(*<*)

theory Paper

imports "../Myhill" "LaTeXsugar"

begin

declare [[show_question_marks = false]]

consts

 REL :: "(string \<times> string) \<Rightarrow> bool"

 UPLUS :: "'a set \<Rightarrow> 'a set \<Rightarrow> (nat \<times> 'a) set"

abbreviation

  "EClass x R \<equiv> R `` {x}"

notation (latex output)

  str_eq_rel ("\<approx>\<^bsub>_\<^esub>") and

  str_eq ("_ \<approx>\<^bsub>_\<^esub> _") and

  Seq (infixr "\<cdot>" 100) and

  Star ("_\<^bsup>\<star>\<^esup>") and

  pow ("_\<^bsup>_\<^esup>" [100, 100] 100) and

  Suc ("_+1" [100] 100) and

  quotient ("_ \<^raw:\ensuremath{\!\sslash\!}> _" [90, 90] 90) and

  REL ("\<approx>") and

  UPLUS ("_ \<^raw:\ensuremath{\uplus}> _" [90, 90] 90) and

  Lam ("\<lambda>'(_')" [100] 100) and 

  Trn ("_, _" [100, 100] 100) and 

  EClass ("\<lbrakk>_\<rbrakk>\<^bsub>_\<^esub>" [100, 100] 100) and

  transition ("_ \<^raw:\ensuremath{\stackrel{\text{>_\<^raw:}}{\Longmapsto}}> _" [100, 100, 100] 100)

(*>*)

section {* Introduction *}

text {*

  Regular languages are an important and well-understood subject in Computer

  Science, with many beautiful theorems and many useful algorithms. There is a

  wide range of textbooks on this subject, many of which are aimed at students

  and contain very detailed ``pencil-and-paper'' proofs

  (e.g.~\cite{Kozen97}). It seems natural to exercise theorem provers by

  formalising these theorems and by verifying formally the algorithms.

  There is however a problem: the typical approach to regular languages is to

  introduce finite automata and then define everything in terms of them.  For

  example, a regular language is normally defined as one whose strings are

  recognised by a finite deterministic automaton. This approach has many

  benefits. Among them is the fact that it is easy to convince oneself that

  regular languages are closed under complementation: one just has to exchange

  the accepting and non-accepting states in the corresponding automaton to

  obtain an automaton for the complement language.  The problem, however, lies with

  formalising such reasoning in a HOL-based theorem prover, in our case

  Isabelle/HOL. Automata are build up from states and transitions that 

  \begin{center}

  \begin{tabular}{ccc}

  \begin{tikzpicture}[scale=0.8]

  %\draw[step=2mm] (-1,-1) grid (1,1);

  \draw[rounded corners=1mm, very thick] (-1.0,-0.3) rectangle (-0.2,0.3);

  \draw[rounded corners=1mm, very thick] ( 0.2,-0.3) rectangle ( 1.0,0.3);

  \node (A) at (-1.0,0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \node (B) at ( 0.2,0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \node (C) at (-0.2, 0.13) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \node (D) at (-0.2,-0.13) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \node (E) at (1.0, 0.2) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \node (F) at (1.0,-0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \node (G) at (1.0,-0.2) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \draw (-0.6,0.0) node {\footnotesize$A_1$};

  \draw ( 0.6,0.0) node {\footnotesize$A_2$};

  \end{tikzpicture}

  & 

  \raisebox{1.1mm}{\bf\Large$\;\;\;\Rightarrow\,\;\;$}

  &

  \begin{tikzpicture}[scale=0.8]

  %\draw[step=2mm] (-1,-1) grid (1,1);

  \draw[rounded corners=1mm, very thick] (-1.0,-0.3) rectangle (-0.2,0.3);

  \draw[rounded corners=1mm, very thick] ( 0.2,-0.3) rectangle ( 1.0,0.3);

