(*<*)
theory Paper
imports "../Myhill_2"
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}"
abbreviation
"Append_rexp2 r_itm r == Append_rexp r r_itm"
abbreviation
"pow" (infixl "\<up>" 100)
where
"A \<up> n \<equiv> A ^^ n"
abbreviation "NULL \<equiv> Zero"
abbreviation "EMPTY \<equiv> One"
abbreviation "CHAR \<equiv> Atom"
abbreviation "ALT \<equiv> Plus"
abbreviation "SEQ \<equiv> Times"
abbreviation "STAR \<equiv> Star"
ML {* @{term "op ^^"} *}
notation (latex output)
str_eq_rel ("\<approx>\<^bsub>_\<^esub>") and
str_eq ("_ \<approx>\<^bsub>_\<^esub> _") and
conc (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
lang ("\<^raw:\ensuremath{\cal{L}}>'(_')" [0] 101) 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) and
Setalt ("\<^raw:\ensuremath{\bigplus}>_" [1000] 999) and
Append_rexp2 ("_ \<^raw:\ensuremath{\triangleleft}> _" [100, 100] 100) and
Append_rexp_rhs ("_ \<^raw:\ensuremath{\triangleleft}> _" [100, 100] 50) and
uminus ("\<^raw:\ensuremath{\overline{>_\<^raw:}}>" [100] 100) and
tag_str_Plus ("tag\<^isub>A\<^isub>L\<^isub>T _ _" [100, 100] 100) and
tag_str_Plus ("tag\<^isub>A\<^isub>L\<^isub>T _ _ _" [100, 100, 100] 100) and
tag_str_Times ("tag\<^isub>S\<^isub>E\<^isub>Q _ _" [100, 100] 100) and
tag_str_Times ("tag\<^isub>S\<^isub>E\<^isub>Q _ _ _" [100, 100, 100] 100) and
tag_str_Star ("tag\<^isub>S\<^isub>T\<^isub>A\<^isub>R _" [100] 100) and
tag_str_Star ("tag\<^isub>S\<^isub>T\<^isub>A\<^isub>R _ _" [100, 100] 100)
lemma meta_eq_app:
shows "f \<equiv> \<lambda>x. g x \<Longrightarrow> f x \<equiv> g x"
by auto
lemma conc_def':
"A \<cdot> B = {s\<^isub>1 @ s\<^isub>2 | s\<^isub>1 s\<^isub>2. s\<^isub>1 \<in> A \<and> s\<^isub>2 \<in> B}"
unfolding conc_def by simp
(* THEOREMS *)
lemma conc_Union_left:
shows "B \<cdot> (\<Union>n. A \<up> n) = (\<Union>n. B \<cdot> (A \<up> n))"
unfolding conc_def by auto
notation (Rule output)
"==>" ("\<^raw:\mbox{}\inferrule{\mbox{>_\<^raw:}}>\<^raw:{\mbox{>_\<^raw:}}>")
syntax (Rule output)
"_bigimpl" :: "asms \<Rightarrow> prop \<Rightarrow> prop"
("\<^raw:\mbox{}\inferrule{>_\<^raw:}>\<^raw:{\mbox{>_\<^raw:}}>")
"_asms" :: "prop \<Rightarrow> asms \<Rightarrow> asms"
("\<^raw:\mbox{>_\<^raw:}\\>/ _")
"_asm" :: "prop \<Rightarrow> asms" ("\<^raw:\mbox{>_\<^raw:}>")
notation (Axiom output)
"Trueprop" ("\<^raw:\mbox{}\inferrule{\mbox{}}{\mbox{>_\<^raw:}}>")
notation (IfThen output)
"==>" ("\<^raw:{\normalsize{}>If\<^raw:\,}> _/ \<^raw:{\normalsize \,>then\<^raw:\,}>/ _.")
syntax (IfThen output)
"_bigimpl" :: "asms \<Rightarrow> prop \<Rightarrow> prop"
("\<^raw:{\normalsize{}>If\<^raw:\,}> _ /\<^raw:{\normalsize \,>then\<^raw:\,}>/ _.")
"_asms" :: "prop \<Rightarrow> asms \<Rightarrow> asms" ("\<^raw:\mbox{>_\<^raw:}> /\<^raw:{\normalsize \,>and\<^raw:\,}>/ _")
"_asm" :: "prop \<Rightarrow> asms" ("\<^raw:\mbox{>_\<^raw:}>")
notation (IfThenNoBox output)
"==>" ("\<^raw:{\normalsize{}>If\<^raw:\,}> _/ \<^raw:{\normalsize \,>then\<^raw:\,}>/ _.")
syntax (IfThenNoBox output)
"_bigimpl" :: "asms \<Rightarrow> prop \<Rightarrow> prop"
("\<^raw:{\normalsize{}>If\<^raw:\,}> _ /\<^raw:{\normalsize \,>then\<^raw:\,}>/ _.")
"_asms" :: "prop \<Rightarrow> asms \<Rightarrow> asms" ("_ /\<^raw:{\normalsize \,>and\<^raw:\,}>/ _")
"_asm" :: "prop \<Rightarrow> asms" ("_")
(*>*)
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 the 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 built up from states and transitions that
need to be represented as graphs, matrices or functions, none
of which can be defined as an inductive datatype.
In case of graphs and matrices, this means we have to build our own
reasoning infrastructure for them, as neither Isabelle/HOL nor HOL4 nor
HOLlight support them with libraries. Even worse, reasoning about graphs and
matrices can be a real hassle in HOL-based theorem provers. Consider for
example the operation of sequencing two automata, say $A_1$ and $A_2$, by
connecting the accepting states of $A_1$ to the initial state of $A_2$:\\[-5.5mm]
%
\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 standard definition for disjoint
union, namely
%
\begin{equation}\label{disjointunion}
@{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}"}
\end{equation}
\noindent
changes the type---the disjoint union is not a set, but a set of pairs.
Using this definition for disjoint union means we do not have a single type for automata
and hence will not be able to state certain properties about \emph{all}
automata, since there is no type quantification available in HOL (unlike in Coq, for example). 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.
Functions are much better supported in Isabelle/HOL, but they still lead to similar
problems as with graphs. Composing, for example, two non-deterministic automata in parallel
requires also the formalisation of disjoint unions. Nipkow \cite{Nipkow98}
dismisses for this the option of using identities, because it leads according to
him to ``messy proofs''. He
opts for a variant of \eqref{disjointunion} using bit lists, but writes
\begin{quote}
\it%
\begin{tabular}{@ {}l@ {}p{0.88\textwidth}@ {}}
`` & All lemmas appear obvious given a picture of the composition of automata\ldots
Yet their proofs require a painful amount of detail.''
\end{tabular}
\end{quote}
\noindent
and
\begin{quote}
\it%
\begin{tabular}{@ {}l@ {}p{0.88\textwidth}@ {}}
`` & If the reader finds the above treatment in terms of bit lists revoltingly
concrete, I cannot disagree. A more abstract approach is clearly desirable.''
\end{tabular}
\end{quote}
\noindent
Moreover, it is not so clear how to conveniently impose a finiteness condition
upon functions in order to represent \emph{finite} automata. The best is
probably to resort to more advanced reasoning frameworks, such as \emph{locales}
or \emph{type classes},
which are \emph{not} available in all HOL-based theorem provers.
Because of these problems to do with representing automata, there seems
to be no substantial formalisation of automata theory and regular languages
carried out in HOL-based theorem provers. Nipkow \cite{Nipkow98} establishes
the link between regular expressions and automata in
the context of lexing. Berghofer and Reiter \cite{BerghoferReiter09}
formalise automata working over
bit strings in the context of Presburger arithmetic.
The only larger formalisations of automata theory
are carried out in Nuprl \cite{Constable00} and in Coq \cite{Filliatre97}.
In this paper, we will not attempt to formalise automata theory in
Isabelle/HOL, but take a different approach to regular
languages. Instead of defining a regular language as one where there exists
an automaton that recognises all strings of the language, we define a
regular language as:
\begin{definition}
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, matrices and functions, 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} and
with an equivalence checker for regular expressions in Isabelle/HOL \cite{KraussNipkow11}.
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:}
There is an extensive literature on regular languages.
