# Theory Pocklington

theory Pocklington
imports Primes
```(*  Title:      HOL/Old_Number_Theory/Pocklington.thy
Author:     Amine Chaieb
*)

section ‹Pocklington's Theorem for Primes›

theory Pocklington
imports Primes
begin

definition modeq:: "nat => nat => nat => bool"    ("(1[_ = _] '(mod _'))")
where "[a = b] (mod p) == ((a mod p) = (b mod p))"

definition modneq:: "nat => nat => nat => bool"    ("(1[_ ≠ _] '(mod _'))")
where "[a ≠ b] (mod p) == ((a mod p) ≠ (b mod p))"

lemma modeq_trans:
"⟦ [a = b] (mod p); [b = c] (mod p) ⟧ ⟹ [a = c] (mod p)"

lemma modeq_sym[sym]:
"[a = b] (mod p) ⟹ [b = a] (mod p)"
unfolding modeq_def by simp

lemma modneq_sym[sym]:
"[a ≠ b] (mod p) ⟹ [b ≠ a] (mod p)"

lemma nat_mod_lemma: assumes xyn: "[x = y] (mod n)" and xy:"y ≤ x"
shows "∃q. x = y + n * q"
using xyn xy unfolding modeq_def using nat_mod_eq_lemma by blast

lemma nat_mod[algebra]: "[x = y] (mod n) ⟷ (∃q1 q2. x + n * q1 = y + n * q2)"
unfolding modeq_def nat_mod_eq_iff ..

(* Lemmas about previously defined terms.                                    *)

lemma prime: "prime p ⟷ p ≠ 0 ∧ p≠1 ∧ (∀m. 0 < m ∧ m < p ⟶ coprime p m)"
(is "?lhs ⟷ ?rhs")
proof-
{assume "p=0 ∨ p=1" hence ?thesis using prime_0 prime_1 by (cases "p=0", simp_all)}
moreover
{assume p0: "p≠0" "p≠1"
{assume H: "?lhs"
{fix m assume m: "m > 0" "m < p"
{assume "m=1" hence "coprime p m" by simp}
moreover
{assume "p dvd m" hence "p ≤ m" using dvd_imp_le m by blast with m(2)
have "coprime p m" by simp}
ultimately have "coprime p m" using prime_coprime[OF H, of m] by blast}
hence ?rhs using p0 by auto}
moreover
{ assume H: "∀m. 0 < m ∧ m < p ⟶ coprime p m"
from prime_factor[OF p0(2)] obtain q where q: "prime q" "q dvd p" by blast
from prime_ge_2[OF q(1)] have q0: "q > 0" by arith
from dvd_imp_le[OF q(2)] p0 have qp: "q ≤ p" by arith
{assume "q = p" hence ?lhs using q(1) by blast}
moreover
{assume "q≠p" with qp have qplt: "q < p" by arith
from H[rule_format, of q] qplt q0 have "coprime p q" by arith
with coprime_prime[of p q q] q have False by simp hence ?lhs by blast}
ultimately have ?lhs by blast}
ultimately have ?thesis by blast}
ultimately show ?thesis  by (cases"p=0 ∨ p=1", auto)
qed

lemma finite_number_segment: "card { m. 0 < m ∧ m < n } = n - 1"
proof-
have "{ m. 0 < m ∧ m < n } = {1..<n}" by auto
thus ?thesis by simp
qed

lemma coprime_mod: assumes n: "n ≠ 0" shows "coprime (a mod n) n ⟷ coprime a n"
using n dvd_mod_iff[of _ n a] by (auto simp add: coprime)

(* Congruences.                                                              *)

lemma cong_mod_01[simp,presburger]:
"[x = y] (mod 0) ⟷ x = y" "[x = y] (mod 1)" "[x = 0] (mod n) ⟷ n dvd x"

lemma cong_sub_cases:
"[x = y] (mod n) ⟷ (if x <= y then [y - x = 0] (mod n) else [x - y = 0] (mod n))"
apply (rule_tac x="q2" in exI)
apply (rule_tac x="q1" in exI, simp)
apply (rule_tac x="q2" in exI)
apply (rule_tac x="q1" in exI, simp)
apply (rule_tac x="q1" in exI)
apply (rule_tac x="q2" in exI, simp)
apply (rule_tac x="q1" in exI)
apply (rule_tac x="q2" in exI, simp)
done

lemma cong_mult_lcancel: assumes an: "coprime a n" and axy:"[a * x = a * y] (mod n)"
shows "[x = y] (mod n)"
proof-
{assume "a = 0" with an axy coprime_0'[of n] have ?thesis by (simp add: modeq_def) }
moreover
{assume az: "a≠0"
{assume xy: "x ≤ y" hence axy': "a*x ≤ a*y" by simp
with axy cong_sub_cases[of "a*x" "a*y" n]  have "[a*(y - x) = 0] (mod n)"
by (simp only: if_True diff_mult_distrib2)
hence th: "n dvd a*(y -x)" by simp
from coprime_divprod[OF th] an have "n dvd y - x"
hence ?thesis using xy cong_sub_cases[of x y n] by simp}
moreover
{assume H: "¬x ≤ y" hence xy: "y ≤ x"  by arith
from H az have axy': "¬ a*x ≤ a*y" by auto
with axy H cong_sub_cases[of "a*x" "a*y" n]  have "[a*(x - y) = 0] (mod n)"
by (simp only: if_False diff_mult_distrib2)
hence th: "n dvd a*(x - y)" by simp
from coprime_divprod[OF th] an have "n dvd x - y"
hence ?thesis using xy cong_sub_cases[of x y n] by simp}
ultimately have ?thesis by blast}
ultimately show ?thesis by blast
qed

lemma cong_mult_rcancel: assumes an: "coprime a n" and axy:"[x*a = y*a] (mod n)"
shows "[x = y] (mod n)"
using cong_mult_lcancel[OF an axy[unfolded mult.commute[of _a]]] .

lemma cong_refl: "[x = x] (mod n)" by (simp add: modeq_def)

lemma eq_imp_cong: "a = b ⟹ [a = b] (mod n)" by (simp add: cong_refl)

lemma cong_commute: "[x = y] (mod n) ⟷ [y = x] (mod n)"

lemma cong_trans[trans]: "[x = y] (mod n) ⟹ [y = z] (mod n) ⟹ [x = z] (mod n)"

lemma cong_add: assumes xx': "[x = x'] (mod n)" and yy':"[y = y'] (mod n)"
shows "[x + y = x' + y'] (mod n)"
proof-
have "(x + y) mod n = (x mod n + y mod n) mod n"
also have "… = (x' mod n + y' mod n) mod n" using xx' yy' modeq_def by simp
also have "… = (x' + y') mod n"
finally show ?thesis unfolding modeq_def .
qed

lemma cong_mult: assumes xx': "[x = x'] (mod n)" and yy':"[y = y'] (mod n)"
shows "[x * y = x' * y'] (mod n)"
proof-
have "(x * y) mod n = (x mod n) * (y mod n) mod n"
by (simp add: mod_mult_left_eq[of x y n] mod_mult_right_eq[of "x mod n" y n])
also have "… = (x' mod n) * (y' mod n) mod n" using xx'[unfolded modeq_def] yy'[unfolded modeq_def] by simp
also have "… = (x' * y') mod n"
by (simp add: mod_mult_left_eq[of x' y' n] mod_mult_right_eq[of "x' mod n" y' n])
finally show ?thesis unfolding modeq_def .
