A PRESENTATION FOR A SUBMONOID OF THE SYMMETRIC INVERSE MONOID Текст научной статьи по специальности «Математика»

URAL MATHEMATICAL JOURNAL, Vol. 9, No. 2, 2023, pp. 175-192

DOI: 10.15826/umj.2023.2.015

A PRESENTATION FOR A SUBMONOID OF THE SYMMETRIC INVERSE MONOID

Apatsara Sareeto

Institute of Mathematics, University of Potsdam, Potsdam, 14476, Germany

[email protected]

Jorg Koppitz

Institute of Mathematics and Informatics, Bulgarian Academy of Sciences, Sofia, 1113, Bulgaria

[email protected]

Abstract: In the present paper, we study a submonoid of the symmetric inverse semigroup In. Specifically, we consider the monoid of all order-, fence-, and parity-preserving transformations of In. While the rank and a set of generators of minimal size for this monoid are already known, we will provide a presentation for this monoid.

Keywords: Symmetric inverse monoid, Order-preserving, Fence-preserving, Presentation.

1. Introduction

Let n be a finite chain with n elements, where n is a positive integer, denoted by n = {1 < 2 < ••• <n}. We denote by PTn the monoid (under composition) of all partial transformations on n. A partial transformation a on the set n is a mapping from a subset A of n into n. The domain (respectively, image or range) of a is denoted by dom(a) (respectively, im(a)). The empty transformation is denoted by e. A transformation a € PTn is called order-preserving if x < y implies xa < ya for all x,y € dom(a). It is worth noting that we write mappings on the right of their arguments and perform composition from left to right. Furthermore, an a € PTn is called a partial injection when a is injective. The set of all partial injections forms a monoid, the symmetric inverse semigroup In, as introduced by Wagner [17]. We denote by POIn the submonoid of In, consisting of all order-preserving partial injections on n. This monoid has already been well-studied (see e.g., [6]).

A non-linear order that is closed to a linear order in some sense is the so-called zig-zag order. The pair (n, is called a zig-zag poset. or fence if

1 2 y ■ ■ ■ ^ n — 1 y n ifnis odd and 1 2 y ■ ■ ■ y n — 1 n ifnis even, respectively.

The definition of the partial order — is self-explanatory. A transformation a € PTn is referred to as fence-preserving if it preserves the partial order —, meaning that for all x,y € dom(a) with x -< y, we have xa ^ ya. The set of fence-preserving transformations on n was initially explored by Currie, Visentin, and Rutkowski. In [2, 14], the authors investigated the number of order-preserving maps of a finite fence. In particular, a formula for the number of order-preserving self-mappings

of a fence was introduced. It is noteworthy that every element of a fence is either minimal or maximal. For all x,y € n with x -< y, we have y £ {x — l,x + 1}. We denote by PFIn the submonoid of In, consisting of all fence-preserving partial injections of n. We denote by IFn the inverse submonoid of PFIn of all regular elements in PFIn. It is easy to see that IFn is the set of all a € PFIn with a-1 € PFIn. It is worth mentioning that several properties of a variety of monoids of fence-preserving transformations were studied [3, 7, 9, 11, 12, 16].

In the present paper, we focus on a submonoid of IOFn = IFnf] POIn. Let a € dom(a) for some a € IOFn. If a + 1 € dom(a) or a — 1 € dom(a) then it is easy to verify that a and aa have the same parity. In other words, a is odd if and only if aa is odd. However, if a — 1 and a + 1 are not in dom(a), then a and aa can have different parity. In order to exclude this case, we require that the image of any a € dom(a) has the same parity as aa. In this context, we refer to a as parity-preserving. In our paper, we consider the monoid IOFn,ar of all parity-preserving transformations of IOFn. Notably, for any a € IOFnar, the inverse partial injection a-1 exists and possesses order-preserving, fence-preserving, and parity-preserving. This observation implies that IOFpar is an inverse submonoid of In, as explained in [15]. Furthermore, in the same paper [15], the authors provided a characterization of the monoid IOFnar:

Proposition 1 [15]. Let p < n and let

Then a € IOF^par if and only if the following four conditions hold:

(i) m1 <m2 < ... < mp;

(ii) d1 and m1 have the same parity;

(iii) di+1 — di = 1 if and only if mi+1 — mi = 1 for all i € {1, ...,p — 1};

(iv) di+1 — dj, is even if and only if mi+1 — mj, is even for all i € {1, ...,p — 1}.

Also in [15], a set of generators of IOF^par of minimal size is given. This leads to the question of a presentation of IOFpar. In this paper, we will exhibit a monoid presentation for IOFpar. A monoid presentation is represented as an ordered pair (X | R), where X is a set, referred to as the alphabet (a set whose elements are called letters), and R is a binary relation on the free monoid generated by X, denoted by X*. A pair (u, v) € X* x X* is represented by u ~ v and is called relation. We state that u & v, for u,v € X*, is a consequence of R if (u,v) € pR, where pR denotes the congruence on X* generated by R. We say that the momoid IOFnar has (monoid) presentation (X | R) if IOFpar is isomorphic to the factor semigroup X*/pR. For a more comprehensive understanding of semigroups, presentations, and standard notation see [1, 8, 10, 13].

Given that IOFpar is a finite monoid, we can always exhibit a presentation for it. A usual method to establish a good presentations is the Guess and Prove Method, which is described by the following theorem, adapted to monoids from Ruskuc (1995, Proposition 3.2.2).

Theorem 1 [13]. Let X be a generating set for IOFnar, let R C X* x X* be a set of relations and let W C X* that the following conditions are satisfied:

1. The generating set X of IOFnar satisfies all the relations from R;

2. For each word w € X*, there exists a word w' € W such that the relation w & w' is a consequence of R;

3. |W| < lIOFPar|.

Then IOFnar is defined by the presentation (X | R).

a

d1 < d2 < • • • < m1 m2 • • •

In the next section, we introduce the alphabet (generating set) denoted as Xn and the binary relation R on Xn. Furthermore, we will demonstrate that Xn fulfills all the relations in R as outlined in Theorem 1, item 1. Following the guidance of item 2 in Theorem 1, we will establish a set of forms, denoted as P, in Section 3. Finally, in the last section, we will provide a proof for item 3 of Theorem 1.

2. The generator and relations

In this section, we will define the alphabet Xn and introduce a binary relation R on X^ We will also demonstrate that the corresponding generating set satisfies all the relations in R. Let Vi be the partial identity with the domain n\{i} for all i € {1,...,??.}. Additionally, let us define

_ /1 ••• i i + 1 i + 2 i + 3 i + 4 ••• n _ \3 ••• i + 2 - - - i + 4 ••• n

and Xi = (Ui)_1 for all i € {1,..., n — 2}. By Proposition 1, it is easy to verify that Ui as well as Xi, i € {1,...,??. — 2}, belong to IOF%ar. In [15], the authors have shown that {vi,v2, ■■■,vn,ui,U'2, ...,un-2,xi,X2, ...,xn-2} is a generating set of IOF%ar. In order to use Theorem 1, we define an alphabet

Xn = {V1 ,V2, ...,Vn ,_1,_2, _n-2 ,X1,X2, ...,Xn-2},

which corresponds to the set of generators of IOFnpar. For w = w1...wm with w1,...,wm € Xn, where m being a positive integer, we write w-1 for the word w-1 = wm...w1.