  \node (A) at (-1.0,0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \node (B) at ( 0.2,0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \node (C) at (-0.2, 0.13) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \node (D) at (-0.2,-0.13) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \node (E) at (1.0, 0.2) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \node (F) at (1.0,-0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \node (G) at (1.0,-0.2) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};

  \draw (C) to [very thick, bend left=45] (B);

  \draw (D) to [very thick, bend right=45] (B);

  \draw (-0.6,0.0) node {\footnotesize$A_1$};

  \draw ( 0.6,0.0) node {\footnotesize$A_2$};

  \end{tikzpicture}

  \end{tabular}

  \end{center}

  \noindent

  On ``paper'' we can define the corresponding graph in terms of the disjoint 

  union of the state nodes. Unfortunately in HOL, the definition for disjoint 

  union, namely 

  @{term "UPLUS A\<^isub>1 A\<^isub>2 \<equiv> {(1, x) | x. x \<in> A\<^isub>1} \<union> {(2, y) | y. y \<in> A\<^isub>2}"}

  \noindent

  changes the type---the disjoint union is not a set, but a set of pairs. 

  Using this definition for disjoint unions means we do not have a single type for automata

  and hence will not be able to state properties about \emph{all}

  automata, since there is no type quantification available in HOL. An

  alternative, which provides us with a single type for automata, is to give every 

  state node an identity, for example a natural

  number, and then be careful to rename these identities apart whenever

  connecting two automata. This results in clunky proofs

  establishing that properties are invariant under renaming. Similarly,

  connecting two automata represented as matrices results in very adhoc

  constructions, which are not pleasant to reason about.

  Because of these problems to do with representing automata, there seems

  to be no substantial formalisation of automata theory and regular languages 

  A language @{text A} is \emph{regular}, provided there is a regular expression that matches all

  strings of @{text "A"}.

  \end{definition}

  \noindent

  The reason is that regular expressions, unlike graphs and matrices, can

  be easily defined as inductive datatype. Consequently a corresponding reasoning 

  infrastructure comes for free. This has recently been exploited in HOL4 with a formalisation

  of regular expression matching based on derivatives \cite{OwensSlind08}.  The purpose of this paper is to

  show that a central result about regular languages---the Myhill-Nerode theorem---can 

  be recreated by only using regular expressions. This theorem gives necessary

  and sufficient conditions for when a language is regular. As a corollary of this

  theorem we can easily establish the usual closure properties, including 

  complementation, for regular languages.\smallskip

  \noindent

  {\bf Contributions:} To our knowledge, our proof of the Myhill-Nerode theorem is the

  first that is based on regular expressions, only. We prove the part of this theorem 

  stating that a regular expression has only finitely many partitions using certain 

  tagging-functions. Again to our best knowledge, these tagging functions have

  not been used before to establish the Myhill-Nerode theorem.

*}

section {* Preliminaries *}

text {*

  Strings in Isabelle/HOL are lists of characters with the \emph{empty string}

  being represented by the empty list, written @{term "[]"}. \emph{Languages}

  are sets of strings. The language containing all strings is written in

  Isabelle/HOL as @{term "UNIV::string set"}. The concatenation of two languages 

  is written @{term "A ;; B"} and a language raised to the power $n$ is written 

  @{term "A \<up> n"}. Their definitions are

  \begin{center}

  @{thm Seq_def[THEN eq_reflection, where A1="A" and B1="B"]}

  \hspace{7mm}

  @{thm pow.simps(1)[THEN eq_reflection, where A1="A"]}

  \hspace{7mm}

  @{thm pow.simps(2)[THEN eq_reflection, where A1="A" and n1="n"]}

  \end{center}

  \noindent

  where @{text "@"} is the usual list-append operation. The Kleene-star of a language @{text A}

  is defined as the union over all powers, namely @{thm Star_def}. In the paper

  we will often make use of the following properties.

