theory Myhill

  imports Main List_Prefix Prefix_subtract Prelude

begin

(*

text {*

     \begin{figure}

    \centering

    \scalebox{0.95}{

    \begin{tikzpicture}[->,>=latex,shorten >=1pt,auto,node distance=1.2cm, semithick]

        \node[state,initial] (n1)                   {$1$};

        \node[state,accepting] (n2) [right = 10em of n1]   {$2$};

        \path (n1) edge [bend left] node {$0$} (n2)

            (n1) edge [loop above] node{$1$} (n1)

            (n2) edge [loop above] node{$0$} (n2)

            (n2) edge [bend left]  node {$1$} (n1)

            ;

    \end{tikzpicture}}

    \caption{An example automaton (or partition)}\label{fig:example_automata}

    \end{figure}

*}

*)

section {* Preliminary definitions *}

types lang = "string set"

text {* 

  Sequential composition of two languages @{text "L1"} and @{text "L2"} 

*}

definition Seq :: "lang \<Rightarrow> lang \<Rightarrow> lang" ("_ ;; _" [100,100] 100)

where 

  "L1 ;; L2 = {s1 @ s2 | s1 s2. s1 \<in> L1 \<and> s2 \<in> L2}"

text {* Transitive closure of language @{text "L"}. *}

inductive_set

  Star :: "string set \<Rightarrow> string set" ("_\<star>" [101] 102)

  for L :: "string set"

where

  start[intro]: "[] \<in> L\<star>"

| step[intro]:  "\<lbrakk>s1 \<in> L; s2 \<in> L\<star>\<rbrakk> \<Longrightarrow> s1@s2 \<in> L\<star>" 

text {* Some properties of operator @{text ";;"}.*}

lemma seq_union_distrib:

  "(A \<union> B) ;; C = (A ;; C) \<union> (B ;; C)"

by (auto simp:Seq_def)

lemma seq_intro:

  "\<lbrakk>x \<in> A; y \<in> B\<rbrakk> \<Longrightarrow> x @ y \<in> A ;; B "

by (auto simp:Seq_def)

lemma seq_assoc:

  "(A ;; B) ;; C = A ;; (B ;; C)"

apply(auto simp:Seq_def)

apply blast

by (metis append_assoc)

lemma star_intro1[rule_format]: "x \<in> lang\<star> \<Longrightarrow> \<forall> y. y \<in> lang\<star> \<longrightarrow> x @ y \<in> lang\<star>"

by (erule Star.induct, auto)

lemma star_intro2: "y \<in> lang \<Longrightarrow> y \<in> lang\<star>"

by (drule step[of y lang "[]"], auto simp:start)

lemma star_intro3[rule_format]: 

  "x \<in> lang\<star> \<Longrightarrow> \<forall>y . y \<in> lang \<longrightarrow> x @ y \<in> lang\<star>"

by (erule Star.induct, auto intro:star_intro2)

lemma star_decom: 

  "\<lbrakk>x \<in> lang\<star>; x \<noteq> []\<rbrakk> \<Longrightarrow>(\<exists> a b. x = a @ b \<and> a \<noteq> [] \<and> a \<in> lang \<and> b \<in> lang\<star>)"

by (induct x rule: Star.induct, simp, blast)

lemma star_decom': 

  "\<lbrakk>x \<in> lang\<star>; x \<noteq> []\<rbrakk> \<Longrightarrow> \<exists>a b. x = a @ b \<and> a \<in> lang\<star> \<and> b \<in> lang"

apply (induct x rule:Star.induct, simp)

apply (case_tac "s2 = []")

apply (rule_tac x = "[]" in exI, rule_tac x = s1 in exI, simp add:start)

apply (simp, (erule exE| erule conjE)+)

by (rule_tac x = "s1 @ a" in exI, rule_tac x = b in exI, simp add:step)

text {* Ardens lemma expressed at the level of language, rather than the level of regular expression. *}

theorem ardens_revised:

  assumes nemp: "[] \<notin> A"

  shows "(X = X ;; A \<union> B) \<longleftrightarrow> (X = B ;; A\<star>)"

proof

  assume eq: "X = B ;; A\<star>"

  have "A\<star> =  {[]} \<union> A\<star> ;; A" 

    by (auto simp:Seq_def star_intro3 star_decom')  

  then have "B ;; A\<star> = B ;; ({[]} \<union> A\<star> ;; A)" 

