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(*<*)
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theory Paper
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imports "../Myhill" "LaTeXsugar"
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begin
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declare [[show_question_marks = false]]
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consts
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REL :: "(string \<times> string) \<Rightarrow> bool"
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UPLUS :: "'a set \<Rightarrow> 'a set \<Rightarrow> (nat \<times> 'a) set"
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abbreviation
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"EClass x R \<equiv> R `` {x}"
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notation (latex output)
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str_eq_rel ("\<approx>\<^bsub>_\<^esub>") and
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str_eq ("_ \<approx>\<^bsub>_\<^esub> _") and
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Seq (infixr "\<cdot>" 100) and
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Star ("_\<^bsup>\<star>\<^esup>") and
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pow ("_\<^bsup>_\<^esup>" [100, 100] 100) and
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Suc ("_+1" [100] 100) and
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quotient ("_ \<^raw:\ensuremath{\!\sslash\!}> _" [90, 90] 90) and
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REL ("\<approx>") and
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UPLUS ("_ \<^raw:\ensuremath{\uplus}> _" [90, 90] 90) and
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L ("L'(_')" [0] 101) and
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Lam ("\<lambda>'(_')" [100] 100) and
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Trn ("_, _" [100, 100] 100) and
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EClass ("\<lbrakk>_\<rbrakk>\<^bsub>_\<^esub>" [100, 100] 100) and
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transition ("_ \<^raw:\ensuremath{\stackrel{\text{>_\<^raw:}}{\Longmapsto}}> _" [100, 100, 100] 100)
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(*>*)
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section {* Introduction *}
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text {*
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Regular languages are an important and well-understood subject in Computer
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Science, with many beautiful theorems and many useful algorithms. There is a
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wide range of textbooks on this subject, many of which are aimed at students
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and contain very detailed ``pencil-and-paper'' proofs
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(e.g.~\cite{Kozen97}). It seems natural to exercise theorem provers by
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formalising these theorems and by verifying formally the algorithms.
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There is however a problem: the typical approach to regular languages is to
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introduce finite automata and then define everything in terms of them. For
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example, a regular language is normally defined as one whose strings are
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recognised by a finite deterministic automaton. This approach has many
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benefits. Among them is the fact that it is easy to convince oneself that
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regular languages are closed under complementation: one just has to exchange
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the accepting and non-accepting states in the corresponding automaton to
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obtain an automaton for the complement language. The problem, however, lies with
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formalising such reasoning in a HOL-based theorem prover, in our case
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Isabelle/HOL. Automata are build up from states and transitions that
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need to be represented as graphs or matrices, neither
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of which can be defined as inductive datatype.\footnote{In some works
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functions are used to represent state transitions, but also they are not
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inductive datatypes.} This means we have to build our own reasoning
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infrastructure for them, as neither Isabelle/HOL nor HOL4 nor HOLlight support
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them with libraries.
