Technische Universit¨At Dresden

TECHNISCHE UNIVERSITATÈ DRESDEN

FakultatÈ Informatik

TUD/ FI98/ 09 - October1998

Daniel Kirsten Technische Berichte Grundlagen der Programmierung Technical Reports Institut fÈur Softwaretechnik I A Connection between the Star Problem ISSN 1430-211X and the Finite Power Property in Trace Monoids

Technische UniversitatÈ Dresden FakultatÈ Informatik D-01062 Dresden Germany

URL: http://www.inf.tu-dresden.de/

A Connection b etween the Star Problem

and the Finite Power Prop erty

in Trace Monoids

Daniel Kirsten

Department of Computer Science

Dresden University of Technology

D Dresden Germany

DanielKirsteninftudresdende

httpwwwinftudresdendedk

Octob er

Abstract

This pap er deals with a connection b etween two decision problems for recognizable trace

languages the star problem and the nite p ower prop erty problem Due to a theorem by

Richomme from we know that b oth problems are decidable in trace monoids

which do not contain a C submonoid It is not known whether the star problem or the

nite p ower prop erty are decidable in the C or in trace monoids containing a C

In this pap er we show a new connection b etween these problems Assume a trace

monoid IM I which is isomorphic to the Cartesian Pro duct of two disjoint trace monoids

IM I and IM I Assume further a recognizable language L in IM I such that

1 1 2 2

every trace in L contains at least one letter in and at least one letter in Then the

1 2

main theorem of this pap er asserts that L is recognizable i L has the nite p ower prop erty

This work has b een supp orted by the p ostgraduate program Sp ecication of discrete pro cesses and

systems of pro cesses by op erational mo dels and logics of the German Research Community Deutsche

Forschungsgemeinsch aft

Contents

Intro duction

Formal Overview

Monoids Languages and Traces

Recognizable Sets

Some Decision Problems for Trace Languages

Main Results

Semigroups and Ideals

Basic Denitions and Notions

Finite Semigroups without Prop er Ideals

Finite Semigroups with Prop er Ideals

A Classication

Pro duct Automata

Denitions

Connections to Subsemigroups and Ideals

Pro of of Theorem

Q do es not have prop er left ideals

Q fullls assertion C

Q fullls assertion D

Q fullls assertion B

Completion of the Pro of

Conclusions and Future Steps

Acknowledgments

Intro duction

Free partially commutative monoids also called trace monoids were intro duced by Cartier

and Foata in In Mazurkiewicz prop osed trace monoids as a p otential mo del

for concurrent pro cesses which marks the b eginning of a systematic study of trace monoids

by mathematicians and theoretical computer scientists see eg the recent surveys

A part of the research in trace theory deals with examinations of wellknown classic results for

free monoids in the framework of traces

One main stream in trace theory is the study of recognizable trace languages which can

b e considered as an extension of the well studied concept of regular languages in free monoids

A ma jor step in this research is Ochmanskys PhD thesis from Some of the results

concerning regular languages in free monoids can b e generalized to recognizable languages in

trace monoids However there is one ma jor dierence The iteration of a recognizable trace

language do es not necessarily yield a recognizable language This fact raises the so called star

problem Given a recognizable language L is L recognizable In general it is not known

whether the star problem is decidable

The nite p ower prop erty problem for short FPPP is related to the star problem Given a

recognizable language L has L the nite p ower prop erty for short FPP ie is there a natural

numb er n such that L L L L

The strategy to achieve partial solutions of these problems by examining connections b etween

them turned out to b e fruitful The main result after a stream of publications is a theorem stated

by Richomme in saying that b oth the star problem and the FPP are decidable in trace

monoids which do not contain a particular submonoid called C It is not known whether

the star problem and the FPP are decidable in trace monoids with a C submonoid Although the

FPP for nite trace languages is obviously decidable decidability of the star problem for nite

trace languages is only known for just a few sp ecial cases

In this pap er we establish a new connection b etween the star problem and the FPP We deal

with trace monoids which are isomorphic to Cartesian Pro ducts of two trace monoids We show

that for a certain class of recognizable languages in Cartesian Pro ducts the iteration of a

language is recognizable i the language has the FPP

The pap er is organized as follows In Section we show some concepts from algebra and

formal language theory In the rst subsection we get familiar with some basic notions from

algebra formal language theory and trace theory In the next subsection we deal with recog

nizable sets and rational sets Then we discuss the two main problems for recognizable trace

languages their connections and solutions as far as known After this discussion I explain the

main results of this pap er

In Section and we deal with concepts which we require to prove the main results

In Section we get familiar with ideal theory which will b e a crucial to ol In Section we deal

with pro duct automata

After these preliminary to ols we prove the main theorems in Section In several subsec

tions we deal with sp ecial cases In the last subsection we summarize these sp ecial cases

Formal Overview

Monoids Languages and Traces

I briey intro duce the basic notions from algebra and trace theory Unless I do not state precise

sources I consider the concepts and notions as wellknown

By IN we denote the set of natural numb ers including zero ie IN f g

We say a set K is a subset of a set L and we denote this by K L i every element of K

FORMAL OVERVIEW

b elongs to L We say K is a proper subset of L and we denote this by K L i every element

of K b elongs to L and additionally there is an element in L which do es not b elong to K

A semigroup S is an algebraic structure consisting of a set S and a binary op eration

S S

such that is asso ciative ie for every p q r S we have p q r p q r We call S

S S S S S

the underlying set and we call the operation or product of S Usually we drop the symb ol

S S

and denote the pro duct by juxtap osition We call a semigroup S nite i S is a nite set

We use the letter S to denote b oth the semigroup and its underlying set

We call a semigroup IM a monoid i IM contains an element such that for every

IM IM

m IM we have m m m We call the neutral element We drop the

IM IM IM IM IM

index at as long as no confusion arises

In the rest of this subsection we assume some semigroup S and some monoid IM

For every natural numb er n we dene the nfold pro duct as follows For every p in S

n n

p yields p itself and further for n p denotes p p For every monoid IM we extend this

denition by claiming that for every p IM we have p

We extend the pro duct to subsets of S If K and L are two subsets of S the set KL contains

all elements k l for some k in K and l in L We extend the nfold pro duct to sets We dene

n n

K K and for n we set K K K For every subset K of IM we dene K f g

For every subset K of S we dene the nonempty iteration K as the union of the sets K

K K For every subset K of IM we additionally dene the iteration of K by K and

dene it by K K f g

n n

For every subset K of S and for every n we abbreviate the union K K by K

n n

For every subset K of IM and for every n we abbreviate K K by K

Assume two semigroups S and S Their Cartesian Pro duct is the semigroup denoted by

and dened in the following way The underlying set is the Cartesian Pro duct of the underlying

sets of S and S We dene the op eration comp onentwise ie for every pair of elements

p p p

1 2 2

The pro ducts p p and p p are the pro ducts in S and S their pro duct yields and

0 0 0

p p p

1 2 2

resp ectively The Cartesian Pro duct of two monoids yields a monoid

Again assume two semigroups S and S We call a function h S S a homomorphism

hl hk l We call a i h preserves the pro duct ie for every k l S we have hk

S S

homomorphism h b etween two monoids IM and IM a monoid homomorphism i h preserves the

