric]
chaieb@26126
   449
    by (rule card_mono[of "{m. 0 < m \<and> m < n}" "{m. 0 < m \<and> m < n \<and> coprime m n}"], auto)
chaieb@26126
   450
qed
chaieb@26126
   451
chaieb@26126
   452
lemma phi_lowerbound_1_strong: assumes n: "n \<ge> 1"
chaieb@26126
   453
  shows "\<phi>(n) \<ge> 1"
chaieb@26126
   454
proof-
chaieb@26126
   455
  let ?S = "{ m. 0 < m \<and> m <= n \<and> coprime m n }"
huffman@30488
   456
  from card_0_eq[of ?S] n have "\<phi> n \<noteq> 0" unfolding phi_alt
chaieb@26126
   457
    apply auto
chaieb@26126
   458
    apply (cases "n=1", simp_all)
chaieb@26126
   459
    apply (rule exI[where x=1], simp)
chaieb@26126
   460
    done
chaieb@26126
   461
  thus ?thesis by arith
chaieb@26126
   462
qed
chaieb@26126
   463
chaieb@26126
   464
lemma phi_lowerbound_1: "2 <= n ==> 1 <= \<phi>(n)"
chaieb@26126
   465
  using phi_lowerbound_1_strong[of n] by auto
chaieb@26126
   466
chaieb@26126
   467
lemma phi_lowerbound_2: assumes n: "3 <= n" shows "2 <= \<phi> (n)"
chaieb@26126
   468
proof-
chaieb@26126
   469
  let ?S = "{ m. 0 < m \<and> m <= n \<and> coprime m n }"
huffman@30488
   470
  have inS: "{1, n - 1} \<subseteq> ?S" using n coprime_plus1[of "n - 1"]
chaieb@26126
   471
    by (auto simp add: coprime_commute)
chaieb@26126
   472
  from n have c2: "card {1, n - 1} = 2" by (auto simp add: card_insert_if)
huffman@30488
   473
  from card_mono[of ?S "{1, n - 1}", simplified inS c2] show ?thesis
chaieb@26126
   474
    unfolding phi_def by auto
chaieb@26126
   475
qed
chaieb@26126
   476
chaieb@26126
   477
lemma phi_prime: "\<phi> n = n - 1 \<and> n\<noteq>0 \<and> n\<noteq>1 \<longleftrightarrow> prime n"
chaieb@26126
   478
proof-
chaieb@26126
   479
  {assume "n=0 \<or> n=1" hence ?thesis by (cases "n=1", simp_all)}
chaieb@26126
   480
  moreover
chaieb@26126
   481
  {assume n: "n\<noteq>0" "n\<noteq>1"
chaieb@26126
   482
    let ?S = "{m. 0 < m \<and> m < n}"
chaieb@26126
   483
    have fS: "finite ?S" by simp
chaieb@26126
   484
    let ?S' = "{m. 0 < m \<and> m < n \<and> coprime m n}"
chaieb@26126
   485
    have fS':"finite ?S'" apply (rule finite_subset[of ?S' ?S]) by auto
chaieb@26126
   486
    {assume H: "\<phi> n = n - 1 \<and> n\<noteq>0 \<and> n\<noteq>1"
huffman@30488
   487
      hence ceq: "card ?S' = card ?S"
chaieb@26126
   488
      using n finite_number_segment[of n] phi_another[OF n(2)] by simp
chaieb@26126
   489
      {fix m assume m: "0 < m" "m < n" "\<not> coprime m n"
wenzelm@32960
   490
        hence mS': "m \<notin> ?S'" by auto
wenzelm@32960
   491
        have "insert m ?S' \<le> ?S" using m by auto
wenzelm@32960
   492
        from m have "card (insert m ?S') \<le> card ?S"
wenzelm@32960
   493
          by - (rule card_mono[of ?S "insert m ?S'"], auto)
wenzelm@32960
   494
        hence False
wenzelm@32960
   495
          unfolding card_insert_disjoint[of "?S'" m, OF fS' mS'] ceq
wenzelm@32960
   496
          by simp }
chaieb@26126
   497
      hence "\<forall>m. 0 <m \<and> m < n \<longrightarrow> coprime m n" by blast
chaieb@26126
   498
      hence "prime n" unfolding prime using n by (simp add: coprime_commute)}
chaieb@26126
   499
    moreover
chaieb@26126
   500
    {assume H: "prime n"
huffman@30488
   501
      hence "?S = ?S'" unfolding prime using n
wenzelm@32960
   502
        by (auto simp add: coprime_commute)
chaieb@26126
   503
      hence "card ?S = card ?S'" by simp
chaieb@26126
   504
      hence "\<phi> n = n - 1" unfolding phi_another[OF n(2)] by simp}
chaieb@26126
   505
    ultimately have ?thesis using n by blast}
chaieb@26126
   506
  ultimately show ?thesis by (cases "n=0") blast+
chaieb@26126
   507
qed
chaieb@26126
   508
chaieb@26126
   509
(* Multiplicativity property.                                                *)
chaieb@26126
   510
chaieb@26126
   511
lemma phi_multiplicative: assumes ab: "coprime a b"
chaieb@26126
   512
  shows "\<phi> (a * b) = \<phi> a * \<phi> b"
chaieb@26126
   513
proof-
huffman@30488
   514
  {assume "a = 0 \<or> b = 0 \<or> a = 1 \<or> b = 1"
chaieb@26126
   515
    hence ?thesis
chaieb@26126
   516
      by (cases "a=0", simp, cases "b=0", simp, cases"a=1", simp_all) }
chaieb@26126
   517
  moreover
chaieb@26126
   518
  {assume a: "a\<noteq>0" "a\<noteq>1" and b: "b\<noteq>0" "b\<noteq>1"
chaieb@26126
   519
    hence ab0: "a*b \<noteq> 0" by simp
chaieb@26126
   520
    let ?S = "\<lambda>k. {m. coprime m k \<and> m < k}"
chaieb@26126
   521
    let ?f = "\<lambda>x. (x mod a, x mod b)"
chaieb@26126
   522
    have eq: "?f ` (?S (a*b)) = (?S a \<times> ?S b)"
chaieb@26126
   523
    proof-
chaieb@26126
   524
      {fix x assume x:"x \<in> ?S (a*b)"
wenzelm@32960
   525
        hence x': "coprime x (a*b)" "x < a*b" by simp_all
wenzelm@32960
   526
        hence xab: "coprime x a" "coprime x b" by (simp_all add: coprime_mul_eq)
wenzelm@32960
   527
        from mod_less_divisor a b have xab':"x mod a < a" "x mod b < b" by auto
wenzelm@32960
   528
        from xab xab' have "?f x \<in> (?S a \<times> ?S b)"
wenzelm@32960
   529
          by (simp add: coprime_mod[OF a(1)] coprime_mod[OF b(1)])}
chaieb@26126
   530
      moreover
chaieb@26126
   531
      {fix x y assume x: "x \<in> ?S a" and y: "y \<in> ?S b"
wenzelm@32960
   532
        hence x': "coprime x a" "x < a" and y': "coprime y b" "y < b" by simp_all
wenzelm@32960
   533
        from chinese_remainder_coprime_unique[OF ab a(1) b(1) x'(1) y'(1)]
wenzelm@32960
   534
        obtain z where z: "coprime z (a * b)" "z < a * b" "[z = x] (mod a)"
wenzelm@32960
   535
          "[z = y] (mod b)" by blast
wenzelm@32960
   536
        hence "(x,y) \<in> ?f ` (?S (a*b))"
wenzelm@32960
   537
          using y'(2) mod_less_divisor[of b y] x'(2) mod_less_divisor[of a x]
wenzelm@32960
   538
          by (auto simp add: image_iff modeq_def)}
chaieb@26126
   539
      ultimately show ?thesis by auto
chaieb@26126
   540
    qed
chaieb@26126
   541
    have finj: "inj_on ?f (?S (a*b))"
chaieb@26126
   542
      unfolding inj_on_def
chaieb@26126
   543
    proof(clarify)
huffman@30488
   544
      fix x y assume H: "coprime x (a * b)" "x < a * b" "coprime y (a * b)"
wenzelm@32960
   545
        "y < a * b" "x mod a = y mod a" "x mod b = y mod b"
huffman@30488
   546
      hence cp: "coprime x a" "coprime x b" "coprime y a" "coprime y b"
wenzelm@32960
   547
        by (simp_all add: coprime_mul_eq)
chaieb@26126
   548
      from chinese_remainder_coprime_unique[OF ab a(1) b(1) cp(3,4)] H
chaieb@26126
   549
      show "x = y" unfolding modeq_def by blast
chaieb@26126
   550
    qed
chaieb@26126
   551
    from card_image[OF finj, unfolded eq] have ?thesis
chaieb@26126
   552
      unfolding phi_alt by simp }
chaieb@26126
   553
  ultimately show ?thesis by auto
chaieb@26126
   554
qed
chaieb@26126
   555
chaieb@26126
   556
(* Fermat's Little theorem / Fermat-Euler theorem.                           *)
chaieb@26126
   557
chaieb@26126
   558
chaieb@26126
   559
lemma nproduct_mod:
chaieb@26126
   560
  assumes fS: "finite S" and n0: "n \<noteq> 0"
chaieb@26126
   561
  shows "[setprod (\<lambda>m. a(m) mod n) S = setprod a S] (mod n)"
chaieb@26126
   562
proof-
chaieb@26126
   563
  have th1:"[1 = 1] (mod n)" by (simp add: modeq_def)
chaieb@26126
   564
  from cong_mult
chaieb@26126
   565
  have th3:"\<forall>x1 y1 x2 y2.
