src/HOL/Library/GCD.thy

removed constant int :: nat => int;
int is now an abbreviation for of_nat :: nat => int

(*  Title:      HOL/GCD.thy
    ID:         $Id$
    Author:     Christophe Tabacznyj and Lawrence C Paulson
    Copyright   1996  University of Cambridge
*)

header {* The Greatest Common Divisor *}

theory GCD
imports Main
begin

text {*
  See \cite{davenport92}.
  \bigskip
*}

consts
  gcd  :: "nat \<times> nat => nat"  -- {* Euclid's algorithm *}

recdef gcd  "measure ((\<lambda>(m, n). n) :: nat \<times> nat => nat)"
  "gcd (m, n) = (if n = 0 then m else gcd (n, m mod n))"

definition
  is_gcd :: "nat => nat => nat => bool" where -- {* @{term gcd} as a relation *}
  "is_gcd p m n = (p dvd m \<and> p dvd n \<and>
    (\<forall>d. d dvd m \<and> d dvd n --> d dvd p))"


lemma gcd_induct:
  "(!!m. P m 0) ==>
    (!!m n. 0 < n ==> P n (m mod n) ==> P m n)
  ==> P (m::nat) (n::nat)"
  apply (induct m n rule: gcd.induct)
  apply (case_tac "n = 0")
   apply simp_all
  done


lemma gcd_0 [simp]: "gcd (m, 0) = m"
  by simp

lemma gcd_non_0: "0 < n ==> gcd (m, n) = gcd (n, m mod n)"
  by simp

declare gcd.simps [simp del]

lemma gcd_1 [simp]: "gcd (m, Suc 0) = 1"
  by (simp add: gcd_non_0)

text {*
  \medskip @{term "gcd (m, n)"} divides @{text m} and @{text n}.  The
  conjunctions don't seem provable separately.
*}

lemma gcd_dvd1 [iff]: "gcd (m, n) dvd m"
  and gcd_dvd2 [iff]: "gcd (m, n) dvd n"
  apply (induct m n rule: gcd_induct)
     apply (simp_all add: gcd_non_0)
  apply (blast dest: dvd_mod_imp_dvd)
  done

text {*
  \medskip Maximality: for all @{term m}, @{term n}, @{term k}
  naturals, if @{term k} divides @{term m} and @{term k} divides
  @{term n} then @{term k} divides @{term "gcd (m, n)"}.
*}

lemma gcd_greatest: "k dvd m ==> k dvd n ==> k dvd gcd (m, n)"
  by (induct m n rule: gcd_induct) (simp_all add: gcd_non_0 dvd_mod)

lemma gcd_greatest_iff [iff]: "(k dvd gcd (m, n)) = (k dvd m \<and> k dvd n)"
  by (blast intro!: gcd_greatest intro: dvd_trans)

lemma gcd_zero: "(gcd (m, n) = 0) = (m = 0 \<and> n = 0)"
  by (simp only: dvd_0_left_iff [symmetric] gcd_greatest_iff)


text {*
  \medskip Function gcd yields the Greatest Common Divisor.
*}

lemma is_gcd: "is_gcd (gcd (m, n)) m n"
  apply (simp add: is_gcd_def gcd_greatest)
  done

text {*
  \medskip Uniqueness of GCDs.
*}

lemma is_gcd_unique: "is_gcd m a b ==> is_gcd n a b ==> m = n"
  apply (simp add: is_gcd_def)
  apply (blast intro: dvd_anti_sym)
  done

lemma is_gcd_dvd: "is_gcd m a b ==> k dvd a ==> k dvd b ==> k dvd m"
  apply (auto simp add: is_gcd_def)
  done


text {*
  \medskip Commutativity
*}

lemma is_gcd_commute: "is_gcd k m n = is_gcd k n m"
  apply (auto simp add: is_gcd_def)
  done

lemma gcd_commute: "gcd (m, n) = gcd (n, m)"
  apply (rule is_gcd_unique)
   apply (rule is_gcd)
  apply (subst is_gcd_commute)
  apply (simp add: is_gcd)
  done

lemma gcd_assoc: "gcd (gcd (k, m), n) = gcd (k, gcd (m, n))"
  apply (rule is_gcd_unique)
   apply (rule is_gcd)
  apply (simp add: is_gcd_def)
  apply (blast intro: dvd_trans)
  done

lemma gcd_0_left [simp]: "gcd (0, m) = m"
  apply (simp add: gcd_commute [of 0])
  done

lemma gcd_1_left [simp]: "gcd (Suc 0, m) = 1"
  apply (simp add: gcd_commute [of "Suc 0"])
  done