qed

lemma cong_exp: "[x = y] (mod n) ⟹ [x^k = y^k] (mod n)"
by (induct k, auto simp add: cong_refl cong_mult)
lemma cong_sub: assumes xx': "[x = x'] (mod n)" and yy': "[y = y'] (mod n)"
and yx: "y <= x" and yx': "y' <= x'"
shows "[x - y = x' - y'] (mod n)"
proof-
{ fix x a x' a' y b y' b'
have "(x::nat) + a = x' + a' ⟹ y + b = y' + b' ⟹ y <= x ⟹ y' <= x'
⟹ (x - y) + (a + b') = (x' - y') + (a' + b)" by arith}
note th = this
from xx' yy' obtain q1 q2 q1' q2' where q12: "x + n*q1 = x'+n*q2"
and q12': "y + n*q1' = y'+n*q2'" unfolding nat_mod by blast+
from th[OF q12 q12' yx yx']
have "(x - y) + n*(q1 + q2') = (x' - y') + n*(q2 + q1')"
thus ?thesis unfolding nat_mod by blast
qed

lemma cong_mult_lcancel_eq: assumes an: "coprime a n"
shows "[a * x = a * y] (mod n) ⟷ [x = y] (mod n)" (is "?lhs ⟷ ?rhs")
proof
assume H: "?rhs" from cong_mult[OF cong_refl[of a n] H] show ?lhs .
next
assume H: "?lhs" hence H': "[x*a = y*a] (mod n)" by (simp add: mult.commute)
from cong_mult_rcancel[OF an H'] show ?rhs  .
qed

lemma cong_mult_rcancel_eq: assumes an: "coprime a n"
shows "[x * a = y * a] (mod n) ⟷ [x = y] (mod n)"
using cong_mult_lcancel_eq[OF an, of x y] by (simp add: mult.commute)

lemma cong_add_lcancel_eq: "[a + x = a + y] (mod n) ⟷ [x = y] (mod n)"

lemma cong_add_rcancel_eq: "[x + a = y + a] (mod n) ⟷ [x = y] (mod n)"

lemma cong_add_rcancel: "[x + a = y + a] (mod n) ⟹ [x = y] (mod n)"

lemma cong_add_lcancel: "[a + x = a + y] (mod n) ⟹ [x = y] (mod n)"

lemma cong_add_lcancel_eq_0: "[a + x = a] (mod n) ⟷ [x = 0] (mod n)"

lemma cong_add_rcancel_eq_0: "[x + a = a] (mod n) ⟷ [x = 0] (mod n)"

lemma cong_imp_eq: assumes xn: "x < n" and yn: "y < n" and xy: "[x = y] (mod n)"
shows "x = y"
using xy[unfolded modeq_def mod_less[OF xn] mod_less[OF yn]] .

lemma cong_divides_modulus: "[x = y] (mod m) ⟹ n dvd m ==> [x = y] (mod n)"
apply (auto simp add: nat_mod dvd_def)
apply (rule_tac x="k*q1" in exI)
apply (rule_tac x="k*q2" in exI)
by simp

lemma cong_0_divides: "[x = 0] (mod n) ⟷ n dvd x" by simp

lemma cong_1_divides:"[x = 1] (mod n) ==> n dvd x - 1"
apply (cases "x≤1", simp_all)
using cong_sub_cases[of x 1 n] by auto

lemma cong_divides: "[x = y] (mod n) ⟹ n dvd x ⟷ n dvd y"
apply (auto simp add: nat_mod dvd_def)
apply (rule_tac x="k + q1 - q2" in exI, simp add: add_mult_distrib2 diff_mult_distrib2)
apply (rule_tac x="k + q2 - q1" in exI, simp add: add_mult_distrib2 diff_mult_distrib2)
done

lemma cong_coprime: assumes xy: "[x = y] (mod n)"
shows "coprime n x ⟷ coprime n y"
proof-
{assume "n=0" hence ?thesis using xy by simp}
moreover
{assume nz: "n ≠ 0"
have "coprime n x ⟷ coprime (x mod n) n"
by (simp add: coprime_mod[OF nz, of x] coprime_commute[of n x])
also have "… ⟷ coprime (y mod n) n" using xy[unfolded modeq_def] by simp
also have "… ⟷ coprime y n" by (simp add: coprime_mod[OF nz, of y])
finally have ?thesis by (simp add: coprime_commute) }
ultimately show ?thesis by blast
qed

lemma cong_mod: "~(n = 0) ⟹ [a mod n = a] (mod n)" by (simp add: modeq_def)

lemma mod_mult_cong: "~(a = 0) ⟹ ~(b = 0)
⟹ [x mod (a * b) = y] (mod a) ⟷ [x = y] (mod a)"

lemma cong_mod_mult: "[x = y] (mod n) ⟹ m dvd n ⟹ [x = y] (mod m)"
apply (auto simp add: nat_mod dvd_def)
apply (rule_tac x="k*q1" in exI)
apply (rule_tac x="k*q2" in exI, simp)
done

(* Some things when we know more about the order.                            *)

lemma cong_le: "y <= x ⟹ [x = y] (mod n) ⟷ (∃q. x = q * n + y)"
using nat_mod_lemma[of x y n]
apply auto
apply (rule_tac x="q" in exI)
apply (rule_tac x="q + q" in exI)
by (auto simp: algebra_simps)

lemma cong_to_1: "[a = 1] (mod n) ⟷ a = 0 ∧ n = 1 ∨ (∃m. a = 1 + m * n)"
proof-
{assume "n = 0 ∨ n = 1∨ a = 0 ∨ a = 1" hence ?thesis
apply (cases "n=0", simp_all add: cong_commute)
apply (cases "n=1", simp_all add: cong_commute modeq_def)
apply arith
apply (cases "a=1")
done }
moreover
{assume n: "n≠0" "n≠1" and a:"a≠0" "a ≠ 1" hence a': "a ≥ 1" by simp
hence ?thesis using cong_le[OF a', of n] by auto }
ultimately show ?thesis by auto
qed

(* Some basic theorems about solving congruences.                            *)

lemma cong_solve: assumes an: "coprime a n" shows "∃x. [a * x = b] (mod n)"
proof-
{assume "a=0" hence ?thesis using an by (simp add: modeq_def)}
moreover
{assume az: "a≠0"
obtain d x y where dxy: "d dvd a" "d dvd n" "a*x = n*y + d" by blast
from an[unfolded coprime, rule_format, of d] dxy(1,2) have d1: "d = 1" by blast
hence "a*x*b = (n*y + 1)*b" using dxy(3) by simp
hence "a*(x*b) = n*(y*b) + b" by algebra
hence "a*(x*b) mod n = (n*(y*b) + b) mod n" by simp
hence "a*(x*b) mod n = b mod n" by (simp add: mod_add_left_eq)
hence "[a*(x*b) = b] (mod n)" unfolding modeq_def .
hence ?thesis by blast}
ultimately  show ?thesis by blast
qed

lemma cong_solve_unique: assumes an: "coprime a n" and nz: "n ≠ 0"
shows "∃!x. x < n ∧ [a * x = b] (mod n)"
proof-
let ?P = "λx. x < n ∧ [a * x = b] (mod n)"
from cong_solve[OF an] obtain x where x: "[a*x = b] (mod n)" by blast
let ?x = "x mod n"
from x have th: "[a * ?x = b] (mod n)"
by (simp add: modeq_def mod_mult_right_eq[of a x n])
from mod_less_divisor[ of n x] nz th have Px: "?P ?x" by simp
{fix y assume Py: "y < n" "[a * y = b] (mod n)"
from Py(2) th have "[a * y = a*?x] (mod n)" by (simp add: modeq_def)
hence "[y = ?x] (mod n)" by (simp add: cong_mult_lcancel_eq[OF an])
with mod_less[OF Py(1)] mod_less_divisor[ of n x] nz
have "y = ?x" by (simp add: modeq_def)}
with Px show ?thesis by blast
qed

lemma cong_solve_unique_nontrivial:
assumes p: "prime p" and pa: "coprime p a" and x0: "0 < x" and xp: "x < p"
shows "∃!y. 0 < y ∧ y < p ∧ [x * y = a] (mod p)"
proof-
from p have p1: "p > 1" using prime_ge_2[OF p] by arith
hence p01: "p ≠ 0" "p ≠ 1" by arith+
from pa have ap: "coprime a p" by (simp add: coprime_commute)
from prime_coprime[OF p, of x] dvd_imp_le[of p x] x0 xp have px:"coprime x p"
from cong_solve_unique[OF px p01(1)]
obtain y where y: "y < p" "[x * y = a] (mod p)" "∀z. z < p ∧ [x * z = a] (mod p) ⟶ z = y" by blast
{assume y0: "y = 0"
with y(2) have th: "p dvd a" by (simp add: cong_commute[of 0 a p])
with p coprime_prime[OF pa, of p] have False by simp}
with y show ?thesis unfolding Ex1_def using neq0_conv by blast
qed
lemma cong_unique_inverse_prime:
assumes p: "prime p" and x0: "0 < x" and xp: "x < p"
shows "∃!y. 0 < y ∧ y < p ∧ [x * y = 1] (mod p)"
using cong_solve_unique_nontrivial[OF p coprime_1[of p] x0 xp] .