chaieb@26126
   566
    [x1 = x2] (mod n) \<and> [y1 = y2] (mod n) \<longrightarrow> [x1 * y1 = x2 * y2] (mod n)"
chaieb@26126
   567
    by blast
chaieb@26126
   568
  have th4:"\<forall>x\<in>S. [a x mod n = a x] (mod n)" by (simp add: modeq_def)
nipkow@28854
   569
  from fold_image_related[where h="(\<lambda>m. a(m) mod n)" and g=a, OF th1 th3 fS, OF th4] show ?thesis unfolding setprod_def by (simp add: fS)
chaieb@26126
   570
qed
chaieb@26126
   571
chaieb@26126
   572
lemma nproduct_cmul:
chaieb@26126
   573
  assumes fS:"finite S"
haftmann@31021
   574
  shows "setprod (\<lambda>m. (c::'a::{comm_monoid_mult})* a(m)) S = c ^ (card S) * setprod a S"
chaieb@26126
   575
unfolding setprod_timesf setprod_constant[OF fS, of c] ..
chaieb@26126
   576
chaieb@26126
   577
lemma coprime_nproduct:
chaieb@26126
   578
  assumes fS: "finite S" and Sn: "\<forall>x\<in>S. coprime n (a x)"
chaieb@26126
   579
  shows "coprime n (setprod a S)"
haftmann@27368
   580
  using fS unfolding setprod_def by (rule finite_subset_induct)
haftmann@27368
   581
    (insert Sn, auto simp add: coprime_mul)
chaieb@26126
   582
chaieb@26126
   583
lemma fermat_little: assumes an: "coprime a n"
chaieb@26126
   584
  shows "[a ^ (\<phi> n) = 1] (mod n)"
chaieb@26126
   585
proof-
chaieb@26126
   586
  {assume "n=0" hence ?thesis by simp}
chaieb@26126
   587
  moreover
chaieb@26126
   588
  {assume "n=1" hence ?thesis by (simp add: modeq_def)}
chaieb@26126
   589
  moreover
chaieb@26126
   590
  {assume nz: "n \<noteq> 0" and n1: "n \<noteq> 1"
chaieb@26126
   591
    let ?S = "{m. coprime m n \<and> m < n}"
chaieb@26126
   592
    let ?P = "\<Prod> ?S"
chaieb@26126
   593
    have fS: "finite ?S" by simp
chaieb@26126
   594
    have cardfS: "\<phi> n = card ?S" unfolding phi_alt ..
chaieb@26126
   595
    {fix m assume m: "m \<in> ?S"
chaieb@26126
   596
      hence "coprime m n" by simp
huffman@30488
   597
      with coprime_mul[of n a m] an have "coprime (a*m) n"
wenzelm@32960
   598
        by (simp add: coprime_commute)}
chaieb@26126
   599
    hence Sn: "\<forall>m\<in> ?S. coprime (a*m) n " by blast
chaieb@26126
   600
    from coprime_nproduct[OF fS, of n "\<lambda>m. m"] have nP:"coprime ?P n"
chaieb@26126
   601
      by (simp add: coprime_commute)
chaieb@26126
   602
    have Paphi: "[?P*a^ (\<phi> n) = ?P*1] (mod n)"
chaieb@26126
   603
    proof-
chaieb@26126
   604
      let ?h = "\<lambda>m. m mod n"
chaieb@26126
   605
      {fix m assume mS: "m\<in> ?S"
wenzelm@32960
   606
        hence "?h m \<in> ?S" by simp}
chaieb@26126
   607
      hence hS: "?h ` ?S = ?S"by (auto simp add: image_iff)
chaieb@26126
   608
      have "a\<noteq>0" using an n1 nz apply- apply (rule ccontr) by simp
chaieb@26126
   609
      hence inj: "inj_on (op * a) ?S" unfolding inj_on_def by simp
huffman@30488
   610
nipkow@28854
   611
      have eq0: "fold_image op * (?h \<circ> op * a) 1 {m. coprime m n \<and> m < n} =
nipkow@28854
   612
     fold_image op * (\<lambda>m. m) 1 {m. coprime m n \<and> m < n}"
nipkow@28854
   613
      proof (rule fold_image_eq_general[where h="?h o (op * a)"])
wenzelm@32960
   614
        show "finite ?S" using fS .
chaieb@26126
   615
      next
wenzelm@32960
   616
        {fix y assume yS: "y \<in> ?S" hence y: "coprime y n" "y < n" by simp_all
wenzelm@32960
   617
          from cong_solve_unique[OF an nz, of y]
wenzelm@32960
   618
          obtain x where x:"x < n" "[a * x = y] (mod n)" "\<forall>z. z < n \<and> [a * z = y] (mod n) \<longrightarrow> z=x" by blast
wenzelm@32960
   619
          from cong_coprime[OF x(2)] y(1)
wenzelm@32960
   620
          have xm: "coprime x n" by (simp add: coprime_mul_eq coprime_commute)
wenzelm@32960
   621
          {fix z assume "z \<in> ?S" "(?h \<circ> op * a) z = y"
wenzelm@32960
   622
            hence z: "coprime z n" "z < n" "(?h \<circ> op * a) z = y" by simp_all
wenzelm@32960
   623
            from x(3)[rule_format, of z] z(2,3) have "z=x"
wenzelm@32960
   624
              unfolding modeq_def mod_less[OF y(2)] by simp}
wenzelm@32960
   625
          with xm x(1,2) have "\<exists>!x. x \<in> ?S \<and> (?h \<circ> op * a) x = y"
wenzelm@32960
   626
            unfolding modeq_def mod_less[OF y(2)] by auto }
wenzelm@32960
   627
        thus "\<forall>y\<in>{m. coprime m n \<and> m < n}.
chaieb@26126
   628
       \<exists>!x. x \<in> {m. coprime m n \<and> m < n} \<and> ((\<lambda>m. m mod n) \<circ> op * a) x = y" by blast
chaieb@26126
   629
      next
wenzelm@32960
   630
        {fix x assume xS: "x\<in> ?S"
wenzelm@32960
   631
          hence x: "coprime x n" "x < n" by simp_all
wenzelm@32960
   632
          with an have "coprime (a*x) n"
wenzelm@32960
   633
            by (simp add: coprime_mul_eq[of n a x] coprime_commute)
wenzelm@32960
   634
          hence "?h (a*x) \<in> ?S" using nz
wenzelm@41541
   635
            by (simp add: coprime_mod[OF nz])}
wenzelm@32960
   636
        thus " \<forall>x\<in>{m. coprime m n \<and> m < n}.