We fix a particular sequence of letters as follows: x^- = xixi+2...xi+2j-2 and _i,j = __i+2...ui+2j-2 for i € {1,...,n — 2}, j € {1,|_(n — i)/2j} and obtain the following sets of words:

Wx = {Xij : i € {1,...,n - 2}, j ^ 1,...,

W-1 = {x-j : Xij € Wx), Wu ={ Uij : i € {1,...,n - 2}, j € {1,...,

n — i 2

n- i

2

Let w be any word of the form w = w\...wm with w\,..., wm € Wx U Wu and m is a positive integer. For k € {1,..., m}, the word wk is of the form

wk i uik ,jk if wk € Wu; k \xik j if wk € Wx

for some ik € {1, ...,n-2}, jk € {1,..., |_(n — i)/2j}. We observe jk = |wk |, i.e. jk is the length of the word wk. We define two sequences 1x, 2x,..., mx and 1u, 2u,..., mu of indicators: for k € {1,..., m} let

kx -

and

ku =

ik + 2KI +2|Wuk|- 2| Wx| if wk € Wu; ik if Wk € Wx

ik + 2|wk|- 2|Wk| +2|Wk| if Wk € Wx; ik if Wk € Wu,

where WS (respectively, W^) means the word ws+i...wm without the letters in {xi,...,xn_2} respectively, in {ui,..., un_2}) for s € {0,1,...,m — 1} and W^1 = Wm = e, where e is the empty word. Let Q0 be the set of all words w = wi...wm with wi,...,wm € Wx U Wu and m being a positive integer such that:

(1q) If wk, wi € Wx then ik + 2jk + 1 <il for k < l < m;

(2q) If wk, wl € Wu then ik + 2jk + 1 <il for k < l < m;

(3q) If wk € Wu then ik + 2jk + 2 < (k + 1)u for k € {1,..., m — 1} and (k + 1)x — kx > 2;

(4q) If wk € Wx then ik + 2jk + 2 < (k + 1)x for k € {1,..., m — 1} and (k + 1)u — ku > 2.

Let now w = wi...wm € Q0 and let w* = Wu)(W,0) . Further, we define recursively a set Aw:

(5q) If mu > mx and mu + 2 < n then Am = {mu + 2,..., n}, if mu < mx and mx + 2 < n then Am = {mx + 2,..., n}, otherwise Am = 0;

(6q) If wk € Wu then Ak = Ak+i U {ik + 2jk + 2,..., (k + 1)u — 1} for k € {1,..., m — 1}, if wk € Wx then Ak = Ak+1 U {ku + 2,..., (k + 1)u — 1)} for k € {1,..., m — 1};

(7q) If 1 € {1x, 1u} then Aw = Ai,

if 1 < 1u < 1x then Aw = A1 U {1,..., 1u — 1},

if 1 < 1x < 1u then Aw = Ai U {1u — 1x + 1,..., 1u — 1}.

For a set A = {¿1 < i2 < ■ ■ ■ < ik} C n, let va = v^v^.-.v^ for some k € {1, ...,n}. Note that means the empty word e. For convenience, we put Vi = e for i > n + 1. Let

Wn = {vAw* :w eQ0,AC Aw} U {vA : A C n}.

On the other hand, we will define now a set of relations. For this, let Wt be the set of all words of the form ui0uil...uilXjx...XjmXjm+1 with the following four properties:

(i) l € {0,..., n — 2}, and m € {0, ...,n — 3};

(ii) io < ii < ■ ■ ■ <ii € {1,..., n — 2};

(iii) ji > j2 > ■ ■ ■ > jm > jm+1 € {1, ..., n — 2};

(iv) if k € {i0,..., ii_i} (respectively, k € {j2, ...,jm+i}) then k + 1,k + 3 / {ii, k + 1,k + 3 / {ji,..,jm}) for all k € {1,..., n — 3}.

let

ii} (respectively,

(E) Xiuj

itions on Xn as follows: for i, j € {1,... ,

viv2vi+3...vj+3 , if i < j, j — i = 2, 3

viv2vj+3...vi+3 , if i > j, i — j = 2, 3

viv2vj+3vj+4, if i > j, i — j = 1;

viv2vj+2vj+3, if i < j, j — i = 1;

viv2vi+3, if i = j;

viv2V,j xi+2, if i < j, j — i > 4;

yiv2V,j+2xi, if i > j, i — j > 4;

22 ~ u2 ~ x2 ~ viv2v3v4v5;

(L1) u2ui « uiu2 « xix2 « x2xi

(L2) u3u2 ^ x2x3 ^ viv2v3v4v5v6;

(L3) uiui ~ viv2ui and xixi ~ v3v4xi, i > 3;

(L4) uiu2 « viv2v3ui and x2xi « v3v4v5xi, i > 4;

(L5) uiui_i & vi+3ui_3ui_i and xi_ixi « vi+3xi_ixi_3, i > 4;

(L6) uiuj & uj_2ui, and xjxi & xj,xj_2, i > j > 3,i — j > 2;

(R1) v2 & vi, i € {1, ...,n};

(R2) ViVj & Vjvi, i,j € {1,n}, i = j;

(R3) v-iUj & Ujv,i and vj,Xj & Xjvi, i € {j + 4, ...,n};

(R4) viUj & Ujvi+2 and vi+2Xj & Xjvi; 1 < i < j;

(R5) viUj & Uj and Xjvi & Xj, i € {j + 1, j + 2, j + 3};

(R6) Ujvi & Uj and vj,Xj & Xj, i € {1, 2, j + 3};

(R7) u\ & x\ & vi...v4;

(R8) u2 & ui-2ui and x2 & xixi-2, i > 3;

(R9) uiui+1 & ui-1ui+1 and xi+1xi & xi+1xi-1, i € {2, ...,n — 5};

(R10) (R11)

(R12)

(R13)

(R14) (R15) (R16) (R17) (R18) (R19)

UiUi+3 & vi+6UiUi+2 and Xi+3Xi & vi+6Xi+2Xi, i < n — 5;

