Automaticity of one-relator semigroups with length less than or equal to three

TL;DR

This paper characterizes when one-relator semigroups with defining relations of length up to three are automatic, prefix-automatic, or biautomatic, providing a complete classification based on the relation structure.

Contribution

It offers a complete characterization of the automaticity properties of small-length one-relator semigroups, identifying precise conditions for automatic, prefix-automatic, and biautomatic cases.

Findings

S is prefix-automatic if u=v not in specific small relation set.

S is not automatic if u=v in the small relation set.

S is biautomatic under certain length and relation conditions.

Abstract

The main results of this paper is to give a complete characterization of the automaticity of one-relator semigroups with length less than or equal to three. Let $S=sgp\langle A|u=v\rangle$ be a semigroup generated by a set $A=\{a_1,a_2,\dots,a_n\},\ n\in \mathbb{N}$ with defining relation $u=v$, where $u,v\in A^*$ and $A^*$ is the free monoid generated by $A$. Such a semigroup is called a one-relator semigroup. Suppose that $|v|\leq|u|\leq3$, where $|u|$ is the length of the word $u$. Suppose that $a,b\in A,\ a\neq b$. Then we have the following: (1) $S$ is prefix-automatic if $u=v\not\in \{aba=ba,\ aab=ba,\ abb=bb\}$. Moreover, if $u=v\in \{aba=ba,\ aab=ba,\ abb=bb\}$ then $S$ is not automatic. (2) $S$ is biautomatic if one of the following holds: (i) $|u|=3,\ |v|=0$, (ii) $|u|=|v|=3$, (iii) $|u|=2$ and $u=v\not\in \{ab=a,\ ab=b\}$. Moreover, if $u=v\in \{ab=a,\ ab=b\}$ then $S$ is not biautomatic.

Equations

$$|\alpha |$$

$$M(L)$$

$$|S(M(L))|$$

$$DFSA$$

$$occ(a,\alpha )$$

$$thenocc(a,\alpha )=3,occ(b,\alpha )=4,$$

$$con(\alpha )$$

$$thencon(\alpha )=\{a,b,c\},$$

$$\mathcal{P}(C)$$

$$Pref(L)$$

$$S^1$$

$$\displaystyle\alpha(t):=\begin{cases}{a_{1}a_{2}\cdots a_{t}}&\mbox{ if $t\leq n$},\\ {a_{1}a_{2}\cdots a_{n}}&\mbox{ if $t>n$},\end{cases}$$

$$\displaystyle\alpha[t]:=\begin{cases}{a_{n-t+1}a_{n-t+2}\cdots a_{n}}&\mbox{ if $t\leq n$},\\ {a_{1}a_{2}\cdots a_{n}}&\mbox{ if $t>n$}.\end{cases}$$

$$\pi :q_1\text{xlongrightarrow}(a_1,u_1)q_2\text{xlongrightarrow}(a_2,u_2)q_3\cdots \text{xlongrightarrow}(a_n,u_n)q_{n+1}$$

$$\Phi (\pi ):=a_1{a}_2\cdots a_n,\Sigma (\pi ):=u_1{u}_2\cdots u_n.$$

$${\eta }_{\mathcal{A}}(X)=\{v\in B^{+}|(\exists u\in X)(\exists t\in T)(q_0\text{xlongrightarrow}{(u,v)}_{+}t)\}.$$

$$(u_1\cdots u_m,v_1\cdots v_n)\mapsto \{\begin{array}{ll}(u_1,v_1)\cdots (u_m,v_n)& \text{if}m=n,\\ (u_1,v_1)\cdots (u_n,v_n)(u_{n+1},\$)\cdots (u_m,\$)& \text{if}m>n,\\ (u_1,v_1)\cdots (u_m,v_m)(\$,v_{m+1})\cdots (\$,v_n)& \text{if}m<n,\end{array}$$

$$(u_1\cdots u_m,v_1\cdots v_n)\mapsto \{\begin{array}{ll}(u_1,v_1)\cdots (u_m,v_n)& \text{if}m=n,\\ (u_1,\$)\cdots (u_{m-n},\$)(u_{m-n+1},v_1)\cdots (u_m,v_n)& \text{if}m>n,\\ (\$,v_1)\cdots (\$,v_{n-m})(u_1,v_{n-m+1})\cdots (u_m,v_n)& \text{if}m<n,\end{array}$$

$${\Delta }_L:=\{(\alpha ,\alpha ){\delta }_A^R|\alpha \in L\}=\{(\alpha ,\alpha ){\delta }_A^L|\alpha \in L\}$$

$$\text{leftidx}{}_a^{\$}L:$$

$$\text{leftidx}{}^{\$}{L}_a:$$

$$\text{leftidx}{}_a{L}^{\$}:$$

$$L_a^{\$}:$$

$$L_{=}^{\prime }:=\{(\alpha ,\beta ){\delta }_A^R|\alpha \in L,\beta \in Pref(L),and\alpha =\beta \}$$

$$a^{rev}:$$

$$L^{rev}:$$

$$U^{rev}:$$

$$(w_1,w_2){\delta }_A^R\in M\Rightarrow ||w_1|-|w_2||\le C,$$

$$M\odot N:=\{(w_1{w}_1^{\prime },w_2{w}_2^{\prime }){\delta }_A^R|(w_1,w_2){\delta }_A^R\in M,(w_1^{\prime },w_2^{\prime }){\delta }_A^R\in N\}$$

$$||\alpha |-|\beta ||\le k$$

$$(w_1,w_2){\delta }_A^L\in M{\delta }_A^L\Rightarrow ||w_1|-|w_2||\le C,$$

$$(w_1^{\prime },w_2^{\prime }){\delta }_A^L\in N{\delta }_A^L\Rightarrow ||w_1^{\prime }|-|w_2^{\prime }||\le C^{\prime },$$

$$M{\odot }^{\prime }N:=\{(w_1{w}_1^{\prime },w_2{w}_2^{\prime }){\delta }_A^L|(w_1,w_2){\delta }_A^L\in M{\delta }_A^L,(w_1^{\prime },w_2^{\prime }){\delta }_A^L\in N{\delta }_A^L\}$$

$$u<v\Rightarrow wu<wv,uw<vw.$$

$$L:=A^{+}-A^{\ast }\{u_i|i\in I\}{A}^{\ast }$$

$$S^1=sg{p}^{+}⟨A,e|R,ea=ae=a,ee=e,a\in A⟩.$$

$$L_{=}^{\prime }=\{(\alpha ,\beta ){\delta }_A^R|\alpha \in L,\beta \in Pref(L),and\alpha =\beta \}$$

$$K_{=}^{\prime }$$

Full text

Automaticity of one-relator semigroups with length less than or equal to three111Supported by the NNSF of China (11571121) and the Science and Technology Program of Guangzhou (201605121833161).

Yuqun Chen222Corresponding author., Haibin Wu, Honglian Xie

School of Mathematical Sciences, South China Normal University

Guangzhou 510631, P. R. China

yqchen@scnu.edu.cn

378238979@qq.com

honglianxie@126.com

Abstract: The main results of this paper is to give a complete characterization of the automaticity of one-relator semigroups with length less than or equal to three. Let $S=sgp\langle A|u=v\rangle$ be a semigroup generated by a set $A=\{a_{1},a_{2},\dots,a_{n}\},\ n\in\mathbb{N}$ with defining relation $u=v$, where $u,v\in A^{*}$ and $A^{*}$ is the free monoid generated by $A$. Such a semigroup is called a one-relator semigroup. Suppose that $|v|\leq|u|\leq 3$, where $|u|$ is the length of the word $u$. Suppose that $a,b\in A,\ a\neq b$. Then we have the following: (1) $S$ is prefix-automatic if $u=v\not\in\{aba=ba,\ aab=ba,\ abb=bb\}$. Moreover, if $u=v\in\{aba=ba,\ aab=ba,\ abb=bb\}$ then $S$ is not automatic. (2) $S$ is biautomatic if one of the following holds: (i) $|u|=3,\ |v|=0$, (ii) $|u|=|v|=3$, (iii) $|u|=2$ and $u=v\not\in\{ab=a,\ ab=b\}$. Moreover, if $u=v\in\{ab=a,\ ab=b\}$ then $S$ is not biautomatic.

Key words: regular language,  automatic semigroup, one-relator semigroup, Gröbner-Shirshov basis, normal form

1 Introduction

The research of automaticity of groups started in the 1980’s. Many scholars engaged in the study of this field and found a lot of research production. For instance, [1, 16, 17, 23, 27].

In the end of 1990’s, the concept of automaticity was generalized to semigroups and monoids. In papers [10, 11, 12, 13, 14], the authors established the basic theory and obtained some results about automatic semigroups.

Not all properties of automatic groups hold for automatic semigroups. For example, the (first order) Dehn function of an automatic group is bounded above by a quadratic function but the (first order) Dehn function of an automatic semigroup may not be bounded above by any primitive recursive function, see [23]. A semigroup is called left-left (left-right, right-left, right-right, resp.) automatic semigroup if there exists $(A,L)$ such that for any a\in A\cup\{\varepsilon\},\ _{a}^{\}L\ (\leftidx{{}^{$}}{L}{a},\ \leftidx{{}{a}}{L}^{$},\ L_{a}^{$},\ resp.) (see Definition [2.2.1](#S2.SS2.Thmdefinition1)) is regular. Examples show that the four types of automatic semigroups are not equivalent, however, the four types of automatic groups are equivalent. An automatic group can be characterized by a geometric property of their Cayley graph, which is intuitively the following: “there is a constant c$such that, if two fellows travel at the same speed by two paths ending at most one edge apart, then the distance between is always less than$c”, see [[16](#bib.bib16)]. This property, called the fellow traveller property, plays an essential role in many of the results obtained so far about automatic groups, but does not characterize automatic semigroups, see [[11](#bib.bib11)]. Therefore, the geometric theory that holds for automatic groups does not hold for automatic semigroups. However, Hoffmann and Thomas [[21](#bib.bib21)] introduced the definition of directed fellow traveller property. They proved that if M$is a left-cancellative monoid finitely generated by$A$and if$L$is a regular subset of the set of all words over$A$that maps onto$M$, then$M$ is automatic if it has the directed fellow traveller property.

Many semigroups, monoids and groups have been proved automatic, such as free groups, free semigroups, braid groups, braid monoids ([16]), divisibility monoids ([24]), plactic monoids ([8]) and Chinese monoids ([9]), and so on.

In recent years, research of the automaticity of semigroups is active, for instance, [8, 9, 10, 11, 12, 13, 14, 15, 18, 19, 20, 21, 22, 29]. Automatcity theory of groups and semigroups have become important in today’s computer algebra.

Suppose semigroup $S$ is generated by a finite set $A$ and $A^{+}$ is the free semigroup (without identity) generated by $A$. The key to decide whether a semigroup $S$ is automatic is to find a regular language $L\subseteq A^{+}$ such that $(A,L)$ is an automatic structure of $S$. We usually choose $L$ to be a normal form of $S$. Generally, it is not easy to find a normal form of a semigroup. However, Gröbner-Shirshov bases theory can help us to solve this problem.

Soppose $S=sgp\langle A|R\rangle$ to be a monoid generated by a set $A$ with defining relations $R$. If the cardinal number of $R$ is 1, we call $S$ a one-relator semigroup. One-relator semigroups is a kind of important semigroups and whether the word problem of one-relator semigroups is solvable is still open.

The paper is organized as follows. In section 2, we review the concepts of regular language, automatic semigroup, biautomatic semigroup, prefix-automatic semigroup, and Gröbner-Shirshov bases for associative algebras. We cite also some results that will be used in the paper. In section 3, we give some characterizations of some automatic (biautomatic, prefix-automatic) semigroups, in particular, a complete characterization of the automaticity of one-relator semigroups with length less than or equal to three is given.

2 Preliminaries

In this section, we give some definitions, notions and mention some results that will be used in the paper.

2.1 Regular language

For any set $A$, we denote $A^{+}$ to be the set of all non-empty words over $A$ and $A^{*}$ to be the set of all words over $A$ including the empty word $\varepsilon$. If $A$ is a generating set of a semigroup $S$, interpreting concatenation is multiplication in $S$. We induce a semigroup epimorphism $\phi:\,A^{+}\rightarrow S$. For convenience, we write $\alpha$ as element $\phi(\alpha)$ in $S$, and write $\alpha\equiv\beta$ when $\alpha,\beta$ are the same element in $A^{*}$, write $\alpha=\beta$ when $\alpha$ and $\beta$ represent the same element of $S$.