chaieb@26126
   637
       ((\<lambda>m. m mod n) \<circ> op * a) x \<in> {m. coprime m n \<and> m < n} \<and>
chaieb@26126
   638
       ((\<lambda>m. m mod n) \<circ> op * a) x = ((\<lambda>m. m mod n) \<circ> op * a) x" by simp
chaieb@26126
   639
      qed
chaieb@26126
   640
      from nproduct_mod[OF fS nz, of "op * a"]
chaieb@26126
   641
      have "[(setprod (op *a) ?S) = (setprod (?h o (op * a)) ?S)] (mod n)"
wenzelm@32960
   642
        unfolding o_def
wenzelm@32960
   643
        by (simp add: cong_commute)
chaieb@26126
   644
      also have "[setprod (?h o (op * a)) ?S = ?P ] (mod n)"
wenzelm@32960
   645
        using eq0 fS an by (simp add: setprod_def modeq_def o_def)
chaieb@26126
   646
      finally show "[?P*a^ (\<phi> n) = ?P*1] (mod n)"
wenzelm@32960
   647
        unfolding cardfS mult_commute[of ?P "a^ (card ?S)"]
wenzelm@32960
   648
          nproduct_cmul[OF fS, symmetric] mult_1_right by simp
chaieb@26126
   649
    qed
chaieb@26126
   650
    from cong_mult_lcancel[OF nP Paphi] have ?thesis . }
chaieb@26126
   651
  ultimately show ?thesis by blast
chaieb@26126
   652
qed
chaieb@26126
   653
chaieb@26126
   654
lemma fermat_little_prime: assumes p: "prime p" and ap: "coprime a p"
chaieb@26126
   655
  shows "[a^ (p - 1) = 1] (mod p)"
chaieb@26126
   656
  using fermat_little[OF ap] p[unfolded phi_prime[symmetric]]
chaieb@26126
   657
by simp
chaieb@26126
   658
chaieb@26126
   659
chaieb@26126
   660
(* Lucas's theorem.                                                          *)
chaieb@26126
   661
chaieb@26126
   662
lemma lucas_coprime_lemma:
chaieb@26126
   663
  assumes m: "m\<noteq>0" and am: "[a^m = 1] (mod n)"
chaieb@26126
   664
  shows "coprime a n"
chaieb@26126
   665
proof-
chaieb@26126
   666
  {assume "n=1" hence ?thesis by simp}
chaieb@26126
   667
  moreover
chaieb@26126
   668
  {assume "n = 0" hence ?thesis using am m exp_eq_1[of a m] by simp}
chaieb@26126
   669
  moreover
chaieb@26126
   670
  {assume n: "n\<noteq>0" "n\<noteq>1"
chaieb@26126
   671
    from m obtain m' where m': "m = Suc m'" by (cases m, blast+)
chaieb@26126
   672
    {fix d
chaieb@26126
   673
      assume d: "d dvd a" "d dvd n"
huffman@30488
   674
      from n have n1: "1 < n" by arith
chaieb@26126
   675
      from am mod_less[OF n1] have am1: "a^m mod n = 1" unfolding modeq_def by simp
chaieb@26126
   676
      from dvd_mult2[OF d(1), of "a^m'"] have dam:"d dvd a^m" by (simp add: m')
chaieb@26126
   677
      from dvd_mod_iff[OF d(2), of "a^m"] dam am1
chaieb@26126
   678
      have "d = 1" by simp }
chaieb@26126
   679
    hence ?thesis unfolding coprime by auto
chaieb@26126
   680
  }
huffman@30488
   681
  ultimately show ?thesis by blast
chaieb@26126
   682
qed
chaieb@26126
   683
chaieb@26126
   684
lemma lucas_weak:
huffman@30488
   685
  assumes n: "n \<ge> 2" and an:"[a^(n - 1) = 1] (mod n)"
chaieb@26126
   686
  and nm: "\<forall>m. 0 <m \<and> m < n - 1 \<longrightarrow> \<not> [a^m = 1] (mod n)"
chaieb@26126
   687
  shows "prime n"
chaieb@26126
   688
proof-
chaieb@26126
   689
  from n have n1: "n \<noteq> 1" "n\<noteq>0" "n - 1 \<noteq> 0" "n - 1 > 0" "n - 1 < n" by arith+
chaieb@26126
   690
  from lucas_coprime_lemma[OF n1(3) an] have can: "coprime a n" .
chaieb@26126
   691
  from fermat_little[OF can] have afn: "[a ^ \<phi> n = 1] (mod n)" .
chaieb@26126
   692
  {assume "\<phi> n \<noteq> n - 1"
chaieb@26126
   693
    with phi_limit_strong[OF n1(1)] phi_lowerbound_1[OF n]
chaieb@26126
   694
    have c:"\<phi> n > 0 \<and> \<phi> n < n - 1" by arith
chaieb@26126
   695
    from nm[rule_format, OF c] afn have False ..}
chaieb@26126
   696
  hence "\<phi> n = n - 1" by blast
chaieb@26126
   697
  with phi_prime[of n] n1(1,2) show ?thesis by simp
chaieb@26126
   698
qed
chaieb@26126
   699
huffman@30488
   700
lemma nat_exists_least_iff: "(\<exists>(n::nat). P n) \<longleftrightarrow> (\<exists>n. P n \<and> (\<forall>m < n. \<not> P m))"
chaieb@26126
   701
  (is "?lhs \<longleftrightarrow> ?rhs")
chaieb@26126
   702
proof
chaieb@26126
   703
  assume ?rhs thus ?lhs by blast
chaieb@26126
   704
next
chaieb@26126
   705
  assume H: ?lhs then obtain n where n: "P n" by blast
chaieb@26126
   706
  let ?x = "Least P"
chaieb@26126
   707
  {fix m assume m: "m < ?x"
chaieb@26126
   708
    from not_less_Least[OF m] have "\<not> P m" .}
chaieb@26126
   709
  with LeastI_ex[OF H] show ?rhs by blast
chaieb@26126
   710
qed
chaieb@26126
   711
huffman@30488
   712
lemma nat_exists_least_iff': "(\<exists>(n::nat). P n) \<longleftrightarrow> (P (Least P) \<and> (\<forall>m < (Least P). \<not> P m))"
chaieb@26126
   713
  (is "?lhs \<longleftrightarrow> ?rhs")
chaieb@26126
   714
proof-
chaieb@26126
   715
  {assume ?rhs hence ?lhs by blast}
huffman@30488
   716
  moreover
chaieb@26126
   717
  { assume H: ?lhs then obtain n where n: "P n" by blast
chaieb@26126
   718
    let ?x = "Least P"
chaieb@26126
   719
    {fix m assume m: "m < ?x"
chaieb@26126
   720
      from not_less_Least[OF m] have "\<not> P m" .}
chaieb@26126
   721
    with LeastI_ex[OF H] have ?rhs by blast}
chaieb@26126
   722
  ultimately show ?thesis by blast
chaieb@26126
   723
qed
huffman@30488
   724
chaieb@26126
   725
lemma power_mod: "((x::nat) mod m)^n mod m = x^n mod m"
chaieb@26126
   726
proof(induct n)
chaieb@26126
   727
  case 0 thus ?case by simp
chaieb@26126
   728
next
huffman@30488
   729
  case (Suc n)
huffman@30488
   730
  have "(x mod m)^(Suc n) mod m = ((x mod m) * (((x mod m) ^ n) mod m)) mod m"
nipkow@30224
   731
    by (simp add: mod_mult_right_eq[symmetric])
chaieb@26126
   732
  also have "\<dots> = ((x mod m) * (x^n mod m)) mod m" using Suc.hyps by simp
chaieb@26126
   733
  also have "\<dots> = x^(Suc n) mod m"
nipkow@30224
   734
    by (simp add: mod_mult_left_eq[symmetric] mod_mult_right_eq[symmetric])
chaieb@26126
   735
  finally show ?case .
chaieb@26126
   736
qed
chaieb@26126
   737
chaieb@26126
   738
lemma lucas:
huffman@30488
   739
  assumes n2: "n \<ge> 2" and an1: "[a^(n - 1) = 1] (mod n)"
chaieb@26126
   740
  and pn: "\<forall>p. prime p \<and> p dvd n - 1 \<longrightarrow> \<not> [a^((n - 1) div p) = 1] (mod n)"
chaieb@26126
   741
  shows "prime n"
chaieb@26126
   742
proof-
chaieb@26126
   743
  from n2 have n01: "n\<noteq>0" "n\<noteq>1" "n - 1 \<noteq> 0" by arith+
chaieb@26126
   744
  from mod_less_divisor[of n 1] n01 have onen: "1 mod n = 1" by simp
huffman@30488
   745
  from lucas_coprime_lemma[OF n01(3) an1] cong_coprime[OF an1]
chaieb@26126
   746
  have an: "coprime a n" "coprime (a^(n - 1)) n" by (simp_all add: coprime_commute)
chaieb@26126
   747
  {assume H0: "\<exists>m. 0 < m \<and> m < n - 1 \<and> [a ^ m = 1] (mod n)" (is "EX m. ?P m")
huffman@30488
   748
    from H0[unfolded nat_exists_least_iff[of ?P]] obtain m where
chaieb@26126
   749
      m: "0 < m" "m < n - 1" "[a ^ m = 1] (mod n)" "\<forall>k <m. \<not>?P k" by blast
huffman@30488
   750
    {assume nm1: "(n - 1) mod m > 0"
huffman@30488
   751
      from mod_less_divisor[OF m(1)] have th0:"(n - 1) mod m < m" by blast
chaieb@26126
   752
      let ?y = "a^ ((n - 1) div m * m)"
chaieb@26126
   753
      note mdeq = mod_div_equality[of "(n - 1)" m]
huffman@30488
   754
      from coprime_exp[OF an(1)[unfolded coprime_commute[of a n]],
wenzelm@32960
   755
        of "(n - 1) div m * m"]
huffman@30488
   756
      have yn: "coprime ?y n" by (simp add: coprime_commute)
huffman@30488
   757
      have "?y mod n = (a^m)^((n - 1) div m) mod n"
wenzelm@32960
   758
        by (simp add: algebra_simps power_mult)
huffman@30488
   759
      also have "\<dots> = (a^m mod n)^((n - 1) div m) mod n"
wenzelm@32960
   760
        using power_mod[of "a^m" n "(n - 1) div m"] by simp
huffman@30488
   761
      also have "\<dots> = 1" using m(3)[unfolded modeq_def onen] onen
wenzelm@32960
   762
        by (simp add: power_Suc0)
huffman@30488
   763
      finally have th3: "?y mod n = 1"  .