(* Forms of the Chinese remainder theorem.                                   *)

lemma cong_chinese:
assumes ab: "coprime a b" and  xya: "[x = y] (mod a)"
and xyb: "[x = y] (mod b)"
shows "[x = y] (mod a*b)"
using ab xya xyb
by (simp add: cong_sub_cases[of x y a] cong_sub_cases[of x y b]
cong_sub_cases[of x y "a*b"])
(cases "x ≤ y", simp_all add: divides_mul[of a _ b])

lemma chinese_remainder_unique:
assumes ab: "coprime a b" and az: "a ≠ 0" and bz: "b≠0"
shows "∃!x. x < a * b ∧ [x = m] (mod a) ∧ [x = n] (mod b)"
proof-
from az bz have abpos: "a*b > 0" by simp
from chinese_remainder[OF ab az bz] obtain x q1 q2 where
xq12: "x = m + q1 * a" "x = n + q2 * b" by blast
let ?w = "x mod (a*b)"
have wab: "?w < a*b" by (simp add: mod_less_divisor[OF abpos])
from xq12(1) have "?w mod a = ((m + q1 * a) mod (a*b)) mod a" by simp
also have "… = m mod a" by (simp add: mod_mult2_eq)
finally have th1: "[?w = m] (mod a)" by (simp add: modeq_def)
from xq12(2) have "?w mod b = ((n + q2 * b) mod (a*b)) mod b" by simp
also have "… = ((n + q2 * b) mod (b*a)) mod b" by (simp add: mult.commute)
also have "… = n mod b" by (simp add: mod_mult2_eq)
finally have th2: "[?w = n] (mod b)" by (simp add: modeq_def)
{fix y assume H: "y < a*b" "[y = m] (mod a)" "[y = n] (mod b)"
with th1 th2 have H': "[y = ?w] (mod a)" "[y = ?w] (mod b)"
from cong_chinese[OF ab H'] mod_less[OF H(1)] mod_less[OF wab]
have "y = ?w" by (simp add: modeq_def)}
with th1 th2 wab show ?thesis by blast
qed

lemma chinese_remainder_coprime_unique:
assumes ab: "coprime a b" and az: "a ≠ 0" and bz: "b ≠ 0"
and ma: "coprime m a" and nb: "coprime n b"
shows "∃!x. coprime x (a * b) ∧ x < a * b ∧ [x = m] (mod a) ∧ [x = n] (mod b)"
proof-
let ?P = "λx. x < a * b ∧ [x = m] (mod a) ∧ [x = n] (mod b)"
from chinese_remainder_unique[OF ab az bz]
obtain x where x: "x < a * b" "[x = m] (mod a)" "[x = n] (mod b)"
"∀y. ?P y ⟶ y = x" by blast
from ma nb cong_coprime[OF x(2)] cong_coprime[OF x(3)]
have "coprime x a" "coprime x b" by (simp_all add: coprime_commute)
with coprime_mul[of x a b] have "coprime x (a*b)" by simp
with x show ?thesis by blast
qed

(* Euler totient function.                                                   *)

definition phi_def: "φ n = card { m. 0 < m ∧ m <= n ∧ coprime m n }"

lemma phi_0[simp]: "φ 0 = 0"
unfolding phi_def by auto

lemma phi_finite[simp]: "finite ({ m. 0 < m ∧ m <= n ∧ coprime m n })"
proof-
have "{ m. 0 < m ∧ m <= n ∧ coprime m n } ⊆ {0..n}" by auto
thus ?thesis by (auto intro: finite_subset)
qed

declare coprime_1[presburger]
lemma phi_1[simp]: "φ 1 = 1"
proof-
{fix m
have "0 < m ∧ m <= 1 ∧ coprime m 1 ⟷ m = 1" by presburger }
thus ?thesis by (simp add: phi_def)
qed

lemma [simp]: "φ (Suc 0) = Suc 0" using phi_1 by simp

lemma phi_alt: "φ(n) = card { m. coprime m n ∧ m < n}"
proof-
{assume "n=0 ∨ n=1" hence ?thesis by (cases "n=0", simp_all)}
moreover
{assume n: "n≠0" "n≠1"
{fix m
from n have "0 < m ∧ m <= n ∧ coprime m n ⟷ coprime m n ∧ m < n"
apply (cases "m = 0", simp_all)
apply (cases "m = 1", simp_all)
apply (cases "m = n", auto)
done }
hence ?thesis unfolding phi_def by simp}
ultimately show ?thesis by auto
qed

lemma phi_finite_lemma[simp]: "finite {m. coprime m n ∧  m < n}" (is "finite ?S")
by (rule finite_subset[of "?S" "{0..n}"], auto)

lemma phi_another: assumes n: "n≠1"
shows "φ n = card {m. 0 < m ∧ m < n ∧ coprime m n }"
proof-
{fix m
from n have "0 < m ∧ m < n ∧ coprime m n ⟷ coprime m n ∧ m < n"
by (cases "m=0", auto)}
thus ?thesis unfolding phi_alt by auto
qed

lemma phi_limit: "φ n ≤ n"
proof-
have "{ m. coprime m n ∧ m < n} ⊆ {0 ..<n}" by auto
with card_mono[of "{0 ..<n}" "{ m. coprime m n ∧ m < n}"]
show ?thesis unfolding phi_alt by auto
qed

lemma stupid[simp]: "{m. (0::nat) < m ∧ m < n} = {1..<n}"
by auto

lemma phi_limit_strong: assumes n: "n≠1"
shows "φ(n) ≤ n - 1"
proof-
show ?thesis
unfolding phi_another[OF n] finite_number_segment[of n, symmetric]
by (rule card_mono[of "{m. 0 < m ∧ m < n}" "{m. 0 < m ∧ m < n ∧ coprime m n}"], auto)
qed

lemma phi_lowerbound_1_strong: assumes n: "n ≥ 1"
shows "φ(n) ≥ 1"
proof-
let ?S = "{ m. 0 < m ∧ m <= n ∧ coprime m n }"
from card_0_eq[of ?S] n have "φ n ≠ 0" unfolding phi_alt
apply auto
apply (cases "n=1", simp_all)
apply (rule exI[where x=1], simp)
done
thus ?thesis by arith
qed

lemma phi_lowerbound_1: "2 <= n ==> 1 <= φ(n)"
using phi_lowerbound_1_strong[of n] by auto

lemma phi_lowerbound_2: assumes n: "3 <= n" shows "2 <= φ (n)"
proof-
let ?S = "{ m. 0 < m ∧ m <= n ∧ coprime m n }"
have inS: "{1, n - 1} ⊆ ?S" using n coprime_plus1[of "n - 1"]
from n have c2: "card {1, n - 1} = 2" by (auto simp add: card_insert_if)
from card_mono[of ?S "{1, n - 1}", simplified inS c2] show ?thesis
unfolding phi_def by auto
qed

lemma phi_prime: "φ n = n - 1 ∧ n≠0 ∧ n≠1 ⟷ prime n"
proof-
{assume "n=0 ∨ n=1" hence ?thesis by (cases "n=1", simp_all)}
moreover
{assume n: "n≠0" "n≠1"
let ?S = "{m. 0 < m ∧ m < n}"
have fS: "finite ?S" by simp
let ?S' = "{m. 0 < m ∧ m < n ∧ coprime m n}"
have fS':"finite ?S'" apply (rule finite_subset[of ?S' ?S]) by auto
{assume H: "φ n = n - 1 ∧ n≠0 ∧ n≠1"
hence ceq: "card ?S' = card ?S"
using n finite_number_segment[of n] phi_another[OF n(2)] by simp
{fix m assume m: "0 < m" "m < n" "¬ coprime m n"
hence mS': "m ∉ ?S'" by auto
have "insert m ?S' ≤ ?S" using m by auto
from m have "card (insert m ?S') ≤ card ?S"
by - (rule card_mono[of ?S "insert m ?S'"], auto)
hence False
unfolding card_insert_disjoint[of "?S'" m, OF fS' mS'] ceq
by simp }
hence "∀m. 0 <m ∧ m < n ⟶ coprime m n" by blast
hence "prime n" unfolding prime using n by (simp add: coprime_commute)}
moreover
{assume H: "prime n"
hence "?S = ?S'" unfolding prime using n
hence "card ?S = card ?S'" by simp
hence "φ n = n - 1" unfolding phi_another[OF n(2)] by simp}
ultimately have ?thesis using n by blast}
ultimately show ?thesis by (cases "n=0") blast+
qed