text {*
  \medskip Multiplication laws
*}

lemma gcd_mult_distrib2: "k * gcd (m, n) = gcd (k * m, k * n)"
    -- {* \cite[page 27]{davenport92} *}
  apply (induct m n rule: gcd_induct)
   apply simp
  apply (case_tac "k = 0")
   apply (simp_all add: mod_geq gcd_non_0 mod_mult_distrib2)
  done

lemma gcd_mult [simp]: "gcd (k, k * n) = k"
  apply (rule gcd_mult_distrib2 [of k 1 n, simplified, symmetric])
  done

lemma gcd_self [simp]: "gcd (k, k) = k"
  apply (rule gcd_mult [of k 1, simplified])
  done

lemma relprime_dvd_mult: "gcd (k, n) = 1 ==> k dvd m * n ==> k dvd m"
  apply (insert gcd_mult_distrib2 [of m k n])
  apply simp
  apply (erule_tac t = m in ssubst)
  apply simp
  done

lemma relprime_dvd_mult_iff: "gcd (k, n) = 1 ==> (k dvd m * n) = (k dvd m)"
  apply (blast intro: relprime_dvd_mult dvd_trans)
  done

lemma gcd_mult_cancel: "gcd (k, n) = 1 ==> gcd (k * m, n) = gcd (m, n)"
  apply (rule dvd_anti_sym)
   apply (rule gcd_greatest)
    apply (rule_tac n = k in relprime_dvd_mult)
     apply (simp add: gcd_assoc)
     apply (simp add: gcd_commute)
    apply (simp_all add: mult_commute)
  apply (blast intro: dvd_trans)
  done


text {* \medskip Addition laws *}

lemma gcd_add1 [simp]: "gcd (m + n, n) = gcd (m, n)"
  apply (case_tac "n = 0")
   apply (simp_all add: gcd_non_0)
  done

lemma gcd_add2 [simp]: "gcd (m, m + n) = gcd (m, n)"
proof -
  have "gcd (m, m + n) = gcd (m + n, m)" by (rule gcd_commute)
  also have "... = gcd (n + m, m)" by (simp add: add_commute)
  also have "... = gcd (n, m)" by simp
  also have  "... = gcd (m, n)" by (rule gcd_commute)
  finally show ?thesis .
qed

lemma gcd_add2' [simp]: "gcd (m, n + m) = gcd (m, n)"
  apply (subst add_commute)
  apply (rule gcd_add2)
  done

lemma gcd_add_mult: "gcd (m, k * m + n) = gcd (m, n)"
  by (induct k) (simp_all add: add_assoc)


text {*
  \medskip Division by gcd yields rrelatively primes.
*}

lemma div_gcd_relprime:
  assumes nz: "a \<noteq> 0 \<or> b \<noteq> 0"
  shows "gcd (a div gcd(a,b), b div gcd(a,b)) = 1"
proof -
  let ?g = "gcd (a, b)"
  let ?a' = "a div ?g"
  let ?b' = "b div ?g"
  let ?g' = "gcd (?a', ?b')"
  have dvdg: "?g dvd a" "?g dvd b" by simp_all
  have dvdg': "?g' dvd ?a'" "?g' dvd ?b'" by simp_all
  from dvdg dvdg' obtain ka kb ka' kb' where
      kab: "a = ?g * ka" "b = ?g * kb" "?a' = ?g' * ka'" "?b' = ?g' * kb'"
    unfolding dvd_def by blast
  then have "?g * ?a' = (?g * ?g') * ka'" "?g * ?b' = (?g * ?g') * kb'" by simp_all
  then have dvdgg':"?g * ?g' dvd a" "?g* ?g' dvd b"
    by (auto simp add: dvd_mult_div_cancel [OF dvdg(1)]
      dvd_mult_div_cancel [OF dvdg(2)] dvd_def)
  have "?g \<noteq> 0" using nz by (simp add: gcd_zero)
  then have gp: "?g > 0" by simp
  from gcd_greatest [OF dvdgg'] have "?g * ?g' dvd ?g" .
  with dvd_mult_cancel1 [OF gp] show "?g' = 1" by simp
qed


text {*
  \medskip Gcd on integers.
*}

definition
  igcd :: "int \<Rightarrow> int \<Rightarrow> int" where
  "igcd i j = int (gcd (nat (abs i), nat (abs j)))"

lemma igcd_dvd1 [simp]: "igcd i j dvd i"
  by (simp add: igcd_def int_dvd_iff)

lemma igcd_dvd2 [simp]: "igcd i j dvd j"
  by (simp add: igcd_def int_dvd_iff)

