(* Title: HOL/Old_Number_Theory/Primes.thy Author: Amine Chaieb, Christophe Tabacznyj and Lawrence C Paulson Copyright 1996 University of Cambridge *) section ‹Primality on nat› theory Primes imports Complex_Main Legacy_GCD begin definition coprime :: "nat => nat => bool" where "coprime m n ⟷ gcd m n = 1" definition prime :: "nat ⇒ bool" where "prime p ⟷ (1 < p ∧ (∀m. m dvd p --> m = 1 ∨ m = p))" lemma two_is_prime: "prime 2" apply (auto simp add: prime_def) apply (case_tac m) apply (auto dest!: dvd_imp_le) done lemma prime_imp_relprime: "prime p ==> ¬ p dvd n ==> gcd p n = 1" apply (auto simp add: prime_def) apply (metis gcd_dvd1 gcd_dvd2) done text ‹ This theorem leads immediately to a proof of the uniqueness of factorization. If @{term p} divides a product of primes then it is one of those primes. › lemma prime_dvd_mult: "prime p ==> p dvd m * n ==> p dvd m ∨ p dvd n" by (blast intro: relprime_dvd_mult prime_imp_relprime) lemma prime_dvd_square: "prime p ==> p dvd m^Suc (Suc 0) ==> p dvd m" by (auto dest: prime_dvd_mult) lemma prime_dvd_power_two: "prime p ==> p dvd m⇧^{2}==> p dvd m" by (rule prime_dvd_square) (simp_all add: power2_eq_square) lemma exp_eq_1:"(x::nat)^n = 1 ⟷ x = 1 ∨ n = 0" by (induct n, auto) lemma exp_mono_lt: "(x::nat) ^ (Suc n) < y ^ (Suc n) ⟷ x < y" by(metis linorder_not_less not_less0 power_le_imp_le_base power_less_imp_less_base) lemma exp_mono_le: "(x::nat) ^ (Suc n) ≤ y ^ (Suc n) ⟷ x ≤ y" by (simp only: linorder_not_less[symmetric] exp_mono_lt) lemma exp_mono_eq: "(x::nat) ^ Suc n = y ^ Suc n ⟷ x = y" using power_inject_base[of x n y] by auto lemma even_square: assumes e: "even (n::nat)" shows "∃x. n⇧^{2}= 4*x" proof- from e have "2 dvd n" by presburger then obtain k where k: "n = 2*k" using dvd_def by auto hence "n⇧^{2}= 4 * k⇧^{2}" by (simp add: power2_eq_square) thus ?thesis by blast qed lemma odd_square: assumes e: "odd (n::nat)" shows "∃x. n⇧^{2}= 4*x + 1" proof- from e have np: "n > 0" by presburger from e have "2 dvd (n - 1)" by presburger then obtain k where "n - 1 = 2 * k" .. hence k: "n = 2*k + 1" using e by presburger hence "n⇧^{2}= 4* (k⇧^{2}+ k) + 1" by algebra thus ?thesis by blast qed lemma diff_square: "(x::nat)⇧^{2}- y⇧^{2}= (x+y)*(x - y)" proof- have "x ≤ y ∨ y ≤ x" by (rule nat_le_linear) moreover {assume le: "x ≤ y" hence "x⇧^{2}≤ y⇧^{2}" by (simp only: numeral_2_eq_2 exp_mono_le Let_def) with le have ?thesis by simp } moreover {assume le: "y ≤ x" hence le2: "y⇧^{2}≤ x⇧^{2}" by (simp only: numeral_2_eq_2 exp_mono_le Let_def) from le have "∃z. y + z = x" by presburger then obtain z where z: "x = y + z" by blast from le2 have "∃z. x⇧^{2}= y⇧^{2}+ z" by presburger then obtain z2 where z2: "x⇧^{2}= y⇧^{2}+ z2" by blast from z z2 have ?thesis by simp algebra } ultimately show ?thesis by blast qed text ‹Elementary theory of divisibility› lemma divides_ge: "(a::nat) dvd b ⟹ b = 0 ∨ a ≤ b" unfolding dvd_def by auto lemma divides_antisym: "(x::nat) dvd y ∧ y dvd x ⟷ x = y" using dvd_antisym[of x y] by auto lemma divides_add_revr: assumes da: "(d::nat) dvd a" and dab:"d dvd (a + b)" shows "d dvd b" proof- from da obtain k where k:"a = d*k" by (auto simp add: dvd_def) from dab obtain k' where k': "a + b = d*k'" by (auto simp add: dvd_def) from k k' have "b = d *(k' - k)" by (simp add : diff_mult_distrib2) thus ?thesis unfolding dvd_def by blast qed declare nat_mult_dvd_cancel_disj[presburger] lemma nat_mult_dvd_cancel_disj'[presburger]: "(m::nat)*k dvd n*k ⟷ k = 0 ∨ m dvd n" unfolding mult.commute[of m k] mult.commute[of n k] by presburger lemma divides_mul_l: "(a::nat) dvd b ==> (c * a) dvd (c * b)" by presburger lemma divides_mul_r: "(a::nat) dvd b ==> (a * c) dvd (b * c)" by presburger lemma divides_cases: "(n::nat) dvd m ==> m = 0 ∨ m = n ∨ 2 * n <= m" by (auto simp add: dvd_def) lemma divides_div_not: "(x::nat) = (q * n) + r ⟹ 0 < r ⟹ r < n ==> ~(n dvd x)" proof(auto simp add: dvd_def) fix k assume H: "0 < r" "r < n" "q * n + r = n * k" from H(3) have r: "r = n* (k -q)" by(simp add: diff_mult_distrib2 mult.commute) {assume "k - q = 0" with r H(1) have False by simp} moreover {assume "k - q ≠ 0" with r have "r ≥ n" by auto with H(2) have False by simp} ultimately show False by blast qed lemma divides_exp: "(x::nat) dvd y ==> x ^ n dvd y ^ n" by (auto simp add: power_mult_distrib dvd_def) lemma divides_exp2: "n ≠ 0 ⟹ (x::nat) ^ n dvd y ⟹ x dvd y" by (induct n ,auto simp add: dvd_def) fun fact :: "nat ⇒ nat" where "fact 0 = 1" | "fact (Suc n) = Suc n * fact n" lemma fact_lt: "0 < fact n" by(induct n, simp_all) lemma fact_le: "fact n ≥ 1" using fact_lt[of n] by simp lemma fact_mono: assumes le: "m ≤ n" shows "fact m ≤ fact n" proof- from le have "∃i. n = m+i" by presburger then obtain i where i: "n = m+i" by blast have "fact m ≤ fact (m + i)" proof(induct m) case 0 thus ?case using fact_le[of i] by simp next case (Suc m) have "fact (Suc m) = Suc m * fact m" by simp have th1: "Suc m ≤ Suc (m + i)" by simp from mult_le_mono[of "Suc m" "Suc (m+i)" "fact m" "fact (m+i)", OF th1 Suc.hyps] show ?case by simp qed thus ?thesis using i by simp qed lemma divides_fact: "1 <= p ⟹ p <= n ==> p dvd fact n" proof(induct n arbitrary: p) case 0 thus ?case by simp next case (Suc n p) from Suc.prems have "p = Suc n ∨ p ≤ n" by presburger moreover {assume "p = Suc n" hence ?case by (simp only: fact.simps dvd_triv_left)} moreover {assume "p ≤ n" with Suc.prems(1) Suc.hyps have th: "p dvd fact n" by simp from dvd_mult[OF th] have ?case by (simp only: fact.simps) } ultimately show ?case by blast qed declare dvd_triv_left[presburger] declare dvd_triv_right[presburger] lemma divides_rexp: "x dvd y ⟹ (x::nat) dvd (y^(Suc n))" by (simp add: dvd_mult2[of x y]) text ‹Coprimality› lemma coprime: "coprime a b ⟷ (∀d. d dvd a ∧ d dvd b ⟷ d = 1)" using gcd_unique[of 1 a b, simplified] by (auto simp add: coprime_def) lemma coprime_commute: "coprime a b ⟷ coprime b a" by (simp add: coprime_def gcd_commute) lemma coprime_bezout: "coprime a b ⟷ (∃x y. a * x - b * y = 1 ∨ b * x - a * y = 1)" using coprime_def gcd_bezout by auto lemma coprime_divprod: "d dvd a * b ⟹ coprime d a ⟹ d dvd b" using relprime_dvd_mult_iff[of d a b] by (auto simp add: coprime_def mult.commute) lemma coprime_1[simp]: "coprime a 1" by (simp add: coprime_def) lemma coprime_1'[simp]: "coprime 1 a" by (simp add: coprime_def) lemma coprime_Suc0[simp]: "coprime a (Suc 0)" by (simp add: coprime_def) lemma coprime_Suc0'[simp]: "coprime (Suc 0) a" by (simp add: coprime_def) lemma gcd_coprime: assumes z: "gcd a b ≠ 0" and a: "a = a' * gcd a b" and b: "b = b' * gcd a b" shows "coprime a' b'" proof- let ?g = "gcd a b" {assume bz: "a = 0" from b bz z a have ?thesis by (simp add: gcd_zero coprime_def)} moreover {assume az: "a≠ 0" from z have z': "?g > 0" by simp from bezout_gcd_strong[OF az, of b] obtain x y where xy: "a*x = b*y + ?g" by blast from xy a b have "?g * a'*x = ?g * (b'*y + 1)" by (simp add: algebra_simps) hence "?g * (a'*x) = ?g * (b'*y + 1)" by (simp add: mult.assoc) hence "a'*x = (b'*y + 1)" by (simp only: nat_mult_eq_cancel1[OF z']) hence "a'*x - b'*y = 1" by simp with coprime_bezout[of a' b'] have ?thesis by auto} ultimately show ?thesis by blast qed lemma coprime_0: "coprime d 0 ⟷ d = 1" by (simp add: coprime_def) lemma coprime_mul: assumes da: "coprime d a" and db: "coprime d b" shows "coprime d (a * b)" proof- from da have th: "gcd a d = 1" by (simp add: coprime_def gcd_commute) from gcd_mult_cancel[of a d b, OF th] db[unfolded coprime_def] have "gcd d (a*b) = 1" by (simp add: gcd_commute) thus ?thesis unfolding coprime_def . qed lemma coprime_lmul2: assumes dab: "coprime d (a * b)" shows "coprime d b" using dab unfolding coprime_bezout apply clarsimp apply (case_tac "d * x - a * b * y = Suc 0 ", simp_all) apply (rule_tac x="x" in exI) apply (rule_tac x="a*y" in exI) apply (simp add: ac_simps) apply (rule_tac x="a*x" in exI) apply (rule_tac x="y" in exI) apply (simp add: ac_simps) done lemma coprime_rmul2: "coprime d (a * b) ⟹ coprime d a" unfolding coprime_bezout apply clarsimp apply (case_tac "d * x - a * b * y = Suc 0 ", simp_all) apply (rule_tac x="x" in exI) apply (rule_tac x="b*y" in exI) apply (simp add: ac_simps) apply (rule_tac x="b*x" in exI) apply (rule_tac x="y" in exI) apply (simp add: ac_simps) done lemma coprime_mul_eq: "coprime d (a * b) ⟷ coprime d a ∧ coprime d b" using coprime_rmul2[of d a b] coprime_lmul2[of d a b] coprime_mul[of d a b] by blast lemma gcd_coprime_exists: assumes nz: "gcd a b ≠ 0" shows "∃a' b'. a = a' * gcd a b ∧ b = b' * gcd a b ∧ coprime a' b'" proof- let ?g = "gcd a b" from gcd_dvd1[of a b] gcd_dvd2[of a b] obtain a' b' where "a = ?g*a'" "b = ?g*b'" unfolding dvd_def by blast hence ab': "a = a'*?g" "b = b'*?g" by algebra+ from ab' gcd_coprime[OF nz ab'] show ?thesis by blast qed lemma coprime_exp: "coprime d a ==> coprime d (a^n)" by(induct n, simp_all add: coprime_mul) lemma coprime_exp_imp: "coprime a b ==> coprime (a ^n) (b ^n)" by (induct n, simp_all add: coprime_mul_eq coprime_commute coprime_exp) lemma coprime_refl[simp]: "coprime n n ⟷ n = 1" by (simp add: coprime_def) lemma coprime_plus1[simp]: "coprime (n + 1) n" apply (simp add: coprime_bezout) apply (rule exI[where x=1]) apply (rule exI[where x=1]) apply simp done lemma coprime_minus1: "n ≠ 0 ==> coprime (n - 1) n" using coprime_plus1[of "n - 1"] coprime_commute[of "n - 1" n] by auto lemma bezout_gcd_pow: "∃x y. a ^n * x - b ^ n * y = gcd a b ^ n ∨ b ^ n * x - a ^ n * y = gcd a b ^ n" proof- let ?g = "gcd a b" {assume z: "?g = 0" hence ?thesis apply (cases n, simp) apply arith apply (simp only: z power_0_Suc) apply (rule exI[where x=0]) apply (rule exI[where x=0]) apply simp done } moreover {assume z: "?g ≠ 0" from gcd_dvd1[of a b] gcd_dvd2[of a b] obtain a' b' where ab': "a = a'*?g" "b = b'*?g" unfolding dvd_def by (auto simp add: ac_simps) hence ab'': "?g*a' = a" "?g * b' = b" by algebra+ from coprime_exp_imp[OF gcd_coprime[OF z ab'], unfolded coprime_bezout, of n] obtain x y where "a'^n * x - b'^n * y = 1 ∨ b'^n * x - a'^n * y = 1" by blast hence "?g^n * (a'^n * x - b'^n * y) = ?g^n ∨ ?g^n*(b'^n * x - a'^n * y) = ?g^n" using z by auto then have "a^n * x - b^n * y = ?g^n ∨ b^n * x - a^n * y = ?g^n" using z ab'' by (simp only: power_mult_distrib[symmetric] diff_mult_distrib2 mult.assoc[symmetric]) hence ?thesis by blast } ultimately show ?thesis by blast qed lemma gcd_exp: "gcd (a^n) (b^n) = gcd a b^n" proof- let ?g = "gcd (a^n) (b^n)" let ?gn = "gcd a b^n" {fix e assume H: "e dvd a^n" "e dvd b^n" from bezout_gcd_pow[of a n b] obtain x y where xy: "a ^ n * x - b ^ n * y = ?gn ∨ b ^ n * x - a ^ n * y = ?gn" by blast from dvd_diff_nat [OF dvd_mult2[OF H(1), of x] dvd_mult2[OF H(2), of y]] dvd_diff_nat [OF dvd_mult2[OF H(2), of x] dvd_mult2[OF H(1), of y]] xy have "e dvd ?gn" by (cases "a ^ n * x - b ^ n * y = gcd a b ^ n", simp_all)} hence th: "∀e. e dvd a^n ∧ e dvd b^n ⟶ e dvd ?gn" by blast from divides_exp[OF gcd_dvd1[of a b], of n] divides_exp[OF gcd_dvd2[of a b], of n] th gcd_unique have "?gn = ?g" by blast thus ?thesis by simp qed lemma coprime_exp2: "coprime (a ^ Suc n) (b^ Suc n) ⟷ coprime a b" by (simp only: coprime_def gcd_exp exp_eq_1) simp lemma division_decomp: assumes dc: "(a::nat) dvd b * c" shows "∃b' c'. a = b' * c' ∧ b' dvd b ∧ c' dvd c" proof- let ?g = "gcd a b" {assume "?g = 0" with dc have ?thesis apply (simp add: gcd_zero) apply (rule exI[where x="0"]) by (rule exI[where x="c"], simp)} moreover {assume z: "?g ≠ 0" from gcd_coprime_exists[OF z] obtain a' b' where ab': "a = a' * ?g" "b = b' * ?g" "coprime a' b'" by blast from gcd_dvd2[of a b] have thb: "?g dvd b" . from ab'(1) have "a' dvd a" unfolding dvd_def by blast with dc have th0: "a' dvd b*c" using dvd_trans[of a' a "b*c"] by simp from dc ab'(1,2) have "a'*?g dvd (b'*?g) *c" by auto hence "?g*a' dvd ?g * (b' * c)" by (simp add: mult.assoc) with z have th_1: "a' dvd b'*c" by simp from coprime_divprod[OF th_1 ab'(3)] have thc: "a' dvd c" . from ab' have "a = ?g*a'" by algebra with thb thc have ?thesis by blast } ultimately show ?thesis by blast qed lemma nat_power_eq_0_iff: "(m::nat) ^ n = 0 ⟷ n ≠ 0 ∧ m = 0" by (induct n, auto) lemma divides_rev: assumes ab: "(a::nat) ^ n dvd b ^n" and n:"n ≠ 0" shows "a dvd b" proof- let ?g = "gcd a b" from n obtain m where m: "n = Suc m" by (cases n, simp_all) {assume "?g = 0" with ab n have ?thesis by (simp add: gcd_zero)} moreover {assume z: "?g ≠ 0" hence zn: "?g ^ n ≠ 0" using n by simp from gcd_coprime_exists[OF z] obtain a' b' where ab': "a = a' * ?g" "b = b' * ?g" "coprime a' b'" by blast from ab have "(a' * ?g) ^ n dvd (b' * ?g)^n" by (simp add: ab'(1,2)[symmetric]) hence "?g^n*a'^n dvd ?g^n *b'^n" by (simp only: power_mult_distrib mult.commute) with zn z n have th0:"a'^n dvd b'^n" by (auto simp add: nat_power_eq_0_iff) have "a' dvd a'^n" by (simp add: m) with th0 have "a' dvd b'^n" using dvd_trans[of a' "a'^n" "b'^n"] by simp hence th1: "a' dvd b'^m * b'" by (simp add: m mult.commute) from coprime_divprod[OF th1 coprime_exp[OF ab'(3), of m]] have "a' dvd b'" . hence "a'*?g dvd b'*?g" by simp with ab'(1,2) have ?thesis by simp } ultimately show ?thesis by blast qed lemma divides_mul: assumes mr: "m dvd r" and nr: "n dvd r" and mn:"coprime m n" shows "m * n dvd r" proof- from mr nr obtain m' n' where m': "r = m*m'" and n': "r = n*n'" unfolding dvd_def by blast from mr n' have "m dvd n'*n" by (simp add: mult.commute) hence "m dvd n'" using relprime_dvd_mult_iff[OF mn[unfolded coprime_def]] by simp then obtain k where k: "n' = m*k" unfolding dvd_def by blast from n' k show ?thesis unfolding dvd_def by auto qed text ‹A binary form of the Chinese Remainder Theorem.› lemma chinese_remainder: assumes ab: "coprime a b" and a:"a ≠ 0" and b:"b ≠ 0" shows "∃x q1 q2. x = u + q1 * a ∧ x = v + q2 * b" proof- from bezout_add_strong[OF a, of b] bezout_add_strong[OF b, of a] obtain d1 x1 y1 d2 x2 y2 where dxy1: "d1 dvd a" "d1 dvd b" "a * x1 = b * y1 + d1" and dxy2: "d2 dvd b" "d2 dvd a" "b * x2 = a * y2 + d2" by blast from gcd_unique[of 1 a b, simplified ab[unfolded coprime_def], simplified] dxy1(1,2) dxy2(1,2) have d12: "d1 = 1" "d2 =1" by auto let ?x = "v * a * x1 + u * b * x2" let ?q1 = "v * x1 + u * y2" let ?q2 = "v * y1 + u * x2" from dxy2(3)[simplified d12] dxy1(3)[simplified d12] have "?x = u + ?q1 * a" "?x = v + ?q2 * b" by algebra+ thus ?thesis by blast qed text ‹Primality› text ‹A few useful theorems about primes› lemma prime_0[simp]: "~prime 0" by (simp add: prime_def) lemma prime_1[simp]: "~ prime 1" by (simp add: prime_def) lemma prime_Suc0[simp]: "~ prime (Suc 0)" by (simp add: prime_def) lemma prime_ge_2: "prime p ==> p ≥ 2" by (simp add: prime_def) lemma prime_factor: assumes n: "n ≠ 1" shows "∃ p. prime p ∧ p dvd n" using n proof(induct n rule: nat_less_induct) fix n assume H: "∀m<n. m ≠ 1 ⟶ (∃p. prime p ∧ p dvd m)" "n ≠ 1" let ?ths = "∃p. prime p ∧ p dvd n" {assume "n=0" hence ?ths using two_is_prime by auto} moreover {assume nz: "n≠0" {assume "prime n" hence ?ths by - (rule exI[where x="n"], simp)} moreover {assume n: "¬ prime n" with nz H(2) obtain k where k:"k dvd n" "k ≠ 1" "k ≠ n" by (auto simp add: prime_def) from dvd_imp_le[OF k(1)] nz k(3) have kn: "k < n" by simp from H(1)[rule_format, OF kn k(2)] obtain p where p: "prime p" "p dvd k" by blast from dvd_trans[OF p(2) k(1)] p(1) have ?ths by blast} ultimately have ?ths by blast} ultimately show ?ths by blast qed lemma prime_factor_lt: assumes p: "prime p" and n: "n ≠ 0" and npm:"n = p * m" shows "m < n" proof- {assume "m=0" with n have ?thesis by simp} moreover {assume m: "m ≠ 0" from npm have mn: "m dvd n" unfolding dvd_def by auto from npm m have "n ≠ m" using p by auto with dvd_imp_le[OF mn] n have ?thesis by simp} ultimately show ?thesis by blast qed lemma euclid_bound: "∃p. prime p ∧ n < p ∧ p <= Suc (fact n)" proof- have f1: "fact n + 1 ≠ 1" using fact_le[of n] by arith from prime_factor[OF f1] obtain p where p: "prime p" "p dvd fact n + 1" by blast from dvd_imp_le[OF p(2)] have pfn: "p ≤ fact n + 1" by simp {assume np: "p ≤ n" from p(1) have p1: "p ≥ 1" by (cases p, simp_all) from divides_fact[OF p1 np] have pfn': "p dvd fact n" . from divides_add_revr[OF pfn' p(2)] p(1) have False by simp} hence "n < p" by arith with p(1) pfn show ?thesis by auto qed lemma euclid: "∃p. prime p ∧ p > n" using euclid_bound by auto lemma primes_infinite: "¬ (finite {p. prime p})" apply(simp add: finite_nat_set_iff_bounded_le) apply (metis euclid linorder_not_le) done lemma coprime_prime: assumes ab: "coprime a b" shows "~(prime p ∧ p dvd a ∧ p dvd b)" proof assume "prime p ∧ p dvd a ∧ p dvd b" thus False using ab gcd_greatest[of p a b] by (simp add: coprime_def) qed lemma coprime_prime_eq: "coprime a b ⟷ (∀p. ~(prime p ∧ p dvd a ∧ p dvd b))" (is "?lhs = ?rhs") proof- {assume "?lhs" with coprime_prime have ?rhs by blast} moreover {assume r: "?rhs" and c: "¬ ?lhs" then obtain g where g: "g≠1" "g dvd a" "g dvd b" unfolding coprime_def by blast from prime_factor[OF g(1)] obtain p where p: "prime p" "p dvd g" by blast from dvd_trans [OF p(2) g(2)] dvd_trans [OF p(2) g(3)] have "p dvd a" "p dvd b" . with p(1) r have False by blast} ultimately show ?thesis by blast qed lemma prime_coprime: assumes p: "prime p" shows "n = 1 ∨ p dvd n ∨ coprime p n" using p prime_imp_relprime[of p n] by (auto simp add: coprime_def) lemma prime_coprime_strong: "prime p ⟹ p dvd n ∨ coprime p n" using prime_coprime[of p n] by auto declare coprime_0[simp] lemma coprime_0'[simp]: "coprime 0 d ⟷ d = 1" by (simp add: coprime_commute[of 0 d]) lemma coprime_bezout_strong: assumes ab: "coprime a b" and b: "b ≠ 1" shows "∃x y. a * x = b * y + 1" proof- from ab b have az: "a ≠ 0" by - (rule ccontr, auto) from bezout_gcd_strong[OF az, of b] ab[unfolded coprime_def] show ?thesis by auto qed lemma bezout_prime: assumes p: "prime p" and pa: "¬ p dvd a" shows "∃x y. a*x = p*y + 1" proof- from p have p1: "p ≠ 1" using prime_1 by blast from prime_coprime[OF p, of a] p1 pa have ap: "coprime a p" by (auto simp add: coprime_commute) from coprime_bezout_strong[OF ap p1] show ?thesis . qed lemma prime_divprod: assumes p: "prime p" and pab: "p dvd a*b" shows "p dvd a ∨ p dvd b" proof- {assume "a=1" hence ?thesis using pab by simp } moreover {assume "p dvd a" hence ?thesis by blast} moreover {assume pa: "coprime p a" from coprime_divprod[OF pab pa] have ?thesis .. } ultimately show ?thesis using prime_coprime[OF p, of a] by blast qed lemma prime_divprod_eq: assumes p: "prime p" shows "p dvd a*b ⟷ p dvd a ∨ p dvd b" using p prime_divprod dvd_mult dvd_mult2 by auto lemma prime_divexp: assumes p:"prime p" and px: "p dvd x^n" shows "p dvd x" using px proof(induct n) case 0 thus ?case by simp next case (Suc n) hence th: "p dvd x*x^n" by simp {assume H: "p dvd x^n" from Suc.hyps[OF H] have ?case .