To our best 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 \<cdot> B"} and a language raised to the power @{text n} is written
@{term "A \<up> n"}. They are defined as usual
\begin{center}
@{thm conc_def'[THEN eq_reflection, where A1="A" and B1="B"]}
\hspace{7mm}
@{thm lang_pow.simps(1)[THEN eq_reflection, where A1="A"]}
\hspace{7mm}
@{thm lang_pow.simps(2)[THEN eq_reflection, where A1="A" and n1="n"]}
\end{center}
\noindent
where @{text "@"} is the 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 make use of the following properties of these constructions.
\begin{proposition}\label{langprops}\mbox{}\\
\begin{tabular}{@ {}ll}
(i) & @{thm star_cases} \\
(ii) & @{thm[mode=IfThen] pow_length}\\
(iii) & @{thm conc_Union_left} \\
\end{tabular}
\end{proposition}
\noindent
In @{text "(ii)"} we use the notation @{term "length s"} for the length of a
string; this property states that if \mbox{@{term "[] \<notin> A"}} then the lengths of
the strings in @{term "A \<up> (Suc n)"} must be longer than @{text n}. We omit
the proofs for these properties, but invite the reader to consult our
formalisation.\footnote{Available at \url{http://www4.in.tum.de/~urbanc/regexp.html}}
The notation in Isabelle/HOL 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 \mbox{@{text "{y | y \<approx> x}"}}.
Central to our proof will be the solution of equational systems
involving equivalence classes of languages. For this we will use Arden's Lemma \cite{Brzozowski64},
which solves equations of the form @{term "X = A \<cdot> X \<union> B"} provided
@{term "[] \<notin> A"}. However we will need the following `reverse'
version of Arden's Lemma (`reverse' in the sense of changing the order of @{term "A \<cdot> X"} to
\mbox{@{term "X \<cdot> A"}}).
\begin{lemma}[Reverse Arden's Lemma]\label{arden}\mbox{}\\
If @{thm (prem 1) arden} then
@{thm (lhs) arden} if and only if
@{thm (rhs) arden}.
\end{lemma}
\begin{proof}
For the right-to-left direction we assume @{thm (rhs) arden} and show
that @{thm (lhs) arden} holds. From Prop.~\ref{langprops}@{text "(i)"}
we have @{term "A\<star> = {[]} \<union> A \<cdot> A\<star>"},
which is equal to @{term "A\<star> = {[]} \<union> A\<star> \<cdot> A"}. Adding @{text B} to both
sides gives @{term "B \<cdot> A\<star> = B \<cdot> ({[]} \<union> A\<star> \<cdot> A)"}, whose right-hand side
is equal to @{term "(B \<cdot> A\<star>) \<cdot> A \<union> B"}. This completes this direction.
For the other direction we assume @{thm (lhs) arden}. By a simple induction
on @{text n}, we can establish the property
\begin{center}
@{text "(*)"}\hspace{5mm} @{thm (concl) arden_helper}
\end{center}
\noindent
Using this property we can show that @{term "B \<cdot> (A \<up> n) \<subseteq> X"} holds for
all @{text n}. From this we can infer @{term "B \<cdot> 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 an element in @{text X}. Since @{thm (prem 1) arden}
we know by Prop.~\ref{langprops}@{text "(ii)"} that
@{term "s \<notin> X \<cdot> (A \<up> Suc k)"} since its length is only @{text k}
(the strings in @{term "X \<cdot> (A \<up> Suc k)"} are all longer).
From @{text "(*)"} it follows then that
@{term s} must be an element in @{term "(\<Union>m\<in>{0..k}. B \<cdot> (A \<up> m))"}. This in turn
implies that @{term s} is in @{term "(\<Union>n. B \<cdot> (A \<up> n))"}. Using Prop.~\ref{langprops}@{text "(iii)"}
this is equal to @{term "B \<cdot> A\<star>"}, as we needed to show.\qed
\end{proof}
\noindent
Regular expressions are defined as the 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
and the language matched by a regular expression is defined as
\begin{center}
\begin{tabular}{c@ {\hspace{10mm}}c}
\begin{tabular}{rcl}
@{thm (lhs) lang.simps(1)} & @{text "\<equiv>"} & @{thm (rhs) lang.simps(1)}\\
@{thm (lhs) lang.simps(2)} & @{text "\<equiv>"} & @{thm (rhs) lang.simps(2)}\\
@{thm (lhs) lang.simps(3)[where a="c"]} & @{text "\<equiv>"} & @{thm (rhs) lang.simps(3)[where a="c"]}\\
\end{tabular}
&
\begin{tabular}{rcl}
@{thm (lhs) lang.simps(4)[where ?r="r\<^isub>1" and ?s="r\<^isub>2"]} & @{text "\<equiv>"} &
@{thm (rhs) lang.simps(4)[where ?r="r\<^isub>1" and ?s="r\<^isub>2"]}\\
@{thm (lhs) lang.simps(5)[where ?r="r\<^isub>1" and ?s="r\<^isub>2"]} & @{text "\<equiv>"} &
@{thm (rhs) lang.simps(5)[where ?r="r\<^isub>1" and ?s="r\<^isub>2"]}\\
@{thm (lhs) lang.simps(6)[where r="r"]} & @{text "\<equiv>"} &
@{thm (rhs) lang.simps(6)[where r="r"]}\\
\end{tabular}
\end{tabular}
\end{center}
Given a finite set of regular expressions @{text rs}, we will make use of the operation of generating
a regular expression that matches the union of all languages of @{text rs}. We only need to know the
existence
of such a regular expression and therefore we use Isabelle/HOL's @{const "fold_graph"} and Hilbert's
@{text "\<epsilon>"} to define @{term "\<Uplus>rs"}. This operation, roughly speaking, folds @{const ALT} over the
set @{text rs} with @{const NULL} for the empty set. We can prove that for a finite set @{text rs}
%
\begin{equation}\label{uplus}
\mbox{@{thm (lhs) folds_alt_simp} @{text "= \<Union> (\<calL> ` rs)"}}
\end{equation}
\noindent
holds, whereby @{text "\<calL> ` rs"} stands for the
image of the set @{text rs} under function @{text "\<calL>"}.
*}
section {* The Myhill-Nerode Theorem, First Part *}
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 a tertiary relation.
\begin{definition}[Myhill-Nerode Relation] Given a language @{text A}, two strings @{text x} and
@{text y} are Myhill-Nerode related provided
\begin{center}
@{thm str_eq_def[simplified str_eq_rel_def Pair_Collect]}
\end{center}
\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. To illustrate this quotient construction, let us give a simple
example: consider the regular language containing just
the string @{text "[c]"}. The relation @{term "\<approx>({[c]})"} partitions @{text UNIV}
into three equivalence classes @{text "X\<^isub>1"}, @{text "X\<^isub>2"} and @{text "X\<^isub>3"}
as follows
\begin{center}
@{text "X\<^isub>1 = {[]}"}\hspace{5mm}
@{text "X\<^isub>2 = {[c]}"}\hspace{5mm}
@{text "X\<^isub>3 = UNIV - {[], [c]}"}
\end{center}
One direction of the Myhill-Nerode theorem establishes
that if there are finitely many equivalence classes, like in the example above, then
the language is regular. In our setting we therefore have to show:
\begin{theorem}\label{myhillnerodeone}
@{thm[mode=IfThen] Myhill_Nerode1}
\end{theorem}
\noindent
To prove this theorem, we first define the set @{term "finals A"} as those equivalence
classes from @{term "UNIV // \<approx>A"} that contain strings of @{text A}, namely
%
\begin{equation}
@{thm finals_def}
\end{equation}
\noindent
In our running example, @{text "X\<^isub>2"} is the only
equivalence class in @{term "finals {[c]}"}.
It is straightforward to show that in general @{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 \mbox{@{term "finals A"}} (which by assumption must be
a finite set), then we can use @{text "\<bigplus>"} to obtain a regular expression
that matches every string in @{text A}.
Our proof of Thm.~\ref{myhillnerodeone} relies on a method that can calculate a
regular expression for \emph{every} equivalence class, not just the ones
in @{term "finals A"}. We
first define the notion of \emph{one-character-transition} between
two equivalence classes
%
\begin{equation}
@{thm transition_def}
\end{equation}
\noindent
which means that if we concatenate the character @{text c} 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
(with the help of a character). In our concrete example we have
@{term "X\<^isub>1 \<Turnstile>c\<Rightarrow> X\<^isub>2"}, @{term "X\<^isub>1 \<Turnstile>d\<Rightarrow> X\<^isub>3"} with @{text d} being any
other character than @{text c}, and @{term "X\<^isub>3 \<Turnstile>d\<Rightarrow> X\<^isub>3"} for any @{text d}.