There are homomorphisms b etween two monoids neutral element ie we have h

which do not preserve the neutral element

We extend the notion of homomorphisms to sets If K is a subset of S we dene hK as

the set of all q S such that for some p K we have hp q We denote the inverse of h

by h We dene it on subsets of S If L is a subset of S h L yields the set of all m S

such that hm L We call h a surjective homomorphism from S to S i hS S We call h

an isomorphism i for every p S the set h fpg is a singleton Then we can regard h as

a homomorphism from S to S We call the semigroups isomorphic i an isomorphism b etween

S and S exists

A sp ecial kind of homomorphisms are the pro jections Assume two semigroups S and S

We dene the projections and from the Cartesian Pro duct to S and S resp ectively

p p S p

q p and we dene For every

q q S q

Assume three semigroups S S and S Assume two homomorphisms g S S and

h S S We denote the composition of g and h by h g It yields a homomorphism from

S to S For every p S h g p yields hg p ie we apply g on p and we apply h on the

result of g p

Assume four semigroups S S S and S Assume two homomorphisms g S S and

g S

h S S We can dene a homomorphism from as follows For every to

h S

2 2

Monoids Languages and Traces

g p

p S g p

If b oth g and h are surjective homomorphisms from S to S yields

q S h q

S g

to is a surjective homomorphism from and S to S resp then As an exercise

S h

g S

you can verify that the homomorphisms h and from to S are identical

h S

By an alphabet we mean a nite set of symb ols We call its elements letters Assume an

alphab et We denote the free monoid over by Its underlying set is the set of all words

strings consisting of letters of the monoid pro duct is the concatenation and the neutral

element is the empty string We call the subsets of the free monoid languages For every word

w in we call the numb er of letters of w the length of w and denote it by jw j

Cartier and Foata intro duced the concept of the free partially commutative monoids

in In Mazurkiewicz considered this concept as a p otential mo del for con

current systems Since then free partially commutative monoids are examined by b oth

mathematicians and theoretical computer scientists For a general overview I recommend the

surveys

Assume an alphab et We call a binary relation I over an independence relation i I is

irreexive and symmetric For every pair of letters a and b with aI b we say that a and b are

independent otherwise a and b are dependent We call the pair I an independence alphabet

We call two words w w in equivalent i we can transform w into w by nitely many

exchanges of indep endent adjacent letters which we denote by w w For instance if a and

c are indep endent letters baacbac bacabac and bcaabca are mutually equivalent words

The relation is an equivalence relation For every word w in we denote by w the

I I

equivalence class of w Moreover is a congruence relation This means for every words

w w w w in with w w and w w we have w w w w Therefore we can

I I I

dene a monoid with the sets w as elements For any words w and w we dene the pro duct

of w and w by w w We denote this monoid by IM I and call it the trace monoid

I I I

over and I Its elements ie the equivalence classes w are called traces and its subsets are

called trace languages or shortly languages The function is a homomorphism from the free

monoid to IM I As long as no confusion arises we omit the index I at

If I is the empty relation over the trace monoid IM I is the free monoid If I is

the largest irreexive relation over ie two letters a and b are indep endent i a and b are

dierent then the trace monoid IM I is the free commutative monoid over

Assume two disjoint indep endence alphab ets I and I We can dene another

indep endenc e alphab et I by merging the alphab ets I and I in the following

way I I I Two letters a and b in are

indep endent i either they do not b elong to the same alphab et or b oth a and b b elong to

resp and we have aI b resp aI b The trace monoid IM I is naturally isomorphic to

IM I

1 1

Consequently we can regard the Cartesian Pro duct of two the Cartesian Pro duct

IM I

2 2

trace monoids as a trace monoid

We dene the trace monoid P by the indep endence alphab et consisting of fa b cg

n o

facg

and I a b b a c b b c The P is isomorphic to the Cartesian Pro duct

fbg

We dene the trace monoid C by the indep endenc e alphab et consisting of fa b c dg

n o

facg

and I I a d d a c d d c The C is isomorphic to

C P

fbdg

Assume an indep endence alphab et I A trace t in IM I is called connected i for

every nonempty traces t and t with t t t there is a letter a o ccurring in t and there is a

letter b o ccurring in t such that a and b are dep endent A trace language L is called connected

i every trace in L is connected A trace in P or C is connected i u or v is the empty

word

Assume some trace monoid IM I Assume a trace language T in IM I such that

for every traces t t T their concatenation t t b elongs to T Then we say that T is

FORMAL OVERVIEW

concatenation closed Moreover T is a semigroup If additionally T then T is a monoid

Consequently it is quite natural to ask for generators of T

Denition Assume a trace monoid IM I and some concatenation closed language T

in IM I such that T The set of generators of T is dened by GenT T n T 2

Consequently the generators of T are the traces in T which cannot b e factorized into some

traces in T

Lemma Assume a trace monoid IM I and a concatenation closed language T in IM I

such that T We have

GenT T and

for every language L IM I with L T we have GenT L 2

Pro of At rst we show assertion We have GenT T Hence we have Gen T T

Because T is concatenation closed we have T T Consequently we have GenT T

We show that T is a subset of GenT by a contradiction Assume that there are traces

in T n GenT Then let t b e a smallest trace in T n GenT ie for every trace t in T

with jt j jtj we have t GenT The trace t do es not b elong to GenT Thus t do es

not b elong to GenT ie t do es not b elong to T n T Consequently t b elongs to T We can

factorize t into t and t in T We have jt j jtj and jt j jtj Hence the traces t and

t b elong to GenT but t do es not b elong to GenT This contradicts that GenT is

concatenation closed

Now we prove assertion by a contradiction Assume a language L such that L yields T

but GenT is not a subset of L Then let t b e a trace in GenT n L The trace t b elongs

to GenT Thus t b elongs to T ie t b elongs to L Because t do es not b elong to L we can

factorize t into two nonempty traces t and t in L Thus t b elongs to L This means t

b elongs to T Consequently t do es not b elong to T n T ie we contradicted that t b elongs

to GenT 2

Recognizable Sets

The concept of recognizability describ es a formal metho d how to use nite machines to deal with

innite ob jects It originates from Mezei and Wright from There are numerous

equivalent denitions I intro duce it as far as we use it in this pap er for a more general overview

I recommend I to ok most of the contents of this section from there

Denition Assume a monoid IM An IM automaton is a triple A Q h F where Q is

a nite monoid h is a homomorphism h IM Q and F is a subset of Q The language of an

