theory Myhill+
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  imports Myhill_2+
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begin+
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+
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section {* Preliminaries *}+
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+
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subsection {* Finite automata and \mht  \label{sec_fa_mh} *}+
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+
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text {*+
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+
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A {\em determinisitc finite automata (DFA)} $M$ is a 5-tuple +
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$(Q, \Sigma, \delta, s, F)$, where:+
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\begin{enumerate}+
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  \item $Q$ is a finite set of {\em states}, also denoted $Q_M$.+
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  \item $\Sigma$ is a finite set of {\em alphabets}, also denoted $\Sigma_M$.+
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  \item $\delta$ is a {\em transition function} of type @{text "Q \<times> \<Sigma> \<Rightarrow> Q"} (a total function),+
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           also denoted $\delta_M$.+
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  \item @{text "s \<in> Q"} is a state called {\em initial state}, also denoted $s_M$.+
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  \item @{text "F \<subseteq> Q"} is a set of states named {\em accepting states}, also denoted $F_M$.+
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\end{enumerate}+
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Therefore, we have $M = (Q_M, \Sigma_M, \delta_M, s_M, F_M)$. Every DFA $M$ can be interpreted as +
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a function assigning states to strings, denoted $\dfa{M}$, the  definition of which is as the following:+
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\begin{equation}+
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\begin{aligned}+
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  \dfa{M}([]) &\equiv s_M \\+
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   \dfa{M}(xa) &\equiv  \delta_M(\dfa{M}(x), a)+
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\end{aligned}+
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\end{equation}+
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A string @{text "x"} is said to be {\em accepted} (or {\em recognized}) by a DFA $M$ if +
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$\dfa{M}(x) \in F_M$. The language recoginzed by DFA $M$, denoted+
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$L(M)$, is defined as:+
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\begin{equation}+
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  L(M) \equiv \{x~|~\dfa{M}(x) \in F_M\}+
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\end{equation}+
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The standard way of specifying a laugage $\Lang$ as {\em regular} is by stipulating that:+
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$\Lang = L(M)$ for some DFA $M$.+
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+
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For any DFA $M$, the DFA obtained by changing initial state to another $p \in Q_M$ is denoted $M_p$, +
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which is defined as:+
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\begin{equation}+
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    M_p \ \equiv\ (Q_M, \Sigma_M, \delta_M, p, F_M) +
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\end{equation}+
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Two states $p, q \in Q_M$ are said to be {\em equivalent}, denoted $p \approx_M q$, iff.+
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\begin{equation}\label{m_eq_def}+
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  L(M_p) = L(M_q)+
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\end{equation}+
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It is obvious that $\approx_M$ is an equivalent relation over $Q_M$. and +
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the partition induced by $\approx_M$ has $|Q_M|$ equivalent classes.+
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By overloading $\approx_M$,  an equivalent relation over strings can be defined:+
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\begin{equation}+
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  x \approx_M y ~ ~ \equiv ~ ~ \dfa{M}(x) \approx_M \dfa{M}(y)+
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\end{equation}+
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It can be proved that the the partition induced by $\approx_M$ also has $|Q_M|$ equivalent classes.+
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It is also easy to show that: if $x \approx_M y$, then $x \approx_{L(M)} y$, and this means+
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$\approx_M$ is a more refined equivalent relation than $\approx_{L(M)}$. Since partition induced by+
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$\approx_M$ is finite, the one induced by $\approx_{L(M)}$ must also be finite, and this is +
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one of the two directions of \mht:+
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\begin{Lem}[\mht , Direction two]+
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  If a language $\Lang$ is regular (i.e. $\Lang = L(M)$ for some DFA $M$), then +
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  the partition induced by $\approx_\Lang$ is finite.+
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\end{Lem}+
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+
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The other direction is:+
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\begin{Lem}[\mht , Direction one]\label{auto_mh_d1}+
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  If  the partition induced by $\approx_\Lang$ is finite, then+
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  $\Lang$ is regular (i.e. $\Lang = L(M)$ for some DFA $M$).+
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\end{Lem}+
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The $M$ we are seeking when prove lemma \ref{auto_mh_d2} can be constructed out of $\approx_\Lang$,+
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denoted $M_\Lang$ and defined as the following:+
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\begin{subequations}+
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\begin{eqnarray}+
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  Q_{M_\Lang} ~ & \equiv & ~ \{ \cls{x}{\approx_\Lang}~|~ x \in \Sigma^* \}\\+
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  \Sigma_{M_\Lang} ~ & \equiv & ~ \Sigma_M \\+
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  \delta_{M_\Lang} ~ & \equiv & ~ \left (\lambda (\cls{x}{\approx_\Lang}, a).  \cls{xa}{\approx_\Lang} \right) \\+
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  s_{M_\Lang} ~ & \equiv & ~ \cls{[]}{\approx_\Lang} \\+
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  F_{M_\Lang} ~ & \equiv & ~ \{ \cls{x}{\approx_\Lang}~|~ x \in \Lang \}+
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\end{eqnarray}+
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\end{subequations}+
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It can be proved that $Q_{M_\Lang}$ is indeed finite and $\Lang = L(M_\Lang)$, so lemma \ref{auto_mh_d1} holds.+
