aa r X i v : . [ m a t h . R A ] A p r FREE OPERATED MONOIDS AND REWRITING SYSTEMS
JIN ZHANG AND XING GAO ∗ A bstract . The construction of bases for quotients is an important problem. In this paper, applyingthe method of rewriting systems, we give a unified approach to construct sections—an alternativename for bases in semigroup theory—for quotients of free operated monoids. As applications, wecapture sections of free ∗ -monoids and free groups, respectively. C ontents
1. Introduction 12. Rewrting systems and sections 22.1. Operated monoids and operated congruences 22.2. Relationship between rewriting systems and sections 63. Applications 83.1. Free ∗ -monoids 93.2. Free groups 12References 161. I ntroduction In 1960, A. G. Kurosh [23] first introduced the concept of algebras with one or more linearoperators. An example of such algebras is the di ff erential algebra led by the algebraic abstractionof di ff erential operator in analysis. Di ff erential algebra is from a purely algebraic viewpoint tostudy di ff erentiation and nonlinear di ff erential equations without using an underlying topology,and has been largely successful in many crucial areas, such as uncoupling of nonlinear systems,classification of singular components, and detection of hidden equations [22, 28]. Another im-portant example of such algebras is the Rota-Baxter algebra (first called Baxter algebra) whichis the algebraic abstraction of integral operator in analysis [17]. Rota-Baxter algebra, originatedfrom probability study [5], is related beautifully to the classical Yang-Baxter equation, as well asto operads, to combinatorics and, through the Hopf algebra framework of Connes and Kreimer,to the renormalization of quantum field theory [1, 3, 4, 11, 12, 19]. Other examples are alsoimportant, such as averaging algebra, Reynolds algebras, Nijenhuis algebras and Leroux’s TDalgebras [8, 26, 25]. Each of the above examples is an algebra with one linear operator, which isnamed operated algebras by Guo [16]. Definition 1.1. An operated monoid (resp. operated k -algebra) is a monoid (resp. k -algebra) U together with a map (resp. k -linear map) P U : U → U , where k is a commutative unitary ring.In that paper [16], Guo constructed the free operated k -algebra on a set. Since the free operated k -algebra as modules is precisely the free k -module with basis the free operated monoid, thecrucial step of Guo’s method is to construct the free operated monoid on a set—the main objectconsidered in this paper, See also [7, 13]. Date : November 9, 2018.2010
Mathematics Subject Classification.
Key words and phrases.
Operated monoids, term-rewriting systems, free ∗ -monoids, free groups.*Corresponding author. ∗ Abstract rewriting system is a branch of theoretical computer science, combining elements oflogic, universal algebra, automated theorem proving and functional programming [2, 27]. Thetheory of convergent rewriting systems is successfully applied to find bases of free di ff erentialtype [18] and free Rota-Baxter type algebras [14], which reveals the power of rewriting systemsin the study of operators.Let us point out that groups can be viewed as operated monoids if one considers the inverseoperator as a map from the group to itself. In the same way, many other important classes ofmonoids such as inverse monoids, I-monoids and ∗ -monoids can also be fitted into the frameworkof operated monoids. All of these examples can be obtained from free operated monoids by takingquotients modulo suitable operated congruences. It is interesting to find bases for quotients.The bases of quotients in semigroup theory are also called sections. In the present paper weobtained a method, in terms of convergent rewriting systems, to give sections for quotients offree operated monoids. Our method is parallel to the famous Composition-Diamond lemma inGr¨obner-Shirshov theory [7]. As applications, we capture sections of free ∗ -monoids and freegroups, respectively.The organisation of this paper is as follows. In Section 2, after reviewing the construction offree operated monoids, we characterize the operated congruence generated by a binary relationon a free operated monoid (Proposition 2.10). Then we associate to each binary relation ona free operated monoid a rewriting system (Definition 2.14). We also establish a relationshipbetween convergent rewriting systems on free operated monoids and sections of quotients of freeoperated monoids. (Theorem 2.21). In Section 3, as applications of our main result, we acquirerespectively sections of free ∗ -monoids (Theorem 3.10) and free groups (Theorem 3.15).2. R ewrting systems and sections In this section, based on rewriting systems on free operated monoids M ( X ), we give an ap-proach to construct sections for quotients of M ( X ).2.1. Operated monoids and operated congruences.
The construction of free operated monoidswas given in [16, 18]. See also [7]. We reproduce that construction here to review the notations.For any set Y , denote by M ( Y ) the free monoid on Y .Let X be a set. We proceed via finite stages M n ( X ) defined recursively to construct the freeoperated monoid M ( X ) on X . The initial stage is M ( X ) : = M ( X ) and M ( X ) : = M ( X ∪ ⌊ M ( X ) ⌋ ),where ⌊ M ( X ) ⌋ : = {⌊ u ⌋ | u ∈ M ( X ) } is a disjoint copy of M ( X ). The inclusion X ֒ → X ∪ ⌊ M ⌋ induces a monomorphism i : M ( X ) = M ( X ) ֒ → M ( X ) = M ( X ∪ ⌊ M ⌋ )of monoids through which we identify M ( X ) with its image in M ( X ).For n >
1, assume inductively that M n ( X ) has been defined and the embedding i n − , n : M n − ( X ) ֒ → M n ( X )has been obtained. Then we define M n + ( X ) : = M (cid:0) X ∪ ⌊ M n ( X ) ⌋ (cid:1) . Since M n ( X ) = M (cid:0) X ∪ ⌊ M n − ( X ) ⌋ (cid:1) is a free monoid, the injection ⌊ M n − ( X ) ⌋ ֒ → ⌊ M n ( X ) ⌋ REE OPERATED MONOIDS AND REWRITING SYSTEMS 3 induces a monoid embedding M n ( X ) = M (cid:0) X ∪ ⌊ M n − ( X ) ⌋ (cid:1) ֒ → M n + ( X ) = M (cid:0) X ∪ ⌊ M n ( X ) ⌋ (cid:1) . Finally we define the monoid M ( X ) : = lim −→ M n = [ n > M n ( X ) , whose elements are called bracketed words on X . Elements w ∈ M n \ M n − are said to have depth n , denoted by dep( w ) = n . Lemma 2.1. ( [17] ) Every bracketed word w , has a unique decomposition w = w · · · w m , where w i , i m, is in X or in ⌊ M ( X ) ⌋ : = {⌊ w ⌋ | w ∈ M ( X ) } . We call | w | : = m the breadth ofw. The following result shows that M ( X ) is the free object in the category of operated monoids. Lemma 2.2. ( [16, 17] ) Let i X : X → M ( X ) be the natural embeddings. Then the triple ( M ( X ) , ⌊ ⌋ , i X ) is the free operated monoid on X, where ⌊ ⌋ : M ( X ) → M ( X ) , w
7→ ⌊ w ⌋ is an operator on M ( X ) . The concept of ⋆ -bracketed words plays a crucial role in the theory of Gr¨obner-Shirshovbases [6]. Definition 2.3.
Let X be a set, ⋆ a symbol not in X and X ⋆ = X ∪ { ⋆ } .( a ) By a ⋆ -bracketed word on X , we mean any bracketed word in M ( X ⋆ ) with exactly oneoccurrence of ⋆ , counting multiplicities. The set of all ⋆ -bracketed words on X is denotedby M ⋆ ( X ).( b ) For q ∈ M ⋆ ( X ) and u ∈ M ( X ), we define q | ⋆ u to be the bracketed word on X obtained byreplacing the symbol ⋆ in q by u , for convenience, denoted it by q | u .( c ) Let u , v ∈ M ( X ). We say that u is a bracketed subword of v , if there exist q ∈ M ⋆ ( X )such that v = q | u .Generally, with ⋆ , ⋆ distinct symbols not in X , set X ⋆ : = X ∪ { ⋆ , ⋆ } .( d ) We define an ( ⋆ , ⋆ ) -bracketed word on X to be a bracketed word in M ( X ⋆ ) with exactlyone occurrence of each of ⋆ i , i = ,
2. The set of all ( ⋆ , ⋆ )-bracketed words on X isdenoted by M ⋆ ,⋆ ( X ).( e ) For q ∈ M ⋆ ,⋆ ( X ) and u , u ∈ M ⋆ ,⋆ ( X ), we define q | u , u : = q | ⋆ u ,⋆ u to be obtained by replacing the letters ⋆ i in q by u i for i = , Remark 2.4. If p | u = p | v with p ∈ M ⋆ ( X ) and u , v ∈ M ( X ), then u = v by the freeness of M ( X ).The concept of operated congruences will be used throughout the paper. Definition 2.5.