We also say that $A$ is an alphabet and call language any subset of $A^{*}$. We will consider regular language, i.e. those languages accepted by finite state automata, see [16], for example. For any words $\alpha,\beta\in A^{*}$, letter $a\in A$, regular language $L$ over $A$, and semigroup $S$, we denote

[TABLE]

Suppose $\alpha\equiv a_{1}a_{2}\cdots a_{n}\in A^{+}$, where each $a_{i}\in A$. We define $\alpha(0):=\varepsilon$ and for any $t\geq 1$,

[TABLE]

A $gsm$ (generalized sequential machine) is a six-tuple $\mathcal{A}=(Q,A,B,\mu,q_{0},T)$ where $Q,A$ and $B$ are finite sets (called the states, the input alphabet and the output alphabet, resp.), $\mu$ is a (partial) function from $Q\times A$ to finite subsets of $Q\times B^{*}$, $q_{0}\in Q$ is the initial state and $T\subseteq Q$ is the set of terminal states. The inclusion $(q^{\prime},u)\in\mu(q,a)$ corresponds to the following situation: if $\mathcal{A}$ is in state $q$ and reads input $a$, then it can move into state $q^{\prime}$ and output $u$. We can interpret $\mathcal{A}$ as a directed labelled graph with vertices $Q$, and an edge $q\xlongrightarrow{(a,u)}q^{\prime}$ for every pair $(q^{\prime},u)\in\mu(q,a)$. For a path

[TABLE]

we define

[TABLE]

For any $q,q^{\prime}\in Q,u\in A^{+}$ and $v\in B^{+}$ we write $q\xlongrightarrow{(u,v)}_{+}q^{\prime}$ to mean that there exists a path $\pi$ from $q$ to $q^{\prime}$ such that $\Phi(\pi)\equiv u$ and $\Sigma(\pi)\equiv v$, and we say that $(u,v)$ is the $label$ of the path. We say that a path is successful if it has the form $q_{0}\xlongrightarrow{(u,v)}_{+}t$ with $t\in T$.

Any $gsm$ $\mathcal{A}$ induces a mapping $\eta_{\mathcal{A}}$ : $\mathcal{P}(A^{+})\xlongrightarrow{}{\mathcal{P}(B^{+})}$ defined by

[TABLE]

An useful result is that if $X$ is regular then so is $\eta_{\mathcal{A}}(X)$, see [14].

Definition 2.1.1**.**

([11]) Let $A$ be an alphabet and \$$ be a new symbol not in A$. Let$A(2,$)=(A\cup{$})\times(A\cup{$})-{($,$)}$. Define the mapping$\delta_{A}^{R}:A^{}\times A^{}\rightarrow(A(2,$))^{*}$ by

[TABLE]

and the mapping \delta_{A}^{L}:A^{*}\times A^{*}\rightarrow(A(2,\))^{*}$ by

[TABLE]

where each $u_{i},\ v_{j}\in A$.

Using the mappings $\delta_{A}^{R}$ and $\delta_{A}^{L}$ defined as above, we can transform the relation into a language over A(2,\)$, which provides a way to consider automata accepting pairs$(\alpha,\beta)$of words with$\alpha,\beta\in A^{+}$ as in the case of automatic groups.

Proposition 2.1.2**.**

([11]) If $L$ is a regular subset of $A^{*}$, then

[TABLE]

is regular over A(2,\)$.

2.2 Automaticity

Definition 2.2.1**.**

([19]) Suppose $S$ is a semigroup with a finite generating set $A$, $L$ is a regular language of $A^{+}$, and $\phi:A^{+}\rightarrow S$ is a homomorphism with $\phi(L)=S$. Let

[TABLE]

If for any a\in A\cup\{\varepsilon\},\ _{a}^{\}L\ (\leftidx{{}^{$}}{L}{a},\ \leftidx{{}{a}}{L}^{$},\ L_{a}^{$},\ resp.$) is regular, then we say semigroup$S$has a left-left (left-right, right-left, right-right, resp.) automatic structure$(A,L)$. If this is the case, we also say that$S$is a **left-left (left-right, right-left, right-right, resp.) automatic semigroup**. If for any$a\in A\cup{\varepsilon}$,$\leftidx{{}{a}}{L^{$}},\ \leftidx{{}^{$}}{L{a}},\ \leftidx{{}{a}}{L^{$}}$and$L{a}^{$}$are all regular, then we say semigroup$S$has a biautomatic structure$(A,L)$and say that$S$ is a biautomatic semigroup.

A right-right automatic semigroup is also called an automatic semigroup.

Definition 2.2.2**.**

([28]) Suppose $S$ is a semigroup with a finite generating set $A$, $L$ is a regular language of $A^{+}$, and $\phi:A^{+}\rightarrow S$ is a homomorphism with $\phi(L)=S$. If $(A,L)$ is an automatic structure for $S$ and

[TABLE]

is regular over A(2,\)$, then we say that$(A,L)$is a prefix-automatic structure for$S$and$S$ is a prefix-automatic semigroup.

For any $\alpha\equiv a_{1}a_{2}\cdots a_{n}\in A^{+}\ (a_{i}\in A)$, $L\subseteq A^{+}$, and $U\subseteq A^{+}\times A^{+}$, we denote

[TABLE]

Lemma 2.2.3**.**

([19])

1. (i)

A semigroup $S$ is left-left automatic if and only if $S^{rev}$ is right-right automatic. 2. (ii)

A semigroup $S$ is left-right automatic if and only if $S^{rev}$ is right-left automatic.

Proposition 2.2.4**.**

([19]) Let $S$ be a semigroup. Then $S$ is (bi-)automatic if and only if $S^{1}$ is (bi-)automatic.

Proposition 2.2.5**.**

([11]) Let $S_{1}$ and $S_{2}$ be semigroups. Then the free product $S_{1}*S_{2}$ of $S_{1}$ and $S_{2}$ is right-right (left-left) automatic if and only if both $S_{1}$ and $S_{2}$ are right-right (left-left) automatic.

Proposition 2.2.6**.**

([14]) Let $A$ be an alphabet and let $M,N$ be regular languages over A(2,\)$. If there exists a constant$C$such that, for any two words$w_{1},w_{2}\in A^{*}$, we have

[TABLE]

then the language

[TABLE]

is regular.

Proposition 2.2.7**.**

([19]) Let $L\subseteq A^{*}\times A^{*}$. Let $k$ be a constant such that

[TABLE]

for all $(\alpha,\beta)\in L$. Then $\{(\alpha,\beta)\delta_{A}^{L}|(\alpha,\beta)\in L\}$ is regular if and only if $\{(\alpha,\beta)\delta_{A}^{R}|(\alpha,\beta)\in L\}$ is regular.

Lemma 2.2.8**.**

Let $A$ be an alphabet and let $M,N$ be regular languages over A(2,\)$. If there exist constants$C$and$C^{\prime}$such that, for any words$w_{1},w_{2},w_{1}^{\prime},w_{2}^{\prime}\in A^{}$, we have*

[TABLE]

[TABLE]

then the language

[TABLE]

is regular.

Proof. Since $(w_{1},w_{2})\delta_{A}^{L}\in M\delta_{A}^{L}\Rightarrow||w_{1}|-|w_{2}||\leq C$ and $(w_{1}^{\prime},w_{2}^{\prime})\delta_{A}^{L}\in N\delta_{A}^{L}\Rightarrow||w_{1}^{\prime}|-|w_{2}^{\prime}||\leq C^{\prime},$ by Proposition 2.2.7, we have $M\delta_{A}^{R}$, $N\delta_{A}^{R}$ are regular. Then we have $M\odot N$ is regular by Proposition 2.2.6. For any $(\alpha,\beta)\delta_{A}^{R}\in M\odot N$, we have $||\alpha|-|\beta||\leq C+C^{\prime}$. Hence, by Proposition 2.2.7, $M\odot^{\prime}N$ is regular.

Proposition 2.2.9**.**

([19]) If $M$ is an automatic monoid and $A$ is any finite generating set for $M$, then there is a regular language $L$ over $A^{+}$ such that $(A,L)$ is an automatic structure for $M$.

Definition 2.2.10**.**

([19]) Let $(A,L)$ be an automatic structure for $S$. We say $(A,L)$ is an automatic structure with uniqueness for $S$ if $L$ maps one-to-one to $S$.

Proposition 2.2.11**.**

([19]) If $S$ is a semigroup with an automatic structure $(A,L)$, then there exists an automatic structure $(A,K)$ with uniqueness for $S$ with $K\subseteq L$.

2.3 Gröbner-Shirshov bases for associative algebras

To show that a semigroup $S$ is automatic, one needs to find an automatic structure $(A,L)$, where $A$ is a generating set of $S$ and $L\subset A^{+}$. After given a generating set $A$ of $S$, one usually takes a normal form, say, $L$, of $S$ to test whether $(A,L)$ is an automatic structure or not. It is known that Gröbner-Shirshov bases theory is a special tool to find formal forms for semigroups.

Let $A$ be a well-ordered set and $F$ be a field. We denote $F\langle A\rangle$ the free associative algebra over $F$ generated by $A$.

A well ordering $<$ on $A^{*}$ is called monomial if for any $u,v,w\in A^{*}$, we have

[TABLE]

A classical example of monomial ordering on $A^{*}$ is the deg-lex ordering, which first compare two words by degree (length) and then by comparing them lexicographically.

Let $A^{*}$ be with a monomial ordering $<$. Then, for any polynomial $f\in F\langle A\rangle$, $f$ has the leading word $\overline{f}$. We call $f$ monic if the coefficient of $\overline{f}$ is 1.

Let $f$ and $g$ be two monic polynomials in $F\langle A\rangle$ and $<$ a monomial ordering on $A^{*}$. Then, there are two kinds of compositions:

$(i)$ If  $w$ is a word such that $w=\bar{f}b=a\bar{g}$ for some $a,\ b\in A^{*}$ with $|\bar{f}|+|\bar{g}|>|w|$, then the polynomial $(f,g)_{w}=fb-ag$ is called the intersection composition of $f$ and $g$ with respect to $w$.

$(ii)$ If $w=\bar{f}=a\bar{g}b$ for some $a,\ b\in A^{*}$, then the polynomial $(f,g)_{w}=f-agb$ is called the inclusion composition of $f$ and $g$ with respect to $w$.

In $(f,g)_{w}$, $w$ is called ambiguity of the composition.

Let $R\subset F\langle A\rangle$ be a monic subset. Then the composition $(f,g)_{w}$ is called trivial modulo $(R,w)$ if $(f,g)_{w}=\sum\alpha_{i}a_{i}s_{i}b_{i}$, where each $\alpha_{i}\in F$, $a_{i},\ b_{i}\in A^{*},\ s_{i}\in R$ and $a_{i}\overline{s_{i}}b_{i}<w$.

A monic set $R\subset F\langle A\rangle$ is called a Gröbner-Shirshov basis with respect to the monomial ordering $<$ if any composition of polynomials in $R$ is trivial modulo $R$ and the corresponding ambiguity.

The following lemma was first proved by Shirshov for free Lie algebras [25, 26] (see also [3, 5]). Bokut [4] specialized the approach of Shirshov to associative algebras (see also Bergman [2]). For commutative algebras, this lemma is known as Buchberger’s Theorem (see [6, 7]).

Lemma 2.3.1**.**

(Composition-Diamond lemma for associative algebras)   Let $<$ be a monomial ordering on $A^{*}$. Let $R\subset F\langle A\rangle$ be a nonempty set of monic polynomials and $Id(R)$ the ideal of $F\langle A\rangle$ generated by $R$. Then the following statements are equivalent:

1. (1)

$R$* is a Gröbner-Shirshov basis in $F\langle A\rangle$.* 2. (2)

$f\in Id(R)\Rightarrow\bar{f}=a\bar{s}b$* for some $s\in R$ and $a,b\in A^{*}$.* 3. (3)

$Irr(R)=\{u\in A^{*}\ |\ u\neq a\bar{s}b,\ s\in R,\ a,b\in A^{*}\}$* is a $F$-basis of the algebra $F\langle A|R\rangle:=F\langle A\rangle/Id(R)$.*

If a subset $R$ of $F\langle A\rangle$ is not a Gröbner-Shirshov basis then one can add all nontrivial compositions of polynomials of $R$ to $R$. Continue this process repeatedly, we finally obtain a Gröbner-Shirshov basis $R^{comp}$ that contains $R$. Such a process is called Shirshov’s algorithm.