  \begin{proposition}\label{langprops}\mbox{}\\

  \begin{tabular}{@ {}ll@ {\hspace{10mm}}ll}

  (i)   & @{thm star_cases}      & (ii)  & @{thm[mode=IfThen] pow_length}\\

  (iii) & @{thm seq_Union_left}  & 

  \end{tabular}

  \end{proposition}

  \noindent

  We omit the proofs of these properties, but invite the reader to consult

  our formalisation.\footnote{Available at ???}

  The notation for the quotient of a language @{text A} according to an 

  equivalence relation @{term REL} is @{term "A // REL"}. We will write 

  @{text "\<lbrakk>x\<rbrakk>\<^isub>\<approx>"} for the equivalence class defined 

  as @{text "{y | y \<approx> x}"}.

  Central to our proof will be the solution of equational systems

  involving sets of languages. For this we will use Arden's lemma \cite{Brzozowski64}

  which solves equations of the form @{term "X = A ;; X \<union> B"} provided

  @{term "[] \<notin> A"}. However we will need the following ``reverse'' 

  version of Arden's lemma.

  \begin{lemma}[Reverse Arden's Lemma]\label{arden}\mbox{}\\

  If @{thm (prem 1) ardens_revised} then

  @{thm (lhs) ardens_revised} has the unique solution

  @{thm (rhs) ardens_revised}.

  \end{lemma}

  \begin{proof}

  For the right-to-left direction we assume @{thm (rhs) ardens_revised} and show

  that @{thm (lhs) ardens_revised} holds. From Prop.~\ref{langprops}@{text "(i)"} 

  we have @{term "A\<star> = {[]} \<union> A ;; A\<star>"},

  which is equal to @{term "A\<star> = {[]} \<union> A\<star> ;; A"}. Adding @{text B} to both 

  sides gives @{term "B ;; A\<star> = B ;; ({[]} \<union> A\<star> ;; A)"}, whose right-hand side

  is equal to @{term "(B ;; A\<star>) ;; A \<union> B"}. This completes this direction. 

  For the other direction we assume @{thm (lhs) ardens_revised}. By a simple induction

  on @{text n}, we can establish the property

  \begin{center}

  @{text "(*)"}\hspace{5mm} @{thm (concl) ardens_helper}

  \end{center}

  \noindent

  Using this property we can show that @{term "B ;; (A \<up> n) \<subseteq> X"} holds for

  all @{text n}. From this we can infer @{term "B ;; A\<star> \<subseteq> X"} using the definition

  of @{text "\<star>"}.

  For the inclusion in the other direction we assume a string @{text s}

  with length @{text k} is element in @{text X}. Since @{thm (prem 1) ardens_revised}

  we know by Prop.~\ref{langprops}@{text "(ii)"} that 

  @{term "s \<notin> X ;; (A \<up> Suc k)"} since its length is only @{text k}

  (the strings in @{term "X ;; (A \<up> Suc k)"} are all longer). 

  From @{text "(*)"} it follows then that

  @{term s} must be element in @{term "(\<Union>m\<in>{0..k}. B ;; (A \<up> m))"}. This in turn

  implies that @{term s} is in @{term "(\<Union>n. B ;; (A \<up> n))"}. Using Prop.~\ref{langprops}@{text "(iii)"} 

  this is equal to @{term "B ;; A\<star>"}, as we needed to show.\qed

  \end{proof}

  \noindent

  Regular expressions are defined as the following inductive datatype

  \begin{center}

  @{text r} @{text "::="}

  @{term NULL}\hspace{1.5mm}@{text"|"}\hspace{1.5mm} 

  @{term EMPTY}\hspace{1.5mm}@{text"|"}\hspace{1.5mm} 

  @{term "CHAR c"}\hspace{1.5mm}@{text"|"}\hspace{1.5mm} 

  @{term "SEQ r r"}\hspace{1.5mm}@{text"|"}\hspace{1.5mm} 

  @{term "ALT r r"}\hspace{1.5mm}@{text"|"}\hspace{1.5mm} 

  @{term "STAR r"}

  \end{center}

  \noindent

  \begin{center}

  \begin{tabular}{c@ {\hspace{10mm}}c}

  \begin{tabular}{rcl}

  @{thm (lhs) L_rexp.simps(1)} & @{text "\<equiv>"} & @{thm (rhs) L_rexp.simps(1)}\\