huffman@30488
   764
      have th2: "[?y * a ^ ((n - 1) mod m) = ?y* 1] (mod n)"
wenzelm@32960
   765
        using an1[unfolded modeq_def onen] onen
wenzelm@32960
   766
          mod_div_equality[of "(n - 1)" m, symmetric]
wenzelm@32960
   767
        by (simp add:power_add[symmetric] modeq_def th3 del: One_nat_def)
chaieb@26126
   768
      from cong_mult_lcancel[of ?y n "a^((n - 1) mod m)" 1, OF yn th2]
huffman@30488
   769
      have th1: "[a ^ ((n - 1) mod m) = 1] (mod n)"  .
huffman@30488
   770
      from m(4)[rule_format, OF th0] nm1
wenzelm@32960
   771
        less_trans[OF mod_less_divisor[OF m(1), of "n - 1"] m(2)] th1
chaieb@26126
   772
      have False by blast }
chaieb@26126
   773
    hence "(n - 1) mod m = 0" by auto
chaieb@26126
   774
    then have mn: "m dvd n - 1" by presburger
chaieb@26126
   775
    then obtain r where r: "n - 1 = m*r" unfolding dvd_def by blast
chaieb@26126
   776
    from n01 r m(2) have r01: "r\<noteq>0" "r\<noteq>1" by - (rule ccontr, simp)+
chaieb@26126
   777
    from prime_factor[OF r01(2)] obtain p where p: "prime p" "p dvd r" by blast
chaieb@26126
   778
    hence th: "prime p \<and> p dvd n - 1" unfolding r by (auto intro: dvd_mult)
chaieb@26126
   779
    have "(a ^ ((n - 1) div p)) mod n = (a^(m*r div p)) mod n" using r
chaieb@26126
   780
      by (simp add: power_mult)
chaieb@26126
   781
    also have "\<dots> = (a^(m*(r div p))) mod n" using div_mult1_eq[of m r p] p(2)[unfolded dvd_eq_mod_eq_0] by simp
chaieb@26126
   782
    also have "\<dots> = ((a^m)^(r div p)) mod n" by (simp add: power_mult)
chaieb@26126
   783
    also have "\<dots> = ((a^m mod n)^(r div p)) mod n" using power_mod[of "a^m" "n" "r div p" ] ..
chaieb@26158
   784
    also have "\<dots> = 1" using m(3) onen by (simp add: modeq_def power_Suc0)
huffman@30488
   785
    finally have "[(a ^ ((n - 1) div p))= 1] (mod n)"
chaieb@26126
   786
      using onen by (simp add: modeq_def)
chaieb@26126
   787
    with pn[rule_format, OF th] have False by blast}
chaieb@26126
   788
  hence th: "\<forall>m. 0 < m \<and> m < n - 1 \<longrightarrow> \<not> [a ^ m = 1] (mod n)" by blast
chaieb@26126
   789
  from lucas_weak[OF n2 an1 th] show ?thesis .
chaieb@26126
   790
qed
chaieb@26126
   791
chaieb@26126
   792
(* Definition of the order of a number mod n (0 in non-coprime case).        *)
chaieb@26126
   793
chaieb@26126
   794
definition "ord n a = (if coprime n a then Least (\<lambda>d. d > 0 \<and> [a ^d = 1] (mod n)) else 0)"
chaieb@26126
   795
chaieb@26126
   796
(* This has the expected properties.                                         *)
chaieb@26126
   797
chaieb@26126
   798
lemma coprime_ord:
huffman@30488
   799
  assumes na: "coprime n a"
chaieb@26126
   800
  shows "ord n a > 0 \<and> [a ^(ord n a) = 1] (mod n) \<and> (\<forall>m. 0 < m \<and> m < ord n a \<longrightarrow> \<not> [a^ m = 1] (mod n))"
chaieb@26126
   801
proof-
chaieb@26126
   802
  let ?P = "\<lambda>d. 0 < d \<and> [a ^ d = 1] (mod n)"
chaieb@26126
   803
  from euclid[of a] obtain p where p: "prime p" "a < p" by blast
chaieb@26126
   804
  from na have o: "ord n a = Least ?P" by (simp add: ord_def)
chaieb@26126
   805
  {assume "n=0 \<or> n=1" with na have "\<exists>m>0. ?P m" apply auto apply (rule exI[where x=1]) by (simp  add: modeq_def)}
chaieb@26126
   806
  moreover
huffman@30488
   807
  {assume "n\<noteq>0 \<and> n\<noteq>1" hence n2:"n \<ge> 2" by arith
chaieb@26126
   808
    from na have na': "coprime a n" by (simp add: coprime_commute)
chaieb@26126
   809
    from phi_lowerbound_1[OF n2] fermat_little[OF na']
chaieb@26126
   810
    have ex: "\<exists>m>0. ?P m" by - (rule exI[where x="\<phi> n"], auto) }
chaieb@26126
   811
  ultimately have ex: "\<exists>m>0. ?P m" by blast
huffman@30488
   812
  from nat_exists_least_iff'[of ?P] ex na show ?thesis
chaieb@26126
   813
    unfolding o[symmetric] by auto
chaieb@26126
   814
qed
chaieb@26126
   815
(* With the special value 0 for non-coprime case, it's more convenient.      *)
chaieb@26126
   816
lemma ord_works:
chaieb@26126
   817
 "[a ^ (ord n a) = 1] (mod n) \<and> (\<forall>m. 0 < m \<and> m < ord n a \<longrightarrow> ~[a^ m = 1] (mod n))"
chaieb@26126
   818
apply (cases "coprime n a")
chaieb@26126
   819
using coprime_ord[of n a]
chaieb@26126
   820
by (blast, simp add: ord_def modeq_def)
chaieb@26126
   821
huffman@30488
   822
lemma ord: "[a^(ord n a) = 1] (mod n)" using ord_works by blast
huffman@30488
   823
lemma ord_minimal: "0 < m \<Longrightarrow> m < ord n a \<Longrightarrow> ~[a^m = 1] (mod n)"
chaieb@26126
   824
  using ord_works by blast
chaieb@26126
   825
lemma ord_eq_0: "ord n a = 0 \<longleftrightarrow> ~coprime n a"
wenzelm@41541
   826
by (cases "coprime n a", simp add: coprime_ord, simp add: ord_def)
chaieb@26126
   827
chaieb@26126
   828
lemma ord_divides:
chaieb@26126
   829
 "[a ^ d = 1] (mod n) \<longleftrightarrow> ord n a dvd d" (is "?lhs \<longleftrightarrow> ?rhs")
chaieb@26126
   830
proof
chaieb@26126
   831
  assume rh: ?rhs
chaieb@26126
   832
  then obtain k where "d = ord n a * k" unfolding dvd_def by blast
chaieb@26126
   833
  hence "[a ^ d = (a ^ (ord n a) mod n)^k] (mod n)"
chaieb@26126
   834
    by (simp add : modeq_def power_mult power_mod)
huffman@30488
   835
  also have "[(a ^ (ord n a) mod n)^k = 1] (mod n)"
huffman@30488
   836
    using ord[of a n, unfolded modeq_def]
chaieb@26158
   837
    by (simp add: modeq_def power_mod power_Suc0)
chaieb@26126
   838
  finally  show ?lhs .