Let $M=sgp\langle A|R\rangle=A^{*}/\rho(R)$ be a monoid with the identity $\varepsilon$ (the empty word), where $\rho(R)$ is the congruence on $A^{*}$ generated by $R$. Then $R$ is also a subset of $F\langle A\rangle$ and we can find a Gröbner-Shirshov basis $R^{comp}$. We also call $R^{comp}$ a Gröbner-Shirshov basis in $M$. $Irr(R^{comp})=\{u\in A^{*}\ |\ u\neq a\overline{f}b,\ a,\ b\in A^{*},\ f\in R^{comp}\}$ is a $F$-basis of $F\langle A|R\rangle$ which is also a set of normal forms of $M$.

Let $sgp^{+}\langle A|R\rangle=A^{+}/\rho(R)$ be a semigroup (possibly without identity) generated by $A$ with defining relations $R$, where $\rho(R)$ is the congruence on $A^{+}$ generated by $R$. If $R=\{u_{i}=v_{i}|i\in I\}\ (u_{i},v_{i}\in A^{+},\ u_{i}>v_{i}$ for any $i\in I$) is a Gröbner-Shirshov basis, then

[TABLE]

is a set of normal forms of $M$ which is also called a normal form of $M$. In particular, if $(A,L)$ is an automatic structure for $M$ then $(A,L)$ is also a prefix-automatic structure for $M$.

3 Main results

In this section, we denote $sgp\langle A|R\rangle$ the monoid generated by $A$ with defining relations $R$ and $sgp^{+}\langle A|R\rangle$ the semigroup (possibly without identity) generated by $A$ with defining relations $R$. Suppose $A$ is a well-ordered set. We use the deg-lex ordering on $A^{*}$ if we mention Gröbner-Shirshov bases. Moreover, if $u=v\in R$, then $u>v$.

Suppose that $S=sgp^{+}\langle A|R\rangle$. Then define

[TABLE]

Clearly, $e$ is the identity of $S^{1}$.

3.1 Some (prefix-)automatic semigroups

Lemma 3.1.1**.**

Let $S=sgp^{+}\langle A|R\rangle$. Suppose $L\subseteq A^{+}$ and $(A,L)$ is a prefix-automatic structure for $S$. Let $B=A\cup\{e\}$ and $K=L\cup\{e\}$. Then $(B,K)$ is a prefix-automatic structure for $S^{1}$.

Proof. It is easy to see that $(B,K)$ is an automatic structure for $S^{1}$, where $e$ maps to the identity of $S^{1}$. Since $(A,L)$ is a prefix-automatic structure for $S$, we have

[TABLE]

is regular. Thus,

[TABLE]

is regular. So, $(B,K)$ is a prefix-automatic structure for $S^{1}$.

Theorem 3.1.2**.**

Let $S=sgp^{+}\langle A|u_{i}=v_{i},1\leq i\leq m\rangle$, where $A=\{a_{1},a_{2},\dots,a_{n}\},\ n\in\mathbb{N}$, $u_{i},v_{i}\in A^{+}$, $i=1,2,\dots,m$ and $\{u_{i}=v_{i}|1\leq i\leq m\}$ is a Gröbner-Shirshov basis.

1. (i)

If $v_{i}(t)\not\equiv u_{j}[t]$ for any $t\geq 1$, $i,j\in\{1,2,\dots,m\}$, then S is prefix-automatic. 2. (ii)

If $v_{i}(t)\not\equiv u_{j}[t]$ and $v_{i}[t]\not\equiv u_{j}(t)$ for any $t\geq 1$, $i,j\in\{1,2,\dots,m\}$, then $S$ is biautomatic.

Proof. Since $\{u_{i}=v_{i}|1\leq i\leq m\}$ is a Gröbner-Shirshov basis, by Lemma 2.3.1, we know

[TABLE]

is a normal form of $S$. Therefore, $L$ maps onto $S$. Obviously, $L$ is a regular language.

If (i) holds, then we show that $(A,L)$ is a prefix-automatic structure for $S$.

Let $W_{j}=\{u_{i}\mid u_{i}[1]=a_{j},1\leq i\leq m\}=\{u_{j_{1}},u_{j_{2}},\dots,u_{j_{s_{j}}}\},\ j=1,2,\dots,n$. By Proposition 2.1.2, L_{=}^{\}=\Delta_{L}$ is regular.

Now, by Proposition 2.1.2, L_{a_{j}}^{\}={(\alpha,\alpha a_{j})\delta_{A}^{R}|\alpha\in L}=\Delta_{L}\cdot{($,a_{j})}$is regular if$W_{j}=\emptyset$ and

[TABLE]

is regular if $W_{j}\neq\emptyset$.

Since $L$ is closed under prefix words, $L_{=}^{\prime}=\Delta_{L}$ is regular.

Hence, $S$ is prefix-automatic.

If (ii) holds, then we prove that $(A,L)$ is a biautomatic structure for $S$.

Let $W^{\prime}_{j}=\{u_{i}\mid u_{i}(1)=a_{j},\ 1\leq i\leq m\}=\{u_{j_{1}},u_{j_{2}},\dots,u_{j_{t_{j}}}\},\ j=1,2,\dots,n$. Then, by Proposition 2.1.2, {}^{\}L_{=}=\Delta_{L}$ is regular. Note that

[TABLE]

is regular if $W^{\prime}_{j}=\emptyset$ and

[TABLE]

is regular if $W^{\prime}_{j}\neq\emptyset$.

Hence, L_{=}^{\},\ ^{$}L_{=},\ L_{a_{j}}^{$}\ and\ ^{$}{a{j}}L\ (j=1,2,\dots,n)$ are regular.

Let $N=\displaystyle{\mathop{max}_{1\leq i\leq m}}\{|u_{i}|-|v_{i}|\}$. For any (\alpha,\beta)\delta_{A}^{R}\in L_{a_{j}}^{\}\ (j=1,2,\dots,n)$, since$||\alpha|-|\beta||\leq N+1, by Proposition [2.2.7](#S2.SS2.Thmdefinition7), we have {}^{$}L_{a_{j}}\ (j=1,2,\dots,n)$is regular. Similarly,${}{a{j}}L^{$}\ (j=1,2,\dots,n)$ is regular.

Hence, $(A,L)$ is a biautomatic structure for $S$.

Now, we consider some one-relator semigroups.

%֤

Theorem 3.1.3**.**

Let $S=sgp^{+}\langle a_{1},a_{2},\dots,a_{n}|a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}=a_{j_{1}}a_{j_{2}}\cdots a_{j_{k}}\rangle$, where $n\in\mathbb{N}$, $k\geq 2$, $a_{i_{l}},a_{j_{l}}\in\{a_{1},a_{2},\dots,a_{n}\}$, $l=1,2,\dots,k$ and $\{a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}=a_{j_{1}}a_{j_{2}}\cdots a_{j_{k}}\}$ is a Gröbner-Shirshov basis. If $S$ satisfies one of the following conditions, then $S$ is biautomatic and prefix-automatic.

1. (1)

For any $t\geq 1,\ (a_{j_{1}}a_{j_{2}}\cdots a_{j_{k}})(t)\not\equiv(a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}})[t]$ and $(a_{j_{1}}a_{j_{2}}\cdots a_{j_{k}})[t]\not\equiv(a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}})(t)$; 2. (2)

$|con(a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}})|=k$; 3. (3)

$k=2$; 4. (4)

$con(a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}})\nsubseteq con(a_{j_{1}}a_{j_{2}}\cdots a_{j_{k}})$; 5. (5)

$a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}\not\equiv ww^{\prime}w$* and $a_{j_{1}}a_{j_{2}}\cdots a_{j_{k}}\not\equiv ss^{\prime}s$ for any $w,s\in\{a_{1},a_{2},\dots,a_{n}\}^{+},w^{\prime},$ $s^{\prime}\in\{a_{1},a_{2},\dots,a_{n}\}^{*}$.*

Proof. Let $A=\{a_{1},a_{2},\dots,a_{n}\}$ and $L=A^{+}-A^{*}\{a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}\}A^{*}$. Since $\{a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}=a_{j_{1}}a_{j_{2}}\cdots a_{j_{k}}\}$ is a Gröbner-Shirshov basis, $L$ is a normal form of $S$.

By Proposition 2.1.2, L_{=}^{\}=\Delta_{L}=^{$}L_{=}$ and

[TABLE]

are regular. Now,

[TABLE]

Denote

[TABLE]

If $W_{R}$ and $W_{L}$ are regular, then L_{a_{i_{k}}}^{\}$and${}^{$}{a{i_{1}}}L$are regular. Since$a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}=a_{j_{1}}a_{j_{2}}\cdots a_{j_{k}}$is homogeneous, for any$(\alpha,\beta)\delta_{A}^{R}\in L_{a_{i}}^{$}$(or$(\alpha,\beta)\delta_{A}^{L}\in$${}^{$}{a{i}}L),\ i=1,2,\dots,n$, we have$||\alpha|-|\beta||\leq 1. Then by Proposition [2.2.7](#S2.SS2.Thmdefinition7), we have {}^{$}L_{a_{i}}$and${}{a{i}}L^{$}$are regular. Therefore,$(A,L)$is a biautomatic structure for$S$. That is to say, to prove$S$is biautomatic, it is sufficient to prove that$W_{R}$and$W_{L}$ are regular.

Now, we prove that $W_{R}$ and $W_{L}$ are regular if $S$ satisfies one of conditions $(1)-(5)$.

Denote $u\equiv a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}$, $v\equiv a_{j_{1}}a_{j_{2}}\cdots a_{j_{k}}$.

Case $1.$ If $S$ satisfies condition $(1)$, then by Theorem 3.1.2, $S$ is biautomatic.

Case $2.$ If there exist $l,l^{\prime}\in\mathbb{N},\ t_{1},t_{2},\dots,t_{l}\in\{1,2,\dots,k-1\}$ and $s_{1},s_{2},\dots,s_{{}_{l^{\prime}}}\in\{1,2,\dots,k-1\}$ such that $v(t_{i})\equiv u[t_{i}],\ v[s_{j}]\equiv u(s_{j})$ for each $i\in\{1,2,\dots,l\}$ and each $j\in\{1,2,\dots,l^{\prime}\}$, then let

[TABLE]

where $p_{11},p_{12},\dots,p_{1l},\dots,p_{c1},p_{c2},\dots,p_{cl}\geq 0,\ c\in\mathbb{N}$, $p_{11}^{\prime},p_{12}^{\prime},\dots,p_{1l^{\prime}}^{\prime},\dots,p_{c^{\prime}1}^{\prime},p_{c^{\prime}2}^{\prime},\dots,$ $p_{c^{\prime}l^{\prime}}^{\prime}\geq 0,\ c^{\prime}\in\mathbb{N}$, and $\alpha^{\prime}\in L-A^{*}\{u(k-t_{1}),u(k-t_{2}),\dots,u(k-t_{l})\}$, $\alpha^{{}^{\prime\prime}}\in L-\{u[k-s_{1}],u[k-s_{2}],\dots,u[k-s_{{}_{l^{\prime}}}]\}A^{*}$. Thus, $\alpha\cdot a_{i_{k}}=\beta$ and $a_{i_{1}}\cdot\sigma=\gamma$. By noting that

[TABLE]

are regular by Proposition 2.2.6 and Lemma 2.2.8, we just need to prove that $\beta,\gamma\in L$.

$(i)$ Suppose $S$ satisfies condition $(2)\ |con(u)|=|u|$.