IM automaton A is dened by LA h F 2

We call Q the underlying monoid of A and the elements of Q the states of A We further call

F the set of accepting states of A and h the homomorphism of A Without loss of generality

we can assume that h is a surjective homomorphism from IM to F If LA L then we

say that A denes L or A is an IMautomaton for L We call a subset L of IM a recognizable

language over IM i there is an IMautomaton A such that L LA We denote the class of

all recognizable languages over IM by RECIM

Denition shows a common way to dene recognizability in arbitrary monoids Following

Courcelle we call the triple Q h F an IMautomaton

Assume a pro duct automation A such that LA is closed under monoid pro duct Then

F is the image of the semigroup LA under the surjective homomorphism h Hence F is a

subsemigroup of Q

The following theorem is a classic one you nd the pro of eg in

Recognizable Sets

Theorem Assume a monoid IM The class RECIM contains the empty set IM itself

and it is closed under union intersection complement and inverse homomorphisms 2

There are monoids containing nite subsets which are not recognizable and monoids in which the

pro duct of two recognizable subsets is not always a recognizable set And further the iteration

of a recognizable set is not always recognizable However we have the following theorem for

trace monoids

Theorem Assume a trace monoid IM I The class REC IM I contains all nite

subsets of IM I and it is closed under monoid pro duct and under iteration of connected

trace languages 2

Recognizabili ty of nite trace languages is obvious The pro of of the closure under monoid

pro duct originates from Fliess Closure under iteration of connected trace languages is

due to Ochmansky Clerbout and Latteux and Metivier In you nd a

recent survey on recognizable trace languages it contains neat little pro ofs of the assertions in

Theorem

Example Assume the free monoids IM fag and IM fbg Let L b e the singleton

n o

a IM

language in the trace monoid The language L is recognizable b ecause it is nite

b IM

o n

n IN Below we show that Further L is not a connected language We have L

L is not recognizable 2

We need a theorem by Mezei concerning recognizable sets in Cartesian Pro ducts It is not

published by the author himself but it is widely known as Mezeis Theorem you nd it in

Theorem Assume two monoids IM and IM A set L is recognizable in i there are a

numb er n recognizable sets L L IM and recognizable sets L L IM such that

L L

2 L

L L

Example continued We show by a contradiction that L is not recognizable Assume

that L is recognizable Then by Theorem L is the union of nitely many Cartesian

j i

a a

b elong to the and Pro ducts Consequently there are two numb ers i j such that

j i

b b

j i

a a

b elong to this Cartesian Pro duct and same Cartesian Pro duct Hence the traces

i j

b b

ie they b elong to L This is a contradiction 2

Let us shortly mention the notion of rational sets Assume some monoid IM The class of rational

subsets of IM is the smallest class which contains the empty set and every singleton subset of IM

and is closed under union monoid pro duct and iteration We have Kleenes classic result

which asserts that in free monoids the recognizable sets and the rational sets coincide

In trace monoids we have just one direction Due to a more general result by McKnight

every recognizable trace language is a rational trace language However there are rational

trace languages which are not recognizable unless the underlying trace monoid is a free monoid

For instance the language L in Example is a rational language which is not recognizable

See for more information on rational trace languages

FORMAL OVERVIEW

Some Decision Problems for Trace Languages

Two decision problems concerning the iteration of recognizable trace languages arise

Star Problem Can we decide whether the iteration of a recognizable trace language

yields a recognizable language

Finite Power Prop erty Problem Can we decide whether a recognizable language has

the nite p ower prop erty ie for a recognizable language L can we decide whether there

is a natural numb er n such that we have L L

By FPP we abbreviate the nite p ower prop erty and by FPPP we abbreviate the nite p ower

prop erty problem If a recognizable language L has the FPP then we have L L L L

for some n IN Hence if L has the FPP then L is recognizable by Theorem and

Below we will see languages L such that L is recognizable but L do es not have the FPP

Although during the recent years many pap ers have dealt with the star problem and

the FPPP only partial results have b een achieved In general b oth problems have remained

unsolved I give a survey ab out their history

The star problem in the free monoid is trivial due to Kleenes Theorem from and it is

decidable in free commutative monoids due to Ginsburg and Spanier Brzozowski

raised the FPPP in the free monoid in and it to ok more than ten years till Simon and

Hashiguchi indep endently showed its decidability In Ochmansky examined

recognizable trace languages in his PhD thesis and stated the star problem During the

eighties Ochmansky Clerbout and Latteux and Metivier indep endently

proved that the iteration of a connected recognizable trace language yields a recognizable trace

language In Ochmansky showed the decidabili ty of the star problem in C for nite

trace languages containing at most one nonconnected trace He used the decidabili ty of

the FPP in free monoids This marks the b eginning of the examination of connections b etween

the FPP and the star problem In Sakarovitch solved the recognizability problem

Theorem Assume a trace monoid IM I We can decide whether a rational language in

IM I is recognizable i IM I do es not contain a P submonoid 2

This theorem includes that the star problem is decidable in trace monoids which do not contain

a P submonoid For a short time one had hop e to solve the star problem One conjectured

that the ab ove theorem can b e generalized to the star problem However just in the same

year Gastin Ochmansky Petit and Rozoy showed the decidability of the star problem

in P Decidability of the FPP in free monoids plays a crucial role in their pro of

During the subsequent years Metivier and Richomme develop ed these ideas They showed

decidabili ty of the FPP for connected trace languages and decidabili ty of the star problem for

trace languages containing at most four traces as well as for nite languages containing at most

two connected traces They showed the following connections b etween the star problem

and the FPP

Theorem Assume an indep endence alphab et I and a letter b with b

IMI

then the FPP is decidable in IM I If the star problem is decidable in

fbg

If b oth the star problem and the FPP are decidable in IM I then b oth problems are

IMI

2 decidable in

fbg

Assertion is a conclusion from the following connection

Main Results

Theorem Assume an indep endence alphab et I and a letter b with b Assume a

recognizable language T in IM I and let K Then the language T has the FPP i

fbg

K is recognizable 2

An obvious conclusion from assertion in Theorem is that if the star problem is decidable in

any trace monoid then so is the FPP Richomme improved the induction in assertion

and showed the following theorem

Theorem Assume a trace monoid IM I If the monoid IM I do es not contain a

Csubmonoid then the star problem and the FPP are decidable 2

The main ideas of the pro of are in the complete pro of is in Please note that the if in

the theorem has just one f

In Richomme tried to prove that the trace monoids with a decidable star problem are

exactly the trace monoids with a decidable FPP However one of the pro ofs in this rep ort

contains an error such that the result was not proved

The subsequent years were designated by stagnation One did not achieve new results and

one ceased the research on the star problem and the FPPP To day esp ecially two questions are

interesting

Are the star problem and the FPP decidable in trace monoids which contain a C sub

monoid

Are the trace monoids with a decidable FPP exactly the trace monoids with a decidable

star problem

Main Results

In this pap er we work on connections b etween the star problem and the FPP As already

mentioned for every recognizable language L the iteration of L is recognizable if L has the FPP

fag

The other direction is not true Let us consider some examples in the trace monoid

fbg

n o

Example Assume the language L Both L and L are recognizable

b ecause L is nite and L is the complete monoid However L do es not have the FPP

For every n IN the language L contains only traces consisting of at most n letters ie