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Even worse, reasoning about graphs and matrices can be a real hassle in HOL-based
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theorem provers. Consider for example the operation of sequencing
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two automata, say $A_1$ and $A_2$, by connecting the
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accepting states of $A_1$ to the initial state of $A_2$:
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\begin{center}
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\begin{tabular}{ccc}
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\begin{tikzpicture}[scale=0.8]
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%\draw[step=2mm] (-1,-1) grid (1,1);
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\draw[rounded corners=1mm, very thick] (-1.0,-0.3) rectangle (-0.2,0.3);
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\draw[rounded corners=1mm, very thick] ( 0.2,-0.3) rectangle ( 1.0,0.3);
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\node (A) at (-1.0,0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\node (B) at ( 0.2,0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\node (C) at (-0.2, 0.13) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\node (D) at (-0.2,-0.13) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\node (E) at (1.0, 0.2) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\node (F) at (1.0,-0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\node (G) at (1.0,-0.2) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\draw (-0.6,0.0) node {\footnotesize$A_1$};
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\draw ( 0.6,0.0) node {\footnotesize$A_2$};
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\end{tikzpicture}
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&
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\raisebox{1.1mm}{\bf\Large$\;\;\;\Rightarrow\,\;\;$}
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&
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\begin{tikzpicture}[scale=0.8]
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%\draw[step=2mm] (-1,-1) grid (1,1);
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\draw[rounded corners=1mm, very thick] (-1.0,-0.3) rectangle (-0.2,0.3);
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\draw[rounded corners=1mm, very thick] ( 0.2,-0.3) rectangle ( 1.0,0.3);
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\node (A) at (-1.0,0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\node (B) at ( 0.2,0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\node (C) at (-0.2, 0.13) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\node (D) at (-0.2,-0.13) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\node (E) at (1.0, 0.2) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\node (F) at (1.0,-0.0) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\node (G) at (1.0,-0.2) [circle, very thick, draw, fill=white, inner sep=0.4mm] {};
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\draw (C) to [very thick, bend left=45] (B);
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\draw (D) to [very thick, bend right=45] (B);
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\draw (-0.6,0.0) node {\footnotesize$A_1$};
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\draw ( 0.6,0.0) node {\footnotesize$A_2$};
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\end{tikzpicture}
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\end{tabular}
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\end{center}
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\noindent
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On ``paper'' we can define the corresponding graph in terms of the disjoint
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union of the state nodes. Unfortunately in HOL, the definition for disjoint
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union, namely
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\begin{center}
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@{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}"}
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\end{center}
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\noindent
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changes the type---the disjoint union is not a set, but a set of pairs.
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Using this definition for disjoint unions means we do not have a single type for automata
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and hence will not be able to state properties about \emph{all}
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automata, since there is no type quantification available in HOL. An
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alternative, which provides us with a single type for automata, is to give every
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state node an identity, for example a natural
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number, and then be careful to rename these identities apart whenever
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connecting two automata. This results in clunky proofs
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establishing that properties are invariant under renaming. Similarly,
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connecting two automata represented as matrices results in very adhoc
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constructions, which are not pleasant to reason about.
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Because of these problems to do with representing automata, there seems
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to be no substantial formalisation of automata theory and regular languages
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carried out in a HOL-based theorem prover. We are only aware of the
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large formalisation of automata theory in Nuprl \cite{Constable00} and
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some smaller formalisations in Coq (for example \cite{Filliatre97}).
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In this paper, we will not attempt to formalise automata theory, but take a completely
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different approach to regular languages. Instead of defining a regular language as one
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where there exists an automaton that recognises all strings of the language, we define
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a regular language as:
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\begin{definition}[A Regular Language]
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A language @{text A} is \emph{regular}, provided there is a regular expression that matches all
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strings of @{text "A"}.
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\end{definition}
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\noindent
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The reason is that regular expressions, unlike graphs and matrices, can
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be easily defined as inductive datatype. Consequently a corresponding reasoning
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infrastructure comes for free. This has recently been exploited in HOL4 with a formalisation
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of regular expression matching based on derivatives \cite{OwensSlind08}. The purpose of this paper is to
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show that a central result about regular languages---the Myhill-Nerode theorem---can
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be recreated by only using regular expressions. This theorem gives necessary
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and sufficient conditions for when a language is regular. As a corollary of this
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theorem we can easily establish the usual closure properties, including
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complementation, for regular languages.\smallskip
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\noindent
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{\bf Contributions:} To our knowledge, our proof of the Myhill-Nerode theorem is the
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first that is based on regular expressions, only. We prove the part of this theorem
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stating that a regular expression has only finitely many partitions using certain
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tagging-functions. Again to our best knowledge, these tagging functions have
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not been used before to establish the Myhill-Nerode theorem.