huffman@30488
   839
next
chaieb@26126
   840
  assume lh: ?lhs
chaieb@26126
   841
  { assume H: "\<not> coprime n a"
chaieb@26126
   842
    hence o: "ord n a = 0" by (simp add: ord_def)
chaieb@26126
   843
    {assume d: "d=0" with o H have ?rhs by (simp add: modeq_def)}
chaieb@26126
   844
    moreover
chaieb@26126
   845
    {assume d0: "d\<noteq>0" then obtain d' where d': "d = Suc d'" by (cases d, auto)
huffman@30488
   846
      from H[unfolded coprime]
huffman@30488
   847
      obtain p where p: "p dvd n" "p dvd a" "p \<noteq> 1" by auto
huffman@30488
   848
      from lh[unfolded nat_mod]
chaieb@26126
   849
      obtain q1 q2 where q12:"a ^ d + n * q1 = 1 + n * q2" by blast
chaieb@26126
   850
      hence "a ^ d + n * q1 - n * q2 = 1" by simp
nipkow@31952
   851
      with dvd_diff_nat [OF dvd_add [OF divides_rexp[OF p(2), of d'] dvd_mult2[OF p(1), of q1]] dvd_mult2[OF p(1), of q2]] d' have "p dvd 1" by simp
chaieb@26126
   852
      with p(3) have False by simp
chaieb@26126
   853
      hence ?rhs ..}
chaieb@26126
   854
    ultimately have ?rhs by blast}
chaieb@26126
   855
  moreover
chaieb@26126
   856
  {assume H: "coprime n a"
chaieb@26126
   857
    let ?o = "ord n a"
chaieb@26126
   858
    let ?q = "d div ord n a"
chaieb@26126
   859
    let ?r = "d mod ord n a"
huffman@30488
   860
    from cong_exp[OF ord[of a n], of ?q]
chaieb@26158
   861
    have eqo: "[(a^?o)^?q = 1] (mod n)"  by (simp add: modeq_def power_Suc0)
chaieb@26126
   862
    from H have onz: "?o \<noteq> 0" by (simp add: ord_eq_0)
chaieb@26126
   863
    hence op: "?o > 0" by simp
chaieb@26126
   864
    from mod_div_equality[of d "ord n a"] lh
chaieb@26126
   865
    have "[a^(?o*?q + ?r) = 1] (mod n)" by (simp add: modeq_def mult_commute)
huffman@30488
   866
    hence "[(a^?o)^?q * (a^?r) = 1] (mod n)"
chaieb@26126
   867
      by (simp add: modeq_def power_mult[symmetric] power_add[symmetric])
chaieb@26126
   868
    hence th: "[a^?r = 1] (mod n)"
nipkow@30224
   869
      using eqo mod_mult_left_eq[of "(a^?o)^?q" "a^?r" n]
chaieb@26126
   870
      apply (simp add: modeq_def del: One_nat_def)
nipkow@30224
   871
      by (simp add: mod_mult_left_eq[symmetric])
chaieb@26126
   872
    {assume r: "?r = 0" hence ?rhs by (simp add: dvd_eq_mod_eq_0)}
chaieb@26126
   873
    moreover
huffman@30488
   874
    {assume r: "?r \<noteq> 0"
chaieb@26126
   875
      with mod_less_divisor[OF op, of d] have r0o:"?r >0 \<and> ?r < ?o" by simp
huffman@30488
   876
      from conjunct2[OF ord_works[of a n], rule_format, OF r0o] th
chaieb@26126
   877
      have ?rhs by blast}
chaieb@26126
   878
    ultimately have ?rhs by blast}
chaieb@26126
   879
  ultimately  show ?rhs by blast
chaieb@26126
   880
qed
chaieb@26126
   881
chaieb@26126
   882
lemma order_divides_phi: "coprime n a \<Longrightarrow> ord n a dvd \<phi> n"
chaieb@26126
   883
using ord_divides fermat_little coprime_commute by simp
huffman@30488
   884
lemma order_divides_expdiff:
chaieb@26126
   885
  assumes na: "coprime n a"
chaieb@26126
   886
  shows "[a^d = a^e] (mod n) \<longleftrightarrow> [d = e] (mod (ord n a))"
chaieb@26126
   887
proof-
huffman@30488
   888
  {fix n a d e
chaieb@26126
   889
    assume na: "coprime n a" and ed: "(e::nat) \<le> d"
chaieb@26126
   890
    hence "\<exists>c. d = e + c" by arith
chaieb@26126
   891
    then obtain c where c: "d = e + c" by arith
chaieb@26126
   892
    from na have an: "coprime a n" by (simp add: coprime_commute)
huffman@30488
   893
    from coprime_exp[OF na, of e]
chaieb@26126
   894
    have aen: "coprime (a^e) n" by (simp add: coprime_commute)
huffman@30488
   895
    from coprime_exp[OF na, of c]
chaieb@26126
   896
    have acn: "coprime (a^c) n" by (simp add: coprime_commute)
chaieb@26126
   897
    have "[a^d = a^e] (mod n) \<longleftrightarrow> [a^(e + c) = a^(e + 0)] (mod n)"
chaieb@26126
   898
      using c by simp
chaieb@26126
   899
    also have "\<dots> \<longleftrightarrow> [a^e* a^c = a^e *a^0] (mod n)" by (simp add: power_add)
chaieb@26126
   900
    also have  "\<dots> \<longleftrightarrow> [a ^ c = 1] (mod n)"
chaieb@26126
   901
      using cong_mult_lcancel_eq[OF aen, of "a^c" "a^0"] by simp
chaieb@26126
   902
    also  have "\<dots> \<longleftrightarrow> ord n a dvd c" by (simp only: ord_divides)
chaieb@26126
   903
    also have "\<dots> \<longleftrightarrow> [e + c = e + 0] (mod ord n a)"
chaieb@26126
   904
      using cong_add_lcancel_eq[of e c 0 "ord n a", simplified cong_0_divides]
chaieb@26126
   905
      by simp
chaieb@26126
   906
    finally have "[a^d = a^e] (mod n) \<longleftrightarrow> [d = e] (mod (ord n a))"
chaieb@26126
   907
      using c by simp }
chaieb@26126
   908
  note th = this
chaieb@26126
   909
  have "e \<le> d \<or> d \<le> e" by arith
chaieb@26126
   910
  moreover
chaieb@26126
   911
  {assume ed: "e \<le> d" from th[OF na ed] have ?thesis .}
chaieb@26126
   912
  moreover
chaieb@26126
   913
  {assume de: "d \<le> e"
chaieb@26126
   914
    from th[OF na de] have ?thesis by (simp add: cong_commute) }
chaieb@26126
   915
  ultimately show ?thesis by blast
chaieb@26126
   916
qed
chaieb@26126
   917
chaieb@26126
   918
(* Another trivial primality characterization.                               *)
chaieb@26126
   919
chaieb@26126
   920
lemma prime_prime_factor:
chaieb@26126
   921
  "prime n \<longleftrightarrow> n \<noteq> 1\<and> (\<forall>p. prime p \<and> p dvd n \<longrightarrow> p = n)"
chaieb@26126
   922
proof-
chaieb@26126
   923
  {assume n: "n=0 \<or> n=1" hence ?thesis using prime_0 two_is_prime by auto}
chaieb@26126
   924
  moreover
chaieb@26126
   925
  {assume n: "n\<noteq>0" "n\<noteq>1"
chaieb@26126
   926
    {assume pn: "prime n"
huffman@30488
   927
chaieb@26126
   928
      from pn[unfolded prime_def] have "\<forall>p. prime p \<and> p dvd n \<longrightarrow> p = n"
wenzelm@32960
   929
        using n
wenzelm@32960
   930
        apply (cases "n = 0 \<or> n=1",simp)
wenzelm@32960
   931
        by (clarsimp, erule_tac x="p" in allE, auto)}
chaieb@26126
   932
    moreover
chaieb@26126
   933
    {assume H: "\<forall>p. prime p \<and> p dvd n \<longrightarrow> p = n"
chaieb@26126
   934
      from n have n1: "n > 1" by arith
chaieb@26126
   935
      {fix m assume m: "m dvd n" "m\<noteq>1"
wenzelm@32960
   936
        from prime_factor[OF m(2)] obtain p where
wenzelm@32960
   937
          p: "prime p" "p dvd m" by blast
wenzelm@32960
   938
        from dvd_trans[OF p(2) m(1)] p(1) H have "p = n" by blast
wenzelm@32960
   939
        with p(2) have "n dvd m"  by simp
nipkow@33657
   940
        hence "m=n"  using dvd_antisym[OF m(1)] by simp }
chaieb@26126
   941
      with n1 have "prime n"  unfolding prime_def by auto }
huffman@30488
   942
    ultimately have ?thesis using n by blast}
huffman@30488
   943
  ultimately       show ?thesis by auto
chaieb@26126
   944
qed
chaieb@26126
   945
chaieb@26126
   946
lemma prime_divisor_sqrt:
chaieb@26126
   947
  "prime n \<longleftrightarrow> n \<noteq> 1 \<and> (\<forall>d. d dvd n \<and> d^2 \<le> n \<longrightarrow> d = 1)"
chaieb@26126
   948
proof-
huffman@30488
   949
  {assume "n=0 \<or> n=1" hence ?thesis using prime_0 prime_1
chaieb@26126
   950
    by (auto simp add: nat_power_eq_0_iff)}
chaieb@26126
   951
  moreover
chaieb@26126
   952
  {assume n: "n\<noteq>0" "n\<noteq>1"
chaieb@26126
   953
    hence np: "n > 1" by arith
chaieb@26126
   954
    {fix d assume d: "d dvd n" "d^2 \<le> n" and H: "\<forall>m. m dvd n \<longrightarrow> m=1 \<or> m=n"
chaieb@26126
   955
      from H d have d1n: "d = 1 \<or> d=n" by blast
chaieb@26126
   956
      {assume dn: "d=n"
wenzelm@41541
   957
        have "n^2 > n*1" using n by (simp add: power2_eq_square)
wenzelm@32960
   958
        with dn d(2) have "d=1" by simp}
chaieb@26126
   959
      with d1n have "d = 1" by blast  }
chaieb@26126
   960
    moreover
chaieb@26126
   961
    {fix d assume d: "d dvd n" and H: "\<forall>d'. d' dvd n \<and> d'^2 \<le> n \<longrightarrow> d' = 1"
chaieb@26126
   962
      from d n have "d \<noteq> 0" apply - apply (rule ccontr) by simp
chaieb@26126
   963
      hence dp: "d > 0" by simp
chaieb@26126
   964
      from d[unfolded dvd_def] obtain e where e: "n= d*e" by blast
chaieb@26126
   965
      from n dp e have ep:"e > 0" by simp
chaieb@26126
   966
      have "d^2 \<le> n \<or> e^2 \<le> n" using dp ep
wenzelm@32960
   967
        by (auto simp add: e power2_eq_square mult_le_cancel_left)