Next we construct an \emph{initial equational system} that
contains an equation for each equivalence class. We first give
an informal description of this construction. 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
where the terms @{text "(Y\<^isub>i\<^isub>j, CHAR c\<^isub>i\<^isub>j)"}
stand for all transitions @{term "Y\<^isub>i\<^isub>j \<Turnstile>c\<^isub>i\<^isub>j\<Rightarrow>
X\<^isub>i"}.
%The intuition behind the equational system is that every
%equation @{text "X\<^isub>i = rhs\<^isub>i"} in this system
%corresponds roughly to a state of an automaton whose name is @{text X\<^isub>i} and its predecessor states
%are the @{text "Y\<^isub>i\<^isub>j"}; the @{text "c\<^isub>i\<^isub>j"} are the labels of the transitions from these
%predecessor states to @{text X\<^isub>i}.
There can only be
finitely many terms of the form @{text "(Y\<^isub>i\<^isub>j, CHAR c\<^isub>i\<^isub>j)"} in a right-hand side
since by assumption there are only finitely many
equivalence classes and only finitely many characters.
The term @{text "\<lambda>(EMPTY)"} in the first equation acts as a marker for the initial state, that
is the equivalence class
containing @{text "[]"}.\footnote{Note that we mark, roughly speaking, the
single `initial' state in the equational system, which is different from
the method by Brzozowski \cite{Brzozowski64}, where he marks the
`terminal' states. We are forced to set up the equational system in our
way, because the Myhill-Nerode relation determines the `direction' of the
transitions---the successor `state' of an equivalence class @{text Y} can
be reached by adding a character to the end of @{text Y}. This is also the
reason why we have to use our reverse version of Arden's Lemma.}
%In our initial equation system there can only be
%finitely many terms of the form @{text "(Y\<^isub>i\<^isub>j, CHAR c\<^isub>i\<^isub>j)"} in a right-hand side
%since by assumption there are only finitely many
%equivalence classes and only finitely many characters.
Overloading the function @{text \<calL>} for the two kinds of terms in the
equational system, we have
\begin{center}
@{text "\<calL>(Y, r) \<equiv>"} %
@{thm (rhs) lang_trm.simps(2)[where X="Y" and r="r", THEN eq_reflection]}\hspace{10mm}
@{thm lang_trm.simps(1)[where r="r", THEN eq_reflection]}
\end{center}
\noindent
and we can prove for @{text "X\<^isub>2\<^isub>.\<^isub>.\<^isub>n"} that the following equations
%
\begin{equation}\label{inv1}
@{text "X\<^isub>i = \<calL>(Y\<^isub>i\<^isub>1, CHAR c\<^isub>i\<^isub>1) \<union> \<dots> \<union> \<calL>(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 = \<calL>(Y\<^isub>1\<^isub>1, CHAR c\<^isub>1\<^isub>1) \<union> \<dots> \<union> \<calL>(Y\<^isub>1\<^isub>p, CHAR c\<^isub>1\<^isub>p) \<union> \<calL>(\<lambda>(EMPTY))"}.
\end{equation}
\noindent
holds. The reason for adding the @{text \<lambda>}-marker to our initial equational system is
to obtain this equation: it only holds with the marker, since none of
the other terms contain the empty string. The point of the initial equational system is
that solving it means we will be able to extract a regular expression for every equivalence class.
Our representation for the equations in Isabelle/HOL are pairs,
where the first component is an equivalence class (a set of strings)
and the second component
is a set of terms. Given a set of equivalence
classes @{text CS}, our initial equational system @{term "Init CS"} is thus
formally defined as
%
\begin{equation}\label{initcs}
\mbox{\begin{tabular}{rcl}
@{thm (lhs) Init_rhs_def} & @{text "\<equiv>"} &
@{text "if"}~@{term "[] \<in> X"}\\
& & @{text "then"}~@{term "{Trn Y (CHAR c) | Y c. Y \<in> CS \<and> Y \<Turnstile>c\<Rightarrow> X} \<union> {Lam EMPTY}"}\\
& & @{text "else"}~@{term "{Trn Y (CHAR c)| Y c. Y \<in> CS \<and> Y \<Turnstile>c\<Rightarrow> X}"}\\
@{thm (lhs) Init_def} & @{text "\<equiv>"} & @{thm (rhs) Init_def}
\end{tabular}}
\end{equation}
*}(*<*)
lemma test:
assumes X_in_eqs: "(X, rhs) \<in> Init (UNIV // \<approx>A)"
shows "X = \<Union> (lang_trm ` rhs)"
using assms l_eq_r_in_eqs by (simp)
(*>*)text {*
\noindent
Because we use sets of terms
for representing the right-hand sides of equations, we can
prove \eqref{inv1} and \eqref{inv2} more concisely as
%
\begin{lemma}\label{inv}
If @{thm (prem 1) test} then @{text "X = \<Union> \<calL> ` rhs"}.
\end{lemma}
\noindent
Our proof of Thm.~\ref{myhillnerodeone} will proceed by transforming the
initial equational system into one in \emph{solved form} maintaining the invariant
in Lem.~\ref{inv}. From the solved form we will be able to read
off the regular expressions.
In order to transform an equational system into solved form, we have two
operations: one that takes an equation of the form @{text "X = rhs"} and removes
any recursive occurrences of @{text X} in the @{text rhs} using our variant of Arden's
Lemma. The other operation takes an equation @{text "X = rhs"}
and substitutes @{text X} throughout the rest of the equational system
adjusting the remaining regular expressions appropriately. To define this adjustment
we define the \emph{append-operation} taking a term and a regular expression as argument
\begin{center}
@{thm Append_rexp.simps(2)[where X="Y" and r="r\<^isub>1" and rexp="r\<^isub>2", THEN eq_reflection]}\hspace{10mm}
@{thm Append_rexp.simps(1)[where r="r\<^isub>1" and rexp="r\<^isub>2", THEN eq_reflection]}
\end{center}
\noindent
We lift this operation to entire right-hand sides of equations, written as
@{thm (lhs) Append_rexp_rhs_def[where rexp="r"]}. With this we can define
the \emph{arden-operation} for an equation of the form @{text "X = rhs"} as:
%
\begin{equation}\label{arden_def}
\mbox{\begin{tabular}{rc@ {\hspace{2mm}}r@ {\hspace{1mm}}l}
@{thm (lhs) Arden_def} & @{text "\<equiv>"}~~\mbox{} & \multicolumn{2}{@ {\hspace{-2mm}}l}{@{text "let"}}\\
& & @{text "rhs' ="} & @{term "rhs - {Trn X r | r. Trn X r \<in> rhs}"} \\
& & @{text "r' ="} & @{term "STAR (\<Uplus> {r. Trn X r \<in> rhs})"}\\
& & \multicolumn{2}{@ {\hspace{-2mm}}l}{@{text "in"}~~@{term "append_rhs_rexp rhs' r'"}}\\
\end{tabular}}
\end{equation}
\noindent
In this definition, we first delete all terms of the form @{text "(X, r)"} from @{text rhs};
then we calculate the combined regular expressions for all @{text r} coming
from the deleted @{text "(X, r)"}, and take the @{const STAR} of it;
finally we append this regular expression to @{text rhs'}. It can be easily seen
that this operation mimics Arden's Lemma on the level of equations. To ensure
the non-emptiness condition of Arden's Lemma we say that a right-hand side is
@{text ardenable} provided
\begin{center}
@{thm ardenable_def}
\end{center}
\noindent
This allows us to prove a version of Arden's Lemma on the level of equations.
\begin{lemma}\label{ardenable}
Given an equation @{text "X = rhs"}.
If @{text "X = \<Union>\<calL> ` rhs"},
@{thm (prem 2) Arden_keeps_eq}, and
@{thm (prem 3) Arden_keeps_eq}, then
@{text "X = \<Union>\<calL> ` (Arden X rhs)"}.