If $\beta\not\in L$, then $u$ is a subword of $\beta$ and $u$ must be of the form

[TABLE]

where $t_{i_{j}}\in\{t_{1},t_{2},\dots,t_{l}\}$, $h\geq 2$ and $s+(k-t_{i_{2}})+(k-t_{i_{3}})+\cdots+(k-t_{i_{h-1}})+s^{\prime}=k$. If $h=2$, then $u\equiv v[t]v[t^{\prime}](s)$, where $0<t,t^{\prime},s<k$. Obviously, $t+s=k$. Since $|con(u)|=|u|$, we have $t^{\prime}\geq t+s=k$ (Otherwise $con(v[t])\cap con(v[t^{\prime}](s))\neq\emptyset$ which contradicts $|con(u)|=|u|$) which contradicts $t^{\prime}<k$. If $h>3$, then $u\equiv v[t_{1}^{\prime}]v[t_{2}^{\prime}]\cdots v[t_{h}^{\prime}](s)$. Obviously, we have $con(v[t_{1}^{\prime}])\cap con(v[t_{2}^{\prime}])\neq\varnothing$, a contradiction. Hence $\beta\in L$.

If $\gamma\not\in L$, then $u$ is a subword of $\gamma$ and $u$ must be of the form

[TABLE]

where $s_{i_{j}}\in\{s_{1},s_{2},\dots,s_{{}_{l^{\prime}}}\}$, $h^{\prime}\geq 2$ and $t+(k-s_{i_{2}})+(k-s_{i_{3}})+\cdots+(k-s_{i_{h^{\prime}-1}})+t^{\prime}=k$. If $h^{\prime}=2$, then $u\equiv v(s)[t]v(s^{\prime})$, where $0<s,t,s^{\prime}<k$. Obviously, $t+s^{\prime}=k$. Since $|con(u)|=|u|$, we have $s\geq s^{\prime}+t=k$ (otherwise $con(v(s^{\prime}))\cap con(v(s)[t])\neq\emptyset$, a contradiction) which contradicts $s<k$. If $h^{\prime}>3$, then $u\equiv v(s_{1}^{\prime})[s]v(s_{2}^{\prime})\cdots v(s_{h}^{\prime})$. Obviously, we have $con(v(s_{2}^{\prime}))\cap con(v(s_{3}^{\prime}))\neq\varnothing$, a contradiction. Hence $\gamma\in L$.

%֤ 3 $(ii)$ Suppose $S$ satisfies condition $(3)\ |u|=2$.

If $\beta\not\in L$, then $u$ is a subword of $\beta$ and $u$ must be of the form $v[1]v[1]$. Suppose $v\equiv ca$. Then $u\equiv aa$. Since $aaa=caa=cca$ and $aaa=aca$, we have $cca=aca$. Since $\{u=v\}$ is a Gröbner-Shirshov basis, we have $cca\equiv aca$. Then $a\equiv c$. This contradicts $u\not\equiv v$. Hence $\beta\in L$.

If $\gamma\not\in L$, then $u$ is a subword of $\gamma$ and $u$ must be of the form $v(1)v(1)$. Suppose $v\equiv ac$. Then $u\equiv aa$. Since $aaa=aca$ and $aaa=aac=acc$, we have $acc\equiv aca$ and so $a\equiv c$, that is, $u\equiv v$, a contradiction. Hence $\gamma\in L$.

%֤ 4 $(iii)$ Suppose $S$ satisfies condition $(4)\ con(u)\nsubseteq con(v)$.

If $\beta\not\in L$ ($\gamma\not\in L$, resp.), then $u$ is a subword of $\beta$ ($\gamma$, resp.) and $u$ must be contained in some subword of $\beta$ of the form $v[k-t_{i_{1}}]v[k-t_{i_{2}}]\cdots v[k-t_{i_{h}}]$ (in some subword of $\gamma$ of the form $v(k-s_{i_{1}})v(k-s_{i_{2}})\cdots v(k-s_{i_{h^{\prime}}})$, resp.). This contradicts $con(u)\nsubseteq con(v)$. Hence $\beta,\gamma\in L$.

%֤ 5 $(iv)$ Suppose S satisfies condition $(5)\ a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}\not\equiv ww^{\prime}w,\ a_{j_{1}}a_{j_{2}}\cdots a_{j_{k}}\not\equiv ss^{\prime}s$ for any $w,s\in\{a_{1},a_{2},\dots,a_{n}\}^{+},w^{\prime},s^{\prime}\in\{a_{1},a_{2},\dots,a_{n}\}^{*}$.

If $\beta\not\in L$, then $u\equiv v[k-t_{i_{1}}][s]v[k-t_{i_{2}}]\cdots v[k-t_{i_{h}}](s^{\prime})$, where $t_{i_{j}}\in\{t_{1},t_{2},\dots,t_{l}\}$, $h\geq 2$, $1\leq s\leq k-t_{i_{1}}$ and $1\leq s^{\prime}\leq k-t_{i_{h}}$.

If $h=2$, then $u\equiv v[k-t][s]v[k-t^{\prime}](s^{\prime})$ and $s+s^{\prime}=k>k-t^{\prime}$. Hence $s>k-t^{\prime}-s^{\prime}$.

Case $1.$ If $k-t^{\prime}\geq s$, then $k-t^{\prime}\geq s>k-t^{\prime}-s^{\prime}$. Hence there exists a prefix $w$ of $v[s]$ such that $u\equiv ww^{\prime}w$ for some $w^{\prime}\in A^{*}$, see Figure $1$. This contradicts condition $(5)$.

Case $2.$ If $k-t^{\prime}<s$, then $k-s=s^{\prime}<t^{\prime}$. Since $u[t^{\prime}]\equiv v(t^{\prime})$, there exists a suffix $v[c]$ of $v[k-t]$ such that $v[c]\equiv v(c)$. Hence, there exists a prefix $w$ of $v(c)$ such that $v\equiv ww^{\prime}w$ (if $c\leq\frac{k}{2},w\equiv v(c)$; if $c\geq\frac{k}{2},w\equiv v(2c-k)$) for some $w^{\prime}\in A^{*}$, see Figure $2$. This is a contradiction.

If $h\geq 3$, then $u\equiv v[k-t_{i_{1}}][s]v[k-t_{i_{2}}]\cdots v[k-t_{i_{h}}](s^{\prime})$. Since $v[k-t_{i_{h-1}}]$ is completely contained in $u$, we have $t_{i_{h-1}}=k-(k-t_{i_{h-1}})>s^{\prime}$. Since $u[t_{i_{h-1}}]\equiv v(t_{i_{h-1}})$, there exists a suffix $v[c]$ of $v[t_{i_{j}}]$ for some $t_{i_{j}}\in\{t_{i_{1}},t_{i_{2}},\dots,t_{i_{h-1}}\}$ such that $v[c]\equiv v(c)$, see Figure $3$. That is $v\equiv ww^{\prime}w$ (if $c\leq\frac{k}{2},w\equiv v(c)$; if $c\geq\frac{k}{2},w\equiv v(2c-k)$) for some $w^{\prime}\in A^{*}$, a contradiction.

Hence $\beta\in L$.

If $\gamma\not\in L$, then $u\equiv v(k-s_{i_{1}})[t]v(k-s_{i_{2}})\cdots v(k-s_{i_{h^{\prime}}})(t^{\prime})$, where $s_{i_{j}}\in\{s_{1},s_{2},\dots,s_{{}_{l^{\prime}}}\}$, $h^{\prime}\geq 2$, $1\leq t\leq k-s_{i_{1}}$ and $1\leq t^{\prime}\leq k-s_{i_{h^{\prime}}}$.

If $h^{\prime}=2$, then $\gamma\equiv v(k-s)[t]v(k-s^{\prime})(t^{\prime})$ and $t+t^{\prime}=k>k-s$. Hence $t^{\prime}>k-s-t$.

Case $1.$ If $k-s\geq t^{\prime}$, then $k-s\geq t^{\prime}>k-s-t$. Hence there exists a suffix $w$ of $v(t^{\prime})$ such that $u\equiv ww^{\prime}w$ for some $w^{\prime}\in A^{*}$, see Figure $4$.

Case $2.$ If $k-s<t^{\prime}$, then $k-t^{\prime}=t<s$. Since $u[s]\equiv v(s)$, there exists a prefix $v(c)$ of $v[k-s^{\prime}]$ such that $v[c]\equiv v(c)$. Hence, there exists a prefix $w$ of $v(c)$ such that $v\equiv ww^{\prime}w$ (if $c\leq\frac{k}{2},w\equiv v(c)$; if $c\geq\frac{k}{2},w\equiv v(2c-k)$) for some $w^{\prime}\in A^{*}$, see Figure $5$.

If $h^{\prime}\geq 3$, then $u\equiv v(k-s_{i_{1}})[t]v(k-s_{i_{2}})\cdots v(k-s_{i_{h^{\prime}}})(t^{\prime})$. Since $v(k-s_{i_{2}})$ is completely contained in $u$, we have $s_{i_{2}}=k-(k-s_{i_{2}})>t$. Since $u(s_{i_{2}})\equiv v[s_{i_{2}}]$, there exists a prefix $v(c)$ of $v(s_{i_{j}})$ for some $s_{i_{j}}\in\{s_{i_{1}},s_{i_{2}},\dots,s_{i_{h^{\prime}-1}}\}$ such that $v(c)\equiv v[c]$, see Figure $6$. That is $v\equiv ww^{\prime}w$ (If $c\leq\frac{k}{2},w\equiv v(c)$; if $c\geq\frac{k}{2},w\equiv v(2c-k)$) for some $w^{\prime}\in A^{*}$.

Both cases contradict condition (5). Hence $\gamma\in L$.

Therefore, $\beta,\gamma\in L$ if $S$ satisfies one of conditions $(2)-(5)$. This shows that $\widetilde{W}_{R}=W_{R}$ and $\widetilde{W}_{L}=W_{L}$. Hence $W_{R}$ and $W_{L}$ are regular. Thus, $S$ is biautomatic.

Since $L$ is closed under prefix words, $S$ is prefix-automatic.

%֤

Theorem 3.1.4**.**

Suppose that $S=sgp\langle a_{1},a_{2},\dots,a_{n}|a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}=\varepsilon\rangle,\ n\in\mathbb{N},\ k\geq 2,\ a_{i_{j}}\in\{a_{1},a_{2},\dots,a_{n}\},\ j=1,2,\dots,k$ and $\{a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}=\varepsilon\}$ is a Gröbner-Shirshov basis. Then $S$ is biautomatic.

Proof. Since $S=sgp\langle a_{1},a_{2},\dots,a_{n}|a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}=\varepsilon\rangle\cong\widetilde{S}=sgp^{+}\langle e,a_{1},a_{2},\dots,a_{n}|a_{i_{1}}a_{i_{2}}$ $\cdots a_{i_{k}}=e,ee=e,a_{j}e=a_{j}=ea_{j},j=1,2,\dots,n\rangle$, we just need to prove that $\widetilde{S}$ is biautomatic.

Let $A=\{e,a_{1},a_{2},\dots,a_{n}\}$, $L=\{e\}\cup(A^{+}-A^{*}\{a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}},e\}A^{*})$. It is easy to see that $\{a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}=e,ee=e,a_{j}e=a_{j}=ea_{j},j=1,2,\dots,n\}$ is also a Gröbner-Shirshov basis in $\widetilde{S}$. Thus, $L$ is a normal form of $\widetilde{S}$. Obviously, $L$ is regular and $L$ maps onto $\widetilde{S}$. We prove that $(A,L)$ is a biautomatic structure for $\widetilde{S}$.

By Proposition 2.1.2, L_{=}^{\}=\Delta_{L}=^{$}L_{=}$ is regular. Note that

[TABLE]

are regular for any $a_{i}\neq a_{i_{k}}$ and $a_{j}\neq a_{i_{1}}$. Also,

[TABLE]

are regular.

For any (\alpha,\beta)\delta_{A}^{R}\in L_{a_{j}}^{\}$$(or\ (\alpha,\beta)\delta_{A}^{L}\in$${}^{$}{a{j}}L)$, since$||\alpha|-|\beta||\leq k, by Proposition [2.2.7](#S2.SS2.Thmdefinition7), we have {}^{$}L_{a_{j}}\ (or\ {a{j}}L^{$})$is regular,$j=1,2,\dots,n$.

Hence $(A,L)$ is a biautomatic structure for $\widetilde{S}$.

Theorem 3.1.5**.**

Suppose that $S=sgp^{+}\langle a_{1},a_{2},\dots,a_{n}|a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}=x\rangle,\ n\in\mathbb{N},\ k\geq 2,\ x,a_{i_{j}}\in\{a_{1},a_{2},\dots,a_{n}\}$, $j=1,2,\dots,k$ and $\{a_{i1}a_{i2}\cdots a_{ik}=x\}$ is a Gröbner-Shirshov basis. Then $S$ is prefix-automatic.