    unfolding Seq_def by simp

  also have "\<dots> = B \<union> B ;; (A\<star> ;; A)"  

    unfolding Seq_def by auto

  also have "\<dots> = B \<union> (B ;; A\<star>) ;; A" 

    by (simp only:seq_assoc)

  finally show "X = X ;; A \<union> B" 

    using eq by blast 

next

  assume eq': "X = X ;; A \<union> B"

  hence c1': "\<And> x. x \<in> B \<Longrightarrow> x \<in> X" 

    and c2': "\<And> x y. \<lbrakk>x \<in> X; y \<in> A\<rbrakk> \<Longrightarrow> x @ y \<in> X" 

    using Seq_def by auto

  show "X = B ;; A\<star>" 

  proof

    show "B ;; A\<star> \<subseteq> X"

    proof-

      { fix x y

        have "\<lbrakk>y \<in> A\<star>; x \<in> X\<rbrakk> \<Longrightarrow> x @ y \<in> X "

          apply (induct arbitrary:x rule:Star.induct, simp)

          by (auto simp only:append_assoc[THEN sym] dest:c2')

      } thus ?thesis using c1' by (auto simp:Seq_def) 

    qed

  next

    show "X \<subseteq> B ;; A\<star>"

    proof-

      { fix x 

        have "x \<in> X \<Longrightarrow> x \<in> B ;; A\<star>"

        proof (induct x taking:length rule:measure_induct)

          fix z

          assume hyps: 

            "\<forall>y. length y < length z \<longrightarrow> y \<in> X \<longrightarrow> y \<in> B ;; A\<star>" 

            and z_in: "z \<in> X"

          show "z \<in> B ;; A\<star>"

          proof (cases "z \<in> B")

            case True thus ?thesis by (auto simp:Seq_def start)

          next

            case False hence "z \<in> X ;; A" using eq' z_in by auto

            then obtain za zb where za_in: "za \<in> X" 

              and zab: "z = za @ zb \<and> zb \<in> A" and zbne: "zb \<noteq> []" 

              using nemp unfolding Seq_def by blast

            from zbne zab have "length za < length z" by auto

            with za_in hyps have "za \<in> B ;; A\<star>" by blast

            hence "za @ zb \<in> B ;; A\<star>" using zab 

              by (clarsimp simp:Seq_def, blast dest:star_intro3)

            thus ?thesis using zab by simp       

          qed

        qed 

      } thus ?thesis by blast

    qed

  qed

qed

text {* The syntax of regular expressions is defined by the datatype @{text "rexp"}. *}

datatype rexp =

  NULL

| EMPTY

| CHAR char

| SEQ rexp rexp

| ALT rexp rexp

| STAR rexp

text {* 

  The following @{text "L"} is an overloaded operator, where @{text "L(x)"} evaluates to 

  the language represented by the syntactic object @{text "x"}.

*}

consts L:: "'a \<Rightarrow> string set"

text {* 

  The @{text "L(rexp)"} for regular expression @{text "rexp"} is defined by the 

  following overloading function @{text "L_rexp"}.

*}

overloading L_rexp \<equiv> "L::  rexp \<Rightarrow> string set"

begin

fun

  L_rexp :: "rexp \<Rightarrow> string set"

where

    "L_rexp (NULL) = {}"

  | "L_rexp (EMPTY) = {[]}"

  | "L_rexp (CHAR c) = {[c]}"

  | "L_rexp (SEQ r1 r2) = (L_rexp r1) ;; (L_rexp r2)"

  | "L_rexp (ALT r1 r2) = (L_rexp r1) \<union> (L_rexp r2)"

  | "L_rexp (STAR r) = (L_rexp r)\<star>"

end

text {*

  To obtain equational system out of finite set of equivalent classes, a fold operation

  on finite set @{text "folds"} is defined. The use of @{text "SOME"} makes @{text "fold"}

  more robust than the @{text "fold"} in Isabelle library. The expression @{text "folds f"}

  makes sense when @{text "f"} is not @{text "associative"} and @{text "commutitive"},

  while @{text "fold f"} does not.  

*}

definition 

  folds :: "('a \<Rightarrow> 'b \<Rightarrow> 'b) \<Rightarrow> 'b \<Rightarrow> 'a set \<Rightarrow> 'b"

where

  "folds f z S \<equiv> SOME x. fold_graph f z S x"

text {* 

  The following lemma assures that the arbitrary choice made by the @{text "SOME"} in @{text "folds"}

  does not affect the @{text "L"}-value of the resultant regular expression. 