\end{lemma}
\noindent
Our @{text ardenable} condition is slightly stronger than needed for applying Arden's Lemma,
but we can still ensure that it holds troughout our algorithm of transforming equations
into solved form. The \emph{substitution-operation} takes an equation
of the form @{text "X = xrhs"} and substitutes it into the right-hand side @{text rhs}.
\begin{center}
\begin{tabular}{rc@ {\hspace{2mm}}r@ {\hspace{1mm}}l}
@{thm (lhs) Subst_def} & @{text "\<equiv>"}~~\mbox{} & \multicolumn{2}{@ {\hspace{-2mm}}l}{@{text "let"}}\\
& & @{text "rhs' ="} & @{term "rhs - {Trn X r | r. Trn X r \<in> rhs}"} \\
& & @{text "r' ="} & @{term "\<Uplus> {r. Trn X r \<in> rhs}"}\\
& & \multicolumn{2}{@ {\hspace{-2mm}}l}{@{text "in"}~~@{term "rhs' \<union> append_rhs_rexp xrhs r'"}}\\
\end{tabular}
\end{center}
\noindent
We again delete first all occurrences of @{text "(X, r)"} in @{text rhs}; we then calculate
the regular expression corresponding to the deleted terms; finally we append this
regular expression to @{text "xrhs"} and union it up with @{text rhs'}. When we use
the substitution operation we will arrange it so that @{text "xrhs"} does not contain
any occurrence of @{text X}.
With these two operations in place, we can define the operation that removes one equation
from an equational systems @{text ES}. The operation @{const Subst_all}
substitutes an equation @{text "X = xrhs"} throughout an equational system @{text ES};
@{const Remove} then completely removes such an equation from @{text ES} by substituting
it to the rest of the equational system, but first eliminating all recursive occurrences
of @{text X} by applying @{const Arden} to @{text "xrhs"}.
\begin{center}
\begin{tabular}{rcl}
@{thm (lhs) Subst_all_def} & @{text "\<equiv>"} & @{thm (rhs) Subst_all_def}\\
@{thm (lhs) Remove_def} & @{text "\<equiv>"} & @{thm (rhs) Remove_def}
\end{tabular}
\end{center}
\noindent
Finally, we can define how an equational system should be solved. For this
we will need to iterate the process of eliminating equations until only one equation
will be left in the system. However, we do not just want to have any equation
as being the last one, but the one involving the equivalence class for
which we want to calculate the regular
expression. Let us suppose this equivalence class is @{text X}.
Since @{text X} is the one to be solved, in every iteration step we have to pick an
equation to be eliminated that is different from @{text X}. In this way
@{text X} is kept to the final step. The choice is implemented using Hilbert's choice
operator, written @{text SOME} in the definition below.
\begin{center}
\begin{tabular}{rc@ {\hspace{4mm}}r@ {\hspace{1mm}}l}
@{thm (lhs) Iter_def} & @{text "\<equiv>"}~~\mbox{} & \multicolumn{2}{@ {\hspace{-4mm}}l}{@{text "let"}}\\
& & @{text "(Y, yrhs) ="} & @{term "SOME (Y, yrhs). (Y, yrhs) \<in> ES \<and> X \<noteq> Y"} \\
& & \multicolumn{2}{@ {\hspace{-4mm}}l}{@{text "in"}~~@{term "Remove ES Y yrhs"}}\\
\end{tabular}
\end{center}
\noindent
The last definition we need applies @{term Iter} over and over until a condition
@{text Cond} is \emph{not} satisfied anymore. This condition states that there
are more than one equation left in the equational system @{text ES}. To solve
an equational system we use Isabelle/HOL's @{text while}-operator as follows:
\begin{center}
@{thm Solve_def}
\end{center}
\noindent
We are not concerned here with the definition of this operator
(see Berghofer and Nipkow \cite{BerghoferNipkow00}), but note that we eliminate
in each @{const Iter}-step a single equation, and therefore
have a well-founded termination order by taking the cardinality
of the equational system @{text ES}. This enables us to prove
properties about our definition of @{const Solve} when we `call' it with
the equivalence class @{text X} and the initial equational system
@{term "Init (UNIV // \<approx>A)"} from
\eqref{initcs} using the principle:
%
\begin{equation}\label{whileprinciple}
\mbox{\begin{tabular}{l}
@{term "invariant (Init (UNIV // \<approx>A))"} \\
@{term "\<forall>ES. invariant ES \<and> Cond ES \<longrightarrow> invariant (Iter X ES)"}\\
@{term "\<forall>ES. invariant ES \<and> Cond ES \<longrightarrow> card (Iter X ES) < card ES"}\\
@{term "\<forall>ES. invariant ES \<and> \<not> Cond ES \<longrightarrow> P ES"}\\
\hline
\multicolumn{1}{c}{@{term "P (Solve X (Init (UNIV // \<approx>A)))"}}
\end{tabular}}
\end{equation}
\noindent
This principle states that given an invariant (which we will specify below)
we can prove a property
@{text "P"} involving @{const Solve}. For this we have to discharge the following
proof obligations: first the
initial equational system satisfies the invariant; second the iteration
step @{text "Iter"} preserves the invariant as long as the condition @{term Cond} holds;
third @{text "Iter"} decreases the termination order, and fourth that
once the condition does not hold anymore then the property @{text P} must hold.
The property @{term P} in our proof will state that @{term "Solve X (Init (UNIV // \<approx>A))"}
returns with a single equation @{text "X = xrhs"} for some @{text "xrhs"}, and
that this equational system still satisfies the invariant. In order to get
the proof through, the invariant is composed of the following six properties:
\begin{center}
\begin{tabular}{@ {}rcl@ {\hspace{-13mm}}l @ {}}
@{text "invariant ES"} & @{text "\<equiv>"} &
@{term "finite ES"} & @{text "(finiteness)"}\\
& @{text "\<and>"} & @{thm (rhs) finite_rhs_def} & @{text "(finiteness rhs)"}\\
& @{text "\<and>"} & @{text "\<forall>(X, rhs)\<in>ES. X = \<Union>\<calL> ` rhs"} & @{text "(soundness)"}\\
& @{text "\<and>"} & @{thm (rhs) distinctness_def}\\
& & & @{text "(distinctness)"}\\
& @{text "\<and>"} & @{thm (rhs) ardenable_all_def} & @{text "(ardenable)"}\\
& @{text "\<and>"} & @{thm (rhs) validity_def} & @{text "(validity)"}\\
\end{tabular}
\end{center}
\noindent
The first two ensure that the equational system is always finite (number of equations
and number of terms in each equation); the third makes sure the `meaning' of the
equations is preserved under our transformations. The other properties are a bit more
technical, but are needed to get our proof through. Distinctness states that every
equation in the system is distinct. @{text Ardenable} ensures that we can always
apply the @{text Arden} operation.
The last property states that every @{text rhs} can only contain equivalence classes
for which there is an equation. Therefore @{text lhss} is just the set containing
the first components of an equational system,
while @{text "rhss"} collects all equivalence classes @{text X} in the terms of the
form @{term "Trn X r"}. That means formally @{thm (lhs) lhss_def}~@{text "\<equiv> {X | (X, rhs) \<in> ES}"}
and @{thm (lhs) rhss_def}~@{text "\<equiv> {X | (X, r) \<in> rhs}"}.
It is straightforward to prove that the initial equational system satisfies the
invariant.
\begin{lemma}\label{invzero}
@{thm[mode=IfThen] Init_ES_satisfies_invariant}
\end{lemma}
\begin{proof}
Finiteness is given by the assumption and the way how we set up the
initial equational system. Soundness is proved in Lem.~\ref{inv}. Distinctness
follows from the fact that the equivalence classes are disjoint. The @{text ardenable}
property also follows from the setup of the initial equational system, as does
validity.\qed
\end{proof}
\noindent
Next we show that @{text Iter} preserves the invariant.
\begin{lemma}\label{iterone}
@{thm[mode=IfThen] iteration_step_invariant[where xrhs="rhs"]}
\end{lemma}
\begin{proof}
The argument boils down to choosing an equation @{text "Y = yrhs"} to be eliminated
and to show that @{term "Subst_all (ES - {(Y, yrhs)}) Y (Arden Y yrhs)"}
preserves the invariant.
We prove this as follows:
\begin{center}
@{text "\<forall> ES."} @{thm (prem 1) Subst_all_satisfies_invariant} implies
@{thm (concl) Subst_all_satisfies_invariant}
\end{center}
\noindent
Finiteness is straightforward, as the @{const Subst} and @{const Arden} operations
keep the equational system finite. These operations also preserve soundness
and distinctness (we proved soundness for @{const Arden} in Lem.~\ref{ardenable}).
The property @{text ardenable} is clearly preserved because the append-operation
cannot make a regular expression to match the empty string. Validity is
given because @{const Arden} removes an equivalence class from @{text yrhs}
and then @{const Subst_all} removes @{text Y} from the equational system.