  @{thm (lhs) L_rexp.simps(2)} & @{text "\<equiv>"} & @{thm (rhs) L_rexp.simps(2)}\\

  @{thm (lhs) L_rexp.simps(3)[where c="c"]} & @{text "\<equiv>"} & @{thm (rhs) L_rexp.simps(3)[where c="c"]}\\

  \end{tabular}

  &

  \begin{tabular}{rcl}

  @{thm (lhs) L_rexp.simps(4)[where ?r1.0="r\<^isub>1" and ?r2.0="r\<^isub>2"]} & @{text "\<equiv>"} &

       @{thm (rhs) L_rexp.simps(4)[where ?r1.0="r\<^isub>1" and ?r2.0="r\<^isub>2"]}\\

  @{thm (lhs) L_rexp.simps(5)[where ?r1.0="r\<^isub>1" and ?r2.0="r\<^isub>2"]} & @{text "\<equiv>"} &

       @{thm (rhs) L_rexp.simps(5)[where ?r1.0="r\<^isub>1" and ?r2.0="r\<^isub>2"]}\\

  @{thm (lhs) L_rexp.simps(6)[where r="r"]} & @{text "\<equiv>"} &

      @{thm (rhs) L_rexp.simps(6)[where r="r"]}\\

  \end{tabular}

  \end{tabular}

  \end{center}

*}

section {* Finite Partitions Imply Regularity of a Language *}

text {*

  The key definition in the Myhill-Nerode theorem is the

  \emph{Myhill-Nerode relation}, which states that w.r.t.~a language two 

  strings are related, provided there is no distinguishing extension in this

  language. This can be defined as:

  \begin{definition}[Myhill-Nerode Relation]\mbox{}\\

  @{thm str_eq_def[simplified str_eq_rel_def Pair_Collect]}

  \end{definition}

  \noindent

  It is easy to see that @{term "\<approx>A"} is an equivalence relation, which

  partitions the set of all strings, @{text "UNIV"}, into a set of disjoint

  equivalence classes. One direction of the Myhill-Nerode theorem establishes 

  that if there are finitely many equivalence classes, then the language is 

  regular. In our setting we therefore have to show:

  \begin{theorem}\label{myhillnerodeone}

  @{thm[mode=IfThen] hard_direction}

  \end{theorem}

  \noindent

  To prove this theorem, we define the set @{term "finals A"} as those equivalence

  classes that contain strings of @{text A}, namely

  %

  \begin{equation} 

  @{thm finals_def}

  \end{equation}

  \noindent

  It is straightforward to show that @{thm lang_is_union_of_finals} and 

  @{thm finals_in_partitions} hold. 

  Therefore if we know that there exists a regular expression for every

  equivalence class in @{term "finals A"} (which by assumption must be

  a finite set), then we can combine these regular expressions with @{const ALT}

  and obtain a regular expression that matches every string in @{text A}.

  We prove Thm.~\ref{myhillnerodeone} by giving a method that can calculate a

  regular expression for \emph{every} equivalence class, not just the ones 

  in @{term "finals A"}. We

  first define a notion of \emph{transition} between equivalence classes

  %

  \begin{equation} 

  @{thm transition_def}

  \end{equation}

  \noindent

  which means that if we concatenate all strings matching the regular expression @{text r} 

  to the end of all strings in the equivalence class @{text Y}, we obtain a subset of 

  @{text X}. Note that we do not define an automaton here, we merely relate two sets

  (w.r.t.~a regular expression). 

  Next we build an equational system that

  contains an equation for each equivalence class. Suppose we have 

  the equivalence classes @{text "X\<^isub>1,\<dots>,X\<^isub>n"}, there must be one and only one that

  contains the empty string @{text "[]"} (since equivalence classes are disjoint).