W & vio + 1vio+2vio + 3Uil Xjl ...Xjm , w = Ui0 Uil ...Ui; Xjl .. .X jm Xjm + i € Wt

with jm+1 = io + 21 — 2m;

w & vio vio+1vio+2vio+3Uii ...Uil Xji ...Xjm , W = Uio Uil ...Uil Xjl ...Xjm Xjra+i € Wt with jm+1 = i0 + 21 — 2m — 1;

w & vio+1vio+2vio+3vio+4Uii ...UilXji ...Xjm , W = UioUii ...UilXji ...XjmXjm+i € Wt with jm+1 = i0 + 21 — 2m + 1;

w & UioUii ...UilXji ...Xjm , w — UioUii ...UiiXji ...XjmXjm +i € Wt with jm+1 < 21 — 2m;

w & Uii ...Uil Xji ...Xjm Xjm+i, w = Uio Uii ...Uil Xji ...Xjm Xjm+i € Wt with io < 2m — 21; v1...viUi,j & v!...vfc+3, i € {1, ...,n — 2}; vfc-i+3...vfc+2X--71 & v!...vfc+3, i € {1,..., n — 2};

viU

i i,j

vfc+2X—j1 & vfc+3Xi-11)j, i € {2, ...,n — 2}.

vfc+3Ui-1,j, i € {2, ...,n — 2};

-1

Lemma 1. The relations frow, R hold as equations in IOFp , when the letters are replaced by the corresponding transformations.

Proof. We show the statement diagrammatically. This method was also used in [4, 5]. We give an example calculation for the relation (R10) uiui+3 & vi+6uiui+2, i < n — 5, in Figures 1 and 2 below. Note we can show Xi+3 Xi && vi+6Xi+2Xi in a similar way. □

By Figures 1 and 2, we have that 3 = ~h+&UiUi+2-

i + 4

...

i + 7

x

+7

i + 4

i

i

n

n

iui+ 3

i+3

5

Next, we will verify consequences of R, which are important by technical reasons.

Lemma 2. (i) For w = ui0ui1...UilXj1...XjmXjm+1 € Wt with jm+i = 2l — 2m, we have

W & V\Ui0 Uii ...Uil Xji ...Xjm .

(ii) For w = ui0ui1 ...UilXj1 ...XjmXjm+1 €Wt with i0 = 2m — 2l, we have

W & Vio + 3Uil ...Uii Xjl ...Xjm Xjm + 1 ■

Proof. (i) We have

(i&4)

Ui0 Ui1 ...Uil Xj1 ...X jm Xjm+1 & Ui0 Ui1 ...Uil Xj1 ...Xjm Xjm + 1-iXjm + 1 .

Suppose jm+i = 2l — 2m > 4. Then

(&)

Ui0 Ui1 ...Uil Xj1 ...Xjm Xjm + 1-iXjm+1 & Uio Ui1 ."Uil Xj1 .-Xjm V jm+1+3X jm+1-iX jm + 1-3 (R4) (Ri4)

& ViUio Ui1 ...Uil Xj1 ... Xjm Xjm+1-iX jm + 1-3 & ViUio Ui1 .-Uil Xj1 .-Xjm .

Suppose jm+i = 2l — 2m < 4, i.e. jm+i = 2. We prove that

Ui0 Ui1 ...Uil Xj1 ...Xjm X jm+1 & ViUio Ui1 ...Uil Xj1 ...Xjm

by using (L1) and (R4)-(R6) in a similar way.

(ii) The proof is similar to (i), by using (R15) and (L5) if io > 4 and (R15), (L1), and (R4)-(R6) if i0 = 2. □

3. Set of forms

In this section, we introduce an algorithm, which transforms any word w € X*n to a word in Wn using R, with other words, we show that for all w € X, there is w' € Wn such that w & w' is a consequence of R. First, the algorithm transforms each w € X*n to a "new" word w'. All these "new" words will be collected in a set. Later, we show that this set belongs to Wn. Let w € X*n\{e}.

• Using (R1)-(R6), we can move any Vi for i € {1,2, ...,n}, at the beginning of the word or we can cancel it. So we obtain w & Vw, where V € {v1, ...,Vn}* and w € {ui,u2,..., un-2,Xi,X2,

• Moreover, we separate the ui's and Xi's for i € {1, ...,n — 2} by (E) and (R1)-(R6). Then vw ttvBC, where v € {vi,..., vn}*, B € {ui, u2, •••, un-2}*, and C € {a?i, x2,..., xn-2}*.

• By (L1)-(L6) and {R1)-{R6), we get TiBC w v'B'C, where v' € {vu..., vn}*, B' € {ui,u2, ...,un-2}*, and C' € {x1,x2, ...,Xn-2}* such that the indices of the letters in the word B' are ascending and in the word C' are descending (reading from the left to the right).

• By (L1), (R7)-(R10), and (R1)-(R6), we replace subwords of B'C' of the form xi+3xi,xi+1xi,x2i,u2i,uiui+3, and uui+i until v'B'C' & v''wi...wp with v'' € {vi,...,vn}* and wi, ...,wp € W-i U Wu such that

if u € var(wi...wp) (respectively, Xj, € var(wi...wp)) then ui+i,ui+3 / var(wi...wp) (respectively, xi+i,xi+3 / var(wi...wp)) for all i € {1, ...,n — 2} and each letter in wi...wp is unique. (*)

Note that this is possible since each of the relations (L1), (R7)-(R10), and (R1)-(R6) does not increase the index of any letter in {ui,u2, ...,un-2,x1,x2, ...,xn-2} in the "new" word.

• Using (R11)-(R15), Lemmas 2, and (R1)-(R6), we remove letters xi and ui, respectively, until one can not more remove a letter xi or ui for i € {1,2, ...,n — 2}. We obtain v''w1...wp & v'''w[ ...w'p, where v"' € {v1, ...,vn}* and w', ...,w' p € W,-1 U Wu. Note that is possible since each of the relations (R11)-(R15) as well as Lemmas 2 only removes letters (and add letters in {v1,...,vn}, respectively).

• We decrease the indices of the letters in {u1,u2, ...,un-2,x1,x2, ...,xn-2} (if possible) by (R16)-(R19) as well as (R1)-(R6) and obtain v'''w1 ...wV & v*B*C* with v* € {vb ..., vn}*, B* € {u1,u2, ...,un-2}*, and C* € {x1, x2, ..., xn-2}*. Note that the indices of the letters in B* (respectively, in C*) are ascending (respectively, are descending).

We repeat all steps. The procedure terminates if the word will not change more in all steps. We obtain v*B*C* ~ VAW\...Wp, where wi,...,wp € W~l U Wu and A Cn such that no Vj (j € A) can be canceled by using (R1)-(R6). This case has to happen since the number of the letters from {u1,u2 , ...,un-2,x1,x2, ...,xn-2 ,v1,...,vn} decreases or is kept and the indices of the ui's and xi's decrease or are kept in each step.

We denote by P the set of all words obtained from w € Xn* by that algorithm. By (*), we obtain immediately from the algorithm.