An equivalence R on M ( X ) is operated congruence if(C1) ( ∀ a , b , c ∈ M ( X )) ( a , b ) ∈ R ⇒ ( ac , bc ) ∈ R ;(C2) ( ∀ a , b , c ∈ M ( X )) ( a , b ) ∈ R ⇒ ( ca , cb ) ∈ R ;(C3) ( ∀ a , b ∈ M ( X )) ( a , b ) ∈ R ⇒ ( ⌊ a ⌋ , ⌊ b ⌋ ) ∈ R . JIN ZHANG AND XING GAO ∗ Let R be a binary relation on M ( X ). There is a unique smallest operated congruence h R i on M ( X ) containing R , which will be described in the following. Define(1) R c : = { ( q | a , q | b ) | q ∈ M ⋆ ( X ) , ( a , b ) ∈ R } . We record some basic properties of R c . For any u ∈ M ( X ), define recursively ⌊ u ⌋ (0) : = u and ⌊ u ⌋ ( k + : = ⌊⌊ u ⌋ ( k ) ⌋ for k > Lemma 2.6. R c is the smallest binary relation containing R and satisfy (C1), (C2) and (C3).Proof. According to the definition of R c , we have R ⊆ R c by choosing q = ⋆ . Let c ∈ M ( X ) and( q | a , q | b ) ∈ R c with q ∈ M ⋆ ( X ) , ( a , b ) ∈ R . Write q : = qc . Then by Eq. (1),(( q | a ) c , ( q | b ) c ) = (( qc ) | a , ( qc ) | b ) = ( q | a , q | b ) ∈ R c . So R c satisfies (C1). By symmetry, R c also satisfies (C2). To prove R c satisfies (C3), let q : = ⌊ q ⌋ ∈ M ⋆ ( X ). Then again from Eq. (1),( ⌊ q | a ⌋ , ⌊ q | b ⌋ ) = ( q | a , q | b ) ∈ R c . Suppose that T is a binary relation containing R and satisfying (C1), (C2) and (C3). Let( q | a , q | b ) ∈ R c with q ∈ M ⋆ ( X ) and ( a , b ) ∈ R . Then ( a , b ) ∈ T by R ⊆ T . Hence ( q | a , q | b ) ∈ T by(C1), (C2) and (C3). (cid:3) For a binary relation R on M ( X ), write R − : = { ( b , a ) | ( a , b ) ∈ R } . Lemma 2.7.
Let R , S be binary relations on M ( X ) . Then ( a ) R ⊆ S ⇒ R c ⊆ S c ;( b ) ( R − ) c = ( R c ) − ;( c ) ( R ∪ S ) c = R c ∪ S c . Proof. ( a ) This follows directly from Eq. (1).( b ) Let ( q | a , q | b ) ∈ ( R − ) c with ( a , b ) ∈ R − and q ∈ M ⋆ ( X ) . Then( b , a ) ∈ R , ( q | b , q | a ) ∈ R c and ( q | a , q | b ) ∈ ( R c ) − , and so ( R − ) c ⊆ ( R c ) − . Conversely, let ( q | a , q | b ) ∈ ( R c ) − with q ∈ M ⋆ ( X ) . Then( q | b , q | a ) ∈ R c , ( b , a ) ∈ R and ( a , b ) ∈ R − , which implies that ( q | a , q | b ) ∈ ( R − ) c and ( R c ) − ⊆ ( R − ) c , as needed.( c ) By Item ( a ), we have R c ⊆ ( R ∪ S ) c and S c ⊆ ( R ∪ S ) c , and thus R c ∪ S c ⊆ ( R ∪ S ) c . Conversely, let ( q | a , q | b ) ∈ ( R ∪ S ) c with ( a , b ) ∈ R ∪ S . Then( a , b ) ∈ R or ( a , b ) ∈ S and so ( q | a , q | b ) ∈ R c or ( q | a , q | b ) ∈ S c . Hence ( q | a , q | b ) ∈ R c ∪ S c and ( R ∪ S ) c ⊆ R c ∪ S c . (cid:3) Lemma 2.8.
Let R be a binary relation on M ( X ) satisfying (C1), (C2) and (C3). Then, so isR n ( = R ◦ R ◦ · · · ◦ R ) for all n > . REE OPERATED MONOIDS AND REWRITING SYSTEMS 5
Proof.
Let ( a , b ) ∈ R n . Then there exist v , v , · · · , v n − in M ( X ) such that( a , v ) , ( v , v ) , · · · , ( v n − , b ) ∈ R . Because R satisfies (C1),(C2) and (C3), it follows that, for all c in M ( X ),( ca , cv ) , ( cv , cv ) , · · · , ( cv n − , cb ) ∈ R , ( ac , v c ) , ( v c , v c ) , · · · , ( v n − c , bc ) ∈ R , ( ⌊ a ⌋ , ⌊ v ⌋ ) , ( ⌊ v ⌋ , ⌊ v ⌋ ) , · · · , ( ⌊ v n − ⌋ , ⌊ b ⌋ ) ∈ R . So ( ca , cb ) , ( ac , ab ) , ( ⌊ a ⌋ , ⌊ b ⌋ ) ∈ R n , as required. (cid:3) We recall the construction of equivalences R e generated by a binary relation R on a set X [21].Write R ∞ : = ∪ n > R n and 1 X : = { ( x , x ) | x ∈ X } . Lemma 2.9. ( [21] ) Let R be a binary relation on a set X. Then R e = [ R ∪ R − ∪ X ] ∞ . Moreprecisely, ( x , y ) ∈ R e if and only if either x = y or, for some n in N , there is a sequence of elementsx = z , z , · · · , z n = y , in which, for each i in { , , · · · , n − } , either ( z i , z i + ) ∈ R or ( z i + , z i ) ∈ R . Now we arrive at the position to give the description of operated congruences generated bybinary relations.
Proposition 2.10.
For every binary relation R on M ( X ) , h R i = ( R c ) e . Proof.
By Lemma 2.9, ( R c ) e is an equivalence containing R c , and so certainly containing R byLemma 2.6. Next we show that ( R c ) e satisfies (C1),(C2) and (C3). From Lemma 2.9,( R c ) e = S ∞ = ∪ n > S n , where S = R c ∪ ( R c ) − ∪ M ( X ) . Using Lemma 2.7 and the fact that 1 M ( X ) = c M ( X ) , we get S = R c ∪ ( R c ) − ∪ M ( X ) = R c ∪ ( R − ) c ∪ c M ( X ) = ( R ∪ R − ∪ M ( X ) ) c , which implies from Lemma 2.6 that S satisfies (C1), (C2) and (C3), and so is S n for all n > R c ) e = S ∞ = ∪ n > S n also satisfies (C1), (C2) and (C3). Moreover since ( R c ) e is already an equivalence, it is an operatedcongruence by Definition 2.5.Finally, suppose T is an arbitrary operated congruence on M ( X ) containing R . Then T c = T byDefinition 2.5 and Eq. (1). So from Lemma 2.7, R c ⊆ T c = T and ( R c ) e ⊆ T e = T . This completes the proof. (cid:3)
Proposition 2.11.
Let R be a binary relation on M ( X ) . Then ( a , b ) ∈ h R i if and only if eithera = b or, for some n in N , there is a sequence of elements a = v , v , · · · , v n = b, in which, foreach i in { , , · · · , n − } , either ( v i , v i + ) ∈ R c or ( v i + , v i ) ∈ R c . Proof.
It follows from Lemmas 2.9 and 2.10. (cid:3)
JIN ZHANG AND XING GAO ∗ Every operated congruence R on M ( X ) has a corresponding quotient structure M ( X ) / R , whoseelements are operated congruence classes for the relation. We end this subsection with an impor-tant concepts used later. Definition 2.12.