lemma igcd_pos: "igcd i j \<ge> 0"
  by (simp add: igcd_def)

lemma igcd0 [simp]: "(igcd i j = 0) = (i = 0 \<and> j = 0)"
  by (simp add: igcd_def gcd_zero) arith

lemma igcd_commute: "igcd i j = igcd j i"
  unfolding igcd_def by (simp add: gcd_commute)

lemma igcd_neg1 [simp]: "igcd (- i) j = igcd i j"
  unfolding igcd_def by simp

lemma igcd_neg2 [simp]: "igcd i (- j) = igcd i j"
  unfolding igcd_def by simp

lemma zrelprime_dvd_mult: "igcd i j = 1 \<Longrightarrow> i dvd k * j \<Longrightarrow> i dvd k"
  unfolding igcd_def
proof -
  assume "int (gcd (nat \<bar>i\<bar>, nat \<bar>j\<bar>)) = 1" "i dvd k * j"
  then have g: "gcd (nat \<bar>i\<bar>, nat \<bar>j\<bar>) = 1" by simp
  from `i dvd k * j` obtain h where h: "k*j = i*h" unfolding dvd_def by blast
  have th: "nat \<bar>i\<bar> dvd nat \<bar>k\<bar> * nat \<bar>j\<bar>"
    unfolding dvd_def
    by (rule_tac x= "nat \<bar>h\<bar>" in exI, simp add: h nat_abs_mult_distrib [symmetric])
  from relprime_dvd_mult [OF g th] obtain h' where h': "nat \<bar>k\<bar> = nat \<bar>i\<bar> * h'"
    unfolding dvd_def by blast
  from h' have "int (nat \<bar>k\<bar>) = int (nat \<bar>i\<bar> * h')" by simp
  then have "\<bar>k\<bar> = \<bar>i\<bar> * int h'" by simp
  then show ?thesis
    apply (subst zdvd_abs1 [symmetric])
    apply (subst zdvd_abs2 [symmetric])
    apply (unfold dvd_def)
    apply (rule_tac x = "int h'" in exI, simp)
    done
qed

lemma int_nat_abs: "int (nat (abs x)) = abs x"  by arith

lemma igcd_greatest:
  assumes "k dvd m" and "k dvd n"
  shows "k dvd igcd m n"
proof -
  let ?k' = "nat \<bar>k\<bar>"
  let ?m' = "nat \<bar>m\<bar>"
  let ?n' = "nat \<bar>n\<bar>"
  from `k dvd m` and `k dvd n` have dvd': "?k' dvd ?m'" "?k' dvd ?n'"
    unfolding zdvd_int by (simp_all only: int_nat_abs zdvd_abs1 zdvd_abs2)
  from gcd_greatest [OF dvd'] have "int (nat \<bar>k\<bar>) dvd igcd m n"
    unfolding igcd_def by (simp only: zdvd_int)
  then have "\<bar>k\<bar> dvd igcd m n" by (simp only: int_nat_abs)
  then show "k dvd igcd m n" by (simp add: zdvd_abs1)
qed

lemma div_igcd_relprime:
  assumes nz: "a \<noteq> 0 \<or> b \<noteq> 0"
  shows "igcd (a div (igcd a b)) (b div (igcd a b)) = 1"
proof -
  from nz have nz': "nat \<bar>a\<bar> \<noteq> 0 \<or> nat \<bar>b\<bar> \<noteq> 0" by simp
  let ?g = "igcd a b"
  let ?a' = "a div ?g"
  let ?b' = "b div ?g"
  let ?g' = "igcd ?a' ?b'"
  have dvdg: "?g dvd a" "?g dvd b" by (simp_all add: igcd_dvd1 igcd_dvd2)
  have dvdg': "?g' dvd ?a'" "?g' dvd ?b'" by (simp_all add: igcd_dvd1 igcd_dvd2)
  from dvdg dvdg' obtain ka kb ka' kb' where
   kab: "a = ?g*ka" "b = ?g*kb" "?a' = ?g'*ka'" "?b' = ?g' * kb'"
    unfolding dvd_def by blast
  then have "?g* ?a' = (?g * ?g') * ka'" "?g* ?b' = (?g * ?g') * kb'" by simp_all
  then have dvdgg':"?g * ?g' dvd a" "?g* ?g' dvd b"
    by (auto simp add: zdvd_mult_div_cancel [OF dvdg(1)]
      zdvd_mult_div_cancel [OF dvdg(2)] dvd_def)
  have "?g \<noteq> 0" using nz by simp
  then have gp: "?g \<noteq> 0" using igcd_pos[where i="a" and j="b"] by arith
  from igcd_greatest [OF dvdgg'] have "?g * ?g' dvd ?g" .
  with zdvd_mult_cancel1 [OF gp] have "\<bar>?g'\<bar> = 1" by simp
  with igcd_pos show "?g' = 1" by simp
qed

text{* LCM *}

definition "lcm = (\<lambda>(m,n). m*n div gcd(m,n))"

definition "ilcm = (\<lambda>i j. int (lcm(nat(abs i),nat(abs j))))"