Having proved the implication above, we can instantiate @{text "ES"} with @{text "ES - {(Y, yrhs)}"}
which matches with our proof-obligation of @{const "Subst_all"}. Since
\mbox{@{term "ES = ES - {(Y, yrhs)} \<union> {(Y, yrhs)}"}}, we can use the assumption
to complete the proof.\qed
\end{proof}
\noindent
We also need the fact that @{text Iter} decreases the termination measure.
\begin{lemma}\label{itertwo}
@{thm[mode=IfThen] iteration_step_measure[simplified (no_asm), where xrhs="rhs"]}
\end{lemma}
\begin{proof}
By assumption we know that @{text "ES"} is finite and has more than one element.
Therefore there must be an element @{term "(Y, yrhs) \<in> ES"} with
@{term "(Y, yrhs) \<noteq> (X, rhs)"}. Using the distinctness property we can infer
that @{term "Y \<noteq> X"}. We further know that @{text "Remove ES Y yrhs"}
removes the equation @{text "Y = yrhs"} from the system, and therefore
the cardinality of @{const Iter} strictly decreases.\qed
\end{proof}
\noindent
This brings us to our property we want to establish for @{text Solve}.
\begin{lemma}
If @{thm (prem 1) Solve} and @{thm (prem 2) Solve} then there exists
a @{text rhs} such that @{term "Solve X (Init (UNIV // \<approx>A)) = {(X, rhs)}"}
and @{term "invariant {(X, rhs)}"}.
\end{lemma}
\begin{proof}
In order to prove this lemma using \eqref{whileprinciple}, we have to use a slightly
stronger invariant since Lem.~\ref{iterone} and \ref{itertwo} have the precondition
that @{term "(X, rhs) \<in> ES"} for some @{text rhs}. This precondition is needed
in order to choose in the @{const Iter}-step an equation that is not \mbox{@{term "X = rhs"}}.
Therefore our invariant cannot be just @{term "invariant ES"}, but must be
@{term "invariant ES \<and> (\<exists>rhs. (X, rhs) \<in> ES)"}. By assumption
@{thm (prem 2) Solve} and Lem.~\ref{invzero}, the more general invariant holds for
the initial equational system. This is premise 1 of~\eqref{whileprinciple}.
Premise 2 is given by Lem.~\ref{iterone} and the fact that @{const Iter} might
modify the @{text rhs} in the equation @{term "X = rhs"}, but does not remove it.
Premise 3 of~\eqref{whileprinciple} is by Lem.~\ref{itertwo}. Now in premise 4
we like to show that there exists a @{text rhs} such that @{term "ES = {(X, rhs)}"}
and that @{text "invariant {(X, rhs)}"} holds, provided the condition @{text "Cond"}
does not holds. By the stronger invariant we know there exists such a @{text "rhs"}
with @{term "(X, rhs) \<in> ES"}. Because @{text Cond} is not true, we know the cardinality
of @{text ES} is @{text 1}. This means @{text "ES"} must actually be the set @{text "{(X, rhs)}"},
for which the invariant holds. This allows us to conclude that
@{term "Solve X (Init (UNIV // \<approx>A)) = {(X, rhs)}"} and @{term "invariant {(X, rhs)}"} hold,
as needed.\qed
\end{proof}
\noindent
With this lemma in place we can show that for every equivalence class in @{term "UNIV // \<approx>A"}
there exists a regular expression.
\begin{lemma}\label{every_eqcl_has_reg}
@{thm[mode=IfThen] every_eqcl_has_reg}
\end{lemma}
\begin{proof}
By the preceding lemma, we know that there exists a @{text "rhs"} such
that @{term "Solve X (Init (UNIV // \<approx>A))"} returns the equation @{text "X = rhs"},
and that the invariant holds for this equation. That means we
know @{text "X = \<Union>\<calL> ` rhs"}. We further know that
this is equal to \mbox{@{text "\<Union>\<calL> ` (Arden X rhs)"}} using the properties of the
invariant and Lem.~\ref{ardenable}. Using the validity property for the equation @{text "X = rhs"},
we can infer that @{term "rhss rhs \<subseteq> {X}"} and because the @{text Arden} operation
removes that @{text X} from @{text rhs}, that @{term "rhss (Arden X rhs) = {}"}.
This means the right-hand side @{term "Arden X rhs"} can only consist of terms of the form @{term "Lam r"}.
So we can collect those (finitely many) regular expressions @{text rs} and have @{term "X = L (\<Uplus>rs)"}.
With this we can conclude the proof.\qed
\end{proof}
\noindent
Lem.~\ref{every_eqcl_has_reg} allows us to finally give a proof for the first direction
of the Myhill-Nerode theorem.
\begin{proof}[of Thm.~\ref{myhillnerodeone}]
By Lem.~\ref{every_eqcl_has_reg} we know that there exists a regular expression for
every equivalence class in @{term "UNIV // \<approx>A"}. Since @{text "finals A"} is
a subset of @{term "UNIV // \<approx>A"}, we also know that for every equivalence class
in @{term "finals A"} there exists a regular expression. Moreover by assumption
we know that @{term "finals A"} must be finite, and therefore there must be a finite
set of regular expressions @{text "rs"} such that
@{term "\<Union>(finals A) = L (\<Uplus>rs)"}.
Since the left-hand side is equal to @{text A}, we can use @{term "\<Uplus>rs"}
as the regular expression that is needed in the theorem.\qed
\end{proof}
*}
section {* Myhill-Nerode, Second Part *}
text {*
We will prove in this section the second part of the Myhill-Nerode
theorem. It can be formulated in our setting as follows:
\begin{theorem}
Given @{text "r"} is a regular expression, then @{thm Myhill_Nerode2}.
\end{theorem}
\noindent
The proof will be by induction on the structure of @{text r}. It turns out
the base cases are straightforward.
\begin{proof}[Base Cases]
The cases for @{const NULL}, @{const EMPTY} and @{const CHAR} are routine, because
we can easily establish that
\begin{center}
\begin{tabular}{l}
@{thm quot_zero_eq}\\
@{thm quot_one_subset}\\
@{thm quot_atom_subset}
\end{tabular}
\end{center}
\noindent
hold, which shows that @{term "UNIV // \<approx>(L r)"} must be finite.\qed
\end{proof}
\noindent
Much more interesting, however, are the inductive cases. They seem hard to solve
directly. The reader is invited to try.
Our proof will rely on some
\emph{tagging-functions} defined over strings. Given the inductive hypothesis, it will
be easy to prove that the \emph{range} of these tagging-functions is finite
(the range of a function @{text f} is defined as @{text "range f \<equiv> f ` UNIV"}).
With this we will be able to infer that the tagging-functions, seen as relations,
give rise to finitely many equivalence classes of @{const UNIV}. Finally we
will show that the tagging-relations are more refined than @{term "\<approx>(L r)"}, which
implies that @{term "UNIV // \<approx>(L r)"} must also be finite (a relation @{text "R\<^isub>1"}
is said to \emph{refine} @{text "R\<^isub>2"} provided @{text "R\<^isub>1 \<subseteq> R\<^isub>2"}).
We formally define the notion of a \emph{tagging-relation} as follows.
\begin{definition}[Tagging-Relation] Given a tagging-function @{text tag}, then two strings @{text x}
and @{text y} are \emph{tag-related} provided
\begin{center}
@{text "x =tag= y \<equiv> tag x = tag y"}\;.
\end{center}
\end{definition}
In order to establish finiteness of a set @{text A}, we shall use the following powerful
principle from Isabelle/HOL's library.
%
\begin{equation}\label{finiteimageD}
@{thm[mode=IfThen] finite_imageD}
\end{equation}
\noindent
It states that if an image of a set under an injective function @{text f} (injective over this set)
is finite, then the set @{text A} itself must be finite. We can use it to establish the following
two lemmas.
\begin{lemma}\label{finone}
@{thm[mode=IfThen] finite_eq_tag_rel}
\end{lemma}
\begin{proof}
We set in \eqref{finiteimageD}, @{text f} to be @{text "X \<mapsto> tag ` X"}. We have
@{text "range f"} to be a subset of @{term "Pow (range tag)"}, which we know must be
finite by assumption. Now @{term "f (UNIV // =tag=)"} is a subset of @{text "range f"},
and so also finite. Injectivity amounts to showing that @{text "X = Y"} under the
assumptions that @{text "X, Y \<in> "}~@{term "UNIV // =tag="} and @{text "f X = f Y"}.