(* Multiplicativity property.                                                *)

lemma phi_multiplicative: assumes ab: "coprime a b"
shows "φ (a * b) = φ a * φ b"
proof-
{assume "a = 0 ∨ b = 0 ∨ a = 1 ∨ b = 1"
hence ?thesis
by (cases "a=0", simp, cases "b=0", simp, cases"a=1", simp_all) }
moreover
{assume a: "a≠0" "a≠1" and b: "b≠0" "b≠1"
hence ab0: "a*b ≠ 0" by simp
let ?S = "λk. {m. coprime m k ∧ m < k}"
let ?f = "λx. (x mod a, x mod b)"
have eq: "?f ` (?S (a*b)) = (?S a × ?S b)"
proof-
{fix x assume x:"x ∈ ?S (a*b)"
hence x': "coprime x (a*b)" "x < a*b" by simp_all
hence xab: "coprime x a" "coprime x b" by (simp_all add: coprime_mul_eq)
from mod_less_divisor a b have xab':"x mod a < a" "x mod b < b" by auto
from xab xab' have "?f x ∈ (?S a × ?S b)"
by (simp add: coprime_mod[OF a(1)] coprime_mod[OF b(1)])}
moreover
{fix x y assume x: "x ∈ ?S a" and y: "y ∈ ?S b"
hence x': "coprime x a" "x < a" and y': "coprime y b" "y < b" by simp_all
from chinese_remainder_coprime_unique[OF ab a(1) b(1) x'(1) y'(1)]
obtain z where z: "coprime z (a * b)" "z < a * b" "[z = x] (mod a)"
"[z = y] (mod b)" by blast
hence "(x,y) ∈ ?f ` (?S (a*b))"
using y'(2) mod_less_divisor[of b y] x'(2) mod_less_divisor[of a x]
by (auto simp add: image_iff modeq_def)}
ultimately show ?thesis by auto
qed
have finj: "inj_on ?f (?S (a*b))"
unfolding inj_on_def
proof(clarify)
fix x y assume H: "coprime x (a * b)" "x < a * b" "coprime y (a * b)"
"y < a * b" "x mod a = y mod a" "x mod b = y mod b"
hence cp: "coprime x a" "coprime x b" "coprime y a" "coprime y b"
from chinese_remainder_coprime_unique[OF ab a(1) b(1) cp(3,4)] H
show "x = y" unfolding modeq_def by blast
qed
from card_image[OF finj, unfolded eq] have ?thesis
unfolding phi_alt by simp }
ultimately show ?thesis by auto
qed

(* Fermat's Little theorem / Fermat-Euler theorem.                           *)

lemma nproduct_mod:
assumes fS: "finite S" and n0: "n ≠ 0"
shows "[setprod (λm. a(m) mod n) S = setprod a S] (mod n)"
proof-
have th1:"[1 = 1] (mod n)" by (simp add: modeq_def)
from cong_mult
have th3:"∀x1 y1 x2 y2.
[x1 = x2] (mod n) ∧ [y1 = y2] (mod n) ⟶ [x1 * y1 = x2 * y2] (mod n)"
by blast
have th4:"∀x∈S. [a x mod n = a x] (mod n)" by (simp add: modeq_def)
from setprod.related [where h="(λm. a(m) mod n)" and g=a, OF th1 th3 fS, OF th4] show ?thesis by (simp add: fS)
qed

lemma nproduct_cmul:
assumes fS:"finite S"
shows "setprod (λm. (c::'a::{comm_monoid_mult})* a(m)) S = c ^ (card S) * setprod a S"
unfolding setprod.distrib setprod_constant[OF fS, of c] ..

lemma coprime_nproduct:
assumes fS: "finite S" and Sn: "∀x∈S. coprime n (a x)"
shows "coprime n (setprod a S)"
using fS by (rule finite_subset_induct)
(insert Sn, auto simp add: coprime_mul)

lemma fermat_little: assumes an: "coprime a n"
shows "[a ^ (φ n) = 1] (mod n)"
proof-
{assume "n=0" hence ?thesis by simp}
moreover
{assume "n=1" hence ?thesis by (simp add: modeq_def)}
moreover
{assume nz: "n ≠ 0" and n1: "n ≠ 1"
let ?S = "{m. coprime m n ∧ m < n}"
let ?P = "∏ ?S"
have fS: "finite ?S" by simp
have cardfS: "φ n = card ?S" unfolding phi_alt ..
{fix m assume m: "m ∈ ?S"
hence "coprime m n" by simp
with coprime_mul[of n a m] an have "coprime (a*m) n"
hence Sn: "∀m∈ ?S. coprime (a*m) n " by blast
from coprime_nproduct[OF fS, of n "λm. m"] have nP:"coprime ?P n"
have Paphi: "[?P*a^ (φ n) = ?P*1] (mod n)"
proof-
let ?h = "λm. (a * m) mod n"

have eq0: "(∏i∈?S. ?h i) = (∏i∈?S. i)"
proof (rule setprod.reindex_bij_betw)
have "inj_on (λi. ?h i) ?S"
proof (rule inj_onI)
fix x y assume "?h x = ?h y"
then have "[a * x = a * y] (mod n)"
moreover assume "x ∈ ?S" "y ∈ ?S"
ultimately show "x = y"
by (auto intro: cong_imp_eq cong_mult_lcancel an)
qed
moreover have "?h ` ?S = ?S"
proof safe
fix y assume "coprime y n" then show "coprime (?h y) n"
by (metis an nz coprime_commute coprime_mod coprime_mul_eq)
next
fix y assume y: "coprime y n" "y < n"
from cong_solve_unique[OF an nz] obtain x where x: "x < n" "[a * x = y] (mod n)"
by blast
then show "y ∈ ?h ` ?S"
using cong_coprime[OF x(2)] coprime_mul_eq[of n a x] an y x
by (intro image_eqI[of _ _ x]) (auto simp add: coprime_commute modeq_def)
qed (insert nz, simp)
ultimately show "bij_betw ?h ?S ?S"
qed
from nproduct_mod[OF fS nz, of "op * a"]
have "[(∏i∈?S. a * i) = (∏i∈?S. ?h i)] (mod n)"
also have "[(∏i∈?S. ?h i) = ?P] (mod n)"
using eq0 fS an by (simp add: setprod_def modeq_def)
finally show "[?P*a^ (φ n) = ?P*1] (mod n)"
unfolding cardfS mult.commute[of ?P "a^ (card ?S)"]
nproduct_cmul[OF fS, symmetric] mult_1_right by simp
qed
from cong_mult_lcancel[OF nP Paphi] have ?thesis . }
ultimately show ?thesis by blast
qed

lemma fermat_little_prime: assumes p: "prime p" and ap: "coprime a p"
shows "[a^ (p - 1) = 1] (mod p)"
using fermat_little[OF ap] p[unfolded phi_prime[symmetric]]
by simp

(* Lucas's theorem.                                                          *)

lemma lucas_coprime_lemma:
assumes m: "m≠0" and am: "[a^m = 1] (mod n)"
shows "coprime a n"
proof-
{assume "n=1" hence ?thesis by simp}
moreover
{assume "n = 0" hence ?thesis using am m exp_eq_1[of a m] by simp}
moreover
{assume n: "n≠0" "n≠1"
from m obtain m' where m': "m = Suc m'" by (cases m, blast+)
{fix d
assume d: "d dvd a" "d dvd n"
from n have n1: "1 < n" by arith
from am mod_less[OF n1] have am1: "a^m mod n = 1" unfolding modeq_def by simp
from dvd_mult2[OF d(1), of "a^m'"] have dam:"d dvd a^m" by (simp add: m')
from dvd_mod_iff[OF d(2), of "a^m"] dam am1
have "d = 1" by simp }
hence ?thesis unfolding coprime by auto
}
ultimately show ?thesis by blast
qed

lemma lucas_weak:
assumes n: "n ≥ 2" and an:"[a^(n - 1) = 1] (mod n)"
and nm: "∀m. 0 <m ∧ m < n - 1 ⟶ ¬ [a^m = 1] (mod n)"
shows "prime n"
proof-
from n have n1: "n ≠ 1" "n≠0" "n - 1 ≠ 0" "n - 1 > 0" "n - 1 < n" by arith+
from lucas_coprime_lemma[OF n1(3) an] have can: "coprime a n" .
from fermat_little[OF can] have afn: "[a ^ φ n = 1] (mod n)" .
