text:
The Battle for D.A. Is Testing Philadelphia's Commitment to Reform
Larry Krasner ended an era of tough-on-crime policies in the DA's office and sparked a nationwide movement. Now voters will decide whether to continue on this path.
Maura Ewing, | May 17, 2021
A protest in Philadelphia in 2020 (Joe Piette/Flickr)
Maura Ewing,
This article originally appeared on The Appeal, which hosted The Political Report project.
The opposition campaign against Philadelphia District Attorney Larry Krasner, fueled heavily by the local police union and a group of former officers, has intensified in the lead-up to Tuesday's primary election. "Our officers have given us carte blanche to spend whatever we need to spend to be able to remove this cancer from the District Attorney's Office," local Fraternal Order of Police president John McNesby told the Philadelphia Inquirer.
Billboards along I-95 call for Krasner to be fired, and earlier this month an FOP-sponsored ice cream truck stationed outside the DA's office offered soft-serve as a reminder to voters of Krasner's "soft on crime" policies. An accompanying truck was adorned with an anti-Krasner poster.
"The FOP is spending a lot of money and throwing their weight around," said Reuben Jones, a prominent Philadelphia activist who is formerly incarcerated. "The ice cream thing sounds so ridiculous but that's the stuff that gets people's attention."
Since taking office in 2018, Krasner—a firebrand not shy of the spotlight—has become a national darling for criminal justice reform. One term isn't enough time to completely overturn a deeply entrenched system, but Krasner has made strides. As a result of his policies, the size of the court-supervised population shrank by more than a third, the number of people who spent at least one night in jail declined by 22 percent over the first year of his cash bail reform, some immigrants have more protections from ICE, and his conviction integrity unit exonerated 20 wrongly convicted prisoners.
This is a sharp turn away from prior administrations. David Rudovsky, a longtime civil and criminal defense lawyer in Philadelphia, said in an email that a succession of Philadelphia chief prosecutors' policies over decades "led to mass incarceration, an insistence on the death penalty in a wide range of cases, a war on drugs that was counter-productive and resulted in high racial disparities of those prosecuted and sentenced, and a pattern of defending almost every conviction regardless of evidence of innocence or violations of the rights of the person convicted."
On Tuesday, Philadelphia voters will decide if they want to continue on the new trajectory that Krasner initiated—primaries in this deeply blue city are typically de facto elections. And the pressure is high; this election is widely seen as a referendum on progressive prosecutors in Philadelphia and nationwide.
The outcome is "critically important" to the national movement for criminal justice reform, says Jamila Hodge, director of the Reshaping Prosecution Program at the Vera Institute of Justice. "DA Krasner is probably one of the most well-known names when it comes to the progressive prosecutorial movement," she said. Should he lose, "It could have a chilling effect on whether or not a person in another jurisdiction decides to run, to challenge the status quo." Essentially, Krasner is so recognizable that his defeat could slow the momentum of the progressive prosecutor movement.
Locally, though, activists are emphasizing the issues at stake over the symbolism of Krasner's campaign.
"As much as it's about Larry, it's not about Larry," said Sean West Damon, an organizer with Free the Ballot, a social justice alliance that supports Krasner. "It's a referendum on the politics of mass incarceration that have led Philadelphia to being one of the most incarcerated cities in the country."
Krasner's bid for re-election comes at a time when the decades-long decline in violent crime in America has been upended by a confluence of factors, including the COVID-19 pandemic, that defies simple explanation. Murder rates spiked in cities across the country. Last year in Philadelphia, 499 people were murdered, a number 40 percent higher than the year before and the highest since 1990. The number of people shot rose by 53 percent in a single year according to the Philadelphia Center for Gun Violence Reporting, and are up 83 percent since 2017, the year before Krasner took office. There is no evidence that Krasner's policies caused the spike in violence, but that may not stop some voters from retreating to the status quo of a punitive, tough-on-crime approach.
During his campaign, Vega has expressed general support for criminal justice reform while asserting that Krasner's policies have harmed public safety. In a debate this month, Vega blamed the bloodshed in recent years on Krasner's low conviction rates for gun cases and his cash bail policies that have led to fewer people in jail.
But he told the Philadelphia Inquirer's editorial board: "I'm not going to reverse any policies." Later, in a tweet, Vega said "that does not mean I don't have my own policies in plans."
The Appeal has reported on policies of Krasner's office that Vega said he would reverse. In an interview with The Appeal: Political Report, he said he would resume charging for drug possession, except marijuana, and rely on traditional diversion courts, whereas Krasner's policy is to drop possession charges if a person attends just one addiction treatment meeting. And Vega said he would not cap probation and parole sentences, a Krasner policy that has drastically reduced the number of people under supervision.
Vega spent most of his career as a prosecutor under the administration of Lynne Abraham, whose hardline tactics and penchant for seeking the death penalty earned her the moniker the "Deadliest DA." Abraham served from 1991 to 2010.
Recently, he has come under fire for his involvement in the civil case for exoneree Anthony Wright, who spent 25 years in prison for a rape and murder he did not commit. Vega was one of two assistant DAs who retried Wright even after DNA evidence cleared him. Throughout his campaign, Vega has distanced himself from the case, saying that he came in at the 11th hour. The Innocence Project rebuked Vega's claim that he had minor involvement.
Rudovsky, who was part of the team that handled the case, said he believes that Vega's participation in the Wright retrial is "reflective of what you would get if he was DA. … Which is, I think, a return to the old regime."
Many decarceral advocates share that view and are campaigning for Krasner to make sure that their hard-fought progress isn't rolled back. Krasner has faced criticism from the left for not going further to reduce the use of cash bail and for continuing to charge some teens as adults—though he does this less frequently than his predecessors.
"I don't want to pretend like everything worked out the way we envisioned," said Jones. "But the one thing I know is that without Krasner in office, we wouldn't have made the progress that we made … the only way we can continue to work is to get him re-elected."
bail reform Larry Krasner Mass incarceration Parole reform Pennsylvania Philadelphia


