(* ID:         $Id$ *)

(* EXTRACT from HOL/ex/Primes.thy*)

theory Primes = Main:

consts

  gcd     :: "nat*nat \<Rightarrow> nat"               (*Euclid's algorithm *)

recdef gcd "measure ((\<lambda>(m,n).n) ::nat*nat \<Rightarrow> nat)"

    "gcd (m, n) = (if n=0 then m else gcd(n, m mod n))"

ML "Pretty.setmargin 64"

ML "IsarOutput.indent := 5"  (*that is, Doc/TutorialI/settings.ML*)

text {*

\begin{quote}

@{thm[display]"dvd_def"}

\rulename{dvd_def}

\end{quote}

*};

(*** Euclid's Algorithm ***)

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

  apply (simp);

  done

lemma gcd_non_0 [simp]: "0<n \<Longrightarrow> gcd(m,n) = gcd (n, m mod n)"

  apply (simp)

  done;

declare gcd.simps [simp del];

(*gcd(m,n) divides m and n.  The conjunctions don't seem provable separately*)

lemma gcd_dvd_both: "(gcd(m,n) dvd m) \<and> (gcd(m,n) dvd n)"

  apply (induct_tac m n rule: gcd.induct)

  apply (case_tac "n=0")

  apply (simp_all)

  apply (blast dest: dvd_mod_imp_dvd)

  done

text {*

@{thm[display] dvd_mod_imp_dvd}

\rulename{dvd_mod_imp_dvd}

@{thm[display] dvd_trans}

\rulename{dvd_trans}

\begin{isabelle}

proof\ (prove):\ step\ 3\isanewline

\isanewline

goal\ (lemma\ gcd_dvd_both):\isanewline

gcd\ (m,\ n)\ dvd\ m\ \isasymand \ gcd\ (m,\ n)\ dvd\ n\isanewline

\ 1.\ \isasymAnd m\ n.\ \isasymlbrakk 0\ <\ n;\ gcd\ (n,\ m\ mod\ n)\ dvd\ n\ \isasymand \ gcd\ (n,\ m\ mod\ n)\ dvd\ (m\ mod\ n)\isasymrbrakk \isanewline

\ \ \ \ \ \ \ \ \ \ \isasymLongrightarrow \ gcd\ (n,\ m\ mod\ n)\ dvd\ m

\end{isabelle}

*};

lemmas gcd_dvd1 [iff] = gcd_dvd_both [THEN conjunct1]

lemmas gcd_dvd2 [iff] = gcd_dvd_both [THEN conjunct2];

text {*

\begin{quote}

@{thm[display] gcd_dvd1}

\rulename{gcd_dvd1}

@{thm[display] gcd_dvd2}

\rulename{gcd_dvd2}

\end{quote}

*};

(*Maximality: for all m,n,k naturals, 

                if k divides m and k divides n then k divides gcd(m,n)*)

lemma gcd_greatest [rule_format]:

      "(k dvd m) \<longrightarrow> (k dvd n) \<longrightarrow> k dvd gcd(m,n)"

  apply (induct_tac m n rule: gcd.induct)

  apply (case_tac "n=0")

  apply (simp_all add: dvd_mod);

  done;

theorem gcd_greatest_iff [iff]: 

        "k dvd gcd(m,n) = (k dvd m \<and> k dvd n)"

  apply (blast intro!: gcd_greatest intro: dvd_trans);

  done;

constdefs

  is_gcd  :: "[nat,nat,nat] \<Rightarrow> bool"        (*gcd as a relation*)

    "is_gcd p m n == p dvd m  \<and>  p dvd n  \<and>

                     (ALL d. d dvd m \<and> d dvd n \<longrightarrow> d dvd p)"

(*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

(*uniqueness of GCDs*)

lemma is_gcd_unique: "\<lbrakk> is_gcd m a b; is_gcd n a b \<rbrakk> \<Longrightarrow> m=n"

  apply (simp add: is_gcd_def);

  apply (blast intro: dvd_anti_sym)

  done

text {*

@{thm[display] dvd_anti_sym}

\rulename{dvd_anti_sym}

\begin{isabelle}

proof\ (prove):\ step\ 1\isanewline

\isanewline

goal\ (lemma\ is_gcd_unique):\isanewline

\isasymlbrakk is_gcd\ m\ a\ b;\ is_gcd\ n\ a\ b\isasymrbrakk \ \isasymLongrightarrow \ m\ =\ n\isanewline