Remark 1. Let w = vAwi1...uim € P and let 1 < k < k' < m. If wk,wk' € Wu then ik + 2|wfc| +2 < ik'. If wk,wk' € Wx then ik' + 2|wk' | + 2 < ik.

Let fix a word w = vAwi1...uim € P. There are a,b € {0, ...,n} with a + b = m, t1,...,ta+b € {1,..., m}, wti ,...,wta € Wu and

wta+i , ...,wta+b € Wx such that

-1 -1

VJ = vAw1...wm = vAwti ...wta wta+i ...wta+b ,

where {wti, ...,wta} = 0 or {wta+i, ...,wta+b} = 0 (i.e. a = 0 or b = 0) is possible. We observe that {u)1,...,u)m} = {wti ,...,wta ,w-a+i ,...,w-a+b} and {¿1, ...,ta,ta+1, ...,ta+b} = {1,...,m}. We define an order on {t1, ...,ta,ta+1, ...,ta+b} by t1 < ■ ■ ■ <ta and ta+b < ■ ■ ■ < ta+1. If a,b > 1, the order between t1, ...,ta and ta+1, ...,ta+b is given by the following rule:

Let k € {1, ... , a} and 1 € {1, ... , b}

if itk + 2|wtfc 1 — 2 + 2|wtfc+i ...wta 1 — 2|w-a + i ...w-a + i-i 1 < ^a + l + 2|w- + i 1 — 2 then tk < ta+l and

if itk +2|wtfc 1 — 2+2|wtfc+i ...wta 1 — 2 1 w—+ i ...w-a + i_i 1 >ita + l + 2|w-a + i 1 — 2 then tk > ta+l .

The case

itk +2|wtfc | — 2 + 2|wtfc + i ...wta | — 21 w—j+i ...w—a + i_i | = ita + l +2|w—a + i | — 2

is not possible, since otherwise we can cancel uitk +2|wtfc|-2 and xit +2|w-i |-2 in w by (R11). Our

next aim is to describe the relationships between ku, (k + 1)u and kx, (k + 1)x for all k € {1, ...,m — 1} for the word w = w1...wm.

Lemma 3. For all k € {1,..., m — 1}, we have ku < (k + 1)u and kx < (k + 1)x.

Proof. Let k € {1,..., m — 1}. Suppose wk, wk+1 € Wu. We obtain ku < (k + 1)u and

kx = ik + 2|wk | +2| Wk | — 2| Wk |,

(k + 1)x = ik+1 + 2|wk+1| + 2|Wuk+1| — 2|Wxk+1|.

By Remark 1, we have ik + 2|wk| +2 < ik+i. This gives

ik + 2|wfc| + 2|Wuk| — 2|WXI < ik+i + 2|Wk| — 2|WXI = ik+i + 2K+i| + 2|WX+i| — 2|WX+i|

(since wk+i € Wu implies 2|WX| = 2|WX+i ). Then kx < (k + 1)x. For the case wk, wk+i € Wx, we can show that ku < (k + 1)u and kx < (k + 1)x in a similar way.

Suppose wk € Wu and wk+i € Wx. First, we will show ku < (k + 1)u. We have ku = ik and

(k + 1)u = ik+i + 2|wk+i I + 2|Wxk+i| — 2|Wuk+i|.

Since k € {ti,..., ta} and k + 1 € {ta+i,..., ta+b}, we obtain

ik + 2|wk| — 2 + 2|Wuk| — 2|WX+1| < ik+i + 2|wk+i|— 2.

Then

ik <ik + 2|wk| < ik+i + 2|wk+i| + 2|Wxk+i| — 2|Wuk+i|

(since wk+i € Wx implies |WX| = |WX+i|). Then ku < (k + 1)u. Moreover, we prove kx < (k + 1)x similarly. The case wk € Wx and wk+i € Wu can be shown in a similar way as above. □

Of course, the next goal should be the proof of w = wi...wm € Q0, i.e. we will show that w satisfies (1q)-(4q).

Lemma 4. We have w = wi...wm € Q0■

Proof. Exactly, w satisfies (1q) and (2q). This is trivially checked by Remark 1. Let k € {1,...,m — 1} and let wk € Wu,wk+i € Wx. This provides k € {ti,...,ta}, k + 1 € {ta+i, ...,ta+b}. We have

ik + 2|wk| — 2 + 2|Wuk| — 2|Wxk+i| < ik+i + 2|wk+i| — 2.

Since wk+i € Wx, we have

2|Wuk | =2|WX+i|.

So

ik + 2|wk| — 2 + 2|WX+i| — 2|Wxk+i| < ik+i + 2|wk+i| — 2.

We observe that

ik + 2|wk| — 2 + 2|WX+i| — 2|Wxk+i| + 1 < ik+i + 2|wk+i| — 2.

If

ik + 2|wk | — 2 + 2|WX+i| — 2|Wxk+i| + 1 = ik+i + 2|wk+i| — 2,

we can cancel uik+2ywki-2,xik+1+2yWk+1i-2 by (R13) in w. This contradicts w € P. Then ik + 2|wk| — 2 + 2|WX+i| — 2|Wxk+i| + 2 < ik+i + 2|wk+i| — 2,

i.e.

ik + 2|wk| + 2 < ik+i + 2|wk+i| — 2|WX+i| + 2|Wxk+i| = (k + 1)u.

Next, to show that (k + 1)x — kx > 2. Lemma 3 gives (k + 1) X kx > 1. If (k + 1)x — kx = 1 then

ik+i — ik — 2|wk | — 2|WX | + 2|Wxk | = 1.

This implies

ik+i +2|wk+i| — 2 = ik + 2^ |— 2 + 21 W£ | — 2|Wxk+i| + 1

since

2|Wxk | =2|Wk+i| +2|Wxk+i|.

We can cancel Uik+2[wk\-2,Xik+1+2iwk+1l-2 in w by (R13). This contradicts w € P. Thus, (k + 1)x — kx > 2. In case wk,wk+i € Wu, by using Remark 1, we easily get

ik + 2^| +2 < (k + 1)u.

To show (k + 1) x — kx > 2, it is routine to calculate directly. Together with Remark 1, we will get that (k + 1)x — kx > 2. Altogether, w satisfies (3q). We prove that w satisfies (4q) in a similar way. Therefore, w € Q0. □

We have shown w € Q0. This leads us to the next step, showing that A C Aw. First, we point out subsets of n, which do not contain any element of A.

Lemma 5. Let q € {1,..., a} and let

p € {itq + 1, ...,kq + 2|wtJ + 1} ilii.

Then p / A.

Proof. Assume p € A. Then

-i -i (R3) -i -i

^ +_ ...w i - - - - - i - i

ta+1

If p € {k + 1, itq + 2, it + 3} n n then

Vpwt1 ...wtq .. wta wta + 1 .. wta + b & wt1 .. .VP wtq ...wta wta+1 . ..wta + b ■

(R5)

VPUitq & Uitq .