Proof. Case 1. If $x\not\equiv a_{ik}$, then by Theorem 3.1.2, S is prefix-automatic.

Case 2. Let $x\equiv a_{ik}$ and $A=\{a_{1},a_{2},\dots,a_{n}\}$. Then $L=A^{+}-A^{*}\{a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}\}A^{*}$ is a normal form of $S$ and L_{=}^{\}=\Delta_{L}$ is regular by Proposition 2.1.2. By noting that

[TABLE]

is also regular for any $a_{j}\neq a_{i_{k}}$ and

[TABLE]

is regular by Propositions 2.1.2 and 2.2.6, $S$ is automatic. Since $L$ is closed under prefix words, $S$ is prefix-automatic.

Noting that the semigroup $S$ in Theorem 3.1.5 may not be biautomatic, the following theorem is an example.

Theorem 3.1.6**.**

Let $S=sgp^{+}\langle a,b|a^{k}b=b\rangle$, where $k\geq 1$. Then $S$ is prefix-automatic but not biautomatic.

Proof. Clearly, $\{a^{k}b=b\}$ is a Gröbner-Shirshov basis. By Theorem 3.1.5, $S$ is prefix-automatic. We now show that there does not exist biautomatic structure for $S$.

Suppose $S$ is biautomatic. Then $S^{1}$ is biautomatic by Proposition 2.2.4. For $B=\{e,a,b\}$, there exists $K\subseteq B^{+}$ such that $(B,K)$ is a biautomatic structure for $S^{1}$ with uniqueness by Propositions 2.2.9 and 2.2.11. So \leftidx{{}^{\}}K_{b}$is regular. For$i,j\in\mathbb{N}$, let$\alpha,\beta\in K$with$\alpha=b^{j}(a^{k})^{i},\beta=b^{j+1}$. Then we have$(\alpha,\beta)\in\leftidx{{}^{$}}K_{b}$ and

[TABLE]

where $\gamma_{j+1},\tau_{11},\dots,\tau_{1k},\dots,\tau_{i1},\dots,\tau_{ik},\eta_{j+2}\in\{e\}^{*},\gamma_{1},\dots,\gamma_{j},\eta_{1},\dots,\eta_{j+1}\in\{e,a\}^{*}$. Denote $\alpha_{1}\equiv\gamma_{j+1}(a\tau_{11}a\tau_{12}\cdots a\tau_{1k})\cdots(a\tau_{i1}a\tau_{i2}\cdots a\tau_{ik})$. Then $ik\leq|\alpha_{1}|\leq N(ik+1)+ik$ and $j+1\leq|\beta|<N(j+2)+j+1$, where $N=S(M(K))$.

Let $ik>N(j+2)+j+1$ and j>|S(M(^{\}K_{b}))|$. Then there will be a loop$(u_{1},u_{2})\delta_{B}^{L}$in$(\alpha_{1},\beta)\delta_{B}^{L}$. Assume$\alpha\equiv w_{1}u_{1}w_{2},\ \beta\equiv w_{1}^{\prime}u_{2}w_{2}^{\prime}$. Since$u_{1}$is a subword of$\alpha_{1}$,$b\not\in con(u_{1})$. If$b\in con(u_{2})$, we have$occ(b,w_{1}u_{1}^{i}w_{2}b)\neq occ(b,w_{1}^{\prime}u_{2}^{i}w_{2}^{\prime})$, so$(w_{1}u_{1}^{i}w_{2},w_{1}^{\prime}u_{2}^{i}w_{2}^{\prime})\delta_{B}^{R}\not\in{\vphantom{K}}^{$}{K}{b}$for$i>1$, a contradiction. Hence$b\not\in con(u{2})$and so$con(u_{2})\subseteq{e,a}$. Therefore$\beta\equiv w_{1}^{\prime}u_{2}w_{2}^{\prime}=w_{1}^{\prime}u_{2}^{k+1}w_{2}^{\prime}\in K$which contradicts the uniqueness of$K$. Thus$S$ is not biautomatic.

Theorem 3.1.7**.**

Let $S=sgp^{+}\langle a_{1},a_{2},\dots,a_{n}|a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}=xy\rangle$, $n\in\mathbb{N},\ k\geq 2$, where $x,y,a_{i_{1}},\dots,a_{i_{k}}\in\{a_{1},a_{2},\dots,a_{n}\}$ and $\{a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}=xy\}$ is a Gröbner-Shirshov basis. If $a_{i_{k-1}}a_{i_{k}}\not\equiv xy$ and $a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}\not\equiv y^{k-1}x$, then $S$ is automatic and $S^{1}$ is prefix-automatic.

Proof. If $x\not\equiv a_{i_{k}}$, then $S$ is prefix-automatic by Theorem 3.1.2. We now suppose $x\equiv a_{i_{k}}$.

Let $A=\{a_{1},a_{2},\dots,a_{n}\}$ and $L=A^{+}-A^{*}\{a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}x\}A^{*}$. Then $L$ is a normal form of $S$ and $L$ is clearly regular.

Case 1. If $x\equiv y$, then $S=sgp^{+}\langle a_{1},a_{2},\dots,a_{n}|a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}x=x^{2}\rangle$ since $a_{i_{k-1}}x\not\equiv xx$ and $a_{i_{k-1}}\not\equiv x$. Clearly, L_{=}^{\}=\Delta_{L}$and$L_{a_{j}}^{$}=\Delta_{L}\cdot{($,a_{j})}$are regular if$a_{j}\not\equiv x$.

Case 1-1. If $a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}x$ is a subword of $a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}$, then

[TABLE]

is regular by Proposition 2.1.2. Hence $S$ is prefix-automatic.

Case 1-2. If $a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}x$ is not a subword of $a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}$, then $a_{i_{1}}\neq x$. Let $B=\{e,a_{1},a_{2},\dots,a_{n}\}$ and $\mathcal{M}=(S,A,\mu,s_{0},F)$ is a DFSA accepting $L$. Denote $m=max\{i|x^{i}$ is a subword of $a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}\}$. Then $m<k-1$.

Let $\mathcal{A}=(Q,A,B,\sigma,q_{0},T)$ be a $gsm$, where $Q=S\times\{0,1,\dots,m\}$ is the set of states, $q_{0}=(s_{0},0)$ the initial state, $T=F\times\{0,1,\dots,m\}$ the terminal states and $\sigma$ the partial function from $Q\times A$ to $\mathcal{P}(Q\times B^{*})$ defined by the following equations

[TABLE]

where $sb:=\mu(s,b)$ for $s\in S,b\in A$.

Let $K=\eta_{\mathcal{A}}(L)\cup\{e\}$. Since $L$ is regular, we have $K$ is regular and maps onto $S^{1}$. Thus K_{=}^{\}=\Delta_{K}=K_{e}^{$}$ is regular. In addition,

[TABLE]

and

[TABLE]

are regular. So $(B,K)$ is an automatic structure for $S^{1}$. Hence $S$ is automatic by Proposition 2.2.4. Since

[TABLE]

is regular, $S^{1}$ is prefix-automatic.

Case 2. If $x\not\equiv y$, then $S=sgp^{+}\langle a_{1},a_{2},\dots,a_{n}|a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}x=xy\rangle$.

Case 2-1. Suppose $a_{i_{1}}\not\equiv y$. Let $\mathcal{M}=(S,A,\mu,s_{0},F)$ be a DFSA accepting $L$.

Case 2-1-1. If $a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}x$ is a subword of $a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}$, then $(A,L)$ is an automatic structure for $S$. It follows that

[TABLE]

are regular. Since $L$ is closed under prefix words, $S$ is prefix-automatic.

Case 2-1-2. If $a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}x$ is not a subword of $a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}$, let $\mathcal{A}=(Q,A,B,\sigma,$ $q_{0},T)$ be a $gsm$ where $Q=S\times\{0,1,\dots,m\},\ B=A\cup\{e\},\ q_{0}=(s_{0},0),\ T=F\times\{0,1,\dots,m\},\ m=max\{i|y^{i}$ is a subword of $a_{i_{1}}a_{i_{2}}\cdots a_{i_{k-1}}\}$ (note that $m\leq k-2)$ and $\sigma$ the partial function from $Q\times A$ to $\mathcal{P}(Q\times B^{*})$ defined by the following equations

[TABLE]

where $sb:=\mu(s,b)$ for $s\in S,b\in A$.

Let $K=\eta_{\mathcal{A}}(L)\cup\{e\}$. Then by the property of $gsm$, $K$ is regular and maps onto $S^{1}$. Thus K_{=}^{\}=\Delta_{K}=K_{e}^{$}$ is regular. In addition,

[TABLE]

are regular, so $(B,K)$ is an automatic structure for $S^{1}$. Hence $S$ is automatic by Proposition 2.2.4. Since

[TABLE]

is regular, $S^{1}$ is prefix-automatic.

Case 2-2. Let $a_{i_{1}}\equiv y$ and $S=sgp^{+}\langle a_{1},a_{2},\dots,a_{n}|y^{t}ux=xy\rangle$ where $|y^{t}ux|=k$ and $u(1)\not\equiv y$. Since $a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}}\not\equiv y^{k-1}x$, we have $u\not\equiv\varepsilon.$

Case 2-2-1. If $y^{t}ux$ is a subword of $y^{t}uy^{t}u$, then $(A,L)$ is an automatic structure for S, where $L=A^{+}-A^{*}\{y^{t}u\}A^{*}$. By noting that

[TABLE]

are regular, $S$ is prefix-automatic since $L$ is closed under prefix words.

Case 2-2-2. If $y^{t}ux$ is not a subword of $y^{t}uy^{t}u$, then let $\mathcal{M}=(S,A,\mu,s_{0},F)$ be a DFSA accepting $L$. Let $m=max\{i|y^{i}$ is a subword of $y^{t}uy^{t}u\}$. Define a $gsm$ $\mathcal{A}=(Q,A,B,\sigma,q_{0},T)$, where $Q=S\times\{0,1,\dots,m\},\ B=A\cup\{e\},\ q_{0}=(s_{0},0),\ T=F\times\{0,1,\dots,m\}$ and $\sigma$ the partial function from $Q\times A$ to $\mathcal{P}(Q\times B^{*})$ defined by the following equations

[TABLE]

where $sb:=\mu(s,b)$ for $s\in S,b\in A$.

Let $K=\eta_{\mathcal{A}}(L)\cup\{e\}$. Then $K$ is regular and maps onto $S^{1}$. Since

[TABLE]

are all regular, $(B,K)$ is an automatic structure for $S^{1}$. Thus $S$ is automatic by Proposition 2.2.4. Since

[TABLE]

is regular, $S^{1}$ is prefix-automatic.

Theorem 3.1.8**.**

Let $S=sgp^{+}\langle a,b|a^{k}b^{l}=b^{l}\rangle$, where $k\geq 1$ and $k+l\geq 1$. Then $S$ is prefix-automatic if and only if $l\leq 1$.

Proof. $(\Leftarrow)$ If $l=0$, then by Theorem 3.1.4, $S$ is prefix-automatic. If $l=1$, then by Theorem 3.1.5, $S$ is prefix-automatic.

$(\Rightarrow)$ Suppose that $S=sgp^{+}\langle a,b|a^{k}b^{l}=b^{l}\rangle$ is automatic for some $l>1$. Then $S^{1}$ is also automatic by Proposition 2.2.4. Let $B=\{e,a,b\}$. Then there exists $K\subseteq B^{+}$ such that $(B,K)$ is an automatic structure for $S^{1}$ with uniqueness by Propositions 2.2.9 and 2.2.11.

For any $s,t\in\mathbb{N}$, there exist $\alpha,\beta\in K$ such that $\alpha=(a^{k})^{s}(a^{k}b)^{t}b^{l-2},\ \beta=b^{t+l-1}$. Then (\alpha,\beta)\in K_{b}^{\}$ and

[TABLE]

where $\tau_{tj},\xi_{i}\in\{e\}^{*},\ \eta_{t},\eta_{t+1}\cdots\eta_{t+l}\in\{e\}^{*},\ \gamma_{i},\tau_{ij}\ (i=1,2,\dots,t-1),\eta_{1},\eta_{2}\cdots\eta_{t-1}\in\{e,a^{k}\}^{*}$.