} with prime_divprod[OF p th] show ?case by blast qed lemma prime_divexp_n: "prime p ⟹ p dvd x^n ⟹ p^n dvd x^n" using prime_divexp[of p x n] divides_exp[of p x n] by blast lemma coprime_prime_dvd_ex: assumes xy: "¬coprime x y" shows "∃p. prime p ∧ p dvd x ∧ p dvd y" proof- from xy[unfolded coprime_def] obtain g where g: "g ≠ 1" "g dvd x" "g dvd y" by blast from prime_factor[OF g(1)] obtain p where p: "prime p" "p dvd g" by blast from g(2,3) dvd_trans[OF p(2)] p(1) show ?thesis by auto qed lemma coprime_sos: assumes xy: "coprime x y" shows "coprime (x * y) (x⇧^{2}+ y⇧^{2})" proof- {assume c: "¬ coprime (x * y) (x⇧^{2}+ y⇧^{2})" from coprime_prime_dvd_ex[OF c] obtain p where p: "prime p" "p dvd x*y" "p dvd x⇧^{2}+ y⇧^{2}" by blast {assume px: "p dvd x" from dvd_mult[OF px, of x] p(3) obtain r s where "x * x = p * r" and "x⇧^{2}+ y⇧^{2}= p * s" by (auto elim!: dvdE) then have "y⇧^{2}= p * (s - r)" by (auto simp add: power2_eq_square diff_mult_distrib2) then have "p dvd y⇧^{2}" .. with prime_divexp[OF p(1), of y 2] have py: "p dvd y" . from p(1) px py xy[unfolded coprime, rule_format, of p] prime_1 have False by simp } moreover {assume py: "p dvd y" from dvd_mult[OF py, of y] p(3) obtain r s where "y * y = p * r" and "x⇧^{2}+ y⇧^{2}= p * s" by (auto elim!: dvdE) then have "x⇧^{2}= p * (s - r)" by (auto simp add: power2_eq_square diff_mult_distrib2) then have "p dvd x⇧^{2}" .. with prime_divexp[OF p(1), of x 2] have px: "p dvd x" . from p(1) px py xy[unfolded coprime, rule_format, of p] prime_1 have False by simp } ultimately have False using prime_divprod[OF p(1,2)] by blast} thus ?thesis by blast qed lemma distinct_prime_coprime: "prime p ⟹ prime q ⟹ p ≠ q ⟹ coprime p q" unfolding prime_def coprime_prime_eq by blast lemma prime_coprime_lt: assumes p: "prime p" and x: "0 < x" and xp: "x < p" shows "coprime x p" proof- {assume c: "¬ coprime x p" then obtain g where g: "g ≠ 1" "g dvd x" "g dvd p" unfolding coprime_def by blast from dvd_imp_le[OF g(2)] x xp have gp: "g < p" by arith from g(2) x have "g ≠ 0" by - (rule ccontr, simp) with g gp p[unfolded prime_def] have False by blast} thus ?thesis by blast qed lemma prime_odd: "prime p ⟹ p = 2 ∨ odd p" unfolding prime_def by auto text ‹One property of coprimality is easier to prove via prime factors.› lemma prime_divprod_pow: assumes p: "prime p" and ab: "coprime a b" and pab: "p^n dvd a * b" shows "p^n dvd a ∨ p^n dvd b" proof- {assume "n = 0 ∨ a = 1 ∨ b = 1" with pab have ?thesis apply (cases "n=0", simp_all) apply (cases "a=1", simp_all) done} moreover {assume n: "n ≠ 0" and a: "a≠1" and b: "b≠1" then obtain m where m: "n = Suc m" by (cases n, auto) from divides_exp2[OF n pab] have pab': "p dvd a*b" . from prime_divprod[OF p pab'] have "p dvd a ∨ p dvd b" . moreover {assume pa: "p dvd a" have pnba: "p^n dvd b*a" using pab by (simp add: mult.commute) from coprime_prime[OF ab, of p] p pa have "¬ p dvd b" by blast with prime_coprime[OF p, of b] b have cpb: "coprime b p" using coprime_commute by blast from coprime_exp[OF cpb] have pnb: "coprime (p^n) b" by (simp add: coprime_commute) from coprime_divprod[OF pnba pnb] have ?thesis by blast } moreover {assume pb: "p dvd b" have pnba: "p^n dvd b*a" using pab by (simp add: mult.commute) from coprime_prime[OF ab, of p] p pb have "¬ p dvd a" by blast with prime_coprime[OF p, of a] a have cpb: "coprime a p" using coprime_commute by blast from coprime_exp[OF cpb] have pnb: "coprime (p^n) a" by (simp add: coprime_commute) from coprime_divprod[OF pab pnb] have ?thesis by blast } ultimately have ?thesis by blast} ultimately show ?thesis by blast qed lemma nat_mult_eq_one: "(n::nat) * m = 1 ⟷ n = 1 ∧ m = 1" (is "?lhs ⟷ ?rhs") proof assume H: "?lhs" hence "n dvd 1" "m dvd 1" unfolding dvd_def by (auto simp add: mult.commute) thus ?rhs by auto next assume ?rhs then show ?lhs by auto qed lemma power_Suc0: "Suc 0 ^ n = Suc 0" unfolding One_nat_def[symmetric] power_one .. lemma coprime_pow: assumes ab: "coprime a b" and abcn: "a * b = c ^n" shows "∃r s. a = r^n ∧ b = s ^n" using ab abcn proof(induct c arbitrary: a b rule: nat_less_induct) fix c a b assume H: "∀m<c. ∀a b. coprime a b ⟶ a * b = m ^ n ⟶ (∃r s. a = r ^ n ∧ b = s ^ n)" "coprime a b" "a * b = c ^ n" let ?ths = "∃r s. a = r^n ∧ b = s ^n" {assume n: "n = 0" with H(3) power_one have "a*b = 1" by simp hence "a = 1 ∧ b = 1" by simp hence ?ths apply - apply (rule exI[where x=1]) apply (rule exI[where x=1]) using power_one[of n] by simp} moreover {assume n: "n ≠ 0" then obtain m where m: "n = Suc m" by (cases n, auto) {assume c: "c = 0" with H(3) m H(2) have ?ths apply simp apply (cases "a=0", simp_all) apply (rule exI[where x="0"], simp) apply (rule exI[where x="0"], simp) done} moreover {assume "c=1" with H(3) power_one have "a*b = 1" by simp hence "a = 1 ∧ b = 1" by simp hence ?ths apply - apply (rule exI[where x=1]) apply (rule exI[where x=1]) using power_one[of n] by simp} moreover {assume c: "c≠1" "c ≠ 0" from prime_factor[OF c(1)] obtain p where p: "prime p" "p dvd c" by blast from prime_divprod_pow[OF p(1) H(2), unfolded H(3), OF divides_exp[OF p(2), of n]] have pnab: "p ^ n dvd a ∨ p^n dvd b" . from p(2) obtain l where l: "c = p*l" unfolding dvd_def by blast have pn0: "p^n ≠ 0" using n prime_ge_2 [OF p(1)] by simp {assume pa: "p^n dvd a" then obtain k where k: "a = p^n * k" unfolding dvd_def by blast from l have "l dvd c" by auto with dvd_imp_le[of l c] c have "l ≤ c" by auto moreover {assume "l = c" with l c have "p = 1" by simp with p have False by simp} ultimately have lc: "l < c" by arith from coprime_lmul2 [OF H(2)[unfolded k coprime_commute[of "p^n*k" b]]] have kb: "coprime k b" by (simp add: coprime_commute) from H(3) l k pn0 have kbln: "k * b = l ^ n" by (auto simp add: power_mult_distrib) from H(1)[rule_format, OF lc kb kbln] obtain r s where rs: "k = r ^n" "b = s^n" by blast from k rs(1) have "a = (p*r)^n" by (simp add: power_mult_distrib) with rs(2) have ?ths by blast } moreover {assume pb: "p^n dvd b" then obtain k where k: "b = p^n * k" unfolding dvd_def by blast from l have "l dvd c" by auto with dvd_imp_le[of l c] c have "l ≤ c" by auto moreover {assume "l = c" with l c have "p = 1" by simp with p have False by simp} ultimately have lc: "l < c" by arith from coprime_lmul2 [OF H(2)[unfolded k coprime_commute[of "p^n*k" a]]] have kb: "coprime k a" by (simp add: coprime_commute) from H(3) l k pn0 n have kbln: "k * a = l ^ n" by (simp add: power_mult_distrib mult.commute) from H(1)[rule_format, OF lc kb kbln] obtain r s where rs: "k = r ^n" "a = s^n" by blast from k rs(1) have "b = (p*r)^n" by (simp add: power_mult_distrib) with rs(2) have ?ths by blast } ultimately have ?ths using pnab by blast} ultimately have ?ths by blast} ultimately show ?ths by blast qed text ‹More useful lemmas.