\ 1.\ \isasymlbrakk m\ dvd\ a\ \isasymand \ m\ dvd\ b\ \isasymand \ (\isasymforall d.\ d\ dvd\ a\ \isasymand \ d\ dvd\ b\ \isasymlongrightarrow \ d\ dvd\ m);\isanewline

\ \ \ \ \ \ \ n\ dvd\ a\ \isasymand \ n\ dvd\ b\ \isasymand \ (\isasymforall d.\ d\ dvd\ a\ \isasymand \ d\ dvd\ b\ \isasymlongrightarrow \ d\ dvd\ n)\isasymrbrakk \isanewline

\ \ \ \ \isasymLongrightarrow \ m\ =\ n

\end{isabelle}

*};

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 

text{*

\begin{isabelle}

proof\ (prove):\ step\ 3\isanewline

\isanewline

goal\ (lemma\ gcd_assoc):\isanewline

gcd\ (gcd\ (k,\ m),\ n)\ =\ gcd\ (k,\ gcd\ (m,\ n))\isanewline

\ 1.\ gcd\ (k,\ gcd\ (m,\ n))\ dvd\ k\ \isasymand \isanewline

\ \ \ \ gcd\ (k,\ gcd\ (m,\ n))\ dvd\ m\ \isasymand \ gcd\ (k,\ gcd\ (m,\ n))\ dvd\ n

\end{isabelle}

*}

lemma gcd_dvd_gcd_mult: "gcd(m,n) dvd gcd(k*m, n)"

  apply (blast intro: dvd_trans);

  done

(*This is half of the proof (by dvd_anti_sym) of*)

lemma gcd_mult_cancel: "gcd(k,n) = 1 \<Longrightarrow> gcd(k*m, n) = gcd(m,n)"

  oops

text{*\noindent

Forward proof material: of, OF, THEN, simplify.

*}

text{*\noindent

SKIP most developments...

*}

(** 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_1 [simp]: "gcd(m,1) = 1"

  apply (simp)

  done

lemma gcd_1_left [simp]: "gcd(1,m) = 1"

  apply (simp add: gcd_commute [of 1])

  done

text{*\noindent

as far as HERE.

*}

text {*

@{thm[display] gcd_1}

\rulename{gcd_1}

@{thm[display] gcd_1_left}

\rulename{gcd_1_left}

*};

text{*\noindent

SKIP THIS PROOF

*}

lemma gcd_mult_distrib2: "k * gcd(m,n) = gcd(k*m, k*n)"

  apply (induct_tac m n rule: gcd.induct)

  apply (case_tac "n=0")

  apply (simp)

  apply (case_tac "k=0")

  apply (simp_all add: mod_geq gcd_non_0 mod_mult_distrib2)

  done

text {*

@{thm[display] gcd_mult_distrib2}

\rulename{gcd_mult_distrib2}

*};

text{*\noindent

of, simplified

*}

lemmas gcd_mult_0 = gcd_mult_distrib2 [of k 1];

lemmas gcd_mult_1 = gcd_mult_0 [simplified];

text {*

@{thm[display] gcd_mult_distrib2 [of _ 1]}

@{thm[display] gcd_mult_0}

\rulename{gcd_mult_0}

@{thm[display] gcd_mult_1}

\rulename{gcd_mult_1}

@{thm[display] sym}

\rulename{sym}

*};

lemmas gcd_mult = gcd_mult_1 [THEN sym];

lemmas gcd_mult = gcd_mult_distrib2 [of k 1, simplified, THEN sym];

      (*better in one step!*)

text {*

more legible

*};

lemma gcd_mult [simp]: "gcd(k, k*n) = k"

  apply (rule gcd_mult_distrib2 [of k 1, simplified, THEN sym])

  done

lemmas gcd_self = gcd_mult [of k 1, simplified];

text {*

Rules handy with THEN

@{thm[display] iffD1}

\rulename{iffD1}

@{thm[display] iffD2}

\rulename{iffD2}

*};

text {*

again: more legible

*};

lemma gcd_self [simp]: "gcd(k,k) = k"

  apply (rule gcd_mult [of k 1, simplified])

  done

lemma relprime_dvd_mult: 