Let X be a set and R an operated congruence on M ( X ). We call W ⊆ M ( X ) a section of R if, for each operated congruence class A of M ( X ) / R , there exists exactly one element w ∈ W such that w ∈ A .2.2. Relationship between rewriting systems and sections.
In this subsection, we first asso-ciate a rewriting system Π S to a binary relation S on M ( X ). A monomial order compatible withall operations is needed. Definition 2.13.
Let X be a set. A monomial order on M ( X ) is a well-ordering on M ( X ) suchthat(2) u < v = ⇒ uw < vw , wu < wv , ⌊ u ⌋ < ⌊ v ⌋ for all u , v , w ∈ M ( X ) . We denote u < v if u v but u , v .The following concepts are adapted from [2, 14]. Definition 2.14.
Let X be a set and a monomial order on M ( X ). Let S be a binary relation on M ( X ).( a ) A term-rewriting system Π S on M ( X ) associated to S is a binary relation of M ( X ), denoteby(3) Π S : = { ( q | t , q | v ) | q ∈ M ⋆ ( X ) , ( t , v ) ∈ S ∪ S − , t > v } . An element in Π S is called a rewriting rule .( b ) Let f , g ∈ M ( X ), we call f rewrites to g in one-step with respect to Π S , if ( f , g ) ∈ Π S .We indicate any such one-step rewriting by f → Π S g .( c ) The reflexive transitive closure of Π S (as a binary relation on M ( X )) will be denoted by ∗ → Π S and we say f rewrites to g with respect to Π S if f ∗ → Π S g . In this case, we call f isa predecessor of g . Denote by P ( g ) the set of all predecessors of g . Note that g ∈ P ( g ).( d ) Two elements f , g ∈ M ( X ) are joinable if there exists h ∈ M ( X ) such that f ∗ → Π S h and g ∗ → Π S h . Denote it by f ↓ Π S g .( e ) The image π ( Π S ) under the first projection map will be denoted by Dom( Π S ). An element f ∈ M ( X ) is irreducible or in normal form if f < Dom( Π S ), that is, no more rewritingrule from Π S can apply to f . Definition 2.15.
The term-rewriting system Π S defined above is called( a ) terminating if there is no infinite chain of one-setp rewriting f → Π S f → Π S f · · · ;( b ) confluent if every fork ( f ∗ → Π S h , f ∗ → Π S g ), we have g ↓ Π S h ;( c ) locally confluent if for every local fork ( f → Π S h , f → Π S g ), we have g ↓ Π S h ;( d ) convergent if it is both terminating and confluent.A well-known result on rewriting systems is Newman’s Lemma [2, Lemma 2.7.2]. Lemma 2.16. (Newman)
A terminating rewriting system is confluent if and only if it is locallyconfluent.
REE OPERATED MONOIDS AND REWRITING SYSTEMS 7
Lemma 2.17.
Let X be a set and a monomial order on M ( X ) . Let S be a binary relation on M ( X ) . The term-rewriting system Π S defined in Eq. (3) is terminating.Proof. Suppose to the contrary that Π S is not terminating. Then there is an infinite chain ofone-step rewritings f → Π S f → Π S f · · · . Let f → Π S g be a one-step rewriting. Then ( f , g ) ∈ Π S and so we can write ( f , g ) = ( q | t , q | v ) with q ∈ M ⋆ ( X ), ( t , v ) ∈ S ∪ S − and t > v . Since is a monomial order, we have f = q | t > q | v = g .Hence f > f > f · · · , contradicting that is a well-order. (cid:3) We are going to capture the relationship between convergent term-rewriting systems on M ( X )and sections of quotients of M ( X ). Let us record three lemmas as a preparation. Lemma 2.18.
Let X be a set and a monomial order on M ( X ) . Let S be a binary relation on M ( X ) . If a ∗ → Π S b with a , b ∈ M ( X ) , then ( a , b ) ∈ h S i .Proof. If a = b , then ( a , b ) ∈ h S i . Suppose a , b . Then there are a , a , · · · , a n ∈ M ( X ) such that a = a → Π S a → Π S · · · → Π S a n = b , which implies that( a , a ) , · · · , ( a n − , a n ) ∈ Π S ⊆ ( S ∪ S − ) c = S c ∪ ( S c ) − ⊆ ( S c ) e = h S i . From the transitivity, we get ( a , b ) = ( a , a n ) ∈ h S i , as required. (cid:3) Lemma 2.19.
Let X be a set and a monomial order on M ( X ) . Let S be a binary relation on M ( X ) . For each operated congruence class A of h S i , we have | A ∩ Irr( S ) | > and A = ∪ a ∈ A ∩ Irr( S ) P ( a ) , where Irr( S ) : = M ( X ) \ Dom( Π S ) is the set of all irreducible elements under Π S .Proof. Let b ∈ A . By Lemma 2.17, there is a ∈ Irr( S ) such that b ∗ → Π S a . From Lemma 2.18, wehave ( b , a ) ∈ h S i and so a ∈ A . Hence a ∈ A ∩ Irr( S ) and | A ∩ Irr( S ) | >
1. Since b ∗ → Π S a , itfollows that b ∈ P ( a ) and thus A ⊆ ∪ a ∈ A ∩ Irr( S ) P ( a ) . Conversely, for every element a ∈ A , we have P ( a ) ⊆ A by Lemma 2.18 and so ∪ a ∈ A ∩ Irr( S ) P ( a ) ⊆ A . This completes the proof. (cid:3)
Lemma 2.20.
Let X be a set and a monomial order on M ( X ) . Let S be a binary relation on M ( X ) . For each operated congruence class A of h S i , if Π S is confluent, then | A ∩ Irr( S ) | = .Proof. Assume that Π S is confluent. By Lemma 2.19, we get | A ∩ Irr( S ) | > . Suppose to thecontrary that | A ∩ Irr( S ) | >
2. Let A ∩ Irr( S ) = { a i | i ∈ I } with | I | >
2. We have two cases toconsider.
Case 1. P ( a i ) ∩ P ( a j ) , ∅ for some i , j ∈ I with i , j . In this case, we can choose b ∈ P ( a i ) ∩ P ( a j ).Then ( b ∗ → Π S a i , b ∗ → Π S a j ) is a fork, which is not joinable by a i , a j ∈ Irr( S ), contradicting that Π S is confluent. JIN ZHANG AND XING GAO ∗ Case 2. P ( a i ) ∩ P ( a j ) = ∅ for all i , j ∈ I with i , j . We claim that ( t i , t j ) , ( t j , t i ) < Π S forall t i ∈ P ( a i ) and t j ∈ P ( a j ). Otherwise, by symmetry, let ( t i , t j ) ∈ Π S for some t i ∈ P ( a i ) and t j ∈ P ( a j ). Then t i → Π S t j ∗ → Π S a j and so t i ∈ P ( a j ) ∩ P ( a i ), contradicting that P ( a i ) ∩ P ( a j ) = ∅ .Since a i and a j are in the same operated congruence class A , we have ( a i , a j ) ∈ h S i . ByLemma 2.11, there is a sequence a i = v , v , · · · , v n = a j with n >
2, in which, for each k in { , , · · · , n − } , either ( v k , v k + ) ∈ S c or ( v k + , v k ) ∈ S c . Note that v k ∈ A , k n . Because a i ∈ Irr( S ), we can take ℓ : = max { k | v k ∈ P ( a i ) , k n } . If ℓ = n , then a j = v n ∈ P ( a i ) and so a j ∈ P ( a i ) ∩ P ( a j ), a contradiction. If 1 ℓ n −
1, then v ℓ ∈ P ( a i ) , v ℓ + < P ( a i ) and v ℓ → Π S v ℓ + . Since v ℓ + ∈ A , by Lemma 2.19, there exists p ∈ I with p , i such that v ℓ + ∗ → Π S a p . Thus v ℓ ∗ → Π S a i and v ℓ → Π S v ℓ + ∗ → Π S a p , but a i , a p ∈ Irr( S ) and a i , a p , contradicting that Π S isconfluent. (cid:3) Now we are ready for our main result of this section.
Theorem 2.21.