L is nite and thus it is dierent from L 2

o n

fag fag

Both L We have L Example We examine the language L

fbg fbg

and L are recognizable Recognizabili ty of L follows eg from Theorem As well as L

the language L do es not p osses the FPP For every n IN L contains only traces in which

the letter a o ccurs at most n times ie L is dierent from L 2

n o

fag

We have L L Example Assume the recognizable language L

fbg

Hence L has the FPP We ask whether there is a language L such that L L and further L

do es not have the FPP We show that such a language L do es not exist Just assume a language

L such that L L

Assume some numb er n The trace b elongs to L ie it b elongs to L We cannot

a a

factorize into some nonempty traces in L Otherwise we could conclude that

n n

b b

a a

contains the letter a more than once Hence b elongs to L Accordingly the trace

b b

a a

b elong to L b elongs to L To sum up for every n the traces and

b b

SEMIGROUPS AND IDEALS

Now assume two numb ers n m The trace b elongs to L We do not know

and in L whether it b elongs to L However we can factorize it into the traces

n n

a a

such that b elongs to L Hence for every n m the trace b elongs to L

m m

b b

Thus we have L L Consequently for every language L such that L L we know

that L has the FPP Note that this statement is not restricted to recognizable languages L 2

By examining many similar examples one makes an observation Assume two disjoint trace

such that L monoids IM and IM Whenever one considers recognizable languages L in

is recognizable but L do es not have the FPP then certain traces play a crucial role Namely

traces which have an empty and a nonempty comp ound ie nonempty traces in and

play a crucial role This leads to Theorem which is the main result of this pap er

Richomme stated it in but the pro of there contains an error Up to now it remained

op en to correct the error or to disprove the theorem

Theorem Assume two disjoint indep endence alphab ets I and I Assume

IM I

1 1

such that every trace in L contains at least one some recognizable language L in

IM I

2 2

letter of and at least one letter of

Then the language L is recognizable i L has the FPP 2

We will prove Theorem as a corollary of the following theorem Its pro of is the main part

of the present pap er

Theorem Assume two disjoint indep endence alphab ets I and I Assume

IM I

1 1

such that every trace t in T some recognizable concatenation closed language T in

IM I

2 2

contains at least one letter of and at least one letter of

Then the set of generators of T has the FPP 2

We close this section by deriving Theorem from Theorem

Pro of of Theorem Assume some recognizable language L in such that every trace

in L contains at least one letter from and at least one letter from If L has the FPP

then L is recognizable b ecause of the closure prop erties of recognizable trace languages in

Theorem and

To show the other direction assume that L is recognizable We examine L It is also

recognizable Further every trace in L contains at least one letter of and at least one letter

of We can use Theorem on L Hence there is a natural numb er l such that we

have GenL L

l l

By Lemma we have GenL L Hence we have GenL L Thus we

l l l

have L L ie L L Consequently we have L L ie L has the FPP 2

Semigroups and Ideals

We deal with some notions on semigroups In the rst subsection we get familiar with ideals

Then we examine nonempty nite semigroups without and with prop er ideals In the last

subsection we use ideal theory to work out a useful classication of nonempty nite semigroups

To understand the rest of the pap er you have to b ecome familiar with the notions of left ideals

and ideals and you have to understand Lemma and Prop osition

Basic Denitions and Notions

Ideal theory originates mainly from Green This subsection contains a suitable adaptation

of a tiny selection of notions and results from ideal theory For more detailed information

I recommend teaching b o oks on semigroups eg rather than b o oks concerning automata

theory I develop ed this subsection in a way that the reader do es not require previous knowledge

in ideal theory

As already mentioned a semigroup is a set together with a binary asso ciative op eration

Assume some semigroup S We call a semigroup H a subsemigroup of S i the underlying set

of H is a subset of S and the op eration of H is the op eration of S restricted to the elements

in H Hence we can regard H as a subset of S which is closed under the op eration of S

We call a subset U of S a left ideal of S i we have SU U Hence a left ideal is a sp ecial

subsemigroup Every semigroup has itself and the empty set as left ideals We call a left ideal

U of S proper i we have U Q The intersection and the union of two left ideals of S

yield left ideals of S

We call a subset J of S an ideal of S i we have JS J and SJ J Hence an ideal is a

sp ecial left ideal We call an ideal J of S proper i we have J Q

For simplicity we develop the following notions just for nite semigroups Assume some

nite semigroup Q Assume some ideal J of Q which is dierent from Q We call a left ideal U

of QJ minimal i we have J U and there is not any left ideal U with J U U Assume

two dierent J minimal left ideals U and V of Q Their intersection contains J Assume J is

prop erly contained in the left ideal U V Then one of the left ideals U or V is not J minimal

b ecause we have J U V U or J U V V Consequently we have U V J

If J we shortly say minimal instead of minimal

We close this subsection with a technical lemma

Lemma Assume a nonempty nite semigroup Q and an ideal J Q Then the union of

all J minimal left ideals yields an ideal of Q 2

Pro of There is at least one left ideal prop erly containing J namely Q itself Hence there is

also some smallest left ideal which contains J prop erly

Let J b e the union of all J minimal left ideals Then J is a left ideal with J J We have

to show J Q J We prove this by showing that for every J minimal left ideal L and for every

element q Q the set J Lq yields J or some J minimal left ideal Just assume J J Lq

Because L is a left ideal we have QL L Thus we have QLq Lq Therefore Lq and

J Lq are left ideals of Q

Now we show by a contradiction that J Lq is J minimal Just assume a left ideal K such

that we have J K J Lq We dene a set K by K fx L j xq K g We show the

prop er inclusions J K L

We have J L and J q J K Hence we have J K

We show that the inclusion J K is prop er There is some p K n J Then p Lq

Hence there is some p L with p p q We have p J b ecause J is an ideal and p p q

do es not b elong to J However p K

The inclusion K L is obvious There is some r J Lq n K Then we have r Lq n J

Thus there is some r L with r q r Then r K ie we have K L

We show that K is a left ideal Just assume some x in K and some y in Q We have y x L

b ecause x b elongs to L which is a left ideal Further we have y xq K b ecause xq b elongs to

the left ideal K Thus we have y x K

Hence the set K is a left ideal with J K L ie L is not J minimal This is a

contradiction such that the assumed left ideal K do es not exist Therefore the set J Lq is a