  *}

lemma folds_alt_simp [simp]:

  "finite rs \<Longrightarrow> L (folds ALT NULL rs) = \<Union> (L ` rs)"

apply (rule set_eq_intro, simp add:folds_def)

apply (rule someI2_ex, erule finite_imp_fold_graph)

by (erule fold_graph.induct, auto)

(* Just a technical lemma. *)

lemma [simp]:

  shows "(x, y) \<in> {(x, y). P x y} \<longleftrightarrow> P x y"

by simp

text {*

  @{text "\<approx>L"} is an equivalent class defined by language @{text "Lang"}.

*}

definition

  str_eq_rel ("\<approx>_" [100] 100)

where

  "\<approx>Lang \<equiv> {(x, y).  (\<forall>z. x @ z \<in> Lang \<longleftrightarrow> y @ z \<in> Lang)}"

text {* 

  Among equivlant clases of @{text "\<approx>Lang"}, the set @{text "finals(Lang)"} singles out 

  those which contains strings from @{text "Lang"}.

*}

definition 

   "finals Lang \<equiv> {\<approx>Lang `` {x} | x . x \<in> Lang}"

text {* 

  The following lemma show the relationshipt between @{text "finals(Lang)"} and @{text "Lang"}.

*}

lemma lang_is_union_of_finals: 

  "Lang = \<Union> finals(Lang)"

proof 

  show "Lang \<subseteq> \<Union> (finals Lang)"

  proof

    fix x

    assume "x \<in> Lang"   

    thus "x \<in> \<Union> (finals Lang)"

      apply (simp add:finals_def, rule_tac x = "(\<approx>Lang) `` {x}" in exI)

      by (auto simp:Image_def str_eq_rel_def)    

  qed

next

  show "\<Union> (finals Lang) \<subseteq> Lang"

    apply (clarsimp simp:finals_def str_eq_rel_def)

    by (drule_tac x = "[]" in spec, auto)

qed

section {* Direction @{text "finite partition \<Rightarrow> regular language"}*}

text {* 

  The relationship between equivalent classes can be described by an

  equational system.

  For example, in equational system \eqref{example_eqns},  $X_0, X_1$ are equivalent 

  classes. The first equation says every string in $X_0$ is obtained either by

  appending one $b$ to a string in $X_0$ or by appending one $a$ to a string in

  $X_1$ or just be an empty string (represented by the regular expression $\lambda$). Similary,

  the second equation tells how the strings inside $X_1$ are composed.

  \begin{equation}\label{example_eqns}

    \begin{aligned}

      X_0 & = X_0 b + X_1 a + \lambda \\

      X_1 & = X_0 a + X_1 b

    \end{aligned}

  \end{equation}

  The summands on the right hand side is represented by the following data type

  @{text "rhs_item"}, mnemonic for 'right hand side item'.

  Generally, there are two kinds of right hand side items, one kind corresponds to

  pure regular expressions, like the $\lambda$ in \eqref{example_eqns}, the other kind corresponds to

  transitions from one one equivalent class to another, like the $X_0 b, X_1 a$ etc.

  *}

datatype rhs_item = 

   Lam "rexp"                           (* Lambda *)

 | Trn "(string set)" "rexp"              (* Transition *)

text {*

  In this formalization, pure regular expressions like $\lambda$ is 

  repsented by @{text "Lam(EMPTY)"}, while transitions like $X_0 a$ is represented by $Trn~X_0~(CHAR~a)$.

  *}

text {*

  The functions @{text "the_r"} and @{text "the_Trn"} are used to extract

  subcomponents from right hand side items.