      "\<lbrakk> gcd(k,n)=1; k dvd (m*n) \<rbrakk> \<Longrightarrow> k dvd m";

  apply (insert gcd_mult_distrib2 [of m k n])

  apply (simp)

  apply (erule_tac t="m" in ssubst);

  apply (simp)

  done

text {*

Another example of "insert"

@{thm[display] mod_div_equality}

\rulename{mod_div_equality}

*};

lemma div_mult_self_is_m: 

      "0<n \<Longrightarrow> (m*n) div n = (m::nat)"

  apply (insert mod_div_equality [of "m*n" n])

  apply (simp)

  done

lemma relprime_dvd_mult_iff: "gcd(k,n)=1 \<Longrightarrow> k dvd (m*n) = k dvd m";

  apply (blast intro: relprime_dvd_mult dvd_trans)

  done

lemma relprime_20_81: "gcd(#20,#81) = 1";

  apply (simp add: gcd.simps)

  done

text {*

Examples of 'OF'

@{thm[display] relprime_dvd_mult}

\rulename{relprime_dvd_mult}

@{thm[display] relprime_dvd_mult [OF relprime_20_81]}

@{thm[display] dvd_refl}

\rulename{dvd_refl}

@{thm[display] dvd_add}

\rulename{dvd_add}

@{thm[display] dvd_add [OF dvd_refl dvd_refl]}

@{thm[display] dvd_add [OF _ dvd_refl]}

*};

lemma "\<lbrakk>(z::int) < #37; #66 < #2*z; z*z \<noteq> #1225; Q(#34); Q(#36)\<rbrakk> \<Longrightarrow> Q(z)";

apply (subgoal_tac "z = #34 \<or> z = #36")

apply blast

apply (subgoal_tac "z \<noteq> #35")

apply arith

apply force

done

text

{*

proof\ (prove):\ step\ 1\isanewline

\isanewline

goal\ (lemma):\isanewline

\isasymlbrakk z\ <\ \#37;\ \#66\ <\ \#2\ *\ z;\ z\ *\ z\ \isasymnoteq \ \#1225;\ Q\ \#34;\ Q\ \#36\isasymrbrakk \ \isasymLongrightarrow \ Q\ z\isanewline

\ 1.\ \isasymlbrakk z\ <\ \#37;\ \#66\ <\ \#2\ *\ z;\ z\ *\ z\ \isasymnoteq \ \#1225;\ Q\ \#34;\ Q\ \#36;\isanewline

\ \ \ \ \ \ \ z\ =\ \#34\ \isasymor \ z\ =\ \#36\isasymrbrakk \isanewline

\ \ \ \ \isasymLongrightarrow \ Q\ z\isanewline

\ 2.\ \isasymlbrakk z\ <\ \#37;\ \#66\ <\ \#2\ *\ z;\ z\ *\ z\ \isasymnoteq \ \#1225;\ Q\ \#34;\ Q\ \#36\isasymrbrakk \isanewline

\ \ \ \ \isasymLongrightarrow \ z\ =\ \#34\ \isasymor \ z\ =\ \#36

proof\ (prove):\ step\ 3\isanewline

\isanewline

goal\ (lemma):\isanewline

\isasymlbrakk z\ <\ \#37;\ \#66\ <\ \#2\ *\ z;\ z\ *\ z\ \isasymnoteq \ \#1225;\ Q\ \#34;\ Q\ \#36\isasymrbrakk \ \isasymLongrightarrow \ Q\ z\isanewline

\ 1.\ \isasymlbrakk z\ <\ \#37;\ \#66\ <\ \#2\ *\ z;\ z\ *\ z\ \isasymnoteq \ \#1225;\ Q\ \#34;\ Q\ \#36;\isanewline

\ \ \ \ \ \ \ z\ \isasymnoteq \ \#35\isasymrbrakk \isanewline

\ \ \ \ \isasymLongrightarrow \ z\ =\ \#34\ \isasymor \ z\ =\ \#36\isanewline

\ 2.\ \isasymlbrakk z\ <\ \#37;\ \#66\ <\ \#2\ *\ z;\ z\ *\ z\ \isasymnoteq \ \#1225;\ Q\ \#34;\ Q\ \#36\isasymrbrakk \isanewline

\ \ \ \ \isasymLongrightarrow \ z\ \isasymnoteq \ \#35

*}

end