  Let us assume @{text "[] \<in> X\<^isub>1"}. We build the following equational system

  \begin{center}

  \begin{tabular}{rcl}

  @{text "X\<^isub>1"} & @{text "="} & @{text "(Y\<^isub>1\<^isub>1, CHAR c\<^isub>1\<^isub>1) + \<dots> + (Y\<^isub>1\<^isub>p, CHAR c\<^isub>1\<^isub>p) + \<lambda>(EMPTY)"} \\

  @{text "X\<^isub>2"} & @{text "="} & @{text "(Y\<^isub>2\<^isub>1, CHAR c\<^isub>2\<^isub>1) + \<dots> + (Y\<^isub>2\<^isub>o, CHAR c\<^isub>2\<^isub>o)"} \\

  & $\vdots$ \\

  @{text "X\<^isub>n"} & @{text "="} & @{text "(Y\<^isub>n\<^isub>1, CHAR c\<^isub>n\<^isub>1) + \<dots> + (Y\<^isub>n\<^isub>q, CHAR c\<^isub>n\<^isub>q)"}\\

  \end{tabular}

  \end{center}

  \noindent

  class containing @{text "[]"}. (Note that we mark, roughly speaking, the

  single ``initial'' state in the equational system, which is different from

  the method by Brzozowski \cite{Brzozowski64}, since for his purposes he needs to mark 

  the ``terminal'' states.) Overloading the function @{text L} for the two kinds of terms in the 

  equational system as follows

  \begin{center}

  @{thm L_rhs_e.simps(2)[where X="Y" and r="r", THEN eq_reflection]}\hspace{10mm}

  @{thm L_rhs_e.simps(1)[where r="r", THEN eq_reflection]}

  \end{center}

  \noindent

  we can prove for @{text "X\<^isub>2\<^isub>.\<^isub>.\<^isub>n"} that the following equations

  %

  \begin{equation}\label{inv1}

  @{text "X\<^isub>i = L(Y\<^isub>i\<^isub>1, CHAR c\<^isub>i\<^isub>1) \<union> \<dots> \<union> L(Y\<^isub>i\<^isub>q, CHAR c\<^isub>i\<^isub>q)"}.

  \end{equation}

  \noindent

  hold. Similarly for @{text "X\<^isub>1"} we can show the following equation

  %

  \begin{equation}\label{inv2}

  @{text "X\<^isub>1 = L(Y\<^isub>i\<^isub>1, CHAR c\<^isub>i\<^isub>1) \<union> \<dots> \<union> L(Y\<^isub>i\<^isub>p, CHAR c\<^isub>i\<^isub>p) \<union> L(\<lambda>(EMPTY))"}.

  \end{equation}

  \noindent

  The reason for adding the @{text \<lambda>}-marker to our equational system is 

  to obtain this equation, which only holds in this form since none of 

  the other terms contain the empty string. 

  Our proof of Thm.~\ref{myhillnerodeone}

  will be by transforming the equational system into a \emph{solved form}

  maintaining the invariants \eqref{inv1} and \eqref{inv2}. From the solved

  form we will be able to read off the regular expressions using our 

  variant of Arden's Lemma (Lem.~\ref{arden}).

*}

section {* Regular Expressions Generate Finitely Many Partitions *}

text {*

  \begin{theorem}

  Given @{text "r"} is a regular expressions, then @{thm rexp_imp_finite}.

  \end{theorem}  

  \begin{proof}

  By induction on the structure of @{text r}. The cases for @{const NULL}, @{const EMPTY}

  and @{const CHAR} are straightforward, because we can easily establish

  \begin{center}

  \begin{tabular}{l}

  @{thm quot_null_eq}\\

  @{thm quot_empty_subset}\\

  @{thm quot_char_subset}

  \end{tabular}

  \end{center}

  \end{proof}

*}

section {* Conclusion and Related Work *}

(*<*)

end

(*>*)