  *}

fun the_r :: "rhs_item \<Rightarrow> rexp"

where "the_r (Lam r) = r"

fun the_Trn:: "rhs_item \<Rightarrow> (string set \<times> rexp)"

where "the_Trn (Trn Y r) = (Y, r)"

text {*

  Every right hand side item @{text "itm"} defines a string set given 

  @{text "L(itm)"}, defined as:

*}

overloading L_rhs_e \<equiv> "L:: rhs_item \<Rightarrow> string set"

begin

  fun L_rhs_e:: "rhs_item \<Rightarrow> string set"

  where

     "L_rhs_e (Lam r) = L r" |

     "L_rhs_e (Trn X r) = X ;; L r"

end

text {*

  The right hand side of every equation is represented by a set of

  items. The string set defined by such a set @{text "itms"} is given

  by @{text "L(itms)"}, defined as:

*}

overloading L_rhs \<equiv> "L:: rhs_item set \<Rightarrow> string set"

begin

   fun L_rhs:: "rhs_item set \<Rightarrow> string set"

   where "L_rhs rhs = \<Union> (L ` rhs)"

end

text {* 

  Given a set of equivalent classses @{text "CS"} and one equivalent class @{text "X"} among

  @{text "CS"}, the term @{text "init_rhs CS X"} is used to extract the right hand side of

  the equation describing the formation of @{text "X"}. The definition of @{text "init_rhs"}

  is:

  *}

definition

  "init_rhs CS X \<equiv>  

      if ([] \<in> X) then 

          {Lam(EMPTY)} \<union> {Trn Y (CHAR c) | Y c. Y \<in> CS \<and> Y ;; {[c]} \<subseteq> X}

      else 

          {Trn Y (CHAR c)| Y c. Y \<in> CS \<and> Y ;; {[c]} \<subseteq> X}"

text {*

  In the definition of @{text "init_rhs"}, the term 

  @{text "{Trn Y (CHAR c)| Y c. Y \<in> CS \<and> Y ;; {[c]} \<subseteq> X}"} appearing on both branches

  describes the formation of strings in @{text "X"} out of transitions, while 

  the term @{text "{Lam(EMPTY)}"} describes the empty string which is intrinsically contained in

  @{text "X"} rather than by transition. This @{text "{Lam(EMPTY)}"} corresponds to 

  the $\lambda$ in \eqref{example_eqns}.

  With the help of @{text "init_rhs"}, the equitional system descrbing the formation of every

  equivalent class inside @{text "CS"} is given by the following @{text "eqs(CS)"}.

  *}

definition "eqs CS \<equiv> {(X, init_rhs CS X) | X.  X \<in> CS}"

(************ arden's lemma variation ********************)

text {* 

  The following @{text "items_of rhs X"} returns all @{text "X"}-items in @{text "rhs"}.

  *}

definition

  "items_of rhs X \<equiv> {Trn X r | r. (Trn X r) \<in> rhs}"

text {* 

  The following @{text "rexp_of rhs X"} combines all regular expressions in @{text "X"}-items

  using @{text "ALT"} to form a single regular expression. 

  It will be used later to implement @{text "arden_variate"} and @{text "rhs_subst"}.

  *}

definition 

  "rexp_of rhs X \<equiv> folds ALT NULL ((snd o the_Trn) ` items_of rhs X)"

text {* 

  The following @{text "lam_of rhs"} returns all pure regular expression items in @{text "rhs"}.

  *}

definition

  "lam_of rhs \<equiv> {Lam r | r. Lam r \<in> rhs}"

text {*

  The following @{text "rexp_of_lam rhs"} combines pure regular expression items in @{text "rhs"}

  using @{text "ALT"} to form a single regular expression. 

  When all variables inside @{text "rhs"} are eliminated, @{text "rexp_of_lam rhs"}

  is used to compute compute the regular expression corresponds to @{text "rhs"}.

  *}

definition

  "rexp_of_lam rhs \<equiv> folds ALT NULL (the_r ` lam_of rhs)"

text {*

  The following @{text "attach_rexp rexp' itm"} attach 

  the regular expression @{text "rexp'"} to

  the right of right hand side item @{text "itm"}.

  *}

fun attach_rexp :: "rexp \<Rightarrow> rhs_item \<Rightarrow> rhs_item"

where

  "attach_rexp rexp' (Lam rexp)   = Lam (SEQ rexp rexp')"

| "attach_rexp rexp' (Trn X rexp) = Trn X (SEQ rexp rexp')"

text {* 

  The following @{text "append_rhs_rexp rhs rexp"} attaches 

  @{text "rexp"} to every item in @{text "rhs"}.

  *}

definition

  "append_rhs_rexp rhs rexp \<equiv> (attach_rexp rexp) ` rhs"

text {*

  With the help of the two functions immediately above, Ardens'

  transformation on right hand side @{text "rhs"} is implemented

  by the following function @{text "arden_variate X rhs"}.