Denote $\alpha_{1}\equiv\gamma_{1}a\gamma_{2}a\cdots\gamma_{sk}a\gamma_{sk+1}$. Then $sk\leq|\alpha_{1}|<N(sk+1)+ks$ and $t+l-1\leq|\beta|<N(t+l)+t+l-1$, where $N=|S(M(K))|$.

Let $sk>N(t+l)+t+l-1$ and t+l-1>|S(M(K_{b}^{\}))|$. Then there is a loop$(u_{1},u_{2})\delta_{B}^{R}$in$(\alpha_{1},\beta)\delta_{B}^{R}$. If$b\in con(u_{2})$, we have$b\not\in con(u_{1})$since$u_{1}$is a subword of$\alpha_{1}$. Suppose$\alpha\equiv w_{1}u_{1}w_{2}$and$\beta\equiv w_{1}^{\prime}u_{2}w_{2}^{\prime}$. Since$(\alpha,\beta)\delta_{B}^{R}\in K_{b}^{$}$and$(w_{1}u_{1}^{2}w_{2},w_{1}^{\prime}u_{2}^{2}w_{2}^{\prime})\delta_{B}^{R}\in K_{b}^{$}$, we have$occ(b,w_{1}u_{1}^{2}w_{2})\leq occ(b,w_{1}^{\prime}u_{2}^{2}w_{2}^{\prime})-2$, a contradiction. Hence$b\not\in con(u_{2})$, so$con(u_{2})\subseteq{e,a}$and$\beta\equiv w_{1}^{\prime}u_{2}w_{2}^{\prime}=w_{1}^{\prime}u_{2}^{k+1}w_{2}^{\prime}\in K$which contradicts the uniqueness of$K$. Thus$S$ is not automatic.

Theorem 3.1.9**.**

Let $S=sgp^{+}\langle a,b\ |\ a^{k}b=ba\rangle\ (k\geq 0)$. Then $S$ is prefix-automatic if and only if $k\leq 1$.

Proof. $(\Leftarrow)$ If $k\leq 1$, then $S$ is prefix-automatic by Theorems 3.1.5 and 3.1.3.

$(\Rightarrow)$ Suppose $k\geq 2$. Since $S$ is automatic, $S^{1}$ is automatic by Proposition 2.2.4. So, there exists an automatic structure $(B,L)$ with uniqueness for $S^{1}$, where $B=\{e,a,b\}$ and $e$ represents the identity of $S^{1}$. Let $N=|S(M(L))|$, N_{=}=|S(M(L_{=}^{\}))|$,$N_{a}=|S(M(L_{a}^{$}))|$,$N_{b}=|S(M(L_{b}^{$}))|$,$\widetilde{N}=max{N,N_{=},N_{a},N_{b}}$.

First we claim that $||\alpha|-|\beta||\leq\widetilde{N}$ for any (\alpha,\beta)\delta_{B}^{R}\in L_{a}^{\}\cup L_{b}^{$}$. Otherwise there exists$(\alpha,\beta)\delta_{B}^{R}\in L_{a}^{$}\cup L_{b}^{$}$such that$||\alpha|-|\beta||>\widetilde{N}$. We can suppose$|\alpha|>|\beta|+\widetilde{N}$and$\alpha\equiv\alpha_{1}\alpha_{2}$with$|\alpha_{1}|=|\beta|$. Then$(\alpha_{2},\varepsilon)\delta_{B}^{R}$contains a subword$u$that can be pumped in$M(L_{a}^{$})$or in$M(L_{b}^{$})$. So either$L_{a}^{$}$or$L_{b}^{$}$contains words of the form$(\widetilde{\alpha}{1}u^{j}\widetilde{\alpha}{2},\beta)\delta_{B}^{R}$with$j\in\mathbb{N}$where$\widetilde{\alpha}{1}u\widetilde{\alpha}{2}\equiv\alpha$. Since$occ(b,\beta)=occ(b,\widetilde{\alpha}{1}u\widetilde{\alpha}{2})$, we have$b\not\in con(u)$. If$a\in con(u)$, then there exists$j\geq 1$such that$occ(a,\widetilde{\alpha}{1}u^{j}\widetilde{\alpha}{2})>occ(a,\gamma)$for any$\gamma\in B^{*}$and$\gamma=\beta$, a contradiction. So$u\equiv e^{|u|}$, which contradicts the uniqueness of$L$.

For $n,i\in\mathbb{N},n,i\geq 1$, let $\beta_{n},\gamma_{n},\beta_{n}^{(i)}\in L$ with $\beta_{n}=a^{k^{n}},\gamma_{n}=b^{n},\beta_{n}^{(i)}=a^{k^{n}}b^{i}$. We have $|\beta|\geq k^{n},|\gamma_{n}|<(n+1)N+n$ and by the above claim

[TABLE]

and so $||\beta_{n}|-|\beta_{n}^{(n)}||\leq n\widetilde{N}$. Hence $|\beta_{n}^{(n)}|\geq|\beta_{n}|-n\widetilde{N}\geq k^{n}-n\widetilde{N}$. On the other hand, since (\gamma_{n},\beta_{n}^{(n)})\in L_{a}^{\}\cup L_{b}^{$}$, by the claim we have$||\gamma_{n}|-|\beta_{(n)}^{n}||\leq\widetilde{N}$. However,$|\beta_{n}^{(n)}|-|\gamma_{n}|>k^{n}-n\widetilde{N}-(n+1)N-n\rightarrow\infty$, that is, there exists some$n\in\mathbb{N},n\geq 1$such that$|\beta_{n}^{(n)}|\geq\widetilde{N}+|\gamma_{n}|$, a contradiction.

This shows that $k\leq 1$.

3.2 Automaticity of semigroups of one-relator of length $\leq 3$

In this section, we prove the following theorem.

Theorem 3.2.1**.**

Let $S=sgp\langle A|u=v\rangle$, where $A=\{a_{1},a_{2},\dots,a_{n}\},\ n\geq 2,\ u,v\in A^{*},\ |v|\leq|u|\leq 3$ and $a,b\in A,\ a\neq b$. Then

1. (1)

$S$* is prefix-automatic if $u=v\not\in\{aba=ba,\ aab=ba,abb=bb\}$. Moreover, if $u=v\in\{aba=ba,\ aab=ba,abb=bb\}$ then $S$ is not automatic.* 2. (2)

$S$* is biautomatic if one of the following holds: (i) $|u|=3,\ |v|=0$, (ii) $|u|=|v|=3$, (iii) $|u|=2$ and $u=v\not\in\{ab=a,\ ab=b\}$. Moreover, if $u=v\in\{ab=a,\ ab=b\}$ then $S$ is not biautomatic.*

We prove Theorem 3.2.1 step by step.

We fix $A=\{a_{1},a_{2},\dots,a_{n}\},\ u,v\in A^{*},\ S=sgp\langle A|u=v\rangle$.

Lemma 3.2.2**.**

Suppose $|u|=|v|=2$. Then $S$ is biautomatic and prefix-automatic.

Proof. Case 1. Suppose $a\not\equiv b,\ a,b\in A$. Then $\{ab=cd\}$ is a Gröbner-Shirshov basis. Since $|u|=2$, $S$ is biautomatic and prefix-automatic by Theorem 3.1.3.

Case 2. Suppose $a\equiv b,c\equiv d$. By Proposition 2.2.5, we just need to prove $\widetilde{S}=sgp^{+}\langle a,c|a^{2}=c^{2}\rangle$ is biautomatic. Since $\{a^{2}=c^{2},ac^{2}=c^{2}a\}$ is a Gröbner-Shirshov basis for $\widetilde{S}$, $L=A^{+}-A^{*}\{a^{2},ac^{2}\}A^{*}$ is a normal form of $S$, where $A=\{a,c\}$. Clearly L_{=}^{\}={\vphantom{L}}^{$}{L}{=}=\Delta{L}$ is regular. Now,

[TABLE]

are regular.

Since $||\alpha|-|\beta||\leq 1$ for (\alpha,\beta)\delta_{A}^{R}\in L_{a}^{\}\cup L_{c}^{$}$and$(\alpha,\beta)\delta_{A}^{L}\in\leftidx{{}{a}^{$}}{L}\cup\leftidx{{}{c}^{$}}L$, we have$\leftidx{{}^{$}}L_{a},\leftidx{{}^{$}}L_{c}$,$\leftidx{{}{a}}L^{$},\leftidx{{}{c}}L^{$}are regular by Proposition [2.2.7](#S2.SS2.Thmdefinition7). Hence\widetilde{S}$ is biautomatic. Thus, S is biautomatic and prefix-automatic.

Lemma 3.2.3**.**

Suppose $|u|=2,|v|\leq 2$. Then $S$ is prefix-automatic. Moreover, $S$ is biautomatic if and only if $u=v\not\in\{ab=a,ab=b\},\ a,b\in A$.

Proof. Case 1. If $|v|=0$, then $S$ is biautomatic and prefix-automatic by Theorem 3.1.4.

By Proposition 2.2.4, in order to prove that $S$ is biautomatic, it suffices to prove $\widetilde{S}=sgp^{+}\langle a_{1},a_{2},\dots,a_{n}|u=v\rangle$ is biautomatic.

Case 2. If $|v|=2$, then $S$ is biautomatic and prefix-automatic by Lemma 3.2.2.

Case 3. If $|v|=1$, suppose $u\equiv ab,v\equiv c$.

Case 3-1. If $a\not\equiv b$, $c\not\equiv a$ and $c\not\equiv b$, then $\{ab=c\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$. So $\widetilde{S}$ is biautomatic and $S$ is prefix-automatic by Theorem 3.1.2.

Case 3-2. If $a\not\equiv b$ and $c\equiv b$, then $\{ab=b\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$ and by Theorem 3.1.6, $\widetilde{S}$ is automatic but not biautomatic, and $S$ is prefix-automatic.

Case 3-3. If $a\not\equiv b$ and $c\equiv a$, then $\widetilde{S}=spg^{+}\langle a,b|ab=a\rangle$ is isomorphic to $sgp^{+}\langle a,b|ab=b\rangle^{rev}$ and by Case 3-2, Lemma 2.2.3 and Theorem 3.1.5, $\widetilde{S}$ is automatic but not biautomatic, and $S$ is prefix-automatic.

Case 3-4. If $a\equiv b$ and $c\not\equiv a$, then by Proposition 2.2.5, it is sufficient to prove $S_{1}=sgp^{+}\langle a,c|aa=c\rangle$ is biautomatic.

Note that $\{a^{2}=c,ac=ca\}$ is a Gröbner-Shirshov basis in $S_{1}$. Let $A_{1}=\{a,c\}$ and $L=\{c^{i}a^{j}|i\geq 0,j\in\{0,1\},i+j\geq 1\}$. Then $L$ is a normal form of $S_{1}$. Clearly, $L$ and L_{=}^{\}=\Delta_{L}$ are regular. Since

[TABLE]

and

[TABLE]

are regular, $S_{1}$ is an automatic semigroup. Let $\widetilde{L}=A^{+}-A^{*}\{aa,ac\}A^{*}$. Then by Proposition 2.2.5, $(A,\widetilde{L})$ is an automatic structure for $\widetilde{S}$.

For any $x\in A\cup\{\varepsilon\}$, (u,v)\delta_{A}^{R}\in L_{x}^{\}$, we have$||u|-|v||\leq 1. So by Proposition [2.2.7](#S2.SS2.Thmdefinition7), {}^{$}L_{x}$is regular. Hence$(A,\widetilde{L})$is a left-right automatic structure for$\widetilde{S}$.

Since {}^{\}{=}L=\Delta{L}=_{=}L^{$}$,

[TABLE]

are regular, $S_{1}$ is a left-left automatic semigroup.

For any $x\in A\cup\{\varepsilon\}$, (u,v)\delta_{A}^{L}\in_{x}^{\}L$, we have$||u|-|v||\leq 1and by Proposition [2.2.7](#S2.SS2.Thmdefinition7),{}_{x}L^{$}$is regular. Hence$(A,\widetilde{L})$is a right-left automatic structure for$\widetilde{S}$.

Therefore, $\widetilde{S}$ is biautomatic and prefix-automatic, and so is $S$.