chaieb@26126
   968
      moreover
chaieb@26126
   969
      {assume h: "d^2 \<le> n"
wenzelm@32960
   970
        from H[rule_format, of d] h d have "d = 1" by blast}
chaieb@26126
   971
      moreover
chaieb@26126
   972
      {assume h: "e^2 \<le> n"
wenzelm@32960
   973
        from e have "e dvd n" unfolding dvd_def by (simp add: mult_commute)
wenzelm@32960
   974
        with H[rule_format, of e] h have "e=1" by simp
wenzelm@32960
   975
        with e have "d = n" by simp}
chaieb@26126
   976
      ultimately have "d=1 \<or> d=n"  by blast}
chaieb@26126
   977
    ultimately have ?thesis unfolding prime_def using np n(2) by blast}
chaieb@26126
   978
  ultimately show ?thesis by auto
chaieb@26126
   979
qed
chaieb@26126
   980
lemma prime_prime_factor_sqrt:
huffman@30488
   981
  "prime n \<longleftrightarrow> n \<noteq> 0 \<and> n \<noteq> 1 \<and> \<not> (\<exists>p. prime p \<and> p dvd n \<and> p^2 \<le> n)"
chaieb@26126
   982
  (is "?lhs \<longleftrightarrow>?rhs")
chaieb@26126
   983
proof-
chaieb@26126
   984
  {assume "n=0 \<or> n=1" hence ?thesis using prime_0 prime_1 by auto}
chaieb@26126
   985
  moreover
chaieb@26126
   986
  {assume n: "n\<noteq>0" "n\<noteq>1"
chaieb@26126
   987
    {assume H: ?lhs
huffman@30488
   988
      from H[unfolded prime_divisor_sqrt] n
wenzelm@41541
   989
      have ?rhs
wenzelm@41541
   990
        apply clarsimp
wenzelm@41541
   991
        apply (erule_tac x="p" in allE)
wenzelm@41541
   992
        apply simp
wenzelm@41541
   993
        done
chaieb@26126
   994
    }
chaieb@26126
   995
    moreover
chaieb@26126
   996
    {assume H: ?rhs
chaieb@26126
   997
      {fix d assume d: "d dvd n" "d^2 \<le> n" "d\<noteq>1"
wenzelm@32960
   998
        from prime_factor[OF d(3)]
wenzelm@32960
   999
        obtain p where p: "prime p" "p dvd d" by blast
wenzelm@32960
  1000
        from n have np: "n > 0" by arith
wenzelm@32960
  1001
        from d(1) n have "d \<noteq> 0" by - (rule ccontr, auto)
wenzelm@32960
  1002
        hence dp: "d > 0" by arith
wenzelm@32960
  1003
        from mult_mono[OF dvd_imp_le[OF p(2) dp] dvd_imp_le[OF p(2) dp]] d(2)
wenzelm@32960
  1004
        have "p^2 \<le> n" unfolding power2_eq_square by arith
wenzelm@32960
  1005
        with H n p(1) dvd_trans[OF p(2) d(1)] have False  by blast}
chaieb@26126
  1006
      with n prime_divisor_sqrt  have ?lhs by auto}
chaieb@26126
  1007
    ultimately have ?thesis by blast }
chaieb@26126
  1008
  ultimately show ?thesis by (cases "n=0 \<or> n=1", auto)
chaieb@26126
  1009
qed
chaieb@26126
  1010
(* Pocklington theorem. *)
chaieb@26126
  1011
chaieb@26126
  1012
lemma pocklington_lemma:
chaieb@26126
  1013
  assumes n: "n \<ge> 2" and nqr: "n - 1 = q*r" and an: "[a^ (n - 1) = 1] (mod n)"
chaieb@26126
  1014
  and aq:"\<forall>p. prime p \<and> p dvd q \<longrightarrow> coprime (a^ ((n - 1) div p) - 1) n"
chaieb@26126
  1015
  and pp: "prime p" and pn: "p dvd n"
chaieb@26126
  1016
  shows "[p = 1] (mod q)"
chaieb@26126
  1017
proof-
chaieb@26126
  1018
  from pp prime_0 prime_1 have p01: "p \<noteq> 0" "p \<noteq> 1" by - (rule ccontr, simp)+
huffman@30488
  1019
  from cong_1_divides[OF an, unfolded nqr, unfolded dvd_def]
chaieb@26126
  1020
  obtain k where k: "a ^ (q * r) - 1 = n*k" by blast
chaieb@26126
  1021
  from pn[unfolded dvd_def] obtain l where l: "n = p*l" by blast
chaieb@26126
  1022
  {assume a0: "a = 0"
chaieb@26126
  1023
    hence "a^ (n - 1) = 0" using n by (simp add: power_0_left)
chaieb@26126
  1024
    with n an mod_less[of 1 n]  have False by (simp add: power_0_left modeq_def)}
chaieb@26126
  1025
  hence a0: "a\<noteq>0" ..
wenzelm@41541
  1026
  from n nqr have aqr0: "a ^ (q * r) \<noteq> 0" using a0 by simp
chaieb@26126
  1027
  hence "(a ^ (q * r) - 1) + 1  = a ^ (q * r)" by simp
chaieb@26126
  1028
  with k l have "a ^ (q * r) = p*l*k + 1" by simp
chaieb@26126
  1029
  hence "a ^ (r * q) + p * 0 = 1 + p * (l*k)" by (simp add: mult_ac)
chaieb@26126
  1030
  hence odq: "ord p (a^r) dvd q"
chaieb@26126
  1031
    unfolding ord_divides[symmetric] power_mult[symmetric] nat_mod  by blast
chaieb@26126
  1032
  from odq[unfolded dvd_def] obtain d where d: "q = ord p (a^r) * d" by blast
huffman@30488
  1033
  {assume d1: "d \<noteq> 1"
chaieb@26126
  1034
    from prime_factor[OF d1] obtain P where P: "prime P" "P dvd d" by blast
chaieb@26126
  1035
    from d dvd_mult[OF P(2), of "ord p (a^r)"] have Pq: "P dvd q" by simp
chaieb@26126
  1036
    from aq P(1) Pq have caP:"coprime (a^ ((n - 1) div P) - 1) n" by blast
chaieb@26126
  1037
    from Pq obtain s where s: "q = P*s" unfolding dvd_def by blast
chaieb@26126
  1038
    have P0: "P \<noteq> 0" using P(1) prime_0 by - (rule ccontr, simp)
chaieb@26126
  1039
    from P(2) obtain t where t: "d = P*t" unfolding dvd_def by blast
chaieb@26126
  1040
    from d s t P0  have s': "ord p (a^r) * t = s" by algebra
chaieb@26126
  1041
    have "ord p (a^r) * t*r = r * ord p (a^r) * t" by algebra
chaieb@26126
  1042
    hence exps: "a^(ord p (a^r) * t*r) = ((a ^ r) ^ ord p (a^r)) ^ t"
chaieb@26126
  1043
      by (simp only: power_mult)
huffman@30488
  1044
    have "[((a ^ r) ^ ord p (a^r)) ^ t= 1^t] (mod p)"
chaieb@26126
  1045
      by (rule cong_exp, rule ord)
huffman@30488
  1046
    then have th: "[((a ^ r) ^ ord p (a^r)) ^ t= 1] (mod p)"
chaieb@26158
  1047
      by (simp add: power_Suc0)
chaieb@26126
  1048
    from cong_1_divides[OF th] exps have pd0: "p dvd a^(ord p (a^r) * t*r) - 1" by simp
chaieb@26126
  1049
    from nqr s s' have "(n - 1) div P = ord p (a^r) * t*r" using P0 by simp
chaieb@26126
  1050
    with caP have "coprime (a^(ord p (a^r) * t*r) - 1) n" by simp
chaieb@26126
  1051
    with p01 pn pd0 have False unfolding coprime by auto}
huffman@30488
  1052
  hence d1: "d = 1" by blast
huffman@30488
  1053
  hence o: "ord p (a^r) = q" using d by simp
chaieb@26126
  1054
  from pp phi_prime[of p] have phip: " \<phi> p = p - 1" by simp
chaieb@26126
  1055
  {fix d assume d: "d dvd p" "d dvd a" "d \<noteq> 1"
chaieb@26126
  1056
    from pp[unfolded prime_def] d have dp: "d = p" by blast
chaieb@26126
  1057
    from n have n12:"Suc (n - 2) = n - 1" by arith
chaieb@26126
  1058
    with divides_rexp[OF d(2)[unfolded dp], of "n - 2"]
chaieb@26126
  1059
    have th0: "p dvd a ^ (n - 1)" by simp
chaieb@26126
  1060
    from n have n0: "n \<noteq> 0" by simp
huffman@30488
  1061
    from d(2) an n12[symmetric] have a0: "a \<noteq> 0"
chaieb@26126
  1062
      by - (rule ccontr, simp add: modeq_def)
wenzelm@41541
  1063
    have th1: "a^ (n - 1) \<noteq> 0" using n d(2) dp a0 by auto
huffman@30488
  1064
    from coprime_minus1[OF th1, unfolded coprime]
chaieb@26126
  1065
      dvd_trans[OF pn cong_1_divides[OF an]] th0 d(3) dp
chaieb@26126
  1066
    have False by auto}
huffman@30488
  1067
  hence cpa: "coprime p a" using coprime by auto
huffman@30488
  1068
  from coprime_exp[OF cpa, of r] coprime_commute
chaieb@26126
  1069
  have arp: "coprime (a^r) p" by blast
chaieb@26126
  1070
  from fermat_little[OF arp, simplified ord_divides] o phip
chaieb@26126
  1071
  have "q dvd (p - 1)" by simp
chaieb@26126
  1072
  then obtain d where d:"p - 1 = q * d" unfolding dvd_def by blast
chaieb@26126
  1073
  from prime_0 pp have p0:"p \<noteq> 0" by -  (rule ccontr, auto)
chaieb@26126
  1074
  from p0 d have "p + q * 0 = 1 + q * d" by simp
chaieb@26126
  1075
  with nat_mod[of p 1 q, symmetric]
chaieb@26126
  1076
  show ?thesis by blast
chaieb@26126
  1077
qed
chaieb@26126
  1078
chaieb@26126
  1079
lemma pocklington:
chaieb@26126
  1080
  assumes n: "n \<ge> 2" and nqr: "n - 1 = q*r" and sqr: "n \<le> q^2"
chaieb@26126
  1081
  and an: "[a^ (n - 1) = 1] (mod n)"
chaieb@26126
  1082
  and aq:"\<forall>p. prime p \<and> p dvd q \<longrightarrow> coprime (a^ ((n - 1) div p) - 1) n"
chaieb@26126
  1083
  shows "prime n"
chaieb@26126
  1084
unfolding prime_prime_factor_sqrt[of n]
chaieb@26126
  1085
proof-
chaieb@26126
  1086
  let ?ths = "n \<noteq> 0 \<and> n \<noteq> 1 \<and> \<not> (\<exists>p. prime p \<and> p dvd n \<and> p\<twosuperior> \<le> n)"
chaieb@26126
  1087
  from n have n01: "n\<noteq>0" "n\<noteq>1" by arith+
chaieb@26126
  1088
  {fix p assume p: "prime p" "p dvd n" "p^2 \<le> n"
chaieb@26126
  1089
    from p(3) sqr have "p^(Suc 1) \<le> q^(Suc 1)" by (simp add: power2_eq_square)
chaieb@26126
  1090
    hence pq: "p \<le> q" unfolding exp_mono_le .