J minimal left ideal 2

SEMIGROUPS AND IDEALS

Finite Semigroups without Prop er Ideals

In this subsection we examine nonempty nite semigroups without prop er ideals Assume some

nonempty nite semigroup Q without prop er ideals We distinguish two cases Q has or Q do es

not have prop er left ideals At rst we deal with the case that Q do es not have prop er left

ideals

Lemma Assume a nonempty nite semigroup Q which has not any prop er left ideal Then

for every elements p p and q in Q the equality pq p q implies p p 2

Pro of We show that a counter example implies the existence of a prop er left ideal Just assume

three elements p p q Q such that pq p q but nevertheless p p We have QQ Q and

thus Q Qq Qq such that Qq is a left ideal Further the pro duct Qq yields a prop er subset

of Q b ecause the result of the pro duct pq o ccurs twice such that at least one element of Q

cannot o ccur in Qq Consequently the set Qq is a prop er left ideal of Q This contradicts the

presumption of the lemma 2

Now we examine nonempty nite semigroups with prop er left ideals but without prop er ideals

Example Assume the semigroup Q with the underlying set fa a a c c cg We dene

the op eration in Q as follows For every two elements p and q in Q the result of the pro duct

pq has the letter ie a or c of p with the index ie or of q For instance we have

a c ac a a a and cc c a cc c This op eration is asso ciative Obviously for two

elements p and q of Q we have q pq q

The semigroup Q do es not have prop er ideals Assume J is a nonempty ideal of Q

Assume an element p J For every q Q the pro duct q p b elongs to J Then we also have

q pq q J Thus J Q

It is immediate that the subsets fa c g fa cg and fa cg are the minimal left ideals of Q

By merging two of these minimal left ideals we obtain three prop er left ideals of Q which are

not minimal 2

Lemma Assume a nonempty nite semigroup Q with at least one prop er left ideal but

without prop er ideals Then Q has two disjoint prop er left ideals U and V with Q U V 2

Pro of We apply Lemma with J The union of all minimal left ideals of Q yields an

ideal of Q Because Q do es not have prop er ideals the union of all minimal left ideals of Q

yields Q itself

Now assume that Q has exactly one minimal left ideal Then this minimal left ideal is Q

itself Thus the semigroup Q do es not have prop er left ideals which is a contradiction Hence

Q has at least two minimal left ideals

Let U b e an minimal left ideal and let V b e the union of all other minimal left ideals of Q

Then the subsets U and V are two disjoint left ideals and their union yields Q 2

Example continued The semigroup Q fullls Lemma eg we cho ose U fa c g

and V fa a c cg 2

By Lemma and we have strong assertions for nite semigroups without prop er ideals

Dep ending on whether a nite semigroup without prop er ideals has a prop er left ideal or not

we can apply either Lemma or Lemma

Finite Semigroups with Prop er Ideals

In this subsection we deal with nite semigroups with prop er ideals

Lemma Assume a nonempty nite semigroup Q with a prop er ideal Then there is a

prop er ideal J of Q such that Q and J fulll one of the following assertions

The set Q n J yields a singleton fr g and r b elongs to J

The set Q n J yields a subsemigroup of Q

There are two prop er left ideals U and V of Q such that U V Q and U V J 2

Note that in assertion three b oth the left ideals U and V are dierent from J Just assume

that U J Then we have J V Q Because J is contained in V we have V Q which

contradicts that V is a prop er left ideal

Pro of Let J b e a prop er ideal of Q such that there is not any ideal J with J J Q Such

an ideal exists b ecause Q is nite and Q has at least one prop er ideal We show that Q and J

fulll assertion provided that they contradict assertion and

Since J is prop er there is some r Q n J Then Q n J fr g implies assertion or

dep ending on whether r J or r r Hence Q n J contains at least two elements

Since Q n J is not a subsemigroup of Q there are p q Q n J with pq J We have

J Qq J J fpg Q n J n fpg q J J q fpq g Q n J n fpg q The sets J q and

fpq g are contained in J such that we have J Qq J Q n J n fpg q

Q n J n fpg J Q n J n fpgq J J Q n J n fpgq J Qq Now we have

We have p Q n J and thus J Q n J n fpg J Q n J Q Therefore we have

J Qq Q Hence we have the prop er inclusion J Qq Q

We show the existence of some left ideal U of Q with J U Q Assume that Qq is not a

subset of J Then the union J Qq yields the desired left ideal U Assume that Qq is a subset

of J Then the set J fq g is the desired left ideal U The inclusion J fq g Q is prop er

since Q n J contains at least two dierent elements

Now we can apply Lemma The union of all J minimal left ideals of Q yields an ideal

This ideal prop erly contains J The only ideal prop erly containing J is Q itself Hence the

union of all J minimal left ideals yields Q itself

Assume there is exactly one J minimal left ideal Then this J minimal left ideal is Q itself

However Q cannot b e a J minimal left ideal b ecause we have shown that there is some left

with J U Q Therefore there are at least two dierent J minimal left ideals ideal U

Now let U b e a J minimal left ideal and let V b e the union of all other J minimal left ideals

Then U and V are the desired left ideals in assertion 2

The assertions in Lemma are not exclusive Assume a nonempty nite semigroup Q with a

prop er ideal J such that Q and J satisfy assertion in Lemma They obviously contradict

assertion two by r J They also contradict assertion three Dep ending on whether r b elongs

to U we either have J U or Q U which is a contradiction However assertions and

are not exclusive

Example Assume the semigroup Q with the underlying set f a bg We dene the op er

ation in Q as follows For every p Q we set p p For p q Q n fg the pro duct pq

yields q Every ideal of Q has to contain Assume that an ideal J contains one of the elements

a or b Then the pro duct JQ yields Q Consequently the set fg is the only prop er ideal of Q

We see that Q and fg fulll assertion two However Q and fg also fulll assertion three

by U f ag and V f bg 2

PRODUCT AUTOMATA

A Classication

We work out a suitable classication of all nonempty nite semigroups We use ideal theory to

distinguish several classes of nonempty nite semigroups

Prop osition Every nonempty nite semigroup Q fullls one of the following assertions

A Q has not any prop er left ideal

B Q has two prop er left ideals U V such that U V Q and U V is an ideal of Q

C Q has an ideal J such that Q n J yields a singleton fr g with r J

D Q has a prop er ideal J and a subsemigroup H such that J H and J H Q 2

Pro of Assume Q do es not have a prop er ideal If Q do es not have a prop er left ideal it fullls

assertion A If Q has a prop er left ideal then Q fullls assertion B by Lemma b ecause

the empty set is an ideal

Now assume Q has a prop er ideal Then by Lemma there is some prop er ideal J of Q

such that J fullls one of the assertions C D or B 2

Assume a nonempty nite semigroup Q Every prop er ideal is also a prop er left ideal Thus

if Q fullls one of the assertions B C or D then Q cannot satisfy assertion A However

the assertions B C and D are not exclusive

Example Consider the semigroup Q with the underlying set f b b c g and the op er

ation given by the following table

b b c

b b b

The op eration can b e understo o d intuitively For n and p p Q we can calculate

the pro duct p p as follows If a c or o ccurs among p p the result has a

n n

Otherwise the result has a b The index of p p ie or is the index of p The semigroup

n n

Q fullls the assertions B C and D

B The semigroup Q has prop er ideals such that we cannot apply Lemma Nevertheless

we can split Q into two disjoint left ideals by U f b c g and V f bg We could also

cho ose prop er left ideals which are not disjoint eg by U f b c g and V f bg

C We simply set J f b bg

D Q satises this assertion by J f c g and H fb bg 2

Pro duct Automata

In this section we deal with a sp ecial kind of automata We adapt the notion of automata from

