
diff -r 1589bf5c1ad8 -r 63bc9f9d96ba Paper/Paper.thy
--- a/Paper/Paper.thy	Tue Feb 08 09:51:27 2011 +0000
+++ b/Paper/Paper.thy	Tue Feb 08 09:51:49 2011 +0000
@@ -151,7 +151,7 @@
   a regular language as:
 
   \begin{definition}[A Regular Language]
-  A language @{text A} is regular, provided there is a regular expression that matches all
+  A language @{text A} is \emph{regular}, provided there is a regular expression that matches all
   strings of @{text "A"}.
   \end{definition}
   
@@ -216,7 +216,7 @@
 
 
   Central to our proof will be the solution of equational systems
-  involving regular expressions. For this we will use Arden's lemma \cite{Brzozowski64}
+  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.
@@ -297,7 +297,7 @@
 section {* Finite Partitions Imply Regularity of a Language *}
 
 text {*
-  The central definition in the Myhill-Nerode theorem is the
+  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:
@@ -311,7 +311,7 @@
   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 have therefore
+  regular. In our setting we therefore have to show:
   
   \begin{theorem}\label{myhillnerodeone}
   @{thm[mode=IfThen] hard_direction}
@@ -326,15 +326,17 @@
   \end{equation}
 
   \noindent
-  It is straightforward to show that @{thm lang_is_union_of_finals} holds. 
+  It is straightforward to show that @{thm lang_is_union_of_finals} and 
+  @{thm finals_included_in_UNIV} 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 just combine these regular expressions with @{const ALT}
+  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 calculating a regular expression for
-  \emph{all} equivalence classes, not just the ones in @{term "finals A"}. We
+  We prove Thm.~\ref{myhillnerodeone} by giving a method that can calculate a
+  regular expression for \emph{every} equivalence classe, not just the ones 
+  in @{term "finals A"}. We
   first define a notion of \emph{transition} between equivalence classes
   %
   \begin{equation} 
@@ -344,14 +346,14 @@
   \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 any automaton here, we merely relate two sets
-  w.r.t.~a regular expression. 
+  @{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 can build the following equational system
+  Let us assume @{text "[] \<in> X\<^isub>1"}. We build the following equational system
   
   \begin{center}
   \begin{tabular}{rcl}
@@ -367,13 +369,13 @@
   @{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
   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}, which marks the ``terminal''
-  states.) Overloading the function @{text L} for the two kinds of terms in the 
+  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(1)[where r="r", THEN eq_reflection]}\hspace{20mm}
-  @{thm L_rhs_e.simps(2)[where X="X" and r="r", THEN eq_reflection]}
+  @{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
@@ -391,9 +393,12 @@
   \end{equation}
 
   \noindent
-  hold. The reason for adding the @{text \<lambda>}-marker to our equational system is 
-  to obtain this equations, which only holds in this form since none of 
-  the other terms contain the empty string. Our proof of Thm.~\ref{myhillnerodeone}
+  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 