{assume "φ n ≠ n - 1"
with phi_limit_strong[OF n1(1)] phi_lowerbound_1[OF n]
have c:"φ n > 0 ∧ φ n < n - 1" by arith
from nm[rule_format, OF c] afn have False ..}
hence "φ n = n - 1" by blast
with phi_prime[of n] n1(1,2) show ?thesis by simp
qed

lemma nat_exists_least_iff: "(∃(n::nat). P n) ⟷ (∃n. P n ∧ (∀m < n. ¬ P m))"
(is "?lhs ⟷ ?rhs")
proof
assume ?rhs thus ?lhs by blast
next
assume H: ?lhs then obtain n where n: "P n" by blast
let ?x = "Least P"
{fix m assume m: "m < ?x"
from not_less_Least[OF m] have "¬ P m" .}
with LeastI_ex[OF H] show ?rhs by blast
qed

lemma nat_exists_least_iff': "(∃(n::nat). P n) ⟷ (P (Least P) ∧ (∀m < (Least P). ¬ P m))"
(is "?lhs ⟷ ?rhs")
proof-
{assume ?rhs hence ?lhs by blast}
moreover
{ assume H: ?lhs then obtain n where n: "P n" by blast
let ?x = "Least P"
{fix m assume m: "m < ?x"
from not_less_Least[OF m] have "¬ P m" .}
with LeastI_ex[OF H] have ?rhs by blast}
ultimately show ?thesis by blast
qed

lemma power_mod: "((x::nat) mod m)^n mod m = x^n mod m"
proof(induct n)
case 0 thus ?case by simp
next
case (Suc n)
have "(x mod m)^(Suc n) mod m = ((x mod m) * (((x mod m) ^ n) mod m)) mod m"
also have "… = ((x mod m) * (x^n mod m)) mod m" using Suc.hyps by simp
also have "… = x^(Suc n) mod m"
finally show ?case .
qed

lemma lucas:
assumes n2: "n ≥ 2" and an1: "[a^(n - 1) = 1] (mod n)"
and pn: "∀p. prime p ∧ p dvd n - 1 ⟶ ¬ [a^((n - 1) div p) = 1] (mod n)"
shows "prime n"
proof-
from n2 have n01: "n≠0" "n≠1" "n - 1 ≠ 0" by arith+
from mod_less_divisor[of n 1] n01 have onen: "1 mod n = 1" by simp
from lucas_coprime_lemma[OF n01(3) an1] cong_coprime[OF an1]
have an: "coprime a n" "coprime (a^(n - 1)) n" by (simp_all add: coprime_commute)
{assume H0: "∃m. 0 < m ∧ m < n - 1 ∧ [a ^ m = 1] (mod n)" (is "EX m. ?P m")
from H0[unfolded nat_exists_least_iff[of ?P]] obtain m where
m: "0 < m" "m < n - 1" "[a ^ m = 1] (mod n)" "∀k <m. ¬?P k" by blast
{assume nm1: "(n - 1) mod m > 0"
from mod_less_divisor[OF m(1)] have th0:"(n - 1) mod m < m" by blast
let ?y = "a^ ((n - 1) div m * m)"
note mdeq = mod_div_equality[of "(n - 1)" m]
from coprime_exp[OF an(1)[unfolded coprime_commute[of a n]],
of "(n - 1) div m * m"]
have yn: "coprime ?y n" by (simp add: coprime_commute)
have "?y mod n = (a^m)^((n - 1) div m) mod n"
also have "… = (a^m mod n)^((n - 1) div m) mod n"
using power_mod[of "a^m" n "(n - 1) div m"] by simp
also have "… = 1" using m(3)[unfolded modeq_def onen] onen
finally have th3: "?y mod n = 1"  .
have th2: "[?y * a ^ ((n - 1) mod m) = ?y* 1] (mod n)"
using an1[unfolded modeq_def onen] onen
mod_div_equality[of "(n - 1)" m, symmetric]
from cong_mult_lcancel[of ?y n "a^((n - 1) mod m)" 1, OF yn th2]
have th1: "[a ^ ((n - 1) mod m) = 1] (mod n)"  .
from m(4)[rule_format, OF th0] nm1
less_trans[OF mod_less_divisor[OF m(1), of "n - 1"] m(2)] th1
have False by blast }
hence "(n - 1) mod m = 0" by auto
then have mn: "m dvd n - 1" by presburger
then obtain r where r: "n - 1 = m*r" unfolding dvd_def by blast
from n01 r m(2) have r01: "r≠0" "r≠1" by - (rule ccontr, simp)+
from prime_factor[OF r01(2)] obtain p where p: "prime p" "p dvd r" by blast
hence th: "prime p ∧ p dvd n - 1" unfolding r by (auto intro: dvd_mult)
have "(a ^ ((n - 1) div p)) mod n = (a^(m*r div p)) mod n" using r
also have "… = (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
also have "… = ((a^m)^(r div p)) mod n" by (simp add: power_mult)
also have "… = ((a^m mod n)^(r div p)) mod n" using power_mod[of "a^m" "n" "r div p" ] ..
also have "… = 1" using m(3) onen by (simp add: modeq_def power_Suc0)
finally have "[(a ^ ((n - 1) div p))= 1] (mod n)"
using onen by (simp add: modeq_def)
with pn[rule_format, OF th] have False by blast}
hence th: "∀m. 0 < m ∧ m < n - 1 ⟶ ¬ [a ^ m = 1] (mod n)" by blast
from lucas_weak[OF n2 an1 th] show ?thesis .
qed

(* Definition of the order of a number mod n (0 in non-coprime case).        *)

definition "ord n a = (if coprime n a then Least (λd. d > 0 ∧ [a ^d = 1] (mod n)) else 0)"

(* This has the expected properties.                                         *)

lemma coprime_ord:
assumes na: "coprime n a"
shows "ord n a > 0 ∧ [a ^(ord n a) = 1] (mod n) ∧ (∀m. 0 < m ∧ m < ord n a ⟶ ¬ [a^ m = 1] (mod n))"
proof-
let ?P = "λd. 0 < d ∧ [a ^ d = 1] (mod n)"
from euclid[of a] obtain p where p: "prime p" "a < p" by blast
from na have o: "ord n a = Least ?P" by (simp add: ord_def)
{assume "n=0 ∨ n=1" with na have "∃m>0. ?P m" apply auto apply (rule exI[where x=1]) by (simp  add: modeq_def)}
moreover
{assume "n≠0 ∧ n≠1" hence n2:"n ≥ 2" by arith
from na have na': "coprime a n" by (simp add: coprime_commute)
from phi_lowerbound_1[OF n2] fermat_little[OF na']
have ex: "∃m>0. ?P m" by - (rule exI[where x="φ n"], auto) }
ultimately have ex: "∃m>0. ?P m" by blast
from nat_exists_least_iff'[of ?P] ex na show ?thesis
unfolding o[symmetric] by auto
qed
(* With the special value 0 for non-coprime case, it's more convenient.      *)
lemma ord_works:
"[a ^ (ord n a) = 1] (mod n) ∧ (∀m. 0 < m ∧ m < ord n a ⟶ ~[a^ m = 1] (mod n))"
apply (cases "coprime n a")
using coprime_ord[of n a]
by (blast, simp add: ord_def modeq_def)

lemma ord: "[a^(ord n a) = 1] (mod n)" using ord_works by blast
lemma ord_minimal: "0 < m ⟹ m < ord n a ⟹ ~[a^m = 1] (mod n)"
using ord_works by blast
lemma ord_eq_0: "ord n a = 0 ⟷ ~coprime n a"

lemma ord_divides:
"[a ^ d = 1] (mod n) ⟷ ord n a dvd d" (is "?lhs ⟷ ?rhs")
proof
assume rh: ?rhs
then obtain k where "d = ord n a * k" unfolding dvd_def by blast
hence "[a ^ d = (a ^ (ord n a) mod n)^k] (mod n)"
by (simp add : modeq_def power_mult power_mod)
also have "[(a ^ (ord n a) mod n)^k = 1] (mod n)"
using ord[of a n, unfolded modeq_def]
by (simp add: modeq_def power_mod power_Suc0)
finally  show ?lhs .