From the assumptions we can obtain @{text "x \<in> X"} and @{text "y \<in> Y"} with
@{text "tag x = tag y"}. Since @{text x} and @{text y} are tag-related, this in
turn means that the equivalence classes @{text X}
and @{text Y} must be equal.\qed
\end{proof}
\begin{lemma}\label{fintwo}
Given two equivalence relations @{text "R\<^isub>1"} and @{text "R\<^isub>2"}, whereby
@{text "R\<^isub>1"} refines @{text "R\<^isub>2"}.
If @{thm (prem 1) refined_partition_finite[where ?R1.0="R\<^isub>1" and ?R2.0="R\<^isub>2"]}
then @{thm (concl) refined_partition_finite[where ?R1.0="R\<^isub>1" and ?R2.0="R\<^isub>2"]}.
\end{lemma}
\begin{proof}
We prove this lemma again using \eqref{finiteimageD}. This time we set @{text f} to
be @{text "X \<mapsto>"}~@{term "{R\<^isub>1 `` {x} | x. x \<in> X}"}. It is easy to see that
@{term "finite (f ` (UNIV // R\<^isub>2))"} because it is a subset of @{term "Pow (UNIV // R\<^isub>1)"},
which is finite by assumption. What remains to be shown is that @{text f} is injective
on @{term "UNIV // R\<^isub>2"}. This is equivalent to showing that two equivalence
classes, say @{text "X"} and @{text Y}, in @{term "UNIV // R\<^isub>2"} are equal, provided
@{text "f X = f Y"}. For @{text "X = Y"} to be equal, we have to find two elements
@{text "x \<in> X"} and @{text "y \<in> Y"} such that they are @{text R\<^isub>2} related.
We know there exists a @{text "x \<in> X"} with \mbox{@{term "X = R\<^isub>2 `` {x}"}}.
From the latter fact we can infer that @{term "R\<^isub>1 ``{x} \<in> f X"}
and further @{term "R\<^isub>1 ``{x} \<in> f Y"}. This means we can obtain a @{text y}
such that @{term "R\<^isub>1 `` {x} = R\<^isub>1 `` {y}"} holds. Consequently @{text x} and @{text y}
are @{text "R\<^isub>1"}-related. Since by assumption @{text "R\<^isub>1"} refines @{text "R\<^isub>2"},
they must also be @{text "R\<^isub>2"}-related, as we need to show.\qed
\end{proof}
\noindent
Chaining Lem.~\ref{finone} and \ref{fintwo} together, means in order to show
that @{term "UNIV // \<approx>(L r)"} is finite, we have to find a tagging-function whose
range can be shown to be finite and whose tagging-relation refines @{term "\<approx>(L r)"}.
Let us attempt the @{const ALT}-case first.
\begin{proof}[@{const "ALT"}-Case]
We take as tagging-function
%
\begin{center}
@{thm tag_str_Plus_def[where A="A" and B="B", THEN meta_eq_app]}
\end{center}
\noindent
where @{text "A"} and @{text "B"} are some arbitrary languages.
We can show in general, if @{term "finite (UNIV // \<approx>A)"} and @{term "finite (UNIV // \<approx>B)"}
then @{term "finite ((UNIV // \<approx>A) \<times> (UNIV // \<approx>B))"} holds. The range of
@{term "tag_str_ALT A B"} is a subset of this product set---so finite. It remains to be shown
that @{text "=tag\<^isub>A\<^isub>L\<^isub>T A B="} refines @{term "\<approx>(A \<union> B)"}. This amounts to
showing
%
\begin{center}
@{term "tag\<^isub>A\<^isub>L\<^isub>T A B x = tag\<^isub>A\<^isub>L\<^isub>T A B y \<longrightarrow> x \<approx>(A \<union> B) y"}
\end{center}
%
\noindent
which by unfolding the Myhill-Nerode relation is identical to
%
\begin{equation}\label{pattern}
@{text "\<forall>z. tag\<^isub>A\<^isub>L\<^isub>T A B x = tag\<^isub>A\<^isub>L\<^isub>T A B y \<and> x @ z \<in> A \<union> B \<longrightarrow> y @ z \<in> A \<union> B"}
\end{equation}
%
\noindent
since both @{text "=tag\<^isub>A\<^isub>L\<^isub>T A B="} and @{term "\<approx>(A \<union> B)"} are symmetric. To solve
\eqref{pattern} we just have to unfold the definition of the tagging-function and analyse
in which set, @{text A} or @{text B}, the string @{term "x @ z"} is.
The definition of the tagging-function will give us in each case the
information to infer that @{text "y @ z \<in> A \<union> B"}.
Finally we
can discharge this case by setting @{text A} to @{term "L r\<^isub>1"} and @{text B} to @{term "L r\<^isub>2"}.\qed
\end{proof}
\noindent
The pattern in \eqref{pattern} is repeated for the other two cases. Unfortunately,
they are slightly more complicated. In the @{const SEQ}-case we essentially have
to be able to infer that
%
\begin{center}
@{text "\<dots>"}@{term "x @ z \<in> A \<cdot> B \<longrightarrow> y @ z \<in> A \<cdot> B"}
\end{center}
%
\noindent
using the information given by the appropriate tagging-function. The complication
is to find out what the possible splits of @{text "x @ z"} are to be in @{term "A \<cdot> B"}
(this was easy in case of @{term "A \<union> B"}). To deal with this complication we define the
notions of \emph{string prefixes}
%
\begin{center}
@{text "x \<le> y \<equiv> \<exists>z. y = x @ z"}\hspace{10mm}
@{text "x < y \<equiv> x \<le> y \<and> x \<noteq> y"}
\end{center}
%
\noindent
and \emph{string subtraction}:
%
\begin{center}
@{text "[] - y \<equiv> []"}\hspace{10mm}
@{text "x - [] \<equiv> x"}\hspace{10mm}
@{text "cx - dy \<equiv> if c = d then x - y else cx"}
\end{center}
%
\noindent
where @{text c} and @{text d} are characters, and @{text x} and @{text y} are strings.
Now assuming @{term "x @ z \<in> A \<cdot> B"} there are only two possible ways of how to `split'
this string to be in @{term "A \<cdot> B"}:
%
\begin{center}
\begin{tabular}{@ {}c@ {\hspace{10mm}}c@ {}}
\scalebox{0.7}{
\begin{tikzpicture}
\node[draw,minimum height=3.8ex] (xa) { $\hspace{3em}@{text "x'"}\hspace{3em}$ };
\node[draw,minimum height=3.8ex, right=-0.03em of xa] (xxa) { $\hspace{0.2em}@{text "x - x'"}\hspace{0.2em}$ };
\node[draw,minimum height=3.8ex, right=-0.03em of xxa] (z) { $\hspace{5em}@{text z}\hspace{5em}$ };
\draw[decoration={brace,transform={yscale=3}},decorate]
(xa.north west) -- ($(xxa.north east)+(0em,0em)$)
node[midway, above=0.5em]{@{text x}};
\draw[decoration={brace,transform={yscale=3}},decorate]
(z.north west) -- ($(z.north east)+(0em,0em)$)
node[midway, above=0.5em]{@{text z}};
\draw[decoration={brace,transform={yscale=3}},decorate]
($(xa.north west)+(0em,3ex)$) -- ($(z.north east)+(0em,3ex)$)
node[midway, above=0.8em]{@{term "x @ z \<in> A \<cdot> B"}};
\draw[decoration={brace,transform={yscale=3}},decorate]
($(z.south east)+(0em,0ex)$) -- ($(xxa.south west)+(0em,0ex)$)
node[midway, below=0.5em]{@{term "(x - x') @ z \<in> B"}};
\draw[decoration={brace,transform={yscale=3}},decorate]
($(xa.south east)+(0em,0ex)$) -- ($(xa.south west)+(0em,0ex)$)
node[midway, below=0.5em]{@{term "x' \<in> A"}};
\end{tikzpicture}}
&
\scalebox{0.7}{
\begin{tikzpicture}
\node[draw,minimum height=3.8ex] (x) { $\hspace{4.8em}@{text x}\hspace{4.8em}$ };
\node[draw,minimum height=3.8ex, right=-0.03em of x] (za) { $\hspace{0.6em}@{text "z'"}\hspace{0.6em}$ };
\node[draw,minimum height=3.8ex, right=-0.03em of za] (zza) { $\hspace{2.6em}@{text "z - z'"}\hspace{2.6em}$ };
\draw[decoration={brace,transform={yscale=3}},decorate]
(x.north west) -- ($(za.north west)+(0em,0em)$)
node[midway, above=0.5em]{@{text x}};
\draw[decoration={brace,transform={yscale=3}},decorate]
($(za.north west)+(0em,0ex)$) -- ($(zza.north east)+(0em,0ex)$)
node[midway, above=0.5em]{@{text z}};
\draw[decoration={brace,transform={yscale=3}},decorate]
($(x.north west)+(0em,3ex)$) -- ($(zza.north east)+(0em,3ex)$)
node[midway, above=0.8em]{@{term "x @ z \<in> A \<cdot> B"}};
\draw[decoration={brace,transform={yscale=3}},decorate]
($(za.south east)+(0em,0ex)$) -- ($(x.south west)+(0em,0ex)$)
node[midway, below=0.5em]{@{text "x @ z' \<in> A"}};
\draw[decoration={brace,transform={yscale=3}},decorate]
($(zza.south east)+(0em,0ex)$) -- ($(za.south east)+(0em,0ex)$)
node[midway, below=0.5em]{@{text "(z - z') \<in> B"}};
\end{tikzpicture}}
\end{tabular}
\end{center}
%
\noindent
Either there is a prefix of @{text x} in @{text A} and the rest is in @{text B} (first picture),
or @{text x} and a prefix of @{text "z"} is in @{text A} and the rest in @{text B} (second picture).