Denition We use ideas from the pro of of Theorem cf In the rst subsection

we get familiar with pro duct automata In the second subsection we examine connections to

ideal theory

Throughout this section we assume two disjoint indep endence alphab ets I and I

We abbreviate the trace monoids IM I and IM I by IM and IM resp ectively

Denitions

We dene pro duct automata

Denition Assume two disjoint trace monoids IM and IM A pro duct automaton A over

is a quintuple P R g h F where

P and R are nonempty nite semigroups

g and h are surjective homomorphisms g IM P h IM R

F is a subset of 2

i h

g P IM

F by automaton We can regard every pro duct automaton P R g h F as an

h R IM

F Denition A pro duct automaton A denes a recognizable language by LA

b elongs to LA i we obtain a pair in F when we apply g This means that a trace t

and h on the rst and second comp ound of t resp ectively

Let us assume that LA is closed under concatenation Then F is a subsemigroup of

Similarly F and F are subsemigroups of P and R resp ectively

We are going to use pro duct automata to prove assertions on recognizable languages in

Therefore we have to show that every recognizable language T in is the language of some

pro duct automaton

Lemma Assume two disjoint trace monoids IM and IM and a recognizable language T

in There is a pro duct automaton for T 2

Pro of By Theorem there is some n IN and recognizable languages T T IM and

T T IM such that

T T

We can b e an automaton for T resp T For i n let P g F resp R h F

i i i i i i

i i

freely assume P P g g R R and h h for any i j n Then T is the

i j i j i j i j

F F

2 language of the pro duct automaton P R g h F with F

F F

Connections to Subsemigroups and Ideals

We examine connections b etween pro duct automata and ideal theory Assume a recognizable

which is closed under concatenation Assume further a pro duct automaton language T in

A P R g h F for T Let us denote F by Q Then Q is a subsemigroup of R We can

easily verify that Q h T T Assume some subset W of Q We dene a

language T by

T f t T j h t W g

The language T is obviously a subset of T Some trace t in b elongs to T i we

W W

g P

have t F

h W

Prop osition Assume some nonempty concatenation closed language T in Assume

a pro duct automaton A P R g h F for T Let Q denote the semigroup F

PROOF OF THEOREM

For W Q the pro duct automaton A P R g h F denes T Therefore

W W

T is recognizable

For every nonempty W Q the language T is nonempty

For every subsemigroup H Q the language T is concatenation closed

For every left ideal U Q the language T is a left ideal of T

For every ideal J Q the language T is an ideal of T 2

Pro of

We can straightforwardly verify that A is a pro duct automaton b ecause A is a pro duct

g g P

automaton For t T we have t F and t Thus we have

h h W

g P

t F Hence T LA

W W

h W

g g P

Conversely let t LA Then we have t F and t Hence t T

h h W

and h t W ie t b elongs to T Consequently LA T

W W W

Because W is nonempty we can assume some r in W This r b elongs to Q ie there is

g p

is a surjection F Because the homomorphism some p in P such that the pair

h r

g p g p

the set is nonempty Further we have T

h r h r

h W T Since W is a subsemigroup of Q its preimage Note that T

h W is a subsemigroup of Then T is a subsemigroup b ecause it is

the intersection of two subsemigroups

Let f b e the restriction of h to T Then f is a surjective homomorphism from T to

Q and T f U Since U is a left ideal of Q so is its preimage T under f

U U

As the previous p oint 2

Pro of of Theorem

In this section we prove Theorem We take over the notions I I IM and IM

from the previous section We abbreviate IM n fg and IM n fg by IM and IM resp ectively

are exactly the traces in The traces in which contain at least one letter in and

at least one letter in

Theorem Assume two disjoint indep endenc e alphab ets I and I Assume

some concatenation closed recognizable language T in

Then the set of generators of T has the FPP 2

Theorem is obviously true if the language T is empty Thus we just need to prove it for

nonempty languages T The general structure of the pro of is the following Just assume a

By Lemma there is a nonempty concatenation closed recognizable language T in

pro duct automaton A P R g h F for T We denote F by Q Because T is nonempty

Q is nonempty We use Prop osition on Q Consequently the pro of of Theorem consists

of four parts In the rst subsection we deal with the case that Q do es not contain prop er left

ideals After that we deal with the cases that Q fullls one of the assertions B C or D in

Prop osition We will do this by an induction on the numb er of elements of Q

In the last subsection we summarize the results to prove Theorem

Q do es not have prop er left ideals

In this subsection we prove the following sp ecial case of Theorem

which is Prop osition Assume a nonempty concatenation closed language T in

recognized by a pro duct automaton P R g h F such that the semigroup F do es not have

prop er left ideals

j F j

Then GenT has the FPP Moreover we have T Gen T 2

I intro duce some notions esp ecially for the pro of of this prop osition We assume a language T

as in Prop osition

Denition Assume some traces t t s in T We call the pair t s a most oblique cut

of t i t t s and for every traces t s T with t t s we have either

j t j j t j or

j t j j t j and j t j j t j 2

Intuitively we can understand the denition as follows We try to factorize t in T into two

traces t and s of T We try to do this in a way that the rst comp ound of t is small but

the second comp ound of t is big

If the trace t is not a generator of T then there is at least one most oblique cut of t A most

oblique cut cannot exist if the trace t is a generator of T

Lemma Assume some traces t t s in T such that t s is a most oblique cut of t

Then the trace t is a generator of T 2

Pro of We prove the lemma by contradiction Just assume that t is not a generator of T