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*}
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section {* Preliminaries *}
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text {*
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Strings in Isabelle/HOL are lists of characters with the \emph{empty string}
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being represented by the empty list, written @{term "[]"}. \emph{Languages}
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are sets of strings. The language containing all strings is written in
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Isabelle/HOL as @{term "UNIV::string set"}. The concatenation of two languages
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is written @{term "A ;; B"} and a language raised to the power $n$ is written
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@{term "A \<up> n"}. Their definitions are
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\begin{center}
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@{thm Seq_def[THEN eq_reflection, where A1="A" and B1="B"]}
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\hspace{7mm}
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@{thm pow.simps(1)[THEN eq_reflection, where A1="A"]}
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\hspace{7mm}
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@{thm pow.simps(2)[THEN eq_reflection, where A1="A" and n1="n"]}
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\end{center}
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\noindent
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where @{text "@"} is the usual list-append operation. The Kleene-star of a language @{text A}
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is defined as the union over all powers, namely @{thm Star_def}. In the paper
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we will often make use of the following properties.
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\begin{proposition}\label{langprops}\mbox{}\\
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\begin{tabular}{@ {}ll@ {\hspace{10mm}}ll}
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(i) & @{thm star_cases} & (ii) & @{thm[mode=IfThen] pow_length}\\
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(iii) & @{thm seq_Union_left} &
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\end{tabular}
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\end{proposition}
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\noindent
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We omit the proofs of these properties, but invite the reader to consult
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our formalisation.\footnote{Available at ???}
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The notation for the quotient of a language @{text A} according to an
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equivalence relation @{term REL} is @{term "A // REL"}. We will write
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@{text "\<lbrakk>x\<rbrakk>\<^isub>\<approx>"} for the equivalence class defined
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as @{text "{y | y \<approx> x}"}.
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Central to our proof will be the solution of equational systems
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involving sets of languages. For this we will use Arden's lemma \cite{Brzozowski64}
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which solves equations of the form @{term "X = A ;; X \<union> B"} provided
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@{term "[] \<notin> A"}. However we will need the following ``reverse''
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version of Arden's lemma.
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\begin{lemma}[Reverse Arden's Lemma]\label{arden}\mbox{}\\
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If @{thm (prem 1) ardens_revised} then
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@{thm (lhs) ardens_revised} has the unique solution
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@{thm (rhs) ardens_revised}.
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\end{lemma}
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\begin{proof}
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For the right-to-left direction we assume @{thm (rhs) ardens_revised} and show
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that @{thm (lhs) ardens_revised} holds. From Prop.~\ref{langprops}@{text "(i)"}
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we have @{term "A\<star> = {[]} \<union> A ;; A\<star>"},
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which is equal to @{term "A\<star> = {[]} \<union> A\<star> ;; A"}. Adding @{text B} to both
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sides gives @{term "B ;; A\<star> = B ;; ({[]} \<union> A\<star> ;; A)"}, whose right-hand side
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is equal to @{term "(B ;; A\<star>) ;; A \<union> B"}. This completes this direction.
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For the other direction we assume @{thm (lhs) ardens_revised}. By a simple induction
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on @{text n}, we can establish the property
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\begin{center}
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@{text "(*)"}\hspace{5mm} @{thm (concl) ardens_helper}
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\end{center}
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\noindent
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Using this property we can show that @{term "B ;; (A \<up> n) \<subseteq> X"} holds for
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all @{text n}. From this we can infer @{term "B ;; A\<star> \<subseteq> X"} using the definition
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of @{text "\<star>"}.
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For the inclusion in the other direction we assume a string @{text s}
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with length @{text k} is element in @{text X}. Since @{thm (prem 1) ardens_revised}
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we know by Prop.~\ref{langprops}@{text "(ii)"} that
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@{term "s \<notin> X ;; (A \<up> Suc k)"} since its length is only @{text k}
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(the strings in @{term "X ;; (A \<up> Suc k)"} are all longer).