chaieb@26126
  1091
    from pocklington_lemma[OF n nqr an aq p(1,2)]  cong_1_divides
chaieb@26126
  1092
    have th: "q dvd p - 1" by blast
chaieb@26126
  1093
    have "p - 1 \<noteq> 0"using prime_ge_2[OF p(1)] by arith
chaieb@26126
  1094
    with divides_ge[OF th] pq have False by arith }
chaieb@26126
  1095
  with n01 show ?ths by blast
chaieb@26126
  1096
qed
chaieb@26126
  1097
chaieb@26126
  1098
(* Variant for application, to separate the exponentiation.                  *)
chaieb@26126
  1099
lemma pocklington_alt:
chaieb@26126
  1100
  assumes n: "n \<ge> 2" and nqr: "n - 1 = q*r" and sqr: "n \<le> q^2"
chaieb@26126
  1101
  and an: "[a^ (n - 1) = 1] (mod n)"
chaieb@26126
  1102
  and aq:"\<forall>p. prime p \<and> p dvd q \<longrightarrow> (\<exists>b. [a^((n - 1) div p) = b] (mod n) \<and> coprime (b - 1) n)"
chaieb@26126
  1103
  shows "prime n"
chaieb@26126
  1104
proof-
chaieb@26126
  1105
  {fix p assume p: "prime p" "p dvd q"
huffman@30488
  1106
    from aq[rule_format] p obtain b where
chaieb@26126
  1107
      b: "[a^((n - 1) div p) = b] (mod n)" "coprime (b - 1) n" by blast
chaieb@26126
  1108
    {assume a0: "a=0"
chaieb@26126
  1109
      from n an have "[0 = 1] (mod n)" unfolding a0 power_0_left by auto
chaieb@26126
  1110
      hence False using n by (simp add: modeq_def dvd_eq_mod_eq_0[symmetric])}
chaieb@26126
  1111
    hence a0: "a\<noteq> 0" ..
chaieb@26126
  1112
    hence a1: "a \<ge> 1" by arith
chaieb@26126
  1113
    from one_le_power[OF a1] have ath: "1 \<le> a ^ ((n - 1) div p)" .
chaieb@26126
  1114
    {assume b0: "b = 0"
huffman@30488
  1115
      from p(2) nqr have "(n - 1) mod p = 0"
wenzelm@32960
  1116
        apply (simp only: dvd_eq_mod_eq_0[symmetric]) by (rule dvd_mult2, simp)
huffman@30488
  1117
      with mod_div_equality[of "n - 1" p]
huffman@30488
  1118
      have "(n - 1) div p * p= n - 1" by auto
chaieb@26126
  1119
      hence eq: "(a^((n - 1) div p))^p = a^(n - 1)"
wenzelm@32960
  1120
        by (simp only: power_mult[symmetric])
chaieb@26126
  1121
      from prime_ge_2[OF p(1)] have pS: "Suc (p - 1) = p" by arith
chaieb@26126
  1122
      from b(1) have d: "n dvd a^((n - 1) div p)" unfolding b0 cong_0_divides .
chaieb@26126
  1123
      from divides_rexp[OF d, of "p - 1"] pS eq cong_divides[OF an] n
chaieb@26126
  1124
      have False by simp}
huffman@30488
  1125
    then have b0: "b \<noteq> 0" ..
huffman@30488
  1126
    hence b1: "b \<ge> 1" by arith
chaieb@26126
  1127
    from cong_coprime[OF cong_sub[OF b(1) cong_refl[of 1] ath b1]] b(2) nqr
chaieb@26126
  1128
    have "coprime (a ^ ((n - 1) div p) - 1) n" by (simp add: coprime_commute)}
huffman@30488
  1129
  hence th: "\<forall>p. prime p \<and> p dvd q \<longrightarrow> coprime (a ^ ((n - 1) div p) - 1) n "
chaieb@26126
  1130
    by blast
chaieb@26126
  1131
  from pocklington[OF n nqr sqr an th] show ?thesis .
chaieb@26126
  1132
qed
chaieb@26126
  1133
chaieb@26126
  1134
(* Prime factorizations.                                                     *)
chaieb@26126
  1135
chaieb@26126
  1136
definition "primefact ps n = (foldr op * ps  1 = n \<and> (\<forall>p\<in> set ps. prime p))"
chaieb@26126
  1137
chaieb@26126
  1138
lemma primefact: assumes n: "n \<noteq> 0"
chaieb@26126
  1139
  shows "\<exists>ps. primefact ps n"
chaieb@26126
  1140
using n
chaieb@26126
  1141
proof(induct n rule: nat_less_induct)
chaieb@26126
  1142
  fix n assume H: "\<forall>m<n. m \<noteq> 0 \<longrightarrow> (\<exists>ps. primefact ps m)" and n: "n\<noteq>0"
chaieb@26126
  1143
  let ?ths = "\<exists>ps. primefact ps n"
huffman@30488
  1144
  {assume "n = 1"
chaieb@26126
  1145
    hence "primefact [] n" by (simp add: primefact_def)
chaieb@26126
  1146
    hence ?ths by blast }
chaieb@26126
  1147
  moreover
chaieb@26126
  1148
  {assume n1: "n \<noteq> 1"
chaieb@26126
  1149
    with n have n2: "n \<ge> 2" by arith
chaieb@26126
  1150
    from prime_factor[OF n1] obtain p where p: "prime p" "p dvd n" by blast
chaieb@26126
  1151
    from p(2) obtain m where m: "n = p*m" unfolding dvd_def by blast
chaieb@26126
  1152
    from n m have m0: "m > 0" "m\<noteq>0" by auto
chaieb@26126
  1153
    from prime_ge_2[OF p(1)] have "1 < p" by arith
chaieb@26126
  1154
    with m0 m have mn: "m < n" by auto
chaieb@26126
  1155
    from H[rule_format, OF mn m0(2)] obtain ps where ps: "primefact ps m" ..