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It can also be proved that $M_\Lang$ is the minimal DFA (therefore unique) which recoginzes $\Lang$.+
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+
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+
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+
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*}+
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+
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subsection {* The objective and the underlying intuition *}+
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+
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text {*+
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  It is now obvious from section \ref{sec_fa_mh} that \mht\ can be established easily when+
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  {\em reglar languages} are defined as ones recognized by finite automata. +
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  Under the context where the use of finite automata is forbiden, the situation is quite different.+
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  The theorem now has to be expressed as:+
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  \begin{Thm}[\mht , Regular expression version]+
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      A language $\Lang$ is regular (i.e. $\Lang = L(\re)$ for some {\em regular expression} $\re$)+
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      iff. the partition induced by $\approx_\Lang$ is finite.+
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  \end{Thm}+
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  The proof of this version consists of two directions (if the use of automata are not allowed):+
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  \begin{description}+
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    \item [Direction one:] +
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      generating a regular expression $\re$ out of the finite partition induced by $\approx_\Lang$,+
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      such that $\Lang = L(\re)$.+
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    \item [Direction two:]+
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          showing the finiteness of the partition induced by $\approx_\Lang$, under the assmption+
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          that $\Lang$ is recognized by some regular expression $\re$ (i.e. $\Lang = L(\re)$).+
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  \end{description}+
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  The development of these two directions consititutes the body of this paper.+
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+
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*}+
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+
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section {* Direction @{text "regular language \<Rightarrow>finite partition"} *}+
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+
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text {*+
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  Although not used explicitly, the notion of finite autotmata and its relationship with +
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  language partition, as outlined in section \ref{sec_fa_mh}, still servers as important intuitive +
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  guides in the development of this paper.+
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  For example, {\em Direction one} follows the {\em Brzozowski algebraic method}+
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  used to convert finite autotmata to regular expressions, under the intuition that every +
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  partition member $\cls{x}{\approx_\Lang}$ is a state in the DFA $M_\Lang$  constructed to prove +
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  lemma \ref{auto_mh_d1} of section \ref{sec_fa_mh}.+
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+
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  The basic idea of Brzozowski method is to extract an equational system out of the +
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  transition relationship of the automaton in question. In the equational system, every+
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  automaton state is represented by an unknown, the solution of which is expected to be +
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  a regular expresion characterizing the state in a certain sense. There are two choices of +
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  how a automaton state can be characterized.  The first is to characterize by the set of +
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  strings leading from the state in question into accepting states. +
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  The other choice is to characterize by the set of strings leading from initial state +
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  into the state in question. For the second choice, the language recognized the automaton +
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  can be characterized by the solution of initial state, while for the second choice, +
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  the language recoginzed by the automaton can be characterized by +
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  combining solutions of all accepting states by @{text "+"}. Because of the automaton +
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  used as our intuitive guide, the $M_\Lang$, the states of which are +
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  sets of strings leading from initial state, the second choice is used in this paper.+
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+
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  Supposing the automaton in Fig \ref{fig_auto_part_rel} is the $M_\Lang$ for some language $\Lang$, +
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  and suppose $\Sigma = \{a, b, c, d, e\}$. Under the second choice, the equational system extracted is:+
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  \begin{subequations}+
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  \begin{eqnarray}+
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    X_0 & = & X_1 \cdot c + X_2 \cdot d + \lambda \label{x_0_def_o} \\+
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    X_1 & = & X_0 \cdot a + X_1 \cdot b + X_2 \cdot d \label{x_1_def_o} \\+
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    X_2 & = & X_0 \cdot b + X_1 \cdot d + X_2 \cdot a \\+
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    X_3 & = & \begin{aligned}+
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                 & X_0 \cdot (c + d + e) + X_1 \cdot (a + e) + X_2 \cdot (b + e) + \\+
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                 & X_3 \cdot (a + b + c + d + e)+
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              \end{aligned}+
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  \end{eqnarray}+
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  \end{subequations} +
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+
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\begin{figure}[h!]+
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\centering+
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\begin{minipage}{0.5\textwidth}+
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\scalebox{0.8}{+
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\begin{tikzpicture}[ultra thick,>=latex]+