next
assume lh: ?lhs
{ assume H: "¬ coprime n a"
hence o: "ord n a = 0" by (simp add: ord_def)
{assume d: "d=0" with o H have ?rhs by (simp add: modeq_def)}
moreover
{assume d0: "d≠0" then obtain d' where d': "d = Suc d'" by (cases d, auto)
from H[unfolded coprime]
obtain p where p: "p dvd n" "p dvd a" "p ≠ 1" by auto
from lh[unfolded nat_mod]
obtain q1 q2 where q12:"a ^ d + n * q1 = 1 + n * q2" by blast
hence "a ^ d + n * q1 - n * q2 = 1" by simp
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
with p(3) have False by simp
hence ?rhs ..}
ultimately have ?rhs by blast}
moreover
{assume H: "coprime n a"
let ?o = "ord n a"
let ?q = "d div ord n a"
let ?r = "d mod ord n a"
from cong_exp[OF ord[of a n], of ?q]
have eqo: "[(a^?o)^?q = 1] (mod n)"  by (simp add: modeq_def power_Suc0)
from H have onz: "?o ≠ 0" by (simp add: ord_eq_0)
hence op: "?o > 0" by simp
from mod_div_equality[of d "ord n a"] lh
have "[a^(?o*?q + ?r) = 1] (mod n)" by (simp add: modeq_def mult.commute)
hence "[(a^?o)^?q * (a^?r) = 1] (mod n)"
hence th: "[a^?r = 1] (mod n)"
using eqo mod_mult_left_eq[of "(a^?o)^?q" "a^?r" n]
apply (simp add: modeq_def del: One_nat_def)
{assume r: "?r = 0" hence ?rhs by (simp add: dvd_eq_mod_eq_0)}
moreover
{assume r: "?r ≠ 0"
with mod_less_divisor[OF op, of d] have r0o:"?r >0 ∧ ?r < ?o" by simp
from conjunct2[OF ord_works[of a n], rule_format, OF r0o] th
have ?rhs by blast}
ultimately have ?rhs by blast}
ultimately  show ?rhs by blast
qed

lemma order_divides_phi: "coprime n a ⟹ ord n a dvd φ n"
using ord_divides fermat_little coprime_commute by simp
lemma order_divides_expdiff:
assumes na: "coprime n a"
shows "[a^d = a^e] (mod n) ⟷ [d = e] (mod (ord n a))"
proof-
{fix n a d e
assume na: "coprime n a" and ed: "(e::nat) ≤ d"
hence "∃c. d = e + c" by arith
then obtain c where c: "d = e + c" by arith
from na have an: "coprime a n" by (simp add: coprime_commute)
from coprime_exp[OF na, of e]
have aen: "coprime (a^e) n" by (simp add: coprime_commute)
from coprime_exp[OF na, of c]
have acn: "coprime (a^c) n" by (simp add: coprime_commute)
have "[a^d = a^e] (mod n) ⟷ [a^(e + c) = a^(e + 0)] (mod n)"
using c by simp
also have "… ⟷ [a^e* a^c = a^e *a^0] (mod n)" by (simp add: power_add)
also have  "… ⟷ [a ^ c = 1] (mod n)"
using cong_mult_lcancel_eq[OF aen, of "a^c" "a^0"] by simp
also  have "… ⟷ ord n a dvd c" by (simp only: ord_divides)
also have "… ⟷ [e + c = e + 0] (mod ord n a)"
using cong_add_lcancel_eq[of e c 0 "ord n a", simplified cong_0_divides]
by simp
finally have "[a^d = a^e] (mod n) ⟷ [d = e] (mod (ord n a))"
using c by simp }
note th = this
have "e ≤ d ∨ d ≤ e" by arith
moreover
{assume ed: "e ≤ d" from th[OF na ed] have ?thesis .}
moreover
{assume de: "d ≤ e"
from th[OF na de] have ?thesis by (simp add: cong_commute) }
ultimately show ?thesis by blast
qed

(* Another trivial primality characterization.                               *)

lemma prime_prime_factor:
"prime n ⟷ n ≠ 1∧ (∀p. prime p ∧ p dvd n ⟶ p = n)"
proof-
{assume n: "n=0 ∨ n=1" hence ?thesis using prime_0 two_is_prime by auto}
moreover
{assume n: "n≠0" "n≠1"
{assume pn: "prime n"

from pn[unfolded prime_def] have "∀p. prime p ∧ p dvd n ⟶ p = n"
using n
apply (cases "n = 0 ∨ n=1",simp)
by (clarsimp, erule_tac x="p" in allE, auto)}
moreover
{assume H: "∀p. prime p ∧ p dvd n ⟶ p = n"
from n have n1: "n > 1" by arith
{fix m assume m: "m dvd n" "m≠1"
from prime_factor[OF m(2)] obtain p where
p: "prime p" "p dvd m" by blast
from dvd_trans[OF p(2) m(1)] p(1) H have "p = n" by blast
with p(2) have "n dvd m"  by simp
hence "m=n"  using dvd_antisym[OF m(1)] by simp }
with n1 have "prime n"  unfolding prime_def by auto }
ultimately have ?thesis using n by blast}
ultimately       show ?thesis by auto
qed

lemma prime_divisor_sqrt:
"prime n ⟷ n ≠ 1 ∧ (∀d. d dvd n ∧ d⇧2 ≤ n ⟶ d = 1)"
proof-
{assume "n=0 ∨ n=1" hence ?thesis using prime_0 prime_1
moreover
{assume n: "n≠0" "n≠1"
hence np: "n > 1" by arith
{fix d assume d: "d dvd n" "d⇧2 ≤ n" and H: "∀m. m dvd n ⟶ m=1 ∨ m=n"
from H d have d1n: "d = 1 ∨ d=n" by blast
{assume dn: "d=n"
have "n⇧2 > n*1" using n by (simp add: power2_eq_square)
with dn d(2) have "d=1" by simp}
with d1n have "d = 1" by blast  }
moreover
{fix d assume d: "d dvd n" and H: "∀d'. d' dvd n ∧ d'⇧2 ≤ n ⟶ d' = 1"
from d n have "d ≠ 0" apply - apply (rule ccontr) by simp
hence dp: "d > 0" by simp
from d[unfolded dvd_def] obtain e where e: "n= d*e" by blast
from n dp e have ep:"e > 0" by simp
have "d⇧2 ≤ n ∨ e⇧2 ≤ n" using dp ep
by (auto simp add: e power2_eq_square mult_le_cancel_left)
moreover
{assume h: "d⇧2 ≤ n"
from H[rule_format, of d] h d have "d = 1" by blast}
moreover
{assume h: "e⇧2 ≤ n"
from e have "e dvd n" unfolding dvd_def by (simp add: mult.commute)
with H[rule_format, of e] h have "e=1" by simp
with e have "d = n" by simp}
ultimately have "d=1 ∨ d=n"  by blast}
ultimately have ?thesis unfolding prime_def using np n(2) by blast}
ultimately show ?thesis by auto
qed
lemma prime_prime_factor_sqrt:
"prime n ⟷ n ≠ 0 ∧ n ≠ 1 ∧ ¬ (∃p. prime p ∧ p dvd n ∧ p⇧2 ≤ n)"
(is "?lhs ⟷?rhs")
proof-
{assume "n=0 ∨ n=1" hence ?thesis using prime_0 prime_1 by auto}
moreover
{assume n: "n≠0" "n≠1"
{assume H: ?lhs
from H[unfolded prime_divisor_sqrt] n
have ?rhs
apply clarsimp
apply (erule_tac x="p" in allE)
apply simp
done
}
moreover
{assume H: ?rhs
{fix d assume d: "d dvd n" "d⇧2 ≤ n" "d≠1"
from prime_factor[OF d(3)]
obtain p where p: "prime p" "p dvd d" by blast
from n have np: "n > 0" by arith
from d(1) n have "d ≠ 0" by - (rule ccontr, auto)
hence dp: "d > 0" by arith
from mult_mono[OF dvd_imp_le[OF p(2) dp] dvd_imp_le[OF p(2) dp]] d(2)
have "p⇧2 ≤ n" unfolding power2_eq_square by arith
with H n p(1) dvd_trans[OF p(2) d(1)] have False  by blast}
with n prime_divisor_sqrt  have ?lhs by auto}
ultimately have ?thesis by blast }
ultimately show ?thesis by (cases "n=0 ∨ n=1", auto)
qed
(* Pocklington theorem. *)

lemma pocklington_lemma:
assumes n: "n ≥ 2" and nqr: "n - 1 = q*r" and an: "[a^ (n - 1) = 1] (mod n)"
and aq:"∀p. prime p ∧ p dvd q ⟶ coprime (a^ ((n - 1) div p) - 1) n"
and pp: "prime p" and pn: "p dvd n"
shows "[p = 1] (mod q)"
proof-
from pp prime_0 prime_1 have p01: "p ≠ 0" "p ≠ 1" by - (rule ccontr, simp)+
from cong_1_divides[OF an, unfolded nqr, unfolded dvd_def]
obtain k where k: "a ^ (q * r) - 1 = n*k" by blast
from pn[unfolded dvd_def] obtain l where l: "n = p*l" by blast
{assume a0: "a = 0"
hence "a^ (n - 1) = 0" using n by (simp add: power_0_left)
with n an mod_less[of 1 n]  have False by (simp add: power_0_left modeq_def)}
hence a0: "a≠0" ..