text:
div.panel_wrapper {
    height: 315px;
    overflow-y: auto;
}

div.panel fieldset {
    margin-bottom: 15px;
}

#images_panel ul.images_list {
    list-style-type: none;
    margin: 0;
    padding: 0;
}

#images_panel ul.images_list li {
    float: left;
    position: relative;
    width: 100px;
    height: 100px;
    margin: 10px;
    padding: 2px;
}

#documents_panel ul.documents_list li {
    padding: 5px 0;
}

#images_panel ul.images_list li a.rm {
    position: absolute;
    top: 0;
    right: 0;
    display: block;
    padding: 0 2px;
    font-size: 12px;
    font-weight: bold;
    text-decoration: none;
    background: #FF6466;
    color: white;
}

#documents_panel ul.documents_list li a.rm {
    display: inline-block;
    padding: 0 2px;
    margin-left: 5px;
    font-size: 11px;
    font-weight: bold;
    text-decoration: none;
    background: #FF6466;
    color: white;
}

.pager a {
    display: inline-block;
    padding: 3px 5px;
    text-decoration: none;
}

.pager a.disabled {
    color: #ccc;
    cursor: default;
}

redpajama_set_name: RedPajamaGithub


text:
\section{Padded multifraction reduction}\label{SSplitRed}

After recalling in Subsection~\ref{SSRed} the definitions that we need for multifraction reduction and the rewrite system~$\RD\MM$, we introduce in Subsection~\ref{SS:PaddedSC} padded versions of semi-convergence and use them to solve the word problem of the enveloping group. Next, we introduce in Subsection~\ref{SSSplitRed} a new rewrite system~$\SD\MM$, a variant of $\RD\MM$ called split reduction, and we connect its semi-convergence with the padded semi-convergence of~$\RD\MM$. Finally, we establish the connection with subword reversing and Property~$\PropH$ in Subsection~\ref{SSPropH}.

\subsection{Multifraction reduction}\label{SSRed}

If $\MM$ is a monoid, we denote by~$\EG\MM$ the enveloping group of~$\MM$, and by~$\can$ the canonical (not necessarily injective) morphism from~$\MM$ to~$\EG\MM$. We say that a finite sequence $\aav = (\aa_1 \wdots \aa_\nn)$ of elements of~$\MM$, also called a \emph{multifraction} on~$\MM$ and denoted by $\aa_1 \sdots \aa_\nn$, \emph{represents} an element~$\gg$ of~$\EG\MM$ if
\begin{equation}\label{E:AltProd}
\smash{\gg = \can(\aa_1) \can(\aa_2)\inv \can(\aa_3) \pdots \can(\aa_\nn)^{(-1)^{\nn - 1}}}
\end{equation}
holds in~$\EG\MM$. In this context, the parameter~$\nn$ is called the \emph{depth} of~$\aav$, denoted by~$\dh\aav$, and the right hand side of~\eqref{E:AltProd} is denoted by~$\can(\aav)$. We use $\FR\MM$ for the family of all multifractions on~$\MM$, and $\One\nn$ for the depth~$\nn$ multifraction with all entries equal to $1$, skipping~$\nn$ when no ambiguity is possible. Our aim is to recognise which multifractions represent~$1$ in~$\EG\MM$.

We collect together the definitions that we need concerning monoids.

\begin{defi}
A monoid~$\MM$ is called a \emph{gcd-monoid} if it is cancellative, $1$ is its only invertible element, and any two elements of~$\MM$ admit a left- and a right-gcd, where, for~$\aa, \bb$ in~$\MM$, we say that $\aa$ \emph{left-divides}~$\bb$, written $\aa \dive \bb$, if we have $\bb = \aa\xx$ for some~$\xx$ in~$\MM$, and that $\dd$ is a \emph{left-gcd} of~$\aa$ and~$\bb$ if $\dd$ left-divides~$\aa$ and~$\bb$ and every common left-divisor of~$\aa$ and~$\bb$ left-divides~$\dd$. Right-division and right-gcd are defined symmetrically, with $\bb = \xx\aa$ replacing $\bb = \aa\xx$.