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  \node[draw,circle,initial] (start) {$X_0$};+
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  \node[draw,circle,accepting] at ($(start) + (3.5cm,2cm)$) (ac1) {$X_1$};+
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  \node[draw,circle,accepting] at ($(start) + (3.5cm,-2cm)$) (ac2) {$X_2$};+
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  \node[draw,circle] at ($(start) + (6.5cm,0cm)$) (ab) {$X_3$};+
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+
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  \path[->] (start) edge [bend left] node [midway, above] {$a$} (ac1);+
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  \path[->] (start) edge [bend right] node [midway, below] {$b$} (ac2);+
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  \path[->] (ac1) edge [loop above] node [midway, above] {$b$} (ac1);+
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  \path[->] (ac2) edge [loop below] node [midway, below] {$a$} (ac2);+
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  \path[->] (ac1) edge [bend right] node [midway, left] {$c$} (ac2);+
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  \path[->] (ac2) edge [bend right] node [midway, right] {$c$} (ac1);+
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  \path[->] (ac1) edge node [midway, sloped, above] {$d$} (start);+
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  \path[->] (ac2) edge node [midway, sloped, above] {$d$} (start);+
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+
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  \path [draw, rounded corners,->,dashed] (start) -- ($(start) + (0cm, 3.7cm)$)+
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         -- ($(ab) + (0cm, 3.7cm)$)  node[midway,above,sloped]{$\Sigma - \{a, b\}$} -- (ab);+
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  \path[->,dashed] (ac1) edge node [midway, above, sloped] {$\Sigma - \{b,c,d\}$} (ab);+
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  \path[->,dashed] (ac2) edge node [midway, below, sloped] {$\Sigma - \{a,c,d\}$} (ab);+
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  \path[->,dashed] (ab) edge [loop right] node [midway, right] {$\Sigma$} (ab);+
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\end{tikzpicture}}+
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\end{minipage}+
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\caption{An example automaton}\label{fig_auto_part_rel}+
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\end{figure}+
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+
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  Every $\cdot$-item on the right side of equations describes some state transtions, except +
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  the $\lambda$ in \eqref{x_0_def_o}, which represents empty string @{text "[]"}. +
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  The reason is that: every state is characterized by the+
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  set of incoming strings leading from initial state. For non-initial state, every such+
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  string can be splitted into a prefix leading into a preceding state and a single character suffix +
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  transiting into from the preceding state. The exception happens at+
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  initial state, where the empty string is a incoming string which can not be splitted. The $\lambda$+
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  in \eqref{x_0_def_o} is introduce to repsent this indivisible string. There is one and only one+
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  $\lambda$ in every equational system such obtained, becasue $[]$ can only be contaied in one+
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  equivalent class (the intial state in $M_\Lang$) and equivalent classes are disjoint. +
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  +
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  Suppose all unknowns ($X_0, X_1, X_2, X_3$) are  solvable, the regular expression charactering +
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  laugnage $\Lang$ is $X_1 + X_2$. This paper gives a procedure+
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  by which arbitrarily picked unknown can be solved. The basic idea to solve $X_i$ is by +
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  eliminating all variables other than $X_i$ from the equational system. If+
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  $X_0$ is the one picked to be solved,  variables $X_1, X_2, X_3$ have to be removed one by +
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  one.  The order to remove does not matter as long as the remaing equations are kept valid.+
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  Suppose $X_1$ is the first one to remove, the action is to replace all occurences of $X_1$ +
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  in remaining equations by the right hand side of its characterizing equation, i.e. +
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  the $ X_0 \cdot a + X_1 \cdot b + X_2 \cdot d$ in \eqref{x_1_def_o}. However, because+
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  of the recursive occurence of $X_1$, this replacement does not really removed $X_1$. Arden's+
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  lemma is invoked to transform recursive equations like \eqref{x_1_def_o} into non-recursive +
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  ones. For example, the recursive equation \eqref{x_1_def_o} is transformed into the follwing+
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  non-recursive one:+
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  \begin{equation}+
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     X_1  =  (X_0 \cdot a + X_2 \cdot d) \cdot b^*+
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  \end{equation}+
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  which, by Arden's lemma, still characterizes $X_1$ correctly. By subsitute +
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  $(X_0 \cdot a + X_2 \cdot d) \cdot b^*$ for all $X_1$ and remove  \eqref{x_1_def_o},+
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  we get:+
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+
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  \begin{subequations}+
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  \begin{eqnarray}+
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    X_0 & = & ((X_0 \cdot a + X_2 \cdot d) \cdot b^*) \cdot c + X_2 \cdot d + \lambda \label{x_0_def_1} \\+
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    X_2 & = & X_0 \cdot b + ((X_0 \cdot a + X_2 \cdot d) \cdot b^*) \cdot d + X_2 \cdot a \\+
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    X_3 & = & \begin{aligned}+
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                 & X_0 \cdot (c + d + e) + ((X_0 \cdot a + X_2 \cdot d) \cdot b^*) \cdot (a + e)\\+
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                 & + X_2 \cdot (b + e) + X_3 \cdot (a + b + c + d + e)+
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              \end{aligned}+
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  \end{eqnarray}+
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  \end{subequations}  +
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 +
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*}+
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+
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end+
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