Case 3-5. If $a\equiv b$ and $c\equiv a$, then by Propositions 2.2.5 and 2.2.7, it is sufficient to prove that $S_{1}=sgp^{+}\langle a|aa=a\rangle$ is biautomatic. Obviously, $S_{1}=\{a\}$ is biautomatic. Therefore, $\widetilde{S}$ is biautomatic and prefix-automatic.

The lemma is proved.

%֤ ʮһ Ρ

Lemma 3.2.4**.**

Suppose $|u|=|v|=3$. Then $S$ is biautomatic and prefix-automatic.

Proof. Let $R=\{u=v\},\ u=abc,\ v=xyz,\ a,b,c,x,y,z\in A$ and $\widetilde{S}=sgp^{+}\langle A|R\rangle$.

$(i)$ Suppose $a,b,c$ are pairwise different. Then $R$ is a Gröbner-Shirshov basis in $\widetilde{S}$. Since $|con(abc)|=|u|$, by Theorem 3.1.3, $S$ is biautomatic and prefix-automatic.

$(ii)$ Suppose $R=\{aac=xyz\}$, where $a\not\equiv c$. Then $R$ is a Gröbner-Shirshov basis in $\widetilde{S}$ and $L=A^{+}-A^{*}\{aac\}A^{*}$ is a normal form of $\widetilde{S}$. Clearly $L$ is regular. Note that

[TABLE]

are regular. Now we prove that L_{c}^{\}$and${}^{$}_{a}L$ are regular.

Case 1. $x\not\equiv c,\ xy\not\equiv ac$. Then

[TABLE]

is regular,

[TABLE]

is regular if $z\equiv a$, and

[TABLE]

is regular if $z\not\equiv a$.

Case 2. $x\not\equiv c,\ xy\equiv ac$. Then

[TABLE]

is regular,

[TABLE]

is regular if $z\equiv a$, and

[TABLE]

is regular if $z\not\equiv a$.

Case 3. $x\equiv c$. Then

[TABLE]

is regular,

[TABLE]

is regular if $z\equiv a$, and

[TABLE]

is regular if $z\not\equiv a$.

Therefore, L_{a_{j}}^{\}$and${}^{$}{a{j}}L$are regular for each$j\in{1,2,\dots,n}$.

For any (\alpha,\beta)\delta_{A}^{R}\in L_{a_{j}}^{\}\ or\ (\alpha,\beta)\delta_{A}^{L}\in$${}^{$}{a{j}}L\ (j=1,2,\dots,n)$, since$||\alpha|-|\beta||\leq 1, by Proposition [2.2.7](#S2.SS2.Thmdefinition7), we have {}^{$}L_{a_{j}}\ (or\ {a{j}}L^{$})$is regular,$j=1,2,\dots,n$. Hence,$\widetilde{S}$ is biautomatic and prefix-automatic.

$(iii)$ Suppose $R=\{abb=xyz\}$, where $a\not\equiv b$. Then $R$ is a Gröbner-Shirshov basis in $\widetilde{S}$ and $L=A^{+}-A^{*}\{abb\}A^{*}$ is a normal form of $\widetilde{S}$. Clearly, $L$,

[TABLE]

are all regular. Now we prove that L_{b}^{\}$and${}^{$}_{a}L$ are regular.

Case 1. $x\not\equiv b$. Then

[TABLE]

is regular,

[TABLE]

is regular if $z\equiv a$,

[TABLE]

is regular if $z\not\equiv a\ and\ z\not\equiv b$ (or $z\equiv b\ and\ y\not\equiv a$), and

[TABLE]

is regular if $z\equiv b\ and\ y\equiv a$.

Case 2. $x\equiv b$ and $xy\not\equiv bb$. Then

[TABLE]

is regular,

[TABLE]

is regular if $z\equiv a$,

[TABLE]

is regular if $z\not\equiv a\ and\ z\not\equiv b$ (or $z\equiv b\ and\ y\not\equiv a$), and

[TABLE]

is regular if $z\equiv b\ and\ y\equiv a$.

Case 3. $x\equiv b$ and $xy\equiv bb$. Then

[TABLE]

is regular,

[TABLE]

is regular if $z\equiv a$, and

[TABLE]

is regular if $z\not\equiv a$.

Therefore, L_{a_{j}}^{\}$and${}^{$}{a{j}}L$are regular for each$j\in{1,2,\dots,n}$.

Hence, $\widetilde{S}$ is biautomatic and prefix-automatic.

$(iv)$ Suppose $R=\{aba=aya\}$, where $a\not\equiv b,y\not\equiv b$. Then $R$ is a Gröbner-Shirshov basis in $S$. Since $con(aba)\nsubseteq con(aya)$, by Theorem 3.1.3, $S$ is biautomatic and prefix-automatic.

$(v)$ Suppose $R=\{aba=xyx\}$, where $a\not\equiv b,x\not\equiv a$.

Case 1. $x\not\equiv b$ or $y\not\equiv a$. Then $\{aba=xyx,abxyx=xyxba\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$ and $L=A^{+}-A^{*}\{aba,abxyx\}A^{*}$ is a normal form of $\widetilde{S}$. Clearly, $L$,

[TABLE]

are all regular.

Therefore, L_{a_{j}}^{\}$and${}^{$}{a{j}}L$are regular for each$j\in{1,2,\dots,n}$.

Hence, $\widetilde{S}$ is biautomatic and prefix-automatic.

Case 2. $xy\equiv ba$. Then $\{aba=bab,ab^{i+1}ab=babba^{i}|i\geq 1\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$ and $L=A^{+}-A^{*}\{aba,\ \{ab\}\{b\}^{+}\{ab\}\}A^{*}$ is a normal form of $\widetilde{S}$. Clearly, $L$,

[TABLE]

are all regular.

Therefore, L_{a_{j}}^{\}$and${}^{$}{a{j}}L$are regular for each$j\in{1,2,\dots,n}$.

Hence, $\widetilde{S}$ is biautomatic and prefix-automatic.

$(vi)$ Suppose $R=\{a^{3}=b^{3}\}$, where $a\not\equiv b$. Then by Proposition 2.2.5, we just need to prove $\widetilde{S}=sgp^{+}\langle a,b|a^{3}=b^{3}\rangle$ is biautomatic and prefix-automatic.

Since $\{a^{3}=b^{3},ab^{3}=b^{3}a\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$, $L=A^{+}-A^{*}\{a^{3},ab^{3}\}A^{*}$ is a normal form of $\widetilde{S}$. Clearly, $L$,

[TABLE]

are regular. Hence, $\widetilde{S}$ is biautomatic and prefix-automatic.

Therefore, $S=sgp\langle a_{1},a_{2},\dots,a_{n}|abc=xyz\rangle$ is biautomatic and prefix-automatic.

%֤ S=sgp¡a,b—aba=ba¿not automatic.

Lemma 3.2.5**.**

Suppose $S=sgp^{+}\langle a,b|aba=ba\rangle$. Then $S$ is not automatic.

Proof. Since $\{ab^{i}a=b^{i}a|i\geq 1\}$ is a Gröbner-Shirshov basis in $S$, we have $L=A^{+}-A^{*}\{a\}\{b\}^{+}\{a\}A^{*}$ is a normal form of $S$, where $A=\{a,b\}$.

Suppose $S$ is automatic. Then, by Proposition 2.2.4, $S^{1}$ is automatic.

Denote $B=\{e,a,b\}$. Then there exists an automatic structure $(B,K)$ for $S^{1}$ with uniqueness.

For any $i,j\geq 1$, there exist $\alpha,\beta\in K$, such that $\alpha=a^{j}b^{i},\beta=b^{i}a$. Since $a^{j}b^{i}\cdot a=b^{i}a$, we have (\alpha,\beta)\delta_{B}^{R}\in K_{a}^{\}$.

Since $\alpha,\beta\in K$, we have

[TABLE]

where $\gamma_{1},\dots,\gamma_{j},\gamma_{1}^{\prime},\dots,\gamma_{i+1}^{\prime}\in\{e\}^{*}$, $\tau_{1},\dots,\tau_{i}\in\{e,a\}^{*}$ and $\tau_{i+1},\tau_{i+2}\in\{e\}^{*}$.

Let $N=|S(M(K))|$, $\alpha_{1}\equiv\gamma_{1}a\gamma_{2}a\cdots\gamma_{j}a$ and $\beta_{1}\equiv\tau_{1}b\tau_{2}b\cdots\tau_{i}b$. Then $j\leq|\alpha_{1}|<jN+j$ and $i\leq|\beta_{1}|<iN+i$.

Choose j>iN+i,\ i>|S(M(K_{a}^{\}))|$. Then there exists a loop$(u,v)\delta_{B}^{R}$in$(\alpha_{1},\beta_{1})\delta_{B}^{R}$. Suppose$\alpha\equiv\alpha^{\prime}u\alpha^{{}^{\prime\prime}},\beta\equiv\beta^{\prime}v\beta^{{}^{\prime\prime}}$.

If $b\not\in con(v)$, then $con(v)\in\{e,a\}$. By the relations in $S$, $\beta^{\prime}\beta^{{}^{\prime\prime}}=\beta$ and $\beta^{\prime}\beta^{{}^{\prime\prime}},\beta\in K$ which contradicts the uniqueness of $K$.

If $b\in con(v)$, we have $occ(b,\alpha^{\prime}\alpha^{\prime\prime}a)\neq occ(b,\beta^{\prime}\beta^{\prime\prime})$ since $b\not\in con(u)$. But (\alpha^{\prime}\alpha^{{}^{\prime\prime}},\beta^{\prime}\beta^{\prime\prime})\delta_{B}^{R}\in K_{a}^{\}$, a contradiction.

Therefor, $S^{1}$ is not automatic.

%֤ η εġ

Lemma 3.2.6**.**

Let $S=sgp\langle A|u=v\rangle$, where $A=\{a_{1},a_{2},\dots,a_{n}\}$ $(n\in\mathbb{N})$, $u,v\in A^{*}$ and $|v|\leq|u|=3$. Then

1. (i)

if $u=v\in\{aba=ba,aab=ba,abb=bb|a,b\in A,a\not\equiv b\}$, then $S$ is not automatic; 2. (ii)

if $u=v\not\in\{aba=ba,aab=ba,abb=bb|a,b\in A,a\not\equiv b\}$, then $S$ is prefix-automatic; 3. (iii)

if $|v|=0$ or 3, then S is biautomatic.

Proof. $(i)$ If $u=v\in\{aba=ba,aab=ba,abb=bb|a,b\in A,\ a\not\equiv b\}$, then by Proposition 2.2.4, Theorems 3.1.8 and 3.1.9, and Lemma 3.2.5, $S=sgp\langle A|u=v\rangle$ is not automatic.

Now we prove $(ii)$ and $(iii)$. Suppose $u=v\not\in\{aba=ba,aab=ba,abb=bb|a,b\in A,a\not\equiv b\}$ and $\widetilde{S}=sgp^{+}\langle A|u=v\rangle$.

$1)$ Suppose $|v|=0$. If $u=v\in\{aaa=\varepsilon,\ aab=\varepsilon,abb=\varepsilon,abc=\varepsilon|a,b\in A,\ a\not\equiv b\}$, then by Theorem 3.1.4, $S$ is biautomatic and prefix-automatic.

Let $\widetilde{S}=sgp^{+}\langle e,a,b|aba=e,ea=ae=a,eb=eb=b,ee=e\rangle$. If $u=aba$, then $\{aba=e,aab=e,ba=ab,ae=a,ea=a,be=b,eb=b,ee=e\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$. Let $B=\{e,a,b\}$, $L=B^{+}-B^{*}\{aba,aab,ba,ae,$ $ea,be,eb,ee\}B^{*}$. Then L is a normal form of $\widetilde{S}$. Then $L_{=}=L_{e}=_{e}L=\Delta_{L}$,

[TABLE]

are regular and so $\widetilde{S}$ is prefix-automatic. Noting that

[TABLE]

are regular, by Proposition 2.2.7, {}^{\}L_{a},\ ^{$}L_{b},\ _{a}L^{$},\ _{b}L^{$}$are regular and hence$\widetilde{S}$is biautomatic. This shows that$S$ is biautomatic and prefix-automatic.