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From @{text "(*)"} it follows then that
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@{term s} must be element in @{term "(\<Union>m\<in>{0..k}. B ;; (A \<up> m))"}. This in turn
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implies that @{term s} is in @{term "(\<Union>n. B ;; (A \<up> n))"}. Using Prop.~\ref{langprops}@{text "(iii)"}
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this is equal to @{term "B ;; A\<star>"}, as we needed to show.\qed
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\end{proof}
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\noindent
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Regular expressions are defined as the following inductive datatype
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\begin{center}
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@{text r} @{text "::="}
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@{term NULL}\hspace{1.5mm}@{text"|"}\hspace{1.5mm}
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@{term EMPTY}\hspace{1.5mm}@{text"|"}\hspace{1.5mm}
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@{term "CHAR c"}\hspace{1.5mm}@{text"|"}\hspace{1.5mm}
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@{term "SEQ r r"}\hspace{1.5mm}@{text"|"}\hspace{1.5mm}
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@{term "ALT r r"}\hspace{1.5mm}@{text"|"}\hspace{1.5mm}
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@{term "STAR r"}
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\end{center}
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\noindent
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The language matched by a regular expression is defined as usual:
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\begin{center}
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\begin{tabular}{c@ {\hspace{10mm}}c}
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\begin{tabular}{rcl}
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@{thm (lhs) L_rexp.simps(1)} & @{text "\<equiv>"} & @{thm (rhs) L_rexp.simps(1)}\\
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@{thm (lhs) L_rexp.simps(2)} & @{text "\<equiv>"} & @{thm (rhs) L_rexp.simps(2)}\\
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@{thm (lhs) L_rexp.simps(3)[where c="c"]} & @{text "\<equiv>"} & @{thm (rhs) L_rexp.simps(3)[where c="c"]}\\
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\end{tabular}
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&
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\begin{tabular}{rcl}
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@{thm (lhs) L_rexp.simps(4)[where ?r1.0="r\<^isub>1" and ?r2.0="r\<^isub>2"]} & @{text "\<equiv>"} &
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@{thm (rhs) L_rexp.simps(4)[where ?r1.0="r\<^isub>1" and ?r2.0="r\<^isub>2"]}\\
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@{thm (lhs) L_rexp.simps(5)[where ?r1.0="r\<^isub>1" and ?r2.0="r\<^isub>2"]} & @{text "\<equiv>"} &
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@{thm (rhs) L_rexp.simps(5)[where ?r1.0="r\<^isub>1" and ?r2.0="r\<^isub>2"]}\\
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@{thm (lhs) L_rexp.simps(6)[where r="r"]} & @{text "\<equiv>"} &
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@{thm (rhs) L_rexp.simps(6)[where r="r"]}\\
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\end{tabular}
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\end{tabular}
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\end{center}
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*}
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section {* Finite Partitions Imply Regularity of a Language *}
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text {*
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The key definition in the Myhill-Nerode theorem is the
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\emph{Myhill-Nerode relation}, which states that w.r.t.~a language two
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strings are related, provided there is no distinguishing extension in this
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language. This can be defined as:
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\begin{definition}[Myhill-Nerode Relation]\mbox{}\\
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@{thm str_eq_def[simplified str_eq_rel_def Pair_Collect]}
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\end{definition}
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\noindent
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It is easy to see that @{term "\<approx>A"} is an equivalence relation, which
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partitions the set of all strings, @{text "UNIV"}, into a set of disjoint
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equivalence classes. One direction of the Myhill-Nerode theorem establishes
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that if there are finitely many equivalence classes, then the language is
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regular. In our setting we therefore have to show:
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\begin{theorem}\label{myhillnerodeone}
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@{thm[mode=IfThen] hard_direction}
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\end{theorem}
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\noindent
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To prove this theorem, we define the set @{term "finals A"} as those equivalence
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classes that contain strings of @{text A}, namely
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%
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\begin{equation}
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@{thm finals_def}
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\end{equation}
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\noindent
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It is straightforward to show that @{thm lang_is_union_of_finals} and
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@{thm finals_in_partitions} hold.