If p = itq + h + t for some h € {2, 4,..., 2|wtq | — 2} and t € {2, 3} then

-i -i -i -i

wt1 ...Vpwtq ...wta wta+1 . ..wta + b = wt1 ...V PUitq Uitq +2^% +2|wtq-2^+1 ta wta +1 ."wta + b

(R3) -1 -1

& wt1 ... Uitq ...V(itq +h+t)Uitq +h..Uitq +2|wtq-2^+1 ...wta wta + 1 ta + b

(R5) -1 -1

& wt1 ... Uitq ...Uitq +h ...Uitq +2 ^q | -2wtq+ 1 ...wta wta+1 "wta + b ,

i.e. we can cancel vp in w using (R3) and (R5), a contradiction. □

Lemma 6. Let p € A and let q € {1,..., a} such that tq = m. If p € {(tq)u + 1,..., (tq + 1)u — 1}

P € {(tq )u + 2|wtq | +2,..., (tq + 1)u — 1} C Aw.

Proof. We have (tq)u = itq. It is a consequence of Lemma 5 that

p € {itq +2|wtq | +2,..., (tq + 1)u — 1}

and by (6q), we have

{itq +2|wtq | +2,..., (tq + 1)u — ^ C Aw

□

Lemma 7. Let p € A, if ta = m and p € {im + 1, ...,n} then p € {mx + 2,..., n} Ç Aw

Proof. Assume p € {im + 1, ...,mx + 1}. We have mx + 1 = ita + 2|Wta | + 1. Then p € {ita + 1,..., ita + 2|wta | + 1}. By Lemma 5, we have p / A. Therefore, p € {mx + 2,..., n} Ç Aw by (5q). □

Lemma 8. Let p € A, then p = (ta+ )u + 1 for all l € {1,..., b}.

Proof. Let l € {1, ...,b}. Assume p = (ta+l)u + 1. Suppose that there exists q € {1, ...,a} with tq > ta+i. Then

-1 -1 (R3) -1 -1

VpWt! ...Wtq ...Wta wta+1 ...wta + b W Wtl -VPWtq ^Wta Wta + 1 -Wta + b (R4) -1 -1

W Wtl ...Wtq ...Wta vp+2\wtq ...wta | W-a + 1 -W-a + b .

Since we have

Suppose tq < ta+l for all q € {1,..., a}. Then we have

(ta+l)U + 1 = Ua + l + 2 1 W-a+1 ...Wta + 1 1 - 2 1 Wtq-Wta 1 +1,

p + 2|Wtq ...Wta 1 = ita + l +2|W-a + i 1 +1.

(ta+l )u + 1 = ita + l + 21 W-a + l ...W-a + i | + 1,

i.e.

-1 -1 (R3) -1 -1 vpwh .. .Wtq ...Wta Wta+1. ..Wta+b « Wt! .. .Wtq .. .Wta vpwta+1. ..Wta+b .

Both cases imply

-1 -1

Wt1 ...Wtq ...Wta vita + i +2K- + 1 ...<+i| + 1Wta+1 ...Wta + b

(R4) -1 -1 -1 (R6) -1 -1

« Wtl ...Wtq ...Wta W- ...v■ .0, -1 ,W, ...Wf , « Wtl ...Wtq ...Wta W- ...W- ,...Wf , ,

t1 ''q ta ta+1 ita + l +2|Wta + l | + 1 ta + l ta + b t1 lq ta ta+1 ta + l ta + b '

i.e. we can cancel vp in W using (R3), (R4), and (R6), a contradiction. □

Lemma 9. Let p€A and let l€{1,...,b} such that ta+l=m. If p€{(ta+l )u+1,..., (ta+l+1)u—1} then

P € {(ta+l)u + 2,..., (ta+l + 1)u — 1} C Aw.

Proof. It is a consequence of Lemma 8 that p € {(ta+l)u + 2,..., (ta+l + 1)u — 1} and by (6q), we have {(ta+l )u + 2, ..., (ta+l + 1)u — 1} C Aw. □

Lemma 10. Let p € A. If ta+1 = m and p € {mu + 1,..., n} then p € {mu + 2,..., n} C Aw.

Proof. Suppose p = mu + 1 = (ta+1)u + 1. By Lemma 8, we have p / A. Therefore, p € {mu + 2,...,n}C Aw by (5q). □

Lemma 11. If 1 < 1x < 1u then p / A for all p € {1,..., 1u — 1x}. Proof. Let p € {1,..., 1u — 1x}. Assume p € A. We observe that

1u - 1x = 2lw—+b...w—+11 — 2|wtl...Wta 1 = 2k

for some positive integer k. We put U = wtl...wta and X = w—1+b ...w-a+1, i.e. 2k = 2|X| — 2|U| and |X| = |U| + k. Let a+ a

Wta + 1 ...Wta + b = Vl...VlUlVlUI + 1...VlUI+k,

where yi, ...,VUI+k € {xi,..., £„-2}. Then

(R4)

Vpwti...wtaVi...V|u|V|u|+i...V|u|+k ~ wti...wtavp+2lwt1 ...wta|Vi...V|u|V|u|+i...V|u|+k.

Using Remark 1, it is routine to calculate that

2|w-1t...w-1, | < iT h1 + 2|w-1,

1 Ta + b Ta+11 la+l I ia + il'

i.e.

(1u - 1x) +2|WT1 ...wta I = 2|w-+b ...w-h | < ita+1 +2|w-+i |.

P +2|wti .. .WTa 1 < iTa + 1 +2|w-a+i |.

This implies

p Wta\ ^ Ha + 1 I -I ^ta+1

Then

(R4)

wti ...Wtavp+2\wt1 ...wta\Vi--y\u\V\u\+i--y\u\+k ~ Wtl ...Wtayi--y\u\Vpy\u\+i--y\u\+k■

Note that 1u — 1x is even and there is i € {2,4,..., 1u — 1x} such that p G {i — 1, i}. If p = i — 1 then

P — 21 V\U\+I—V\U\+î/2— 11 = 1.

If p = i then

P — 2|y \ u\+l...y \ U \ +i/2— i1 =2.

Thus,

wti ...wtayi...y\u\vpy\u\+i...y\u\+k

(R4)

~ wti ...wta yl...y \U \y\U\+l...Vp—2 \ y\u\ + i-y\u\+H2-l \y \U \ +i/2...y\U\+(lu —lx )/2

= wti ...wta yl...y \ u \ y \ u\+l...Vpy \ u \ +i/2...y \ U\+(lu—lx )/2

(where p G {1, 2})

(R6)

« wti...wtayl...y\u\y\u\+l...y\u\+i/2...y\U\+(lu—lx)/2, i.e. we can cancel vp in w using (R4) and (R6), a contradiction. □

Lemma 12. Let p € A with p € {1,..., 1u - 1}. If 1 < 1u < 1x then p € {1,..., 1u - 1} C Aw and if 1 < 1x < 1u then p € {1u — 1x + 1,..., 1u — 1} C Aw.