› lemma prime_product: assumes "prime (p * q)" shows "p = 1 ∨ q = 1" proof - from assms have "1 < p * q" and P: "⋀m. m dvd p * q ⟹ m = 1 ∨ m = p * q" unfolding prime_def by auto from ‹1 < p * q› have "p ≠ 0" by (cases p) auto then have Q: "p = p * q ⟷ q = 1" by auto have "p dvd p * q" by simp then have "p = 1 ∨ p = p * q" by (rule P) then show ?thesis by (simp add: Q) qed lemma prime_exp: "prime (p^n) ⟷ prime p ∧ n = 1" proof(induct n) case 0 thus ?case by simp next case (Suc n) {assume "p = 0" hence ?case by simp} moreover {assume "p=1" hence ?case by simp} moreover {assume p: "p ≠ 0" "p≠1" {assume pp: "prime (p^Suc n)" hence "p = 1 ∨ p^n = 1" using prime_product[of p "p^n"] by simp with p have n: "n = 0" by (simp only: exp_eq_1 ) simp with pp have "prime p ∧ Suc n = 1" by simp} moreover {assume n: "prime p ∧ Suc n = 1" hence "prime (p^Suc n)" by simp} ultimately have ?case by blast} ultimately show ?case by blast qed lemma prime_power_mult: assumes p: "prime p" and xy: "x * y = p ^ k" shows "∃i j. x = p ^i ∧ y = p^ j" using xy proof(induct k arbitrary: x y) case 0 thus ?case apply simp by (rule exI[where x="0"], simp) next case (Suc k x y) from Suc.prems have pxy: "p dvd x*y" by auto from prime_divprod[OF p pxy] have pxyc: "p dvd x ∨ p dvd y" . from p have p0: "p ≠ 0" by - (rule ccontr, simp) {assume px: "p dvd x" then obtain d where d: "x = p*d" unfolding dvd_def by blast from Suc.prems d have "p*d*y = p^Suc k" by simp hence th: "d*y = p^k" using p0 by simp from Suc.hyps[OF th] obtain i j where ij: "d = p^i" "y = p^j" by blast with d have "x = p^Suc i" by simp with ij(2) have ?case by blast} moreover {assume px: "p dvd y" then obtain d where d: "y = p*d" unfolding dvd_def by blast from Suc.prems d have "p*d*x = p^Suc k" by (simp add: mult.commute) hence th: "d*x = p^k" using p0 by simp from Suc.hyps[OF th] obtain i j where ij: "d = p^i" "x = p^j" by blast with d have "y = p^Suc i" by simp with ij(2) have ?case by blast} ultimately show ?case using pxyc by blast qed lemma prime_power_exp: assumes p: "prime p" and n:"n ≠ 0" and xn: "x^n = p^k" shows "∃i. x = p^i" using n xn proof(induct n arbitrary: k) case 0 thus ?case by simp next case (Suc n k) hence th: "x*x^n = p^k" by simp {assume "n = 0" with Suc have ?case by simp (rule exI[where x="k"], simp)} moreover {assume n: "n ≠ 0" from prime_power_mult[OF p th] obtain i j where ij: "x = p^i" "x^n = p^j"by blast from Suc.hyps[OF n ij(2)] have ?case .} ultimately show ?case by blast qed lemma divides_primepow: assumes p: "prime p" shows "d dvd p^k ⟷ (∃ i. i ≤ k ∧ d = p ^i)" proof assume H: "d dvd p^k" then obtain e where e: "d*e = p^k" unfolding dvd_def apply (auto simp add: mult.commute) by blast from prime_power_mult[OF p e] obtain i j where ij: "d = p^i" "e=p^j" by blast from prime_ge_2[OF p] have p1: "p > 1" by arith from e ij have "p^(i + j) = p^k" by (simp add: power_add) hence "i + j = k" using power_inject_exp[of p "i+j" k, OF p1] by simp hence "i ≤ k" by arith with ij(1) show "∃i≤k. d = p ^ i" by blast next {fix i assume H: "i ≤ k" "d = p^i" hence "∃j. k = i + j" by arith then obtain j where j: "k = i + j" by blast hence "p^k = p^j*d" using H(2) by (simp add: power_add) hence "d dvd p^k" unfolding dvd_def by auto} thus "∃i≤k. d = p ^ i ⟹ d dvd p ^ k" by blast qed lemma coprime_divisors: "d dvd a ⟹ e dvd b ⟹ coprime a b ⟹ coprime d e" by (auto simp add: dvd_def coprime) lemma mult_inj_if_coprime_nat: "inj_on f A ⟹ inj_on g B ⟹ ALL a:A. ALL b:B. coprime (f a) (g b) ⟹ inj_on (%(a,b). f a * g b::nat) (A × B)" apply(auto simp add:inj_on_def) apply(metis coprime_def dvd_triv_left gcd_proj2_if_dvd_nat gcd_semilattice_nat.inf_commute relprime_dvd_mult) apply(metis coprime_commute coprime_divprod dvd.neq_le_trans dvd_triv_right) done declare power_Suc0[simp del] end