(* ilcm_dvd12 are needed later *)
lemma lcm_dvd1: 
  assumes mpos: " m >0"
  and npos: "n>0"
  shows "m dvd (lcm(m,n))"
proof-
  have "gcd(m,n) dvd n" by simp
  then obtain "k" where "n = gcd(m,n) * k" using dvd_def by auto
  then have "m*n div gcd(m,n) = m*(gcd(m,n)*k) div gcd(m,n)" by (simp add: mult_ac)
  also have "\<dots> = m*k" using mpos npos gcd_zero by simp
  finally show ?thesis by (simp add: lcm_def)
qed

lemma lcm_dvd2: 
  assumes mpos: " m >0"
  and npos: "n>0"
  shows "n dvd (lcm(m,n))"
proof-
  have "gcd(m,n) dvd m" by simp
  then obtain "k" where "m = gcd(m,n) * k" using dvd_def by auto
  then have "m*n div gcd(m,n) = (gcd(m,n)*k)*n div gcd(m,n)" by (simp add: mult_ac)
  also have "\<dots> = n*k" using mpos npos gcd_zero by simp
  finally show ?thesis by (simp add: lcm_def)
qed

lemma ilcm_dvd1: 
assumes anz: "a \<noteq> 0" 
  and bnz: "b \<noteq> 0"
  shows "a dvd (ilcm a b)"
proof-
  let ?na = "nat (abs a)"
  let ?nb = "nat (abs b)"
  have nap: "?na >0" using anz by simp
  have nbp: "?nb >0" using bnz by simp
  from nap nbp have "?na dvd lcm(?na,?nb)" using lcm_dvd1 by simp
  thus ?thesis by (simp add: ilcm_def dvd_int_iff)
qed


lemma ilcm_dvd2: 
assumes anz: "a \<noteq> 0" 
  and bnz: "b \<noteq> 0"
  shows "b dvd (ilcm a b)"
proof-
  let ?na = "nat (abs a)"
  let ?nb = "nat (abs b)"
  have nap: "?na >0" using anz by simp
  have nbp: "?nb >0" using bnz by simp
  from nap nbp have "?nb dvd lcm(?na,?nb)" using lcm_dvd2 by simp
  thus ?thesis by (simp add: ilcm_def dvd_int_iff)
qed

lemma zdvd_self_abs1: "(d::int) dvd (abs d)"
by (case_tac "d <0", simp_all)

lemma zdvd_self_abs2: "(abs (d::int)) dvd d"
by (case_tac "d<0", simp_all)

lemma zdvd_abs1: "((d::int) dvd t) = ((abs d) dvd t)"
 by (cases "d < 0") simp_all

(* lcm a b is positive for positive a and b *)

lemma lcm_pos: 
  assumes mpos: "m > 0"
  and npos: "n>0"
  shows "lcm (m,n) > 0"

proof(rule ccontr, simp add: lcm_def gcd_zero)
assume h:"m*n div gcd(m,n) = 0"
from mpos npos have "gcd (m,n) \<noteq> 0" using gcd_zero by simp
hence gcdp: "gcd(m,n) > 0" by simp
with h
have "m*n < gcd(m,n)"
  by (cases "m * n < gcd (m, n)") (auto simp add: div_if[OF gcdp, where m="m*n"])
moreover 
have "gcd(m,n) dvd m" by simp
 with mpos dvd_imp_le have t1:"gcd(m,n) \<le> m" by simp
 with npos have t1:"gcd(m,n)*n \<le> m*n" by simp
 have "gcd(m,n) \<le> gcd(m,n)*n" using npos by simp
 with t1 have "gcd(m,n) \<le> m*n" by arith
ultimately show "False" by simp
qed

lemma ilcm_pos: 
  assumes apos: " 0 < a"
  and bpos: "0 < b" 
  shows "0 < ilcm  a b"
proof-
  let ?na = "nat (abs a)"
  let ?nb = "nat (abs b)"
  have nap: "?na >0" using apos by simp
  have nbp: "?nb >0" using bpos by simp
  have "0 < lcm (?na,?nb)" by (rule lcm_pos[OF nap nbp])
  thus ?thesis by (simp add: ilcm_def)
qed


end