  After this transformation, the recursive occurent of @{text "X"}

  in @{text "rhs"} will be eliminated, while the 

  string set defined by @{text "rhs"} is kept unchanged.

  *}

definition 

  "arden_variate X rhs \<equiv> 

        append_rhs_rexp (rhs - items_of rhs X) (STAR (rexp_of rhs X))"

(*********** substitution of ES *************)

text {* 

  Suppose the equation defining @{text "X"} is $X = xrhs$,

  the purpose of @{text "rhs_subst"} is to substitute all occurences of @{text "X"} in

  @{text "rhs"} by @{text "xrhs"}.

  A litte thought may reveal that the final result

  should be: first append $(a_1 | a_2 | \ldots | a_n)$ to every item of @{text "xrhs"} and then

  union the result with all non-@{text "X"}-items of @{text "rhs"}.

 *}

definition 

  "rhs_subst rhs X xrhs \<equiv> 

        (rhs - (items_of rhs X)) \<union> (append_rhs_rexp xrhs (rexp_of rhs X))"

text {*

  Suppose the equation defining @{text "X"} is $X = xrhs$, the follwing

  @{text "eqs_subst ES X xrhs"} substitute @{text "xrhs"} into every equation

  of the equational system @{text "ES"}.

  *}

definition

  "eqs_subst ES X xrhs \<equiv> {(Y, rhs_subst yrhs X xrhs) | Y yrhs. (Y, yrhs) \<in> ES}"

text {*

  The computation of regular expressions for equivalent classes is accomplished

  using a iteration principle given by the following lemma.

  *}

lemma wf_iter [rule_format]: 

  fixes f

  assumes step: "\<And> e. \<lbrakk>P e; \<not> Q e\<rbrakk> \<Longrightarrow> (\<exists> e'. P e' \<and>  (f(e'), f(e)) \<in> less_than)"

  shows pe:     "P e \<longrightarrow> (\<exists> e'. P e' \<and>  Q e')"

proof(induct e rule: wf_induct 

           [OF wf_inv_image[OF wf_less_than, where f = "f"]], clarify)

  fix x 

  assume h [rule_format]: 

    "\<forall>y. (y, x) \<in> inv_image less_than f \<longrightarrow> P y \<longrightarrow> (\<exists>e'. P e' \<and> Q e')"

    and px: "P x"

  show "\<exists>e'. P e' \<and> Q e'"

  proof(cases "Q x")

    assume "Q x" with px show ?thesis by blast

  next

    assume nq: "\<not> Q x"

    from step [OF px nq]

    obtain e' where pe': "P e'" and ltf: "(f e', f x) \<in> less_than" by auto

    show ?thesis

    proof(rule h)

      from ltf show "(e', x) \<in> inv_image less_than f" 

	by (simp add:inv_image_def)

    next

      from pe' show "P e'" .

    qed

  qed

qed

text {*

  The @{text "P"} in lemma @{text "wf_iter"} is an invaiant kept throughout the iteration procedure.

  The particular invariant used to solve our problem is defined by function @{text "Inv(ES)"},

  an invariant over equal system @{text "ES"}.

  Every definition starting next till @{text "Inv"} stipulates a property to be satisfied by @{text "ES"}.

*}

text {* 

  Every variable is defined at most onece in @{text "ES"}.

  *}

definition 

  "distinct_equas ES \<equiv> 

            \<forall> X rhs rhs'. (X, rhs) \<in> ES \<and> (X, rhs') \<in> ES \<longrightarrow> rhs = rhs'"

text {* 

  Every equation in @{text "ES"} (represented by @{text "(X, rhs)"}) is valid, i.e. @{text "(X = L rhs)"}.

  *}

definition 

  "valid_eqns ES \<equiv> \<forall> X rhs. (X, rhs) \<in> ES \<longrightarrow> (X = L rhs)"

text {*

  The following @{text "rhs_nonempty rhs"} requires regular expressions occuring in transitional 

  items of @{text "rhs"} does not contain empty string. This is necessary for

  the application of Arden's transformation to @{text "rhs"}.

  *}

definition 

  "rhs_nonempty rhs \<equiv> (\<forall> Y r. Trn Y r \<in> rhs \<longrightarrow> [] \<notin> L r)"

text {*

  The following @{text "ardenable ES"} requires that Arden's transformation is applicable

  to every equation of equational system @{text "ES"}.