chaieb@26126
  1156
    from ps m p(1) have "primefact (p#ps) n" by (simp add: primefact_def)
chaieb@26126
  1157
    hence ?ths by blast}
chaieb@26126
  1158
  ultimately show ?ths by blast
chaieb@26126
  1159
qed
chaieb@26126
  1160
huffman@30488
  1161
lemma primefact_contains:
chaieb@26126
  1162
  assumes pf: "primefact ps n" and p: "prime p" and pn: "p dvd n"
chaieb@26126
  1163
  shows "p \<in> set ps"
chaieb@26126
  1164
  using pf p pn
chaieb@26126
  1165
proof(induct ps arbitrary: p n)
chaieb@26126
  1166
  case Nil thus ?case by (auto simp add: primefact_def)
chaieb@26126
  1167
next
chaieb@26126
  1168
  case (Cons q qs p n)
huffman@30488
  1169
  from Cons.prems[unfolded primefact_def]
chaieb@26126
  1170
  have q: "prime q" "q * foldr op * qs 1 = n" "\<forall>p \<in>set qs. prime p"  and p: "prime p" "p dvd q * foldr op * qs 1" by simp_all
chaieb@26126
  1171
  {assume "p dvd q"
chaieb@26126
  1172
    with p(1) q(1) have "p = q" unfolding prime_def by auto
chaieb@26126
  1173
    hence ?case by simp}
chaieb@26126
  1174
  moreover
chaieb@26126
  1175
  { assume h: "p dvd foldr op * qs 1"
huffman@30488
  1176
    from q(3) have pqs: "primefact qs (foldr op * qs 1)"
chaieb@26126
  1177
      by (simp add: primefact_def)
chaieb@26126
  1178
    from Cons.hyps[OF pqs p(1) h] have ?case by simp}
chaieb@26126
  1179
  ultimately show ?case using prime_divprod[OF p] by blast
chaieb@26126
  1180
qed
chaieb@26126
  1181
haftmann@37602
  1182
lemma primefact_variant: "primefact ps n \<longleftrightarrow> foldr op * ps 1 = n \<and> list_all prime ps"
haftmann@37602
  1183
  by (auto simp add: primefact_def list_all_iff)
chaieb@26126
  1184
chaieb@26126
  1185
(* Variant of Lucas theorem.                                                 *)
chaieb@26126
  1186
chaieb@26126
  1187
lemma lucas_primefact:
huffman@30488
  1188
  assumes n: "n \<ge> 2" and an: "[a^(n - 1) = 1] (mod n)"
huffman@30488
  1189
  and psn: "foldr op * ps 1 = n - 1"
chaieb@26126
  1190
  and psp: "list_all (\<lambda>p. prime p \<and> \<not> [a^((n - 1) div p) = 1] (mod n)) ps"
chaieb@26126
  1191
  shows "prime n"
chaieb@26126
  1192
proof-
chaieb@26126
  1193
  {fix p assume p: "prime p" "p dvd n - 1" "[a ^ ((n - 1) div p) = 1] (mod n)"
huffman@30488
  1194
    from psn psp have psn1: "primefact ps (n - 1)"
chaieb@26126
  1195
      by (auto simp add: list_all_iff primefact_variant)
chaieb@26126
  1196
    from p(3) primefact_contains[OF psn1 p(1,2)] psp
chaieb@26126
  1197
    have False by (induct ps, auto)}
chaieb@26126
  1198
  with lucas[OF n an] show ?thesis by blast
chaieb@26126
  1199
qed
chaieb@26126
  1200
chaieb@26126
  1201
(* Variant of Pocklington theorem.                                           *)
chaieb@26126
  1202
chaieb@26126
  1203
lemma mod_le: assumes n: "n \<noteq> (0::nat)" shows "m mod n \<le> m"
chaieb@26126
  1204
proof-
chaieb@26126
  1205
    from mod_div_equality[of m n]
huffman@30488
  1206
    have "\<exists>x. x + m mod n = m" by blast
chaieb@26126
  1207
    then show ?thesis by auto
chaieb@26126
  1208
qed
huffman@30488
  1209
chaieb@26126
  1210
chaieb@26126
  1211
lemma pocklington_primefact:
chaieb@26126
  1212
  assumes n: "n \<ge> 2" and qrn: "q*r = n - 1" and nq2: "n \<le> q^2"
huffman@30488
  1213
  and arnb: "(a^r) mod n = b" and psq: "foldr op * ps 1 = q"
chaieb@26126
  1214
  and bqn: "(b^q) mod n = 1"
chaieb@26126
  1215
  and psp: "list_all (\<lambda>p. prime p \<and> coprime ((b^(q div p)) mod n - 1) n) ps"
chaieb@26126
  1216
  shows "prime n"
chaieb@26126
  1217
proof-
chaieb@26126
  1218
  from bqn psp qrn
chaieb@26126
  1219
  have bqn: "a ^ (n - 1) mod n = 1"
huffman@30488
  1220
    and psp: "list_all (\<lambda>p. prime p \<and> coprime (a^(r *(q div p)) mod n - 1) n) ps"  unfolding arnb[symmetric] power_mod
nipkow@29667
  1221
    by (simp_all add: power_mult[symmetric] algebra_simps)
chaieb@26126
  1222
  from n  have n0: "n > 0" by arith
chaieb@26126
  1223
  from mod_div_equality[of "a^(n - 1)" n]
chaieb@26126
  1224
    mod_less_divisor[OF n0, of "a^(n - 1)"]
huffman@30488
  1225
  have an1: "[a ^ (n - 1) = 1] (mod n)"
chaieb@26126
  1226
    unfolding nat_mod bqn
chaieb@26126
  1227
    apply -
chaieb@26126
  1228
    apply (rule exI[where x="0"])
chaieb@26126
  1229
    apply (rule exI[where x="a^(n - 1) div n"])
nipkow@29667
  1230
    by (simp add: algebra_simps)
chaieb@26126
  1231
  {fix p assume p: "prime p" "p dvd q"
chaieb@26126
  1232
    from psp psq have pfpsq: "primefact ps q"
chaieb@26126
  1233
      by (auto simp add: primefact_variant list_all_iff)
huffman@30488
  1234
    from psp primefact_contains[OF pfpsq p]
chaieb@26126
  1235
    have p': "coprime (a ^ (r * (q div p)) mod n - 1) n"
chaieb@26126
  1236
      by (simp add: list_all_iff)
chaieb@26126
  1237
    from prime_ge_2[OF p(1)] have p01: "p \<noteq> 0" "p \<noteq> 1" "p =Suc(p - 1)" by arith+
huffman@30488
  1238
    from div_mult1_eq[of r q p] p(2)
chaieb@26126
  1239
    have eq1: "r* (q div p) = (n - 1) div p"
chaieb@26126
  1240
      unfolding qrn[symmetric] dvd_eq_mod_eq_0 by (simp add: mult_commute)
chaieb@26126
  1241
    have ath: "\<And>a (b::nat). a <= b \<Longrightarrow> a \<noteq> 0 ==> 1 <= a \<and> 1 <= b" by arith
chaieb@26126
  1242
    from n0 have n00: "n \<noteq> 0" by arith
chaieb@26126
  1243
    from mod_le[OF n00]
chaieb@26126
  1244
    have th10: "a ^ ((n - 1) div p) mod n \<le> a ^ ((n - 1) div p)" .
chaieb@26126
  1245
    {assume "a ^ ((n - 1) div p) mod n = 0"
chaieb@26126
  1246
      then obtain s where s: "a ^ ((n - 1) div p) = n*s"
wenzelm@32960
  1247
        unfolding mod_eq_0_iff by blast
chaieb@26126
  1248
      hence eq0: "(a^((n - 1) div p))^p = (n*s)^p" by simp
chaieb@26126
  1249
      from qrn[symmetric] have qn1: "q dvd n - 1" unfolding dvd_def by auto
chaieb@26126
  1250
      from dvd_trans[OF p(2) qn1] div_mod_equality'[of "n - 1" p]
huffman@30488
  1251
      have npp: "(n - 1) div p * p = n - 1" by (simp add: dvd_eq_mod_eq_0)
chaieb@26126
  1252
      with eq0 have "a^ (n - 1) = (n*s)^p"
wenzelm@32960
  1253
        by (simp add: power_mult[symmetric])
chaieb@26126
  1254
      hence "1 = (n*s)^(Suc (p - 1)) mod n" using bqn p01 by simp
wenzelm@28668
  1255
      also have "\<dots> = 0" by (simp add: mult_assoc)
chaieb@26126
  1256
      finally have False by simp }
huffman@30488
  1257
      then have th11: "a ^ ((n - 1) div p) mod n \<noteq> 0" by auto
huffman@30488
  1258
    have th1: "[a ^ ((n - 1) div p) mod n = a ^ ((n - 1) div p)] (mod n)"
huffman@30488
  1259
      unfolding modeq_def by simp
chaieb@26126
  1260
    from cong_sub[OF th1 cong_refl[of 1]]  ath[OF th10 th11]
chaieb@26126
  1261
    have th: "[a ^ ((n - 1) div p) mod n - 1 = a ^ ((n - 1) div p) - 1] (mod n)"
huffman@30488
  1262
      by blast
huffman@30488
  1263
    from cong_coprime[OF th] p'[unfolded eq1]
chaieb@26126
  1264
    have "coprime (a ^ ((n - 1) div p) - 1) n" by (simp add: coprime_commute) }
chaieb@26126
  1265
  with pocklington[OF n qrn[symmetric] nq2 an1]
huffman@30488
  1266
  show ?thesis by blast
chaieb@26126
  1267
qed
chaieb@26126
  1268
chaieb@26126
  1269
end