Then there are two traces t and t in T such that t t t We can factorize t into t

a b a b a

and t s The traces t and t s b elong to T Further t contains prop erly less letters

b a b a

than t since t b This contradicts that t s is a most oblique cut 2

We can factorize some trace t in T into generators by successive most oblique cuts We factorize

t into a generator t and a trace s in T Then we factorize s by a most oblique cut and so on

until a most oblique cut yields two generators This iterative factorization terminates b ecause

the remaining part of t b ecomes prop erly shorter in every most oblique cut

Pro of of Prop osition Assume some trace t in T We denote by Q the semigroup F

We show that a factorization of t by successive most oblique cuts yields a factorization of t into

at most jQj generators of T

We factorize t into generators of T by successive most oblique cuts We obtain a natural

numb er n and generators t t of T such that t t yields t For every i in n

n n

the pair t t t is a most oblique cut of the trace t t

i i n i n

We intro duce two notions for lucidity For every i in n we dene u t and

i i

For every i in n we have hv Q b ecause the v t ie we have t

i i i i

traces t t b elong to T

We show by a contradiction that n jQj Assume n jQj

By hv v hv hv Q for i n and n jQj we get the existence

i n i n

of i j n such that hv v hv v

i n j n

Then hv hv v hv v hv v hv v Since Q do es not

i Q i n i n i j Q j n

have prop er left ideals we can apply Lemma and get hv hv v

i i j

PROOF OF THEOREM

g u g u

i i

By t T we have F Because of hv hv v we get F

i i i j

h v h v v

i i j

u u g u

i+1 i

T Similarly t t T implies and thus F By hv v

i n i n

v v h v v

i+1 i j

u u g u u

n n

i+1 i+1

hv v we have F and therefore T

j n

v v h v v

n n

j +1 j +1

u u u

i i+1

Thus are a factorization of t t into two traces from T Since and

i n

v v v v

i j j +1

u u

i i

j t we obtain j t t t is a most oblique cut of t and

i i i n

v v v v

i j i j

j t j Hence jv v j jv j Because v is a prex of v v we have jv v j jv j

i i j i i i j i j i

Consequently v v This is a contradiction b ecause every trace in T contains at least

i j

one letter from

Finally our assumption n jQj lead to a contradiction Hence we have n jQj 2

The metho d of most oblique cuts is a very suitable metho d to prove Theorem in the case

that the semigroup Q do es not have prop er left ideals Let us consider an example

Example Assume the free monoids over singletons IM fag and IM fbg Consider

o n o n

a a

The language T fullls every n m in the language T

b b

presumption of Theorem However we cannot prove that GenT has the FPP by factori

zations with most oblique cuts For every n the application of successive most oblique cuts

a a a

factorizes the trace into ie we obtain n generators Hence the numb er of

b b b

generators which we obtain by successive most oblique cuts is unlimited 2

Q fullls assertion C

We prove the remaining cases of Theorem by an induction on the numb er of elements in Q

In the case that Q is a singleton we already know by Prop osition that Theorem is true

for T b ecause the singleton semigroup do es not have prop er left ideals

Assume some natural numb er n By induction we presume that Theorem is true

for languages T if there is a pro duct automaton P R g h F for T such that the numb er

of elements of F ie Q is prop erly smaller than n Then we show that Theorem is

true if Q has n elements and Q fullls one of the assertions B C or D in Prop osition

We do this by a decomp osition of Q and T into subsemigroups left ideals and ideals

We p erform the rst induction step We start with the case that Q fullls assertion C in

Prop osition b ecause this is the most simple one

Prop osition Let n Assume that Theorem holds for every nonempty concatena

R g h F which is recognized by a pro duct automaton P in tion closed language T

with j F j n Let P R g h F b e a pro duct automaton for a language T such that

T is a nonempty concatenation closed language in

j F j n and

there is an r F such that r r and F n fr g is an ideal

Then GenT has the FPP 2

Pro of For simplicity let Q F and J F n fr g We use Prop osition to examine

the language T f t T j h t J g The traces in T n T are exactly these traces t in T

J J

with h t r

The ideal J of Q is a subsemigroup of Q b ecause every ideal is a subsemigroup By assertion

three of Prop osition the language T is concatenation closed J

Q fullls assertion D

By assertion one of Prop osition we know that T is recognizable Moreover the pro duct

P P

automaton A P R g h F denes T We see that F yields J We have

J J

P P

j F j jQj ie we have j F j n By the inductive hyp othesis we have

J J

an l IN such that T GenT

J J J

By assertion ve of Prop osition we know that T is an ideal of T Further we can

show that for every t t T the concatenation t t b elongs to T If one of the traces t

and t b elongs to T then the trace t t b elongs to T b ecause T is an ideal If b oth t and

J J J

t do not b elong to T then we have h t h t r We have h t t

h t h t r J Hence the trace t t b elongs to T ie TT T

Q J J

Let l l We show T Gen T For this let t T

Case The trace t do es not b elong to T

Because TT T we have t TT Hence t is a generator of T Thus t GenT

Case The trace t is a generator of T

Assume we can factorize t into more than three generators of T Then we can also

factorize t into exactly four traces of T Thus there are traces t t T such that

t t t We have t t T and t t T Consequently t is not a generator of T

J J J

This contradicts the case presumption Hence t GenT

Case The trace t b elongs to T

Because of T GenT we have a k l and t t T such that t t t

J J J k J k

By case we can factorize every trace among t t into three or less generators of T

Hence we can factorize t into l l or less generators of T ie t GenT

We have T GenT ie the set GenT has the FPP 2

Q fullls assertion D

We deal with the case that we can split Q into a prop er ideal and a subsemigroup

Prop osition Let n Assume that Theorem holds for every nonempty concatena

which is recognized by a pro duct automaton P R g h F tion closed language T in

with j F j n Let P R g h F b e a pro duct automaton for a language T such that

T is a nonempty concatenation closed language in

j F j n and

there are a prop er ideal J and a subsemigroup H in F with J H F and

J H

Then GenT has the FPP 2

Pro of We denote by Q the semigroup F We examine the languages T and T They

J H

are nonempty subsets of T They are disjoint and their union yields T Both T and T are

J H

recognizable and concatenation closed Further the language T is an ideal of T There are two

l l

J H

natural numb ers l and l such that T GenT and T GenT

J H J J H H

Let l l l l We show T GenT Assume some trace t in T

H J J

Case t is a generator of T

Assume that there are t t T such that t t t If one of the traces t or t b elongs to

the ideal T then t b elongs to T Hence b oth t and t b elong to T This contradicts

J J H

that t is a generator of T Thus t is a generator of T H

PROOF OF THEOREM

Case t is a trace of T

We can factorize t into l or less generators of T By case such a factorization is a

H H

factorization of t into generators of T Thus t GenT

Case t is a generator of T

The trace t is not necessarily a generator of T We examine factorizations of t as follows

Assume three traces t t t T fg such that

t t t t t T fg t T fg and t T fg

H J H

There are traces t t t which fulll these conditions namely t t t t

We cho ose a triple t t t such that the numb er of letters of t is minimal

We can apply case to the traces t and t The traces t and t are either or they

b elong to T such that we can factorize each of t and t into l or less generators of T

H H

We deal with t Assume t is not the empty trace By a contradiction we can show that

t is a generator of T Just assume t t T with t t t We distinguish four cases

t T and t T

J J

The language T is an ideal of T such that t t t t T Thus t t and t t form

J J

a factorization of t into two traces of T This contradicts that t is a generator of T