Proof. If 1 < 1u < 1x then {1,..., 1u — 1} C Aw by (7q). If 1 < 1x < 1u, it is a consequence

of Lemma 11 that p € {1u — 1x + 1,..., 1u — 1} and by (7q), we have {1u — 1x + 1,..., 1u — 1} C Aw.

□

Lemma 13. We have (tq)u / A for all q € {1,..., a}. Proof. Let q € {1,..., a}. We have

Wtq = Uitq Uitq + 2^Uitq + 2|wtq |-2

and (tq)u = itq. Assume (tq)u € A. If itq > 2 then

-1 -1 (R3) -1 -1 vitq Wt1 . . .Wtq ...Wta Wta + 1 . ..Wta + b « Wt1 . ..vitq Uhq Uitq +2 .. Uitq +2| wtq |-2Wtq+1 ...Wta Wta+1 . ^t^b

(R18) -1 -1

« Wt1 ...vitq +2|wtq | +1 Uitq-1Uitq + 1 -^'itq +2^tq |-3Wtq+1 ..Wta Wta+1 .-Wta + b .

If itq = 1 then q = 1 and

-1 -1 -1 -1

vit1 Wt1 Wt2 ...Wta Wta + 1 ...Wta + b = v1U1U3...U1+2\wtl |-2Wt2.-Wta Wta+1 .-Wta + b

(R16) -1 -1

« v1v2 ...V1+2|wt1 | + 1Wt2 ...Wta Wta + 1 ...Wta + b .

We observe that we can replace several letters in W by letters with decreasing index by (R18) and the letters u1,u3, ...,u1+2|wt11-2 were canceled in W by (R16), respectively, a contradiction. □

Lemma 14. We have (ta+l)u / A for all l € {1,..., b}.

Proof. Let l € {1, ...,b}. Now assume that (ta+l)u € A. We will have the following two cases. In the first case, we suppose that there exists q € {1,..., a} with tq > ta+l and, of course, for the trivial second case is supposed tq < ta+l for all q € {1, ...,a}. Using (R3) and (R4) in the first case and (R4) in the second case, together with a few tedious calculations, both cases imply

-1 -1 -1 -1

v(t ,) wti ...wtq...wt„ w— ...w— , & wti ...wt„v- ,0i -i -i w ...w— ,.

(ta + l)u ti <'q ta ta+i ta + b ti ta ita + l +2|wt +_ ...wt | ta+i ta + b

ta+l ta+1 ta+l

It is routine to calculate that

-1 -1 (R4) -1 -1 -1 Wt1 ...Wta vit+ +2|w-a1+1 ...W-a1+i |Wta+1 ...Wta + b « W1 "Wta Wta+1 ^it^ +2^+ |Wta + lta + b"

If it +l + 2|w-1 | > 3 then

ta+l ta+l

-1 -1 -1 Wtl ...Wta W- , ...v■ ,ol„„-1 |W- ...W- ,

t1 ta ta+1 it„ + l +2|wt | ta + l ta + b

a+l a+l

_ -1 -1 -1 = Wt1 ...Wta Wta + 1 ...vita + l + 2^+ | Xita + l +2K, + i M^a + l +2|wta + l |-4...Xita + l Wta + l + 1 "^ta + b

(R19) -1 -1 -1

« Wt1 ...Wta Wta+1 ...vita + i +2|w-a1+l | + 1Xita + l +2|wta + l |-3Xita + l +2|wta + l |-5^Xita + l-1Wta + l + 1 a + b '

If it +, + 2IW-1 | = 3 then Wt-1 = X1. Thus,

ta + l ' I ta + l ta + b 1 '

-1 -1

Wt1 ...Wta vita + t +2|w-a1+1 ...W-+ |^a+1 "Wa + b (R4) -1 -1 (R17) -1 -1

« w'1 ...Wta Wta + 1 ...Wta + b-1 v3 X1 « Wt1 -Wta Wta + 1 -Wta + b-1 v1v2v3v4.

We observe that we can replace several letters in W by letters with decreasing index by (R19) and the letter x1 can be canceled in W by (R17), respectively, a contradiction. □

If we summarize the previous lemmas, then we obtain:

Lemma 15. We have A C Aw.

Proof. Let p € A. Then it is easy to verify that p € {1,..., 1u} or p € {ku + 1,..., (k + 1)u} for some k € {1,...., m — 1} or p € {mu + 1,..., n}. Suppose that p € {ku + 1,..., (k + 1)u — 1} for some k € {1, ...,m — 1}. Lemmas 13 and 14 show that ku / A. Then we can conclude that p € Aw by Lemmas 6 and 9. Suppose p € {mu + 1, ...,n}. Then we can conclude that p € Aw by Lemmas 7 and 10. Finally, we suppose that p € {1,..., 1u — 1}. Then we can conclude that p € Aw by Lemma 12. Eventually, we have p € Aw for all p € A. Therefore, A C Aw. □

Lemmas 4 and 15 prove that w = VAwii...uim € Wn. Consequently, we have: Proposition 2. P C Wn.

By the definition of the set P and Proposition 2, it is proved: Corollary 1. Let w € X,,. Then there is w' € P C Wn with w ~ w .

4. A presentation for IOFpar

In this section, we exhibit a presentation for IOFpar. Concerning the results from the previous sections, it remains to show that |Wn| < |IOF,,ar|. For this, we construct a word wa, for all a € IOFnar, in the following way. Let

a = (di < d2 < ••• < dA €lOFpar\{e} \mi m2 • • • mpJ

for a positive integer p < n. There are a unique l € {0,1, ...,p — 1} and a unique set {ri,..., rl} C {1, ...,p — 1} such that (i)-(iii) are satisfied:

(i) ri < ... < ri;

(ii) dn+i — dn = mn+i — mn for i € {1,..., l};

(iii) di+i — di = mi+i — mi for i € {1, ...,p — 1}\{ri, ...,ri}.

Note that l = 0 means {ri,..., ri} = 0. Further, we put ri+i = p. For i € {1,..., l}, we define

wi =

xmri ,((mr.+i-m,ri )-(dr. + i-dri ))/2 if mn+i — mn > dn+i — dn; Udr. ,((dri+i-dri)-(mri + i-mri))/2 if mri+i — mn < dn+i — dn.

Obviously, we have w-i € WxU Wu for all i € {1,..., l}. If mp = dp then we put wi+i = e. If mp = dp, we define additionally

w K(d,-mp)/2 if dp >mp;

wi+i =

[udp,(mp-dp)/2 if dp < mp.

Clearly, wi+i € Wx U Wu. We consider the word

w = wi...wi+i.