$2)$ Suppose $|v|=1$. If $u=v\in\{abc=x,aab=x,abb=x,aba=a,aaa=a|a,b,c,x\in A,a\not\equiv b,a\not\equiv c,b\not\equiv c\}$, then $\{u=v\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$. Then by Theorem 3.1.5, $S$ is prefix-automatic.

If $u\equiv aba,\ v\equiv x$, where $x\in A-\{a\}$, then $\{aba=x,abx=xba\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$ and so $L=A^{+}-A^{*}\{aba,abx\}A^{*}$ is a normal form of $\widetilde{S}$. Clearly, $L$,

[TABLE]

are regular.

Therefore, L_{a_{j}}^{\}$are regular for each$j\in{1,2,\dots,n}$. Hence$\widetilde{S}$is automatic. Thus$S$ is prefix-automatic.

If $u\equiv a^{3},\ v\equiv x$, where $x\in A-\{a\}$, then $\{aaa=x,ax=xa\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$ and $L=A^{+}-A^{*}\{aaa,ax\}A^{*}$ is a normal form of $\widetilde{S}$. Clearly, $L$,

[TABLE]

are regular.

Therefore, L_{a_{j}}^{\}$are regular for each$j\in{1,2,\dots,n}$. Hence$\widetilde{S}$ is prefix-automatic.

$3)$ Suppose $|v|=2$.

Case 1. $\widetilde{S}=sgp^{+}\langle A|abc=xy\rangle$, where $a,b,c,x,y\in A$ and $a,b,c$ are pairwise different. Obviously, $\{abc=xy\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$.

Case 1-1. If $xy\not\equiv bc$, by Theorem 3.1.7, $\widetilde{S}$ is prefix-automatic.

Case 1-2. If $xy\equiv bc$, by Proposition 2.2.5, it is sufficient to prove $\widetilde{S}=sgp^{+}\langle a,b,c|abc=bc\rangle$ is prefix-automatic.

Since $\{abc=bc\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$, we have $L=B^{+}-B^{*}\{abc\}B^{*}$ is a normal form of $\widetilde{S}$, where $B=\{a,b,c\}$. Clearly,

[TABLE]

are regular. Hence $\widetilde{S}$ is prefix-automatic.

Case 2. $\widetilde{S}=sgp^{+}\langle A|aab=xy\rangle$, where $a,b,x,y\in A$ and $a\not\equiv b$. Obviously, $\{aab=xy\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$.

Case 2-1. If $xy\not\equiv ab$, we have $v\not\equiv ba$ since $u=v\not\in\{aba=ba,aab=ba,abb=bb|a,b\in A,a\not\equiv b\}$. By Theorem 3.1.7, $S$ is prefix-automatic.

Case 2-2. If $xy\equiv ab$, by Proposition 2.2.5, it is sufficient to prove $\widetilde{S}=sgp^{+}\langle a,b|aab=ab\rangle$ is prefix-automatic.

Since $\{aab=ab\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$, we have $L=B^{+}-B^{*}\{aab\}B^{*}$ is a normal form of $\widetilde{S}$, where $B=\{a,b\}$. Clearly, $L$, L_{=}^{\}=\Delta_{L}$,$L_{a}^{$}={(\alpha,\alpha a)\delta_{B}^{R}|\alpha\in L}$and$L_{b}^{$}={(\alpha,\alpha b)\delta_{B}^{R}|\alpha\in L-B^{}{aa}}\cup{(\alpha a^{i+2},\alpha ab)\delta_{B}^{R}|\alpha\in L-B^{}{a},i\geq 0}$are all regular. Hence$\widetilde{S}$ is prefix-automatic.

Case 3. $\widetilde{S}=sgp^{+}\langle A|abb=xy\rangle$, where $a,b,x,y\in A$ and $a\not\equiv b$. Obviously, $\{abb=xy\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$.

Since $u=v\not\in\{aba=ba,aab=ba,abb=bb|a,b\in A,a\not\equiv b\}$, we have $xy\not\equiv bb$. By Theorem 3.1.7, $S$ is prefix-automatic.

Case 4. $\widetilde{S}=sgp^{+}\langle A|aba=xy\rangle$, where $a,b,x,y\in A$ and $a\not\equiv b$.

Case 4-1. If $xy\equiv aa$, then $\{aba=aa\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$. By Theorem 3.1.7, $S$ is prefix-automatic.

Case 4-2. If $xy\equiv ab$, then $\{ab^{i}a=ab^{i}|i\geq 1\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$. By Proposition 2.2.5, it is sufficient to prove $\widetilde{S}=sgp^{+}\langle a,b|aba=ab\rangle$ is prefix-automatic.

Let $B=\{a,b\}$, $L=B^{+}-B^{*}\{a\}\{b\}^{+}\{a\}B^{*}$. Then $L$ is a normal form of $\widetilde{S}$. It is clear that $L$, L_{=}^{\}=\Delta_{L}$,

[TABLE]

and L_{b}^{\}={(\alpha,\alpha b)\delta_{B}^{R}|\alpha\in L}$are regular. Hence$\widetilde{S}$ is prefix-automatic.

Case 4-3. If $x\equiv a,\ y\not\equiv a$ and $y\not\equiv b$, then $\{ay^{i}ba=ay^{i+1}|i\geq 0\}$ is a Gröbner-Shirshov basis in $S$. By Proposition 2.2.5, it is sufficient to prove $\widetilde{S}=sgp^{+}\langle a,b,y|aba=ay\rangle$ is prefix-automatic.

Let $B=\{a,b,y\}$, $L=B^{+}-B^{*}\{a\}\{y\}^{*}\{ba\}B^{*}$. Then $L$ is a normal form of $\widetilde{S}$. Clearly

[TABLE]

are regular. Hence $\widetilde{S}$ is prefix-automatic.

Case 4-4. If $xy\equiv ba$, then $u=v\in\{aba=ba,aab=ba,abb=bb|a,b\in A,a\neq b\}$ and hence $S$ is not automatic by $(i)$.

Case 4-5. If $xy\equiv bb$, then $\{aba=bb,ab^{3}=b^{3}a\}$ is a Gröbner-Shirshov basis in $\widetilde{S}$. By Proposition 2.2.5, it is sufficient to prove $\widetilde{S}=sgp^{+}\langle a,b|aba=bb\rangle$ is prefix-automatic.

Let $B=\{a,b\}$, $L=B^{+}-B^{*}\{aba,ab^{3}\}B^{*}$. Then $L$ is a normal form of $\widetilde{S}$. Clearly,

[TABLE]

are regular. Hence $\widetilde{S}$ is prefix-automatic.

Case 4-6. If $x\equiv b,\ y\not\equiv a$ and $y\not\equiv b$, then $\{aba=by,ab^{2}y=byba\}$ is a Gröbner-Shirshov basis in $S$. By Proposition 2.2.5, it is sufficient to prove $\widetilde{S}=sgp^{+}\langle a,b,y|aba=by\rangle$ is prefix-automatic.

Let $B=\{a,b,y\}$, $L=B^{+}-B^{*}\{aba,ab^{2}y\}B^{*}$. Then $L$ is a normal form of $\widetilde{S}$. Obviously,

[TABLE]

are regular. Hence $\widetilde{S}$ is prefix-automatic.

Case 4-7. If $x\not\equiv a,\ x\not\equiv b$ and $y\equiv a$, then $\{abx^{i}a=x^{i+1}a|i\geq 0\}$ is a Gröbner-Shirshov basis in $S$. By Proposition 2.2.5, it is sufficient to prove $\widetilde{S}=sgp^{+}\langle a,b,x|aba=xa\rangle$ is prefix-automatic.

Let $B=\{a,b,x\}$, $L=B^{+}-B^{*}\{ab\}\{x\}^{*}\{a\}B^{*}$. Then $L$ is a normal form of $\widetilde{S}$. Clearly, L_{=}^{\}=\Delta_{L}$,

[TABLE]

L_{b}^{\}={(\alpha,\alpha b)\delta_{B}^{R}|\alpha\in L}}$and$L_{x}^{$}={(\alpha,\alpha x)\delta_{B}^{R}|\alpha\in L}}$are all regular. Hence$\widetilde{S}$ is prefix-automatic.

Case 4-8. If $x\not\equiv a,\ x\not\equiv b$ and $y\not\equiv a,x$, then $\{aba=xy,abxy=xyba\}$ is a Gröbner-Shirshov basis in $S$. Let $B=\{a,b,x,y\}$ if $y\not\equiv b$ (Otherwise, let $B=\{a,b,x\}$). By Proposition 2.2.5, it is sufficient to prove $\widetilde{S}=sgp^{+}\langle B|aba=xy\rangle$ is prefix-automatic. Note that $L=B^{+}-B^{*}\{aba,abxy\}B^{*}$ is a normal form of $\widetilde{S}$. Clearly,

[TABLE]

are all regular. Hence $\widetilde{S}$ is prefix-automatic.

Case 4-9. If $x\not\equiv a,\ x\not\equiv b$ and $y\equiv x$, then $\{aba=xx,abx^{2}=x^{2}ba\}$ is a Gröbner-Shirshov basis in $S$. By Proposition 2.2.5, it is sufficient to prove $\widetilde{S}=sgp^{+}\langle a,b,x|aba=x^{2}\rangle$ is prefix-automatic.

Let $B=\{a,b,x\}$, $L=B^{+}-B^{*}\{aba,abx^{2}\}B^{*}$. Then $L$ is a normal form of $\widetilde{S}$. Noting that

[TABLE]

are regular, $\widetilde{S}$ is prefix-automatic.

Case 5. $S=sgp^{+}\langle A|a^{3}=xy\rangle$, where $a,x,y\in A$.

Case 5-1. If $x\not\equiv a$ and $y\not\equiv a$, then $\{a^{3}=xy,axy=xya\}$ is a Gröbner-Shirshov basis in $S$. Let $B=\{a,x,y\}$ if $x\not\equiv y$ (Otherwise, let $B=\{a,x\}$). By Proposition 2.2.5, it is sufficient to prove $\widetilde{S}=sgp^{+}\langle B|aaa=xy\rangle$ is prefix-automatic.

Note that $L=B^{+}-B^{*}\{aaa,axy\}B^{*}$ is a normal form of $\widetilde{S}$. Clearly,

[TABLE]

are regular. Hence $\widetilde{S}$ is prefix-automatic.

Case 5-2. If $x\not\equiv a$ and $y\equiv a$, then $\{a^{3}=xa,ax^{i}a=x^{i}a^{2}|i\geq 1\}$ is a Gröbner-Shirshov basis in $S$. By Proposition 2.2.5, it is sufficient to prove $\widetilde{S}=sgp^{+}\langle a,x|aaa=xa\rangle$ is prefix-automatic.

Let $B=\{a,x\}$, $L=B^{+}-B^{*}\{aaa\}B^{*}-B^{*}\{a\}\{x\}^{+}\{a\}B^{*}$. Then $L$ is a normal form of $\widetilde{S}$. Clearly,

[TABLE]

are regular. Hence $\widetilde{S}$ is prefix-automatic.

Case 5-3. If $x\equiv a$ and $y\not\equiv a$, then $\{a^{3}=ay,ay^{i}a=a^{2}y^{i}|i\geq 1\}$ is a Gröbner-Shirshov basis in $S$. By Proposition 2.2.5, it is sufficient to prove $\widetilde{S}=sgp^{+}\langle a,y|aaa=ay\rangle$ is prefix-automatic.

Let $B=\{a,y\}$, $L=(B^{+}-B^{*}\{aaa\}B^{*})-B^{*}\{a\}\{y\}^{+}\{a\}B^{*}$. Then $L$ is a normal form of $\widetilde{S}$. By noting that

[TABLE]

are regular, $\widetilde{S}$ is prefix-automatic.

Case 5-4. If $x\equiv a$ and $y\equiv a$, then $\{a^{3}=a^{2}\}$ is a Gröbner-Shirshov basis in $S$. By Proposition 2.2.5, it is sufficient to prove $\widetilde{S}=sgp^{+}\langle a\mid aaa=aa\rangle$ is prefix-automatic. Since $\widetilde{S}$ is a finite semigroup, we have $\widetilde{S}$ is prefix-automatic.

$4)$ Suppose $|v|=3$. Then $S=sgp\langle A\mid u=v\rangle$ is biautomatic and prefix-automatic by Lemma 3.2.4.

Theorem 3.2.1 follows from Lemmas 3.2.3 and 3.2.6.