Let S be a binary relation on M ( X ) , and let be a monomial order on M ( X ) and Π S the term-rewriting system with respect to . Then Irr( S ) is a section of h S i if and only if Π S isconvergent.Proof. ( ⇐ ) Suppose Π S is convergent. Then in view of Lemma 2.20, every operated congruenceclass intersects with Irr( S ) exactly one element. So Irr( S ) is a section of M ( X ) / h S i .( ⇒ ) Suppose to the contrary that Π S is not convergent. Since Π S is terminating by Lemma 2.17, Π S is not confluent. Then there exists a fork ( t ∗ → Π S v , t ∗ → Π S v ) which is not joinable. ByLemma 2.17, v ∗ → Π S u and v ∗ → Π S u for some u , u ∈ Irr( S ) , u , u . which implies that t ∗ → Π S u and t ∗ → Π S u . By Lemma 2.18, we have( t , u ) , ( t , u ) ∈ h S i and so ( u , u ) ∈ h S i . Hence u and u in a same operated congruence class and they are in normal form, contradictingthat Irr( S ) is a section of M ( X ) / h S i . (cid:3)
3. A pplications
Inverse monoids appear in a range of contexts, for example, they can be employed in the studyof partial symmetries [24]. U -monoids are natural generalizations of inverse monoids. Definition 3.1. A U -monoid is a monoid G equipped with a (unary) operator ◦ such that ( u ◦ ) ◦ = u for all u ∈ G .The following are two classes of U -monoids. Definition 3.2. ( a ) A ∗ -monoid is a U -monoid G satisfying the axiom ( uv ) ◦ = v ◦ u ◦ for all u , v ∈ G . Such an operator is called an involution, and typically denoted by ∗ .( b ) A group is a ∗ -monoid G satisfying the axiom u ∗ u = = uu ∗ for all u ∈ G .In this section, as applications of Theorem 2.21, we construct respectively sections of free ∗ -monoids and free groups, which are viewed as quotients of free operated monoids. The monomialorder given in [18, Lem. 5.3] will be used throughout this section. REE OPERATED MONOIDS AND REWRITING SYSTEMS 9
Free ∗ -monoids. This subsection is spent to construct sections of free ∗ -monoids. Recallthat M ( X ) is the free operated monoid on X . As a special case of a well-known result in universalalgebra [10, Prop.1.3.6], the free ∗ -monoid on a set X is the quotient of M ( X ) by the operatedcongruence h S i , where(4) S = { φ ( w ) : = ( ⌊⌊ w ⌋⌋ , w ) , ψ ( u , v ) : = ( ⌊ uv ⌋ , ⌊ v ⌋⌊ u ⌋ ) , ω : = ( ⌊ ⌋ , | w ∈ M ( X ) , u , v ∈ M ( X ) \ { }} . Before we go on to obtain a section of the free ∗ -monoid, we recall a monomial order on M ( X ) from [18]. Let be a well-ordering on a set X . It can be extended to a well-ordering on M ( X ) = lim −→ M n ( X ) by recursively defining a well-ordering n , on M n : = M n ( X ) for each n > n =
0, we have M = M ( X ). In this case, we obtain a well-ordering by taking the shortlexorder sl on M ( X ) induced by with the convention that 1 sl u for all u ∈ M ( X ) \{ } . Suppose n has been defined on M n : = M ( X ⊔ ⌊ M n − ⌋ ) for an n >
0. Denote by deg X ( u ) the number of x ∈ X in u with repetition. Then n induces( a ) a well-ordering ′ n on ⌊ M n ⌋ by ⌊ u ⌋ < ′ n ⌊ v ⌋ ⇐⇒ u < n v ;( b ) a well-ordering ′′ n on X ⊔ ⌊ M n ⌋ ;( c ) a well-ordering ′′′ n on X ⊔ ⌊ M n ⌋ by u < ′′′ n v ⇐⇒ ( either deg X ( u ) < deg X ( v )or deg X ( u ) = deg X ( v ) and u < ′′ n v . ( d ) the shortlex well-ordering n + on M n + = M ( X ⊔ ⌊ M n ⌋ ) induced by ′′′ n .The orders n are compatible with the direct system { M n } n > and hence induces a well-ordering,still denoted by , on M ( X ) = lim −→ M n . Lemma 3.3. [18, Lem. 5.3]
The order on M ( X ) defined above is a monomial order. Using this monomial order, we have w < ⌊⌊ w ⌋⌋ , ⌊ v ⌋⌊ u ⌋ < ⌊ uv ⌋ , < ⌊ ⌋ for all w ∈ M ( X ) , and u , v ∈ M ( X ) \ { } . For simplicity, if α is an element of a binary relation, we denote α and R ( α ) by the domain andimage of α , respectively. For example, φ ( w ) = ⌊⌊ w ⌋⌋ , R ( φ ( w )) = w , ψ ( u , v ) = ⌊ uv ⌋ , R ( ϕ ( u )) = ⌊ v ⌋⌊ u ⌋ , and ω = ⌊ ⌋ , R ( ω ) = . In the remainder of this paper, if q | ⌊⌊ x ⌋⌋ → Π S q | x , we will indicate such rewriting step in more detailby q | ⌊⌊ x ⌋⌋ → φ q | x . Similar notations will be used for → ϕ and → ω .The following concept is finer than bracketed subwords, containing the information of place-ments [29]. Definition 3.4.
Let X be a set, and let w ∈ M ( X ) be such that(5) q | u = w = q | u for some u , u ∈ M ( X ) , q , q ∈ M ⋆ ( X ) . The two placements ( u , q ) and ( u , q ) are called( a ) separated if there exist p ∈ M ⋆ ,⋆ ( X ) such that q | ⋆ = p | ⋆ , u , q | ⋆ = p | u , ⋆ , and w = p | u , u ;( b ) nested if there exists q ∈ M ⋆ ( X ) such that either q = q | q or q = q | q ;( c ) intersecting if there exist q ∈ M ⋆ ( X ) and a , b , c ∈ M ( X ) \{ } such that w = q | abc and either(i) q = q | ⋆ c and q = q | a ⋆ ; or ∗ (ii) q = q | a ⋆ and q = q | ⋆ c . Remark 3.5. ( a ) Suppose the placements ( u , q ) and ( u , q ) are nested. If q = q | q , then q | q | u = q | u = w = q | u . By Remark 2.4, we have q | u = u , i.e., u is a bracketedsubword of u . Similarly, if q = q | q , then q | u = u , i.e., u is a bracketed subword of u .( b ) Suppose the placements ( u , q ) and ( u , q ) are intersecting. If q = q | ⋆ c and q = q | a ⋆ ,then q | u c = q | u = w = q | abc and q | au = q | u = w = q | abc . Again by Remark 2.4, u c = abc and au = abc . So u = ab and u = bc . Similarly, if q = q | a ⋆ and q = q | ⋆ c , then u = ab and u = bc . Lemma 3.6. [29, Thm. 4.11]
Let w ∈ M ( X ) . For any two placements ( u , q ) and ( u , q ) in w,exactly one of the following is true :( a ) ( u , q ) and ( u , q ) are separated ;( b ) ( u , q ) and ( u , q ) are nested ;( c ) ( u , q ) and ( u , q ) are intersecting. Lemma 3.7.
Let w ∈ M ( X ) . For any two placements ( u , q ) and ( u , q ) in w, if the breadth | u | is 1, then ( u , q ) and ( u , q ) cann’t be intersecting.Proof. Suppose the placements ( u , q ) and ( u , q ) are intersecting. From Definition 3.4 ( c ), thereare q ∈ M ⋆ ( X ) and a , b , c ∈ M ( X ) \{ } such that w = q | abc and either q = q | ⋆ c and q = q | a ⋆ , or q = q | a ⋆ and q = q | ⋆ c . For the former case, we have u = ab and u = bc by Remark 3.5; for thelater case, we get u = bc and u = ab . Since a , b , c ∈ M ( X ) \ { } , it follows that | a | , | b | , | c | > | u | > | u | = (cid:3) Lemma 3.8.