From this word, we construct a new word wa by arranging the subwords s € Wx in reverse order at the end, replacing s by s-i. In other words, we consider the word

w*a = wsi ...wsa w-ai+i .. .w-a + b

such that Ws1 ,---,w.Sa € Wu, WSa+1 ,'",WSa + b € Wx and

{Ws-i,••• , wsa , WSa+1 , •••, Wsa + b } — {w1, •••, Wa+b },

where si < ... < sa, sa+b < ••• < s<i+i, and a,b Gn U {0} with

\l if dp — mp; a + b — <

I l + 1 if dp — mp •

For convenience, a — 0 means wt — w_\, •••w-\ and b — 0 means wt — ws- •••wSa. Now, we add

' a sa+1 sa + b a s1 Sa 1

recursively letters from the set {v1, •••,vn} C Xn to the word w*, obtaining new words A0, X1, •••, Xp.

(1) For dp < n - 2:

(1.1) if mp < dp then Ao — Vdp+2---Vnw*a;

(1.2) if n - 1 > mp > dp then Ao — Vmp^^Vriwa;

(1.3) if mp — dp then Ao — vmp+1:Vnw*a;

otherwise A0 — w*

(2) If dp — mp — n — 1 then A0 — vnwO*. Otherwise A0 — w,

,p — Hip — Ib — ± unen /\(J — VnUJa- vymcinioc /\o — uja-

(3) For k € {2, •••,p}:

(3.1) if 2 < mk — mk-1 — dk — dfc_1 then Ap_fc+1 — vdk_1+l•••vdk-1\p-k;

(3.2) if 2 <mk — mk_1 < dk — dk_1 then Ap_k+1 — vdk_1Ap_k;

(3.3) if mk — mk_1 > dk — dk_1 > 2 then Ap_k+1 — Vdk_1+2•••Vdk_1Ap;

otherwise Ap_k+1 — Ap_k.

(4) If d1 — 1 or m1 — 1 then Ap — Ap_1.

(5) If 1 < d1 < m1 then Ap — v1•••vdl_1Ap_1.

(6) If 1 <m1 <d1 then Ap — Vdl_ml+l•••Vdl_1Ap_1.

The word \p induces a set A = {a € n : va is a letter in Ap} and it is easy to verify that p £ A for all p € dom(a). We put wa — Ap. The word wa has the form wa — vaw*

Our next aim is to present the relationship between cardinality of Wn and IOFnar. This leads us to assume the existence of a map / : IOF%ar\{e} —>• Wn\{vn}, where f(a) = wa for all a € IOFnar\{e}. We start by constructing the transformation aVAW* for any vaw* € Wn, different from vjt■ Let vaw* € Wn\{vn}. We have w € Qo,A C Aw, and there are w\, ...,wm € Wu U Wx such that w — wi^^wm for some positive integer m. For k € {^••^m}, we define ak — ku + 2 and bk — ik + 2jk + 2, whenever wk € Wx. On the other hand, we define ak — ik + 2jk + 2 and bk — kx + 2, whenever wk € Wu. It is easy to verify that am — bm. We put

1 + 1u — min{1u, 1x}-1u a1_2u ••• am_1^:mu am-n

a'VA'W* 1 + 1x — min{1u, 1x}-1x b1^2x ••• bm_l•••mx b

For convenience, we also give

d1 d,2 ■ ■ ■ dp

•••n

a

VAWr

p

m1 m2 ■ ■ ■ mp

for some positive integer p < n. In the following, we show that aVAW* is well-defined in the sense that the construction of aVAw* gives a transformation.

Lemma 16. aVAW* is well-defined.

■i

Proof. Let k € {1,..., m — 1}. Suppose wk,wk+1 € Wu. We have

ku = ik, kx = ik + 2|wfc | + 2| Wk | — 2| Wk |, (k + 1)u = ik+1, (k + 1)x = ik+1 + 2|wk+i| + 2|Wuk+:| — 2|Wak+1|,

and ak = ik + 2jk + 2,bk = kx + 2. Then

(k + 1)u — ak = ik+i — (ik + 2jk + 2), (k + 1)x — bk = ik+i + 2|wk+i| + 2|Wuk+1| — 2|Wxk+1| — kx — 2 = ik+1 + 2|wk+1| + 2|Wuk+1| — 2| Wxk+1| — ik — 2|wk | — 2lWk | + 2|Wk | — 2 = ik+1 — ik — 2jk — 2 = ik+1 — (ik + 2jk + 2).

Therefore, (k + 1)u — ak = (k + 1)x — bk.

For the rest cases (wk € Wu and wk+1 € Wx, wk € Wx and wk+1 € Wu as well as wk,wk+1 € Wx), a proof similar as above will eventually show that (k + 1)u — ak = (k + 1)x — bk. Furthermore, suppose dp = mp. Let k € {1, ...,m} and wk € Wu. We have

ak — ku = ik + 2jk + 2 — ku = ik + 2jk + 2 — ik = 2jk + 2, bk kx — kx I 2 kx — 2.

Thus, ak — ku = bk — kx.

For the case wk € Wx, we can show ak — ku = bk — kx in the same way.

Continuously, suppose dp = mp. By the previous part of the proof, we have ak — ku = bk — kx for all k € {1, ...,m — 1}. Moreover, we observe that dp / {am, ...,n} and mp / {bm, ...,n} because n — am = n — bm. This implies dp = mu and mp = mx. By any of the above, we can conclude that aVAw* is well-defined. □

The proof of Lemma 16 shows (k + 1)u — ak = (k + 1)x — bk for all k € {1,..., m — 1}. Then ak — ku = bk — kx for all k € {1,...,m}, whenever dp = mp, and ak — ku = bk — kx for all k € {1, ...,m — 1} and dp = mu,mp = mx, whenever dp = mp. Furthermore, observing by trivial calculation, ak — ku > 2 and bk — kx > 2. Therefore, if there exists i € {1,...,p — 1}, where di+1 — di = mi+1 — mi, then di € {1u,..., (m — 1)u}(U{mu}), mi € {1x,..., (m — 1)x}(U{mx}) and we put ku = drk,kx = mrfc for all k € {1, ...,m — 1}(U{m}) (we put rm = p, whenever dp = mp). This gives the unique set {r1,...,rm} as required by the definition of wav . Moreover, we need to show that aVAw* € IOFnar\{e} by checking (i)-(iv) of Proposition 1. We will now show that aVAw* € IOFnar as well as wa = v^w*. This gives the tools to calculate that |Wn| < IIOFpar|.

Lemma 17. avAw * € IOFpar\{e}.

Proof. Clearly, aVAw* = e. We will prove that aVAw* satisfies the conditions (i)-(iv) in Proposition 1. We observe that d1 < d2 < ■ ■ ■ < dp and m1 < m2 < ■ ■ ■ < mp by definition of aVAw*. We have 1u — d1 = 1x — m1, i.e. 1u — 1x = d1 — m1. By the definition of ku and kx, for k € {1, ...,m}, we observe that 1u — 1x is even, i.e. d1 — m1 is even. Thus, d1 and m1 have the same parity.