In both cases we have to show that @{term "y @ z \<in> A \<cdot> B"}. For this we use the
following tagging-function
%
\begin{center}
@{thm tag_str_Times_def[where ?L1.0="A" and ?L2.0="B", THEN meta_eq_app]}
\end{center}
\noindent
with the idea that in the first split we have to make sure that @{text "(x - x') @ z"}
is in the language @{text B}.
\begin{proof}[@{const SEQ}-Case]
If @{term "finite (UNIV // \<approx>A)"} and @{term "finite (UNIV // \<approx>B)"}
then @{term "finite ((UNIV // \<approx>A) \<times> (Pow (UNIV // \<approx>B)))"} holds. The range of
@{term "tag_str_SEQ A B"} is a subset of this product set, and therefore finite.
We have to show injectivity of this tagging-function as
%
\begin{center}
@{term "\<forall>z. tag_str_SEQ A B x = tag_str_SEQ A B y \<and> x @ z \<in> A \<cdot> B \<longrightarrow> y @ z \<in> A \<cdot> B"}
\end{center}
%
\noindent
There are two cases to be considered (see pictures above). First, there exists
a @{text "x'"} such that
@{text "x' \<in> A"}, @{text "x' \<le> x"} and @{text "(x - x') @ z \<in> B"} hold. We therefore have
%
\begin{center}
@{term "(\<approx>B `` {x - x'}) \<in> ({\<approx>B `` {x - x'} |x'. x' \<le> x \<and> x' \<in> A})"}
\end{center}
%
\noindent
and by the assumption about @{term "tag_str_SEQ A B"} also
%
\begin{center}
@{term "(\<approx>B `` {x - x'}) \<in> ({\<approx>B `` {y - y'} |y'. y' \<le> y \<and> y' \<in> A})"}
\end{center}
%
\noindent
That means there must be a @{text "y'"} such that @{text "y' \<in> A"} and
@{term "\<approx>B `` {x - x'} = \<approx>B `` {y - y'}"}. This equality means that
@{term "(x - x') \<approx>B (y - y')"} holds. Unfolding the Myhill-Nerode
relation and together with the fact that @{text "(x - x') @ z \<in> B"}, we
have @{text "(y - y') @ z \<in> B"}. We already know @{text "y' \<in> A"}, therefore
@{term "y @ z \<in> A \<cdot> B"}, as needed in this case.
Second, there exists a @{text "z'"} such that @{term "x @ z' \<in> A"} and @{text "z - z' \<in> B"}.
By the assumption about @{term "tag_str_SEQ A B"} we have
@{term "\<approx>A `` {x} = \<approx>A `` {y}"} and thus @{term "x \<approx>A y"}. Which means by the Myhill-Nerode
relation that @{term "y @ z' \<in> A"} holds. Using @{text "z - z' \<in> B"}, we can conclude also in this case
with @{term "y @ z \<in> A \<cdot> B"}. We again can complete the @{const SEQ}-case
by setting @{text A} to @{term "L r\<^isub>1"} and @{text B} to @{term "L r\<^isub>2"}.\qed
\end{proof}
\noindent
The case for @{const STAR} is similar to @{const SEQ}, but poses a few extra challenges. When
we analyse the case that @{text "x @ z"} is an element in @{term "A\<star>"} and @{text x} is not the
empty string, we
have the following picture:
%
\begin{center}
\scalebox{0.7}{
\begin{tikzpicture}
\node[draw,minimum height=3.8ex] (xa) { $\hspace{4em}@{text "x'\<^isub>m\<^isub>a\<^isub>x"}\hspace{4em}$ };
\node[draw,minimum height=3.8ex, right=-0.03em of xa] (xxa) { $\hspace{0.5em}@{text "x - x'\<^isub>m\<^isub>a\<^isub>x"}\hspace{0.5em}$ };
\node[draw,minimum height=3.8ex, right=-0.03em of xxa] (za) { $\hspace{2em}@{text "z\<^isub>a"}\hspace{2em}$ };
\node[draw,minimum height=3.8ex, right=-0.03em of za] (zb) { $\hspace{7em}@{text "z\<^isub>b"}\hspace{7em}$ };
\draw[decoration={brace,transform={yscale=3}},decorate]
(xa.north west) -- ($(xxa.north east)+(0em,0em)$)
node[midway, above=0.5em]{@{text x}};
\draw[decoration={brace,transform={yscale=3}},decorate]
(za.north west) -- ($(zb.north east)+(0em,0em)$)
node[midway, above=0.5em]{@{text z}};
\draw[decoration={brace,transform={yscale=3}},decorate]
($(xa.north west)+(0em,3ex)$) -- ($(zb.north east)+(0em,3ex)$)
node[midway, above=0.8em]{@{term "x @ z \<in> A\<star>"}};
\draw[decoration={brace,transform={yscale=3}},decorate]
($(za.south east)+(0em,0ex)$) -- ($(xxa.south west)+(0em,0ex)$)
node[midway, below=0.5em]{@{text "(x - x'\<^isub>m\<^isub>a\<^isub>x) @ z\<^isub>a \<in> A"}};
\draw[decoration={brace,transform={yscale=3}},decorate]
($(xa.south east)+(0em,0ex)$) -- ($(xa.south west)+(0em,0ex)$)
node[midway, below=0.5em]{@{term "x'\<^isub>m\<^isub>a\<^isub>x \<in> A\<star>"}};
\draw[decoration={brace,transform={yscale=3}},decorate]
($(zb.south east)+(0em,0ex)$) -- ($(zb.south west)+(0em,0ex)$)
node[midway, below=0.5em]{@{term "z\<^isub>b \<in> A\<star>"}};
\draw[decoration={brace,transform={yscale=3}},decorate]
($(zb.south east)+(0em,-4ex)$) -- ($(xxa.south west)+(0em,-4ex)$)
node[midway, below=0.5em]{@{term "(x - x'\<^isub>m\<^isub>a\<^isub>x) @ z \<in> A\<star>"}};
\end{tikzpicture}}
\end{center}
%
\noindent
We can find a strict prefix @{text "x'"} of @{text x} such that @{term "x' \<in> A\<star>"},
@{text "x' < x"} and the rest @{term "(x - x') @ z \<in> A\<star>"}. For example the empty string
@{text "[]"} would do.