Let S be the binary relation given in Eq. (4) and α, β ∈ S . Suppose q | α = q | β forsome q , q ∈ M ⋆ ( X ) . If the placements ( α, q ) and ( β, q ) are separated, then q | R ( α ) ↓ Π S q | R ( β ) .Proof. In view of Definition 3.4 ( a ), there exists p ∈ M ⋆ ,⋆ ( X ) such that q | ⋆ = p | ⋆ , β and q | ⋆ = p | α, ⋆ , whence q | R ( α ) = p | R ( α ) , β → Π S p | R ( α ) , R ( β ) , q | R ( β ) = p | α, R ( β ) → Π S p | R ( α ) , R ( β ) . So we conclude that q | R ( α ) ↓ Π S q | R ( β ) . (cid:3) Lemma 3.9.
Let S be the binary relation given in Eq. (4) and α, β ∈ S . Suppose q | α = q | β forsome q , q ∈ M ⋆ ( X ) . If the placements ( α, q ) and ( β, q ) are nested, then q | R ( α ) ↓ Π S q | R ( β ) . Proof.
Suppose the two placements ( α, q ) and ( β, q ) are nested. According to the choice of α and β , we have the following cases to consider. Case 1. α = φ ( u ) = ⌊⌊ u ⌋⌋ and β = φ ( v ) = ⌊⌊ v ⌋⌋ . By symmetry, we may assume that q = q | q forsome q ∈ M ⋆ ( X ). Then by Remark 3.5, q | ⌊⌊ u ⌋⌋ = ⌊⌊ v ⌋⌋ , i.e., ⌊⌊ u ⌋⌋ is a bracketed subword of ⌊⌊ v ⌋⌋ . Subcase 1.1. ⌊⌊ u ⌋⌋ = ⌊⌊ v ⌋⌋ . Then α = β , q = ⋆ and q = q . Hence q | R ( α ) = q | R ( β ) and q | R ( α ) ↓ Π S q | R ( β ) trivially. Subcase 1.2. ⌊⌊ u ⌋⌋ = ⌊ v ⌋ . Then v = ⌊ u ⌋ , q | ⌊⌊ u ⌋⌋ = ⌊⌊ v ⌋⌋ = ⌊⌊⌊ u ⌋⌋⌋ = ⌊ ⋆ ⌋| ⌊⌊ u ⌋⌋ and q = ⌊ ⋆ ⌋ . Hence q | R ( α ) = q | u = ( q | q ) | u = q | q | u = q | ⌊ ⋆ ⌋| u = q | ⌊ u ⌋ = q | v = q | R ( β ) and so q | R ( α ) ↓ Π S q | R ( β ) trivially. REE OPERATED MONOIDS AND REWRITING SYSTEMS 11
Subcase 1.3. ⌊⌊ u ⌋⌋ is a bracketed subword of v . Then there exists p ∈ M ⋆ ( X ) such that v = p | ⌊⌊ u ⌋⌋ and q = ⌊⌊ p ⌋⌋ , which implies q | R ( α ) = q | u = ( q | q ) | u = q | q | u = q | ⌊⌊ p ⌋⌋| u = q | ⌊⌊ p | u ⌋⌋ → φ q | p | u , q | R ( β ) = q | v = q | p | ⌊⌊ u ⌋⌋ → φ q | p | u . Consequently q | R ( α ) ↓ Π S q | R ( β ) . Case 2. α = ψ ( u ′ , v ′ ) = ⌊ u ′ v ′ ⌋ and β = ψ ( u , v ) = ⌊ uv ⌋ . By symmetry, we may assume that q = q | q for some q ∈ M ⋆ ( X ). Then by Remark 3.5, q | ⌊ u ′ v ′ ⌋ = ⌊ uv ⌋ , i.e., ⌊ u ′ v ′ ⌋ is a bracketed subword of ⌊ uv ⌋ . Subcase 2.1. ⌊ u ′ v ′ ⌋ = ⌊ uv ⌋ . This is similar to Subcase 1.1. Subcase 2.2. ⌊ u ′ v ′ ⌋ is a bracketed subword of u or v . By symmetry, we may assume ⌊ u ′ v ′ ⌋ is abracketed subword of u . i.e., there exists p ∈ M ⋆ ( X ) such that u = p | ⌊ u ′ v ′ ⌋ and q = ⌊ pv ⌋ . Whence q | R ( α ) = q | ⌊ v ′ ⌋⌊ u ′ ⌋ = ( q | q ) | ⌊ v ′ ⌋⌊ u ′ ⌋ = q | q | ⌊ v ′⌋⌊ u ′⌋ = q | ⌊ pv ⌋| ⌊ v ′⌋⌊ u ′⌋ = q | ⌊ p | ⌊ v ′⌋⌊ u ′⌋ v ⌋ → ψ q | ⌊ v ⌋⌊ p | ⌊ v ′⌋⌊ u ′⌋ ⌋ , q | R ( β ) = q | ⌊ v ⌋⌊ u ⌋ = q | ⌊ v ⌋⌊ p ⌋| ⌊ u ′ v ′⌋ → ψ q | ⌊ v ⌋⌊ p ⌋| ⌊ v ′⌋⌊ u ′⌋ = q | ⌊ v ⌋⌊ p | ⌊ v ′⌋⌊ u ′⌋ ⌋ . Hence q | R ( α ) ↓ Π S q | R ( β ) . Case 3. α = φ ( w ) = ⌊⌊ w ⌋⌋ and β = ψ ( u , v ) = ⌊ uv ⌋ . Subcase 3.1. q = q | q for some q ∈ M ⋆ ( X ). By Remark 3.5, q | ⌊⌊ w ⌋⌋ = ⌊ uv ⌋ , that is, ⌊⌊ w ⌋⌋ is abracketed subword of ⌊ uv ⌋ . Subcase 3.1.1. ⌊⌊ w ⌋⌋ = ⌊ uv ⌋ . Then ⌊ w ⌋ = uv and so u = v =
1, which contradicts fromEq. (6) that β ∈ S . Subcase 3.1.2. ⌊⌊ w ⌋⌋ is a bracketed subword of uv . By symmetry, we can assume that ⌊⌊ w ⌋⌋ is abracketed subword of u , i.e., there exists p ∈ M ⋆ ( X ) such that u = p | ⌊⌊ w ⌋⌋ and q = ⌊ pv ⌋ . So q | R ( α ) = q | w = ( q | q ) | w = q | q | w = q | ⌊ pv ⌋| w = q | ⌊ p | w v ⌋ → ψ q | ⌊ v ⌋⌊ p | w ⌋ , q | R ( β ) = q | ⌊ v ⌋⌊ u ⌋ = q | ⌊ v ⌋⌊ p | ⌊⌊ w ⌋⌋ ⌋ → φ q | ⌊ v ⌋⌊ p | w ⌋ . Thus q | R ( α ) ↓ Π S q | R ( β ) . Subcase 3.2. q = q | q for some q ∈ M ⋆ ( X ). By Remark 3.5, ⌊⌊ w ⌋⌋ = q | ⌊ uv ⌋ , i.e., ⌊ uv ⌋ is abracketed subword of ⌊⌊ w ⌋⌋ . If ⌊ uv ⌋ = ⌊⌊ w ⌋⌋ , similar to Case 3.1.1, we get q | R ( α ) ↓ Π S q | R ( β ) .Suppose ⌊ uv ⌋ , ⌊⌊ w ⌋⌋ . Subcase 3.2.1. ⌊ uv ⌋ = ⌊ w ⌋ . Then uv = w , q | ⌊ uv ⌋ = ⌊⌊ w ⌋⌋ = ⌊⌊ uv ⌋⌋ = ⌊ ⋆ ⌋| ⌊ uv ⌋ and q = ⌊ ⋆ ⌋ . Consequently q | R ( α ) = q | w = q | uv and q | R ( β ) = q | ⌊ v ⌋⌊ u ⌋ = ( q | q ) | ⌊ v ⌋⌊ u ⌋ = q | q | ⌊ v ⌋⌊ u ⌋ = q | ⌊ ⋆ ⌋| ⌊ v ⌋⌊ u ⌋ = q | ⌊⌊ v ⌋⌊ u ⌋⌋ → ψ q | ⌊⌊ u ⌋⌋⌊⌊ v ⌋⌋ → φ q | u ⌊⌊ v ⌋⌋ → φ q | uv . Hence q | R ( α ) ↓ Π S q | R ( β ) . Subcase 3.2.2. ⌊ uv ⌋ is a bracketed subword of w , i.e., there exists p ∈ M ⋆ ( X ) such that w = p | ⌊ uv ⌋ and q = ⌊⌊ p ⌋⌋ . Thus q | R ( α ) = q | w = q | p | ⌊ uv ⌋ → ψ q | p | ⌊ v ⌋⌊ u ⌋ , q | R ( β ) = q | ⌊ v ⌋⌊ u ⌋ = ( q | q ) | ⌊ v ⌋⌊ u ⌋ = q | q | ⌊ v ⌋⌊ u ⌋ = q | ⌊⌊ p ⌋⌋| ⌊ v ⌋⌊ u ⌋ = q | ⌊⌊ p | ⌊ v ⌋⌊ u ⌋ ⌋⌋ → φ q | p | ⌊ v ⌋⌊ u ⌋ , and so q | R ( α ) ↓ Π S q | R ( β ) . Case 4. α = φ ( w ) = ⌊⌊ w ⌋⌋ and β = ω = ⌊ ⌋ . Subcase 4.1. q = q | q for some q ∈ M ⋆ ( X ). Then by Remark 3.5, q | ⌊⌊ w ⌋⌋ = ⌊ ⌋ , which impliesthat ⌊⌊ w ⌋⌋ is a bracketed subword of ⌊ ⌋ , a contradiction by comparing the depth. ∗ Subcase 4.2. q = q | q for some q ∈ M ⋆ ( X ). By Remark 3.5, ⌊⌊ w ⌋⌋ = q | ⌊ ⌋ , i.e., ⌊ ⌋ is a bracketedsubword of ⌊⌊ w ⌋⌋ . Note that ⌊ ⌋ , ⌊⌊ w ⌋⌋ . Subcase 4.2.1. ⌊ ⌋ = ⌊ w ⌋ . Then w = , q | ⌊ ⌋ = ⌊⌊ w ⌋⌋ = ⌊ ⋆ ⌋| ⌊ ⌋ and q = ⌊ ⋆ ⌋ . Hence q | R ( α ) = q | w = q | , q | R ( β ) = q | = ( q | q ) | = q | ⌊ ⋆ ⌋| = q | ⌊ ⌋ → ω q | , and so q | R ( α ) ↓ Π S q | R ( β ) . Subcase 4.2.2. ⌊ ⌋ is a bracketed subword of w , i.e., there exists p ∈ M ⋆ ( X ) such that w = p | ⌊ ⌋ and q = ⌊⌊ p ⌋⌋ . Then q | R ( α ) = q | w = q | p | ⌊ ⌋ → ω q | p | , q | R ( β ) = q | = ( q | q ) | = q | ⌊⌊ p ⌋⌋| → φ q | p | . Therefore q | R ( α ) ↓ Π S q | R ( β ) . Case 5. α = ψ ( u , v ) = ⌊ uv ⌋ and β = ω = ⌊ ⌋ . This is similar to Case 4. Case 6. α = ω = ⌊ ⌋ and β = ω = ⌊ ⌋ . This case is trivial since α and β are equal.This completes the proof. (cid:3) Now we arrive at our first main result of this section.
Theorem 3.10.
Let X be a set and S the binary relation given in Eq. (4). With the monomialorder given in [18] , we have ( a ) the term-rewriting system Π S is convergent. ( b ) the set Irr( S ) = M ( X ) \ Dom( Π S ) is a section of M ( X ) / h S i .Proof. ( a ) Since is a monomial order on M ( X ), Π S is terminating by Lemma 2.17. So weare left to prove that Π S is confluent. From Lemma 2.16, it su ffi ces to show that Π S is locallyconfluent. Let q | R ( α ) Π S ← q | α = w = q | β → Π S q | R ( β ) be an arbitrary local fork, where q , q ∈ M ⋆ ( X ) , α, β ∈ S . From Eq. (4), both of the breadth of α and β are 1, and so the placements ( α, q ) and ( β, q ) cann’t be intersecting by Lemma 3.7. If theplacements ( α, q ) and ( β, q ) are separated and nested, then q | R ( α ) ↓ Π S q | R ( β ) by Lemmas 3.12and 3.9.( b ) It follows from Item ( a ) and Theorem 2.21. (cid:3) Remark 3.11.
It is well known [24] that ∗ -monoid is also called monoid with involution, and thefree monoid with involution ∗ on a set X is the free monoid M ( X ∪ X ∗ ), where X ∗ : = { x ∗ | x ∈ X } is a disjoint copy of X . The Irr( S ) obtained in Theorem 3.10 is precisely the set M ( X ∪ X ∗ ) if weidentify ⌊ x ⌋ with x ∗ for each x ∈ X .3.2. Free groups.
Again as a special case of a well-known result in universal algebra [10,Prop.1.3.6], the free group on a set X is the quotient of M ( X ) by the operated congruence h S i ,where(6) S : = { ( ⌊⌊ w ⌋⌋ , w ) , ( ⌊ uv ⌋ , ⌊ v ⌋⌊ u ⌋ ) , ( ⌊ w ⌋ w , , ( w ⌊ w ⌋ , | w ∈ M ( X ) , u , v ∈ M ( X ) \ { }} . In this subsection, we turn to construct a section of the free group on a set W . Write φ ( w ) : = ( ⌊⌊ w ⌋⌋ , w ) , ψ ( u , v ) : = ( ⌊ uv ⌋ , ⌊ v ⌋⌊ u ⌋ ) , ϕ ( w ) : = ( ⌊ w ⌋ w , , χ ( w ) : = ( w ⌊ w ⌋ , , REE OPERATED MONOIDS AND REWRITING SYSTEMS 13 where w ∈ M ( X ) , u , v ∈ M ( X ) \ { } . Note that if w =
1, then w ⌊ w ⌋ = ⌊ ⌋ = ⌊ ⌋ . So ( ⌊ ⌋ , ∈ S .Here again under the monomial order given in [18], we have w < ⌊⌊ w ⌋⌋ , ⌊ v ⌋⌊ u ⌋ < ⌊ uv ⌋ , < ⌊ w ⌋ w , < w ⌊ w ⌋ for w ∈ M ( X ) , u , v ∈ M ( X ) \ { } . Lemma 3.12.
Let S be the binary relation given in Eq. (6) and α, β ∈ S . Suppose q | α = q | β forsome q , q ∈ M ⋆ ( X ) . If the placements ( α, q ) and ( β, q ) are separated, then q | R ( α ) ↓ Π S q | R ( β ) .Proof. It is parallel to the proof of Lemma 3.12, because the proof of Lemma 3.12 does notdepend on the concrete expressions of α and β . (cid:3) Lemma 3.13.