Let di+1 — dj, = 1 for some i € {1, ...,p — 1}. Then dj, € dom(a)\{1u,..., mu} implies mi+1 — m-i = di+1 — di = 1.

Let mi+1 — mi = 1 for some i € {1,...,p — 1}. Then mi € im(a)\{1x,..., mx} implies di+1 — di = mi+1 — mi = 1.

Let di+1 — di is even. Suppose di+1 — di = mi+1 — mi. This gives di = ku and mi = kx for some k € {1,..., m — 1}. By the definition of ku and kx, we observe that ku — kx is even.

Moreover, (k + 1)u — di+1 = (k + 1)x — mi+1 since (k + 1)u — (k + 1)x is even, we have di+1 — mi+1 is even. Then di+1, di and di, mi as well as di+1,mi+1 have the same parity. This implies that mi+1,mi have the same parity, i.e. mi+1 — mi is even. Conversely, we can prove similarly that, if mi+1 — mi is even then di+1 — di is even. By Proposition 1, we get aVAW* € IOFnar. □

We can construct f (aVAW*) = wa^w*, where wa^w* = vAw*a ^ with w = wi...wm for wi, ...,wm € Wu U Wx and Ac.n. We will prove that / is surject.ive in the next lemma.

Lemma 18. Let vaw* € Wn\{t>7i}. Then there is a € IOF%ar\{e} with vaw* = wa.

Proof. We have wa * = v¿Wl *, where W = Wi...Wm with Wi, ...,Wm € Wu U Wx and

VAw A aVAw* ' ' '

A C n. First, our goal is to show that w = w. Suppose dp = mv and let k € {1, ...,m} such that bk - kx > ak - ku. By the definition of Wk, we have Wk = xkx,((bk-kx)-(ak-kn))/2 and kx = ik. Then

(bk - kx) - (a.fc - ku) _ ik + 2jk + 2 - ik - ku - 2 + ku _ . 2 ~ 2 ~

i.e. Wk = Xik,jk = Wk. For the case bk — kx < ak — ku, we can prove that Wk = Wk in a similar way. This gives Wi...Wm = Wi...Wm.

Suppose dp = mp. We have ak — ku = bk — kx for all k € {1,...,m — 1} and by a similar proof as above, we have Wi...Wm-i = Wi...Wm-i. If mp < dp then Wm = xmp¿dp_mp)/2 and mp = mx = im. Then

dp "mp mu mx im + 2jm im

2 ~~ 2 ~~ 2 ~~ •?m'

i.e. Wm = ximjm = Wm. For the case mp > dp, we can prove Wm = Wm in a similar way. Thus, Wi...Wm-iWm = Wi...Wm-iWm. Then w = W, i.e. w* = W*n *. The next goal is to show that

' aVAw*

A = A.

1) To show that A C A: let a € A. We have A C Aw since vaw* € Wn. Therefore, we have the following cases: a € {am,..., n} = Ai or a € {ak,..., (k + 1)u — 1} = A2 for some k € {1,..., m — 1} or

a € {1 + 1u — min{1u, 1x},..., 1u — 1} = A3.

If a € Ai and mp = dp then a € A since (1.1) and (1.2), respectively. If a € Ai and a € {dp + 1, ...,n} with mp = dp then a € A since (1.3) and (2), respectively.

Suppose a € A2 with a € {ak,..., drk+i — 1}. If 2 < drk+i — drk < mrk+i — mrk then wk € Wx. Note that ak = ku + 2 = drk + 2. Thus, a € A since (3.3). If 2 < mrk+i — mrk < drk+i — drk then Wk € Wu. Note

drk+i — ak = mrk+i — bk, bk = kx + 2, ak = ak — bk + bk = drk+i — mrk+i + kx + 2 = drk+i — mrk+i + mrk + 2.

Thus, a € A since (3.2).

Suppose a € A3. If 1 < di < mi and a € {1,..., di — 1} then a € A since (5). If 1 < mi < di and a € {di — mi + 1,..., 1u — 1} then a € A since (6) (note that 1u — 1x = di — mi).

Suppose a € Ai U A2 U A3 and there exists s € {2, ...,p} such that ds — ds-i = ms — ms-i > 2 with a € {ds-i + 1,..., ds — 1}. Then a € A since (3.1). By any of the above, we have A C A.

2) To show that A C A: let

A1 — {1 + 1u — min{1u, 1x}, •••, 1u — 1}, A2 — {a1, •••, 2u — 1} U {a2, •••, 3u — 1} U ••• U {am_1, •••, mv, — 1},

A3 — {am, -^n}-

Because A C Aw, we have A C A1 U A2 U A3 and A n {d1, ...,dp} — 0. This implies A C A1 U A2 U A3\{d1, •••, dp}. Conversely, we have A1 U A2 U A3\{d1, •••, dp} C A by the definition of aVAW*. Thus, A — A1 U A2 U A3\{d1, •••, dp}.

Let a € A. By the definition of A, we can observe that a — di for all i € {1, •••,p}. Suppose a is given by (1.1) or (1.2) or (1.3) or (2). Then a € A3\{d1, •••,dp}. Suppose a is given by (3.1). Then a € A1 U A2 U A3\{d1, •••, dp}.

Suppose a is given by (3.2), i.e. a € {ds — ms + ms_1 + 2, •••, ds — 1} for some s € {2, •••,p}. We have already shown that there is k € {1, •••,m — 1} such that ds — ms + ms_1 + 2 — ak. Then a € A2\{d1, •••,dp}.

Suppose a is given by (3.3). Then a € A2\{d1, •••, dp}. Suppose a is given by (5). Then a € A1\{d1, •••,dp}.

Suppose a is given by (6). Then a € A1\{d1, •••, dp} (note that d1 — m1 — 1u — 1x). Therefore, we have a € A, i.e. A C A.

By 1) and 2), we get A — A. This implies vaw* — vph* — wa . □

Lemma 18 establishes that f is surjective, which implies |Wn| < IIOFn,ar|. We will now adjust our alphabet and relations to meet the requirements of Theorem 1. As mentioned previously, Xn = {s : s € Xn} is a generating set for the monoid IOF%ar. Building on the insights from Lemma 1, we can conclude that Xn satisfies all the relations from R = {si s2 : si s2 € R}.

Corollary 1 further shows that for any w € Xn, there exists a corresponding w' € Wn, for which w ~ w' is a consequence of R. This implies that R C Xn x Xn and that Wn C Xn meet the conditions 1-3 in Theorem 1. We now possess all the necessary items to conclude our main result.

Theorem 2. (Xn | R) is a monoid presentation for IOF%ar.

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