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(*  Title:      HOL/GCD.thy
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    ID:         $Id$
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    Author:     Christophe Tabacznyj and Lawrence C Paulson
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    Copyright   1996  University of Cambridge
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Builds on Integ/Parity mainly because that contains recdef, which we
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need, but also because we may want to include gcd on integers in here
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as well in the future.
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*)
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header {* The Greatest Common Divisor *}
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theory GCD
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imports Parity
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begin
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text {*
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  See \cite{davenport92}.
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  \bigskip
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*}
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consts
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  gcd  :: "nat \<times> nat => nat"  -- {* Euclid's algorithm *}
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recdef gcd  "measure ((\<lambda>(m, n). n) :: nat \<times> nat => nat)"
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  "gcd (m, n) = (if n = 0 then m else gcd (n, m mod n))"
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constdefs
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  is_gcd :: "nat => nat => nat => bool"  -- {* @{term gcd} as a relation *}
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  "is_gcd p m n == p dvd m \<and> p dvd n \<and>
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    (\<forall>d. d dvd m \<and> d dvd n --> d dvd p)"
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lemma gcd_induct:
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  "(!!m. P m 0) ==>
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    (!!m n. 0 < n ==> P n (m mod n) ==> P m n)
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  ==> P (m::nat) (n::nat)"
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  apply (induct m n rule: gcd.induct)
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  apply (case_tac "n = 0")
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   apply simp_all
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  done
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lemma gcd_0 [simp]: "gcd (m, 0) = m"
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  apply simp
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  done
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lemma gcd_non_0: "0 < n ==> gcd (m, n) = gcd (n, m mod n)"
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  apply simp
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  done
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declare gcd.simps [simp del]
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lemma gcd_1 [simp]: "gcd (m, Suc 0) = 1"
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  apply (simp add: gcd_non_0)
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  done
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text {*
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  \medskip @{term "gcd (m, n)"} divides @{text m} and @{text n}.  The
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  conjunctions don't seem provable separately.
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*}
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lemma gcd_dvd1 [iff]: "gcd (m, n) dvd m"
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  and gcd_dvd2 [iff]: "gcd (m, n) dvd n"
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  apply (induct m n rule: gcd_induct)
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   apply (simp_all add: gcd_non_0)
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  apply (blast dest: dvd_mod_imp_dvd)
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  done
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text {*
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  \medskip Maximality: for all @{term m}, @{term n}, @{term k}
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  naturals, if @{term k} divides @{term m} and @{term k} divides
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  @{term n} then @{term k} divides @{term "gcd (m, n)"}.
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*}
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lemma gcd_greatest: "k dvd m ==> k dvd n ==> k dvd gcd (m, n)"
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  apply (induct m n rule: gcd_induct)
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   apply (simp_all add: gcd_non_0 dvd_mod)
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  done
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lemma gcd_greatest_iff [iff]: "(k dvd gcd (m, n)) = (k dvd m \<and> k dvd n)"
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  apply (blast intro!: gcd_greatest intro: dvd_trans)
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  done
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lemma gcd_zero: "(gcd (m, n) = 0) = (m = 0 \<and> n = 0)"
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  by (simp only: dvd_0_left_iff [THEN sym] gcd_greatest_iff)
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text {*
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  \medskip Function gcd yields the Greatest Common Divisor.
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*}
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lemma is_gcd: "is_gcd (gcd (m, n)) m n"
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  apply (simp add: is_gcd_def gcd_greatest)
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  done
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text {*
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  \medskip Uniqueness of GCDs.
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*}
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lemma is_gcd_unique: "is_gcd m a b ==> is_gcd n a b ==> m = n"
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  apply (simp add: is_gcd_def)
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  apply (blast intro: dvd_anti_sym)
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  done
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lemma is_gcd_dvd: "is_gcd m a b ==> k dvd a ==> k dvd b ==> k dvd m"
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  apply (auto simp add: is_gcd_def)
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  done
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text {*
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  \medskip Commutativity
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*}
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lemma is_gcd_commute: "is_gcd k m n = is_gcd k n m"
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  apply (auto simp add: is_gcd_def)
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  done
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lemma gcd_commute: "gcd (m, n) = gcd (n, m)"
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  apply (rule is_gcd_unique)
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   apply (rule is_gcd)
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  apply (subst is_gcd_commute)
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  apply (simp add: is_gcd)
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  done
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lemma gcd_assoc: "gcd (gcd (k, m), n) = gcd (k, gcd (m, n))"
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  apply (rule is_gcd_unique)
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   apply (rule is_gcd)
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  apply (simp add: is_gcd_def)
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  apply (blast intro: dvd_trans)
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  done
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lemma gcd_0_left [simp]: "gcd (0, m) = m"
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  apply (simp add: gcd_commute [of 0])
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  done
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lemma gcd_1_left [simp]: "gcd (Suc 0, m) = 1"
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  apply (simp add: gcd_commute [of "Suc 0"])
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  done
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text {*
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  \medskip Multiplication laws
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*}
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lemma gcd_mult_distrib2: "k * gcd (m, n) = gcd (k * m, k * n)"
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    -- {* \cite[page 27]{davenport92} *}
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  apply (induct m n rule: gcd_induct)
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   apply simp
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  apply (case_tac "k = 0")
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   apply (simp_all add: mod_geq gcd_non_0 mod_mult_distrib2)
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  done
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lemma gcd_mult [simp]: "gcd (k, k * n) = k"
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  apply (rule gcd_mult_distrib2 [of k 1 n, simplified, symmetric])
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  done
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lemma gcd_self [simp]: "gcd (k, k) = k"
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  apply (rule gcd_mult [of k 1, simplified])
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  done
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lemma relprime_dvd_mult: "gcd (k, n) = 1 ==> k dvd m * n ==> k dvd m"
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  apply (insert gcd_mult_distrib2 [of m k n])
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  apply simp
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  apply (erule_tac t = m in ssubst)
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  apply simp
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  done
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lemma relprime_dvd_mult_iff: "gcd (k, n) = 1 ==> (k dvd m * n) = (k dvd m)"
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  apply (blast intro: relprime_dvd_mult dvd_trans)
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  done
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lemma gcd_mult_cancel: "gcd (k, n) = 1 ==> gcd (k * m, n) = gcd (m, n)"
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  apply (rule dvd_anti_sym)
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   apply (rule gcd_greatest)
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    apply (rule_tac n = k in relprime_dvd_mult)
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     apply (simp add: gcd_assoc)
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     apply (simp add: gcd_commute)
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    apply (simp_all add: mult_commute)
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  apply (blast intro: dvd_trans)
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  done
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text {* \medskip Addition laws *}
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lemma gcd_add1 [simp]: "gcd (m + n, n) = gcd (m, n)"
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  apply (case_tac "n = 0")
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   apply (simp_all add: gcd_non_0)
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  done
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lemma gcd_add2 [simp]: "gcd (m, m + n) = gcd (m, n)"
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proof -
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  have "gcd (m, m + n) = gcd (m + n, m)" by (rule gcd_commute) 
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  also have "... = gcd (n + m, m)" by (simp add: add_commute)
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  also have "... = gcd (n, m)" by simp
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  also have  "... = gcd (m, n)" by (rule gcd_commute) 
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  finally show ?thesis .
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qed
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lemma gcd_add2' [simp]: "gcd (m, n + m) = gcd (m, n)"
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  apply (subst add_commute)
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  apply (rule gcd_add2)
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  done
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lemma gcd_add_mult: "gcd (m, k * m + n) = gcd (m, n)"
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  apply (induct k)
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   apply (simp_all add: add_assoc)
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  done
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end
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