J J

T T and t t

J H

Then t t T b ecause T is a subsemigroup The trace t contains prop erly less

H H

letters than t Consequently the traces t t t and t contradict the choice of t

t and t ab ove

t T and t T

J H

Similar to the previous case the traces t t and t t contradict the choice of the

traces t t and t

t T and t T

H H

The traces t t and t t contradict the choice of t t and t

Thus the trace t is either or a generator of T We can factorize t and t into l or

less generators of T Therefore t GenT

Case t is a trace in T

We can factorize t into l or less generators of T By case we can factorize every

J J

l l l

H J J

generator of T into l generators of T Consequently t GenT

J H

l l l

H J J

ie the set GenT has the FPP 2 We have T GenT

Q fullls assertion B

Just one case remains The semigroup Q has two left ideals U and V such that their union

yields Q and their intersection yields some ideal of Q This case is the most involved one

Prop osition Let n Assume that Theorem holds for every nonempty concatena

which is recognized by a pro duct automaton P R g h F tion closed language T in

with j F j n Let P R g h F b e a pro duct automaton for a language T such that

T is a nonempty concatenation closed language in

j F j n and

Q fullls assertion B

there are two prop er left ideals U and V such that U V F and U V yields an

ideal of F

Then GenT has the FPP 2

Pro of As in the previous pro ofs we set Q F For lucidity we set J U V

As in the previous pro ofs the languages T and T are nonempty recognizable and con

U V

catenation closed subsets of T Further T and T are left ideals of T and T is an ideal of T

U V J

l l

V U

and T GenT We have two natural numb ers l and l such that T Gen T

V V U V U U

Provided that J is nonempty there is some natural numb er l such that T GenT

J J J

If J then T is also empty

We have U V Q and U V J For every t T we have h t U or h t V

Thus T T T Further for every t T we have h t J i h t U and

U V

h t V Hence we have T T T

U V J

Let l l l l maxl l l We show T GenT Assume some trace t in T

J U V U V J

Case t is a generator of T Further t do es not b elong to T

U J

Then t cannot b elong to T Furthermore if we factorize t into some traces in T not any

factor do es b elong to the ideal T Otherwise t would b elong to T Consequently if we

J J

factorize t into some traces in T the factors either b elong to T or T but they do not

U V

b elong to T T

U V

The trace t is not necessarily a generator of T In the case that t is a generator of T we are

done So assume that t is not a generator of T There is a trace x T and a generator y

of T with xy t

Assume y b elongs to the left ideal T Then xy t also b elongs to T This is a

V V

contradiction Hence y T Assume x also b elongs to T Then xy t is not a

U U

generator of T Hence we have x T and y T

U V U

We factorize the trace x into l or less generators of T There are a k l and generators

V V V

x x of T such that x x x

k V k

We show by a contradiction that the traces x x are generators of T Just assume

and some natural numb er i with i k such that x can b e factorized into two traces x

i i

x in T Assume that x b elongs to T Then x b elongs to T which is a contradiction

U U

i i i

b elongs to T Then x is not a generator of T T Now assume that x Hence x

V V V

i i i

Thus we have x T and x T

U V

i i

However this yields a contradiction We factorize t into the traces x x x and

i i

x x x y Both factors b elong to T b ecause x and y b elong to the left ideal T

U U

i i i

Hence we can factorize t into two traces from T which is a contradiction

The assumption that some trace among x x is not a generator of T yields a contra

diction Thus we have by x x y a factorization of t into generators of T Hence

t GenT

Case t is a trace in T n T

U J

There are a k l and generators t t of T such that t t t The generators

U k U k

t t cannot b elong to T By case we have t t GenT Because

k J k

l l l

U V U

k l we have t GenT

Case t is a trace in T n T

We can deal with the traces in T n T as we dealt with the traces in T n T We obtain

V J U J

l l l

U V V

T n T GenT

V J

l l maxl l

U V U V

Hence we have T n T GenT J

CONCLUSIONS AND FUTURE STEPS

Case t is a generator of T

We use the triple factorization metho d from case in the pro of of Prop osition Assume

three traces t t t in T fg such that we have

fg and t T fg t t t t t t T n T

J J

There are traces t t t which fulll these conditions eg t resp ectively We cho ose

a triple t t t which fullls the ab ove conditions such that the numb er of letters of t is

minimal We can apply case on the traces t and t if they are not empty

We deal with t If t is the empty trace we have t t t Then we can factorize t into

l l maxl l or less generators of T If t is a generator of T then we need one

U V U V

more generator to factorize t into generators of T

We deal with the case that t is not a generator of T Then we can factorize t into a

generator t of T and some trace t in T Assume that t T We distinguish two cases

dep ending on whether we have t t T or not If we have t t T then can factorize

J J

t into the traces t t and t t in T which contradicts that t is a generator of T If we

J J

have t t T n T then we have a factorization of t into t t t and t This contradicts

the choice of t t t ab ove Consequently t T

To sum up we have a factorization of t into t t t and t The trace t is a generator

of T We can apply case on t t and t

l l maxl l

U V U V

Thus t GenT

Case t is a trace in T

We can factorize t into l or less generators of T and apply case to every generator

J J

l l l maxl l l

J U V U V J

Consequently t Gen T

l l l maxl l l

J U V U V J

We have T GenT ie the set GenT has the FPP 2

Completion of the Pro of

Pro of of Theorem The theorem is obviously true if T is the empty set As a conclusion

of Prop osition Theorem holds for every nonempty concatenation closed language

if there is a pro duct automaton P R g h F for T such that F is a singleton T

Assume some natural numb er n Assume that Theorem is true for every concate

if there is a pro duct automaton P R g h F for T with nation closed language T

j F j n

dened by a pro duct automaton Now let T b e a concatenation closed language in

P R g h F with F n Then by Prop osition the semigroup F satises one of

the assertions A B C or D such that we can apply one of the Prop ositions

or resp ectively 2

Conclusions and Future Steps

By proving Theorem we corrected an error in Provided that this rep ort do es not

contain further errors its main result is true ie the trace monoids with a decidable star

problem are exactly the trace monoids with a decidable FPP

Opp osed to the previously known connections b etween the star problem and the FPP in

Theorem and we have by Theorem a connection b etween the star problem and

the FPP in one and the same trace monoid

It raises the following decision problem which is a sp ecial case of the star problem Assume

two disjoint trace monoids IM I and IM I and a language L as in Theorem

Can we decide whether L is recognizable ie whether L has the FPP provided that b oth the

star problem and the FPP are decidable in b oth IM I and IM I

Acknowledgments

I acknowledge the helpful discussions with Manfred Droste and Dietrich Kuske at the

Institute of Algebra I esp ecially acknowledge Manfred Drostes lecture Algebraic Theory of

Automata in which I came in tough with ideal theory and the helpful teaching b o ok I thank

to Dietrich Kuske for reading a preliminary version of this pap er and giving many useful hints

I thank to Gwenael Richomme for bringing my attention to this particular problem

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