from n nqr have aqr0: "a ^ (q * r) ≠ 0" using a0 by simp
hence "(a ^ (q * r) - 1) + 1  = a ^ (q * r)" by simp
with k l have "a ^ (q * r) = p*l*k + 1" by simp
hence "a ^ (r * q) + p * 0 = 1 + p * (l*k)" by (simp add: ac_simps)
hence odq: "ord p (a^r) dvd q"
unfolding ord_divides[symmetric] power_mult[symmetric] nat_mod  by blast
from odq[unfolded dvd_def] obtain d where d: "q = ord p (a^r) * d" by blast
{assume d1: "d ≠ 1"
from prime_factor[OF d1] obtain P where P: "prime P" "P dvd d" by blast
from d dvd_mult[OF P(2), of "ord p (a^r)"] have Pq: "P dvd q" by simp
from aq P(1) Pq have caP:"coprime (a^ ((n - 1) div P) - 1) n" by blast
from Pq obtain s where s: "q = P*s" unfolding dvd_def by blast
have P0: "P ≠ 0" using P(1) prime_0 by - (rule ccontr, simp)
from P(2) obtain t where t: "d = P*t" unfolding dvd_def by blast
from d s t P0  have s': "ord p (a^r) * t = s" by algebra
have "ord p (a^r) * t*r = r * ord p (a^r) * t" by algebra
hence exps: "a^(ord p (a^r) * t*r) = ((a ^ r) ^ ord p (a^r)) ^ t"
by (simp only: power_mult)
have "[((a ^ r) ^ ord p (a^r)) ^ t= 1^t] (mod p)"
by (rule cong_exp, rule ord)
then have th: "[((a ^ r) ^ ord p (a^r)) ^ t= 1] (mod p)"
from cong_1_divides[OF th] exps have pd0: "p dvd a^(ord p (a^r) * t*r) - 1" by simp
from nqr s s' have "(n - 1) div P = ord p (a^r) * t*r" using P0 by simp
with caP have "coprime (a^(ord p (a^r) * t*r) - 1) n" by simp
with p01 pn pd0 have False unfolding coprime by auto}
hence d1: "d = 1" by blast
hence o: "ord p (a^r) = q" using d by simp
from pp phi_prime[of p] have phip: " φ p = p - 1" by simp
{fix d assume d: "d dvd p" "d dvd a" "d ≠ 1"
from pp[unfolded prime_def] d have dp: "d = p" by blast
from n have n12:"Suc (n - 2) = n - 1" by arith
with divides_rexp[OF d(2)[unfolded dp], of "n - 2"]
have th0: "p dvd a ^ (n - 1)" by simp
from n have n0: "n ≠ 0" by simp
from d(2) an n12[symmetric] have a0: "a ≠ 0"
by - (rule ccontr, simp add: modeq_def)
have th1: "a^ (n - 1) ≠ 0" using n d(2) dp a0 by auto
from coprime_minus1[OF th1, unfolded coprime]
dvd_trans[OF pn cong_1_divides[OF an]] th0 d(3) dp
have False by auto}
hence cpa: "coprime p a" using coprime by auto
from coprime_exp[OF cpa, of r] coprime_commute
have arp: "coprime (a^r) p" by blast
from fermat_little[OF arp, simplified ord_divides] o phip
have "q dvd (p - 1)" by simp
then obtain d where d:"p - 1 = q * d" unfolding dvd_def by blast
from prime_0 pp have p0:"p ≠ 0" by -  (rule ccontr, auto)
from p0 d have "p + q * 0 = 1 + q * d" by simp
with nat_mod[of p 1 q, symmetric]
show ?thesis by blast
qed

lemma pocklington:
assumes n: "n ≥ 2" and nqr: "n - 1 = q*r" and sqr: "n ≤ q⇧2"
and an: "[a^ (n - 1) = 1] (mod n)"
and aq:"∀p. prime p ∧ p dvd q ⟶ coprime (a^ ((n - 1) div p) - 1) n"
shows "prime n"
unfolding prime_prime_factor_sqrt[of n]
proof-
let ?ths = "n ≠ 0 ∧ n ≠ 1 ∧ ¬ (∃p. prime p ∧ p dvd n ∧ p⇧2 ≤ n)"
from n have n01: "n≠0" "n≠1" by arith+
{fix p assume p: "prime p" "p dvd n" "p⇧2 ≤ n"
from p(3) sqr have "p^(Suc 1) ≤ q^(Suc 1)" by (simp add: power2_eq_square)
hence pq: "p ≤ q" unfolding exp_mono_le .
from pocklington_lemma[OF n nqr an aq p(1,2)]  cong_1_divides
have th: "q dvd p - 1" by blast
have "p - 1 ≠ 0"using prime_ge_2[OF p(1)] by arith
with divides_ge[OF th] pq have False by arith }
with n01 show ?ths by blast
qed

(* Variant for application, to separate the exponentiation.                  *)
lemma pocklington_alt:
assumes n: "n ≥ 2" and nqr: "n - 1 = q*r" and sqr: "n ≤ q⇧2"
and an: "[a^ (n - 1) = 1] (mod n)"
and aq:"∀p. prime p ∧ p dvd q ⟶ (∃b. [a^((n - 1) div p) = b] (mod n) ∧ coprime (b - 1) n)"
shows "prime n"
proof-
{fix p assume p: "prime p" "p dvd q"
from aq[rule_format] p obtain b where
b: "[a^((n - 1) div p) = b] (mod n)" "coprime (b - 1) n" by blast
{assume a0: "a=0"
from n an have "[0 = 1] (mod n)" unfolding a0 power_0_left by auto
hence False using n by (simp add: modeq_def dvd_eq_mod_eq_0[symmetric])}
hence a0: "a≠ 0" ..
hence a1: "a ≥ 1" by arith
from one_le_power[OF a1] have ath: "1 ≤ a ^ ((n - 1) div p)" .