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Therefore if we know that there exists a regular expression for every
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equivalence class in @{term "finals A"} (which by assumption must be
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a finite set), then we can combine these regular expressions with @{const ALT}
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and obtain a regular expression that matches every string in @{text A}.
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We prove Thm.~\ref{myhillnerodeone} by giving a method that can calculate a
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regular expression for \emph{every} equivalence class, not just the ones
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in @{term "finals A"}. We
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first define a notion of \emph{transition} between equivalence classes
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%
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\begin{equation}
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@{thm transition_def}
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\end{equation}
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\noindent
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which means that if we concatenate all strings matching the regular expression @{text r}
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to the end of all strings in the equivalence class @{text Y}, we obtain a subset of
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@{text X}. Note that we do not define an automaton here, we merely relate two sets
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(w.r.t.~a regular expression).
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Next we build an equational system that
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contains an equation for each equivalence class. Suppose we have
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the equivalence classes @{text "X\<^isub>1,\<dots>,X\<^isub>n"}, there must be one and only one that
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contains the empty string @{text "[]"} (since equivalence classes are disjoint).
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Let us assume @{text "[] \<in> X\<^isub>1"}. We build the following equational system
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\begin{center}
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\begin{tabular}{rcl}
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@{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)"} \\
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@{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)"} \\
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& $\vdots$ \\
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@{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)"}\\
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\end{tabular}
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\end{center}
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75
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\noindent
+ − 368
where the pairs @{text "(Y\<^isub>i\<^isub>j, r\<^isub>i\<^isub>j)"} stand for all transitions
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@{term "Y\<^isub>i\<^isub>j \<Turnstile>r\<^isub>i\<^isub>j\<Rightarrow> X\<^isub>i"}. The term @{text "\<lambda>(EMPTY)"} acts as a marker for the equivalence
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class containing @{text "[]"}. (Note that we mark, roughly speaking, the
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single ``initial'' state in the equational system, which is different from
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the method by Brzozowski \cite{Brzozowski64}, since for his purposes he needs to mark
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the ``terminal'' states.) Overloading the function @{text L} for the two kinds of terms in the
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equational system as follows
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\begin{center}
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@{thm L_rhs_e.simps(2)[where X="Y" and r="r", THEN eq_reflection]}\hspace{10mm}
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@{thm L_rhs_e.simps(1)[where r="r", THEN eq_reflection]}
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\end{center}
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\noindent
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we can prove for @{text "X\<^isub>2\<^isub>.\<^isub>.\<^isub>n"} that the following equations
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%
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\begin{equation}\label{inv1}
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@{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)"}.
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\end{equation}
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\noindent
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hold. Similarly for @{text "X\<^isub>1"} we can show the following equation
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%
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\begin{equation}\label{inv2}
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@{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))"}.
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\end{equation}
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\noindent
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The reason for adding the @{text \<lambda>}-marker to our equational system is
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to obtain this equation, which only holds in this form since none of
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the other terms contain the empty string.
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Our proof of Thm.~\ref{myhillnerodeone}
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will be by transforming the equational system into a \emph{solved form}
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maintaining the invariants \eqref{inv1} and \eqref{inv2}. From the solved
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form we will be able to read off the regular expressions using our
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variant of Arden's Lemma (Lem.~\ref{arden}).
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54
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*}
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section {* Regular Expressions Generate Finitely Many Partitions *}
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text {*
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54
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\begin{theorem}
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Given @{text "r"} is a regular expressions, then @{thm rexp_imp_finite}.
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\end{theorem}
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\begin{proof}
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By induction on the structure of @{text r}. The cases for @{const NULL}, @{const EMPTY}
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and @{const CHAR} are straightforward, because we can easily establish
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\begin{center}
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\begin{tabular}{l}
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@{thm quot_null_eq}\\
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@{thm quot_empty_subset}\\
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@{thm quot_char_subset}
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\end{tabular}
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\end{center}
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\end{proof}
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*}
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54
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section {* Conclusion and Related Work *}
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24
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(*<*)
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end
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(*>*)