Let S be the binary relation given in Eq. (6) and α, β ∈ S . Suppose q | α = q | β forsome q , q ∈ M ⋆ ( X ) . If the placements ( α, q ) and ( β, q ) are intersecting, then q | R ( α ) ↓ Π S q | R ( β ) .Proof. Note that φ ( w ) = ⌊⌊ w ⌋⌋ , ψ ( u , v ) = ⌊ uv ⌋ and | φ ( u ) | = | ψ ( u , v ) | = w ∈ M ( X ) , u , v ∈ M ( X ) \ { } . It follows from Lemma 3.7 that α, β ∈ { ϕ ( u ) , χ ( u ) | u ∈ M ( X ) } . Let w : = q | α = q | β .Then by Definition 3.4( c ), there are q ∈ M ⋆ ( X ) and a , b , c ∈ M ( X ) \ { } such that w = q | abc .Depending on the forms of α and β , there are three cases to consider. Case 1. α = ϕ ( u ) = ⌊ u ⌋ u and β = ϕ ( v ) = ⌊ v ⌋ v for some u , v ∈ M ( X ). By symmetry, we mayassume q = q | ⋆ c and q = q | a ⋆ . Using Remark 3.5, we get ⌊ u ⌋ u = ab and ⌊ v ⌋ v = bc . Since | b | > ⌊ v ⌋ is a bracketed subword of b . Suppose v = v v and b = ⌊ v ⌋ v . Then c = v . Similarly from ⌊ u ⌋ u = ab , we can assume a = ⌊ u ⌋ u and b = u with u = u u . Then u = b = ⌊ v ⌋ v = ⌊ v v ⌋ v and a = ⌊ u ⌋ u = ⌊ u u ⌋ u = ⌊ u ⌊ v v ⌋ v ⌋ u and so q | R ( α ) = q | = q | ⋆ c | = q | c = q | v , q | R ( β ) = q | = q | a ⋆ | = q | a = q | ⌊ u ⌊ v v ⌋ v ⌋ u → ψ q | ⌊⌊ v v ⌋ v ⌋⌊ u ⌋ u → ϕ q | ⌊⌊ v v ⌋ v ⌋ → ψ q | ⌊⌊ v ⌋⌊ v ⌋ v ⌋ → ϕ q | ⌊⌊ v ⌋⌋ → φ q | v , which implies q | R ( α ) ↓ Π S q | R ( β ) . Case 2. α = χ ( u ) = u ⌊ u ⌋ and β = χ ( v ) = v ⌊ v ⌋ for some u , v ∈ M ( X ). This is similar to Case 1. Case 3. α = ϕ ( u ) = ⌊ u ⌋ u and β = χ ( v ) = v ⌊ v ⌋ , or α = χ ( u ) = u ⌊ u ⌋ and β = ϕ ( v ) = ⌊ v ⌋ v for some u , v ∈ M ( X ). By symmetry, it su ffi ces to consider the former case. Then according toDefinition 3.4( c ), we have two subcases to consider. Subcase 3.1. q = q | ⋆ c and q = q | a ⋆ . From Remark 3.5, ⌊ u ⌋ u = ab and v ⌊ v ⌋ = bc . With a similarargument to Case 1, we can assume b = v , c = v ⌊ v v ⌋ with v = v v , a = ⌊ u u ⌋ u , b = u with u = u u . Thus q | R ( α ) = q | = q | ⋆ c | = q | c = q | v ⌊ v v ⌋ → ψ q | v ⌊ v ⌋⌊ v ⌋ → χ q | ⌊ v ⌋ , q | R ( β ) = q | = q | a ⋆ | = q | a = q | ⌊ u u ⌋ u → ψ q | ⌊ u ⌋⌊ u ⌋ u → ϕ q | ⌊ u ⌋ = q | ⌊ v ⌋ , and so q | R ( α ) ↓ Π S q | R ( β ) . Subcase 3.2. q = q | a ⋆ and q = q | ⋆ c . From Remark 3.5, ⌊ u ⌋ u = bc and v ⌊ v ⌋ = ab . Again similarto Case 1, we may suppose b = ⌊ u u ⌋ u and c = u with u = u u . ∗ Then a ⌊ u u ⌋ u = ab = v ⌊ v ⌋ . If u , , then ⌊ v ⌋ is a bracketed subword of u and a ⌊ u u ⌋ is abracketed subword of v . So we getdep( ⌊ v ⌋ ) dep( u ) < dep( a ⌊ u u ⌋ ) dep( v ) , a contradiction. So u = , u = u , b = ⌊ u u ⌋ = ⌊ u ⌋ , c = u , a ⌊ u ⌋ = ab = v ⌊ v ⌋ , which implies that a = v , ⌊ u ⌋ = ⌊ v ⌋ and u = v . Thus q | R ( α ) = q | = q | a ⋆ | = q | a = q | v , q | R ( β ) = q | = q | ⋆ c | = q | c = q | u = q | v , and so q | R ( α ) ↓ Π S q | R ( β ) . (cid:3) Lemma 3.14.
Let S be the binary relation given in Eq. (6) and α, β ∈ S . Suppose q | α = q | β forsome q , q ∈ M ⋆ ( X ) . If the placements ( α, q ) and ( β, q ) are nested, then q | R ( α ) ↓ Π S q | R ( β ) .Proof. By Definition 3.4( c ), we can assume q | α = w = q | β for some w ∈ M ( X ) . From Eq. (6),there are four choices for each α and β . In view of symmetry, there are ten pairs of α and β to consider. If α, β ∈ { φ ( w ) , ψ ( u , v ) | w ∈ M ( X ) , u , v ∈ M ( X ) \ { }} , the result follows fromLemma 3.9. So three cases have been done and we are left to consider the following seven cases. Case 1. α = φ ( u ) = ⌊⌊ u ⌋⌋ and β = ϕ ( v ) = ⌊ v ⌋ v . Since either α is subword of β or β is subword of α , we have the following two subcases. Subcase 1.1. q = q | q for some q ∈ M ⋆ ( X ) . By Remark 3.5, ⌊⌊ u ⌋⌋ = q | ⌊ v ⌋ v , i.e., ⌊ v ⌋ v is a bracketedsubword of ⌊⌊ u ⌋⌋ . Note that ⌊ v ⌋ v , ⌊⌊ u ⌋⌋ , ⌊ u ⌋ by comparing the breadth. So ⌊ v ⌋ v is a bracketedsubword of u , i.e., there exists p ∈ M ⋆ ( X ) such that u = p | ⌊ v ⌋ v and q = ⌊⌊ p ⌋⌋ . Thus q | R ( α ) = q | u = q | p | ⌊ v ⌋ v → ϕ q | p | , q | R ( β ) = q | = q | q | = q | ⌊⌊ p ⌋⌋| = q | ⌊⌊ p | ⌋⌋ → φ q | p | , and so q | R ( α ) ↓ Π S q | R ( β ) . Subcase 1.2. q = q | q for some q ∈ M ⋆ ( X ) . Then by Remark 3.5, q | ⌊⌊ u ⌋⌋ = ⌊ v ⌋ v , i.e., ⌊⌊ u ⌋⌋ is abracketed subword of ⌊ v ⌋ v . Note that ⌊⌊ u ⌋⌋ , ⌊ v ⌋ v . So there are two points to consider. Subcase 1.2.1. ⌊⌊ u ⌋⌋ = ⌊ v ⌋ . Then v = ⌊ u ⌋ . Since q | ⌊⌊ u ⌋⌋ = ⌊ v ⌋ v = ⌊⌊ u ⌋⌋⌊ u ⌋ , we have q = ⋆ ⌊ u ⌋ . Thus q | R ( α ) = q | u = q | q | u = q | u ⌊ u ⌋ → χ q | and q | R ( β ) = q | and so q | R ( α ) ↓ Π S q | R ( β ) . Subcase 1.2.2. ⌊⌊ u ⌋⌋ is a bracketed subword of v , i.e., there exists p ∈ M ⋆ ( X ) such that v = p | ⌊⌊ u ⌋⌋ .Then q = ⌊ p ⌋ v or q = ⌊ v ⌋ p . If q = ⌊ p ⌋ v , then q | R ( α ) = q | u = q | q | u = q | ⌊ p | u ⌋ v = q | ⌊ p | u ⌋ p | ⌊⌊ u ⌋⌋ → φ q | ⌊ p | u ⌋ p | u → ϕ q | . If q = ⌊ v ⌋ p , then q | R ( α ) = q | u = q | q | u = q | ⌊ v ⌋ p | u = q | ⌊ p | ⌊⌊ u ⌋⌋ ⌋ p | u → φ q | ⌊ p | u ⌋ p | u → ϕ q | . Note that q | R ( β ) = q | . So we conclude q | R ( α ) ↓ Π S q | R ( β ) . Case 2. α = φ ( u ) = ⌊⌊ u ⌋⌋ and β = χ ( v ) = v ⌊ v ⌋ . This is similar to Case 1.