{assume b0: "b = 0"
from p(2) nqr have "(n - 1) mod p = 0"
apply (simp only: dvd_eq_mod_eq_0[symmetric]) by (rule dvd_mult2, simp)
with mod_div_equality[of "n - 1" p]
have "(n - 1) div p * p= n - 1" by auto
hence eq: "(a^((n - 1) div p))^p = a^(n - 1)"
by (simp only: power_mult[symmetric])
from prime_ge_2[OF p(1)] have pS: "Suc (p - 1) = p" by arith
from b(1) have d: "n dvd a^((n - 1) div p)" unfolding b0 cong_0_divides .
from divides_rexp[OF d, of "p - 1"] pS eq cong_divides[OF an] n
have False by simp}
then have b0: "b ≠ 0" ..
hence b1: "b ≥ 1" by arith
from cong_coprime[OF cong_sub[OF b(1) cong_refl[of 1] ath b1]] b(2) nqr
have "coprime (a ^ ((n - 1) div p) - 1) n" by (simp add: coprime_commute)}
hence th: "∀p. prime p ∧ p dvd q ⟶ coprime (a ^ ((n - 1) div p) - 1) n "
by blast
from pocklington[OF n nqr sqr an th] show ?thesis .
qed

(* Prime factorizations.                                                     *)

definition "primefact ps n = (foldr op * ps  1 = n ∧ (∀p∈ set ps. prime p))"

lemma primefact: assumes n: "n ≠ 0"
shows "∃ps. primefact ps n"
using n
proof(induct n rule: nat_less_induct)
fix n assume H: "∀m<n. m ≠ 0 ⟶ (∃ps. primefact ps m)" and n: "n≠0"
let ?ths = "∃ps. primefact ps n"
{assume "n = 1"
hence "primefact [] n" by (simp add: primefact_def)
hence ?ths by blast }
moreover
{assume n1: "n ≠ 1"
with n have n2: "n ≥ 2" by arith
from prime_factor[OF n1] obtain p where p: "prime p" "p dvd n" by blast
from p(2) obtain m where m: "n = p*m" unfolding dvd_def by blast
from n m have m0: "m > 0" "m≠0" by auto
from prime_ge_2[OF p(1)] have "1 < p" by arith
with m0 m have mn: "m < n" by auto
from H[rule_format, OF mn m0(2)] obtain ps where ps: "primefact ps m" ..
from ps m p(1) have "primefact (p#ps) n" by (simp add: primefact_def)
hence ?ths by blast}
ultimately show ?ths by blast
qed

lemma primefact_contains:
assumes pf: "primefact ps n" and p: "prime p" and pn: "p dvd n"
shows "p ∈ set ps"
using pf p pn
proof(induct ps arbitrary: p n)
case Nil thus ?case by (auto simp add: primefact_def)
next
case (Cons q qs p n)
from Cons.prems[unfolded primefact_def]
have q: "prime q" "q * foldr op * qs 1 = n" "∀p ∈set qs. prime p"  and p: "prime p" "p dvd q * foldr op * qs 1" by simp_all
{assume "p dvd q"
with p(1) q(1) have "p = q" unfolding prime_def by auto
hence ?case by simp}
moreover
{ assume h: "p dvd foldr op * qs 1"
from q(3) have pqs: "primefact qs (foldr op * qs 1)"
from Cons.hyps[OF pqs p(1) h] have ?case by simp}
ultimately show ?case using prime_divprod[OF p] by blast
qed

lemma primefact_variant: "primefact ps n ⟷ foldr op * ps 1 = n ∧ list_all prime ps"
by (auto simp add: primefact_def list_all_iff)

(* Variant of Lucas theorem.                                                 *)

lemma lucas_primefact:
assumes n: "n ≥ 2" and an: "[a^(n - 1) = 1] (mod n)"
and psn: "foldr op * ps 1 = n - 1"
and psp: "list_all (λp. prime p ∧ ¬ [a^((n - 1) div p) = 1] (mod n)) ps"
shows "prime n"
proof-
{fix p assume p: "prime p" "p dvd n - 1" "[a ^ ((n - 1) div p) = 1] (mod n)"
from psn psp have psn1: "primefact ps (n - 1)"
by (auto simp add: list_all_iff primefact_variant)
from p(3) primefact_contains[OF psn1 p(1,2)] psp
have False by (induct ps, auto)}
with lucas[OF n an] show ?thesis by blast
qed

(* Variant of Pocklington theorem.                                           *)

lemma mod_le: assumes n: "n ≠ (0::nat)" shows "m mod n ≤ m"
proof-
from mod_div_equality[of m n]
have "∃x. x + m mod n = m" by blast
then show ?thesis by auto
qed

lemma pocklington_primefact:
assumes n: "n ≥ 2" and qrn: "q*r = n - 1" and nq2: "n ≤ q⇧2"
and arnb: "(a^r) mod n = b" and psq: "foldr op * ps 1 = q"
and bqn: "(b^q) mod n = 1"
and psp: "list_all (λp. prime p ∧ coprime ((b^(q div p)) mod n - 1) n) ps"
shows "prime n"
proof-
from bqn psp qrn
have bqn: "a ^ (n - 1) mod n = 1"
and psp: "list_all (λp. prime p ∧ coprime (a^(r *(q div p)) mod n - 1) n) ps"  unfolding arnb[symmetric] power_mod
from n  have n0: "n > 0" by arith
from mod_div_equality[of "a^(n - 1)" n]
mod_less_divisor[OF n0, of "a^(n - 1)"]
have an1: "[a ^ (n - 1) = 1] (mod n)"
unfolding nat_mod bqn
apply -
apply (rule exI[where x="0"])
apply (rule exI[where x="a^(n - 1) div n"])
{fix p assume p: "prime p" "p dvd q"
from psp psq have pfpsq: "primefact ps q"
by (auto simp add: primefact_variant list_all_iff)
from psp primefact_contains[OF pfpsq p]
have p': "coprime (a ^ (r * (q div p)) mod n - 1) n"
from prime_ge_2[OF p(1)] have p01: "p ≠ 0" "p ≠ 1" "p =Suc(p - 1)" by arith+
from div_mult1_eq[of r q p] p(2)
have eq1: "r* (q div p) = (n - 1) div p"
unfolding qrn[symmetric] dvd_eq_mod_eq_0 by (simp add: mult.commute)
have ath: "⋀a (b::nat). a <= b ⟹ a ≠ 0 ==> 1 <= a ∧ 1 <= b" by arith
from n0 have n00: "n ≠ 0" by arith
from mod_le[OF n00]
have th10: "a ^ ((n - 1) div p) mod n ≤ a ^ ((n - 1) div p)" .
{assume "a ^ ((n - 1) div p) mod n = 0"
then obtain s where s: "a ^ ((n - 1) div p) = n*s"
unfolding mod_eq_0_iff by blast
hence eq0: "(a^((n - 1) div p))^p = (n*s)^p" by simp
from qrn[symmetric] have qn1: "q dvd n - 1" unfolding dvd_def by auto
from dvd_trans[OF p(2) qn1] div_mod_equality'[of "n - 1" p]
have npp: "(n - 1) div p * p = n - 1" by (simp add: dvd_eq_mod_eq_0)
with eq0 have "a^ (n - 1) = (n*s)^p"
hence "1 = (n*s)^(Suc (p - 1)) mod n" using bqn p01 by simp
also have "… = 0" by (simp add: mult.assoc)
finally have False by simp }
then have th11: "a ^ ((n - 1) div p) mod n ≠ 0" by auto
have th1: "[a ^ ((n - 1) div p) mod n = a ^ ((n - 1) div p)] (mod n)"
unfolding modeq_def by simp
from cong_sub[OF th1 cong_refl[of 1]]  ath[OF th10 th11]
have th: "[a ^ ((n - 1) div p) mod n - 1 = a ^ ((n - 1) div p) - 1] (mod n)"
by blast
from cong_coprime[OF th] p'[unfolded eq1]
have "coprime (a ^ ((n - 1) div p) - 1) n" by (simp add: coprime_commute) }
with pocklington[OF n qrn[symmetric] nq2 an1]
show ?thesis by blast
qed

end
```