There are potentially many such prefixes, but there can only be finitely many of them (the
string @{text x} is finite). Let us therefore choose the longest one and call it
@{text "x'\<^isub>m\<^isub>a\<^isub>x"}. Now for the rest of the string @{text "(x - x'\<^isub>m\<^isub>a\<^isub>x) @ z"} we
know it is in @{term "A\<star>"}. By definition of @{term "A\<star>"}, we can separate
this string into two parts, say @{text "a"} and @{text "b"}, such that @{text "a \<in> A"}
and @{term "b \<in> A\<star>"}. Now @{text a} must be strictly longer than @{text "x - x'\<^isub>m\<^isub>a\<^isub>x"},
otherwise @{text "x'\<^isub>m\<^isub>a\<^isub>x"} is not the longest prefix. That means @{text a}
`overlaps' with @{text z}, splitting it into two components @{text "z\<^isub>a"} and
@{text "z\<^isub>b"}. For this we know that @{text "(x - x'\<^isub>m\<^isub>a\<^isub>x) @ z\<^isub>a \<in> A"} and
@{term "z\<^isub>b \<in> A\<star>"}. To cut a story short, we have divided @{term "x @ z \<in> A\<star>"}
such that we have a string @{text a} with @{text "a \<in> A"} that lies just on the
`border' of @{text x} and @{text z}. This string is @{text "(x - x'\<^isub>m\<^isub>a\<^isub>x) @ z\<^isub>a"}.
In order to show that @{term "x @ z \<in> A\<star>"} implies @{term "y @ z \<in> A\<star>"}, we use
the following tagging-function:
%
\begin{center}
@{thm tag_str_Star_def[where ?L1.0="A", THEN meta_eq_app]}\smallskip
\end{center}
\begin{proof}[@{const STAR}-Case]
If @{term "finite (UNIV // \<approx>A)"}
then @{term "finite (Pow (UNIV // \<approx>A))"} holds. The range of
@{term "tag_str_STAR A"} is a subset of this set, and therefore finite.
Again we have to show injectivity of this tagging-function as
%
\begin{center}
@{term "\<forall>z. tag_str_STAR A x = tag_str_STAR A y \<and> x @ z \<in> A\<star> \<longrightarrow> y @ z \<in> A\<star>"}
\end{center}
%
\noindent
We first need to consider the case that @{text x} is the empty string.
From the assumption we can infer @{text y} is the empty string and
clearly have @{term "y @ z \<in> A\<star>"}. In case @{text x} is not the empty
string, we can divide the string @{text "x @ z"} as shown in the picture
above. By the tagging-function we have
%
\begin{center}
@{term "\<approx>A `` {(x - x'\<^isub>m\<^isub>a\<^isub>x)} \<in> ({\<approx>A `` {x - x'} |x'. x' < x \<and> x' \<in> A\<star>})"}
\end{center}
%
\noindent
which by assumption is equal to
%
\begin{center}
@{term "\<approx>A `` {(x - x'\<^isub>m\<^isub>a\<^isub>x)} \<in> ({\<approx>A `` {y - y'} |y'. y' < y \<and> y' \<in> A\<star>})"}
\end{center}
%
\noindent
and we know that we have a @{term "y' \<in> A\<star>"} and @{text "y' < y"}
and also know @{term "(x - x'\<^isub>m\<^isub>a\<^isub>x) \<approx>A (y - y')"}. Unfolding the Myhill-Nerode
relation we know @{term "(y - y') @ z\<^isub>a \<in> A"}. We also know that @{term "z\<^isub>b \<in> A\<star>"}.
Therefore @{term "y' @ ((y - y') @ z\<^isub>a) @ z\<^isub>b \<in> A\<star>"}, which means
@{term "y @ z \<in> A\<star>"}. As the last step we have to set @{text "A"} to @{term "L r"} and
complete the proof.\qed
\end{proof}
*}
section {* Conclusion and Related Work *}
text {*
In this paper we took the view that a regular language is one where there
exists a regular expression that matches all of its strings. Regular
expressions can conveniently be defined as a datatype in HOL-based theorem
provers. For us it was therefore interesting to find out how far we can push
this point of view. We have established in Isabelle/HOL both directions
of the Myhill-Nerode theorem.
%
\begin{theorem}[The Myhill-Nerode Theorem]\mbox{}\\
A language @{text A} is regular if and only if @{thm (rhs) Myhill_Nerode}.
\end{theorem}
%
\noindent
Having formalised this theorem means we
pushed our point of view quite far. Using this theorem we can obviously prove when a language
is \emph{not} regular---by establishing that it has infinitely many
equivalence classes generated by the Myhill-Nerode relation (this is usually
the purpose of the pumping lemma \cite{Kozen97}). We can also use it to
establish the standard textbook results about closure properties of regular
languages. Interesting is the case of closure under complement, because
it seems difficult to construct a regular expression for the complement
language by direct means. However the existence of such a regular expression
can be easily proved using the Myhill-Nerode theorem since
%
\begin{center}
@{term "s\<^isub>1 \<approx>A s\<^isub>2"} if and only if @{term "s\<^isub>1 \<approx>(-A) s\<^isub>2"}
\end{center}
%
\noindent
holds for any strings @{text "s\<^isub>1"} and @{text
"s\<^isub>2"}. Therefore @{text A} and the complement language @{term "-A"} give rise to the same
partitions. Proving the existence of such a regular expression via automata
using the standard method would
be quite involved. It includes the
steps: regular expression @{text "\<Rightarrow>"} non-deterministic automaton @{text
"\<Rightarrow>"} deterministic automaton @{text "\<Rightarrow>"} complement automaton @{text "\<Rightarrow>"}
regular expression.
While regular expressions are convenient in formalisations, they have some
limitations. One is that there seems to be no method of calculating a
minimal regular expression (for example in terms of length) for a regular
language, like there is
for automata. On the other hand, efficient regular expression matching,
without using automata, poses no problem \cite{OwensReppyTuron09}.
For an implementation of a simple regular expression matcher,
whose correctness has been formally established, we refer the reader to
Owens and Slind \cite{OwensSlind08}.
Our formalisation consists of 780 lines of Isabelle/Isar code for the first
direction and 460 for the second, plus around 300 lines of standard material about
regular languages. While this might be seen as too large to count as a
concise proof pearl, this should be seen in the context of the work done by
Constable at al \cite{Constable00} who formalised the Myhill-Nerode theorem
in Nuprl using automata. They write that their four-member team needed
something on the magnitude of 18 months for their formalisation. The
estimate for our formalisation is that we needed approximately 3 months and
this included the time to find our proof arguments. Unlike Constable et al,
who were able to follow the proofs from \cite{HopcroftUllman69}, we had to
find our own arguments. So for us the formalisation was not the
bottleneck. It is hard to gauge the size of a formalisation in Nurpl, but
from what is shown in the Nuprl Math Library about their development it
seems substantially larger than ours. The code of ours can be found in the
Mercurial Repository at
\mbox{\url{http://www4.in.tum.de/~urbanc/regexp.html}}.
Our proof of the first direction is very much inspired by \emph{Brzozowski's
algebraic method} used to convert a finite automaton to a regular
expression \cite{Brzozowski64}. The close connection can be seen by considering the equivalence
classes as the states of the minimal automaton for the regular language.
However there are some subtle differences. Since we identify equivalence
classes with the states of the automaton, then the most natural choice is to
characterise each state with the set of strings starting from the initial
state leading up to that state. Usually, however, the states are characterised as the
strings starting from that state leading to the terminal states. The first
choice has consequences about how the initial equational system is set up. We have
the $\lambda$-term on our `initial state', while Brzozowski has it on the
terminal states. This means we also need to reverse the direction of Arden's
Lemma.
We briefly considered using the method Brzozowski presented in the Appendix
of~\cite{Brzozowski64} in order to prove the second direction of the
Myhill-Nerode theorem. There he calculates the derivatives for regular
expressions and shows that for every language there can be only
finitely many of them %derivations
(if regarded equal modulo ACI). We could
have used as tagging-function the set of derivatives of a regular expression
with respect to a language. Using the fact that two strings are
Myhill-Nerode related whenever their derivative is the same, together with
the fact that there are only finitely such derivatives
would give us a similar argument as ours. However it seems not so easy to
calculate the set of derivatives modulo ACI. Therefore we preferred our
direct method of using tagging-functions. This
is also where our method shines, because we can completely side-step the
standard argument \cite{Kozen97} where automata need to be composed, which
as stated in the Introduction is not so easy to formalise in a
HOL-based theorem prover. However, it is also the direction where we had to
spend most of the `conceptual' time, as our proof-argument based on tagging-functions
is new for establishing the Myhill-Nerode theorem. All standard proofs
of this direction use %proceed by
arguments over automata.\\[-6mm]%\medskip
%
%\noindent
%{\bf Acknowledgements:} We are grateful for the comments we received from Larry
%Paulson and the referees of the paper.
*}
(*<*)
end
(*>*)