Case 3. α = ψ ( u , u ) = ⌊ u u ⌋ and β = ϕ ( v ) = ⌊ v ⌋ v . Again since either α is subword of β or β issubword of α , there are two subcases to consider. Subcase 3.1. q = q | q for some q ∈ M ( X ). Then by Remark 3.5, q | ⌊ u u ⌋ = ⌊ v ⌋ v , i.e., ⌊ u u ⌋ is abracketed subword of ⌊ v ⌋ v . Note that ⌊ v ⌋ v , ⌊ u u ⌋ by comparing the breadth. REE OPERATED MONOIDS AND REWRITING SYSTEMS 15
Subcase 3.1.1. ⌊ u u ⌋ = ⌊ v ⌋ . Then v = u u , q | ⌊ u u ⌋ = ⌊ v ⌋ v = ⌊ u u ⌋ u u and q = ⋆ u u and so q | R ( α ) = q | ⌊ u ⌋⌊ u ⌋ = q | q | ⌊ u ⌋⌊ u ⌋ = q | ⋆ u u | ⌊ u ⌋⌊ u ⌋ = q | ⌊ u ⌋⌊ u ⌋ u u → ϕ q | ⌊ u ⌋ u → ϕ q | . Since q | R ( β ) = q | , we get q | R ( α ) ↓ Π S q | R ( β ) . Subcase 3.1.2. ⌊ u u ⌋ is a bracketed subword of v , i.e., there exists p ∈ M ⋆ ( X ) such that v = p | ⌊ u u ⌋ . Then q = ⌊ p ⌋ v or q = ⌊ v ⌋ p . If q = ⌊ p ⌋ v , then q | R ( α ) = q | ⌊ u ⌋⌊ u ⌋ = q | q | ⌊ u ⌋⌊ u ⌋ = q | ⌊ p | ⌊ u ⌋⌊ u ⌋ ⌋ v = q | ⌊ p | ⌊ u ⌋⌊ u ⌋ ⌋ p | ⌊ u u ⌋ → ψ q | ⌊ p | ⌊ u ⌋⌊ u ⌋ ⌋ p | ⌊ u ⌋⌊ u ⌋ → ϕ q | , If q = ⌊ v ⌋ p , then q | R ( α ) = q | ⌊ u ⌋⌊ u ⌋ = q | q | ⌊ u ⌋⌊ u ⌋ = q | ⌊ v ⌋ p | ⌊ u ⌋⌊ u ⌋ = q | ⌊ p | ⌊ u u ⌋ ⌋ p | ⌊ u ⌋⌊ u ⌋ → ψ q | ⌊ p | ⌊ u ⌋⌊ u ⌋ ⌋ p | ⌊ u ⌋⌊ u ⌋ → ϕ q | . Since q | R ( β ) = q | , we conclude that q | R ( α ) ↓ Π S q | R ( β ) . Subcase 3.2. q = q | q for some q ∈ M ⋆ ( X ). Then by Remark 3.5, q | ⌊ v ⌋ v = ⌊ u u ⌋ , i.e., ⌊ v ⌋ v is abracketed subword of ⌊ u u ⌋ . Note that ⌊ v ⌋ v , ⌊ u u ⌋ . Thus we can assume ⌊ v ⌋ v is a bracketedsubword of u u . Subcase 3.2.1. ⌊ v ⌋ v is a bracketed subword of u or u . By symmetry, we only need to considerthe former. Then there exists p ∈ M ⋆ ( X ) such that u = p | ⌊ v ⌋ v and q = ⌊ pu ⌋ . Consequently, q | R ( α ) = q | ⌊ u ⌋⌊ u ⌋ = q | ⌊ u ⌋⌊ p | ⌊ v ⌋ v ⌋ → ϕ q | ⌊ u ⌋⌊ p | ⌋ , q | R ( β ) = q | = q | q | = q | ⌊ p | u ⌋ → ψ q | ⌊ u ⌋⌊ p | ⌋ , and so q | R ( α ) ↓ Π S q | R ( β ) . Subcase 3.2.2. ⌊ v ⌋ v is neither a bracketed subword of u nor a bracketed subword of u . Then v = v v , u = u ′ ⌊ v ⌋ v and u = v u ′ for some v , v , u ′ , u ′ ∈ M ( X ) with v ,
1. Consequently, ⌊ u ′ ⋆ u ′ ⌋| ⌊ v ⌋ v = ⌊ u ′ ⌊ v ⌋ vu ′ ⌋ = ⌊ u ′ ⌊ v ⌋ v v u ′ ⌋ = ⌊ u u ⌋ = q | ⌊ v ⌋ v and ⌊ u ′ ⋆ u ′ ⌋ = q . Thus q | R ( α ) = q | ⌊ u ⌋⌊ u ⌋ = q | ⌊ v u ′ ⌋⌊ u ′ ⌊ v v ⌋ v ⌋ → ψ q | ⌊ v u ′ ⌋⌊ u ′ ⌊ v ⌋⌊ v ⌋ v ⌋ → ϕ q | ⌊ v u ′ ⌋⌊ u ′ ⌊ v ⌋⌋ → ψ q | ⌊ u ′ ⌋⌊ v ⌋⌊ u ′ ⌊ v ⌋⌋ → ψ q | ⌊ u ′ ⌋⌊ v ⌋⌊⌊ v ⌋⌋⌊ u ′ ⌋ → φ q | ⌊ u ′ ⌋⌊ v ⌋ v ⌊ u ′ ⌋ → ϕ q | ⌊ u ′ ⌋⌊ u ′ ⌋ , q | R ( β ) = q | = q | q | = q | ⌊ u ′ u ′ ⌋ → ψ q | ⌊ u ′ ⌋⌊ u ′ ⌋ , and so q | R ( α ) ↓ Π S q | R ( β ) . Case 4. α = ψ ( u , u ) = ⌊ u u ⌋ and β = χ ( v ) = v ⌊ v ⌋ . This is similar to Case 3.
Case 5. α = ϕ ( u ) = ⌊ u ⌋ u and β = ϕ ( v ) = ⌊ v ⌋ v . By symmetry, we may assume q = q | q forsome q ∈ M ⋆ ( X ). From Remark 3.5, q | ⌊ u ⌋ u = ⌊ v ⌋ v , i.e., ⌊ u ⌋ u is a bracketed subword of ⌊ v ⌋ v . If ⌊ u ⌋ u = ⌊ v ⌋ v , then u = v , α = β , q = q and so q | R ( α ) ↓ Π S q | R ( β ) . Suppose ⌊ u ⌋ u , ⌊ v ⌋ v . Since ⌊ u ⌋ u , ⌊ v ⌋ , ⌊ u ⌋ u is a bracketed subword of v , i.e., there exists p ∈ M ⋆ ( X ) such that v = p | ⌊ u ⌋ u .Then q = ⌊ p ⌋ v or q = ⌊ v ⌋ p . If q = ⌊ p ⌋ v , then q | R ( α ) = q | = q | q | = q | ⌊ p ⌋| v = q | ⌊ p | ⌋ p | ⌊ u ⌋ u → ϕ q | ⌊ p | ⌋ p | → ϕ q | . If q = ⌊ p ⌋ v , then q | R ( α ) = q | = q | q | = q | ⌊ v ⌋ p | = q | ⌊ p | ⌊ u ⌋ u ⌋ p | → ϕ q | ⌊ p | ⌋ p | → ϕ q | . Since q | R ( β ) = q | , we conclude that q | R ( α ) ↓ Π S q | R ( β ) . ∗ Case 6. α = χ ( u ) = u ⌊ u ⌋ and β = χ ( v ) = v ⌊ v ⌋ . This is similar to Case 5.
Case 7. α = ϕ ( u ) = ⌊ u ⌋ u and β = χ ( v ) = v ⌊ v ⌋ . This is also similar to Case 5. (cid:3)
Theorem 3.15.
Let X be a set and S the binary relation given in Eq. (6). With the monomialorder given in [18] , we have ( a ) the term-rewriting system Π S is convergent. ( b ) the set Irr( S ) = M ( X ) \ Dom( Π S ) is a section of the free group M ( X ) / h S i .Proof. ( a ) With a similar argument to the proof of Theorem 3.10( a ), the result follows fromLemmas 3.12, 3.13 and 3.14.( b ) This part follows from Item ( a ) and Theorem 3.10. (cid:3) Remark 3.16.
It is well known that reduced words are elements in the free group on a set X [20].The set Irr( S ) = M ( X ) \ Dom( Π S ) obtained in Theorem 3.15, of course, coincides with the set ofreduced words. Acknowledgements : The authors are supported by the National Natural Science Foundation ofChina (No. 11771191), the Fundamental Research Funds for the Central Universities (No. lzujbky-2017-162), and the Natural Science Foundation of Gansu Province (Grant No. 17JR5RA175) andShandong Province (No. ZR2016AM02).We thank the anonymous referee for valuable suggestions helping to improve the paper.R eferences [1] M. Aguiar, On the associative analog of Lie bialgebras,
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