An \emph{atom} in~$\MM$ is an element that is not expressible as $\aa\bb$ with $\aa, \bb \not=1$. A right- (\resp left-) \emph{basic element} is one that is obtained from atoms using the operation~$\under$ (\resp $/$) defined by $\aa (\aa {\under} \bb) = \aa \lcm \bb$ (\resp $(\aa / \bb)\bb = \aa \lcmt \bb$).

The monoid~$\MM$ is called \emph{noetherian} if there is no infinite descending sequence with respect to proper left- or right-division, and \emph{strongly noetherian} if there exists a map $\lambda$ from $\MM \setminus \{1\}$ to the positive integers satisfying $\lambda(\aa\bb) \ge \lambda(\aa) + \lambda(\bb)$.
\end{defi}

Throughout the paper, we restrict our attention to gcd-monoids: Artin--Tits monoids are typical examples. In a gcd-monoid, any two elements~$\aa, \bb$ that admit a common right-multiple(\resp left-multiple) admit a least one, the \emph{right-lcm} (\resp \emph{left-lcm}) of~$\aa$ and~$\bb$, denoted by~$\aa \lcm \bb$ (\resp $\aa \lcmt \bb$).

If $\MM$ is a gcd-monoid, a family~$\RD\MM$ of rewrite rules on~$\FR\MM$ is defined in~\cite{Dit}: for $\aav, \bbv$ in~$\FR\MM$, and for $\ii \ge 1$ and $\xx \in \MM$, we write $\aav \act \Red{\ii, \xx} = \bbv$ if we have $\dh\bbv = \dh\aav$, $\bb_\kk = \aa_\kk$ for $\kk \not= \ii - 1, \ii, \ii + 1$, and there exists~$\xx'$ satisfying
$$\begin{array}{lccc}
\text{for $\ii$ even:}
&\bb_{\ii-1} = \aa_{\ii-1} \xx',
&\xx \bb_\ii = \aa_\ii \xx' = \xx \lcm \aa_\ii,
&\xx \bb_{\ii+1} = \aa_{\ii+1},\\
\text{for $\ii \ge 3$ odd:\qquad}
&\bb_{\ii-1} = \xx' \aa_{\ii-1},
&\bb_\ii \xx = \xx' \aa_\ii = \xx \lcmt \aa_\ii,
&\bb_{\ii+1} \xx = \aa_{\ii+1},\\
\text{for $\ii = 1$:}
&&\bb_\ii \xx = \aa_\ii,
&\bb_{\ii+1} \xx = \aa_{\ii+1}.
\end{array}$$
We say that $\aav$ \emph{reduces} to~$\bbv$ in one step, and write $\aav \rd \bbv$, if $\aav \act \Red{\ii, \xx} = \bbv$ holds for some~$\ii$ and some $\xx \not= 1$. We use $\rds$ for the reflexive--transitive closure of~$\rd$.

By~\cite[Lemma~3.8 and Cor.~3.20]{Dit}, $\aav \rds \bbv$ implies that $\aav$ and~$\bbv$ represent the same element in~$\EG\MM$, and, conversely, the relation~$\can(\aav) = \can(\bbv)$ is essentially the equivalence relation generated by~$\rd$ (up to deleting trivial final entries). Furthermore, whenever the monoid~$\MM$ is noetherian, $\RRR$-reduction is terminating for~$\MM$.

By~\cite[Prop.~4.16]{Dit} and~\cite[Prop.~3.2]{Diu}, $\RRR$-reduction is locally confluent for~$\MM$ if and only if $\MM$ satisfies the \emph{$3$-Ore condition}, meaning that any three elements of~$\MM$ that pairwise admit a common right- (\resp left-) multiple admit a global one. In this case, $\RRR$-reduction is convergent, implying that a multifraction~$\aav$ represents~$1$ in~$\EG\MM$ if and only if $\aav \rds \one$ is satisfied. In the case of Artin--Tits monoids, this exactly corresponds to type~FC.

When $\MM$ does not satisfy the $3$-Ore condition, $\RRR$-reduction is not confluent for~$\MM$, and no convergence result can be expected. However, a condition weaker than convergence can be used to solve the word problem:

\begin{defi}[\cite{Diu}]
We say that $\RRR$-reduction is \emph{semi-convergent for~$\MM$} if
$\aav \rds\nobreak \one$ holds for every multifraction~$\aav$ that
represents~$1$ in~$\EG\MM$.
\end{defi}

Convergence implies semi-convergence, but explicit examples of gcd-monoids~$\MM$ for which $\RRR$-reduction is semi-convergent but not convergent are known~\cite{Div}. However:

\begin{prop}[{\cite[Prop.~3.16]{Diu}}]\label{P:WP}
 If $\MM$ is a strongly noetherian gcd-monoid with finitely many basic elements for which $\RRR$-reduction is semi-convergent, then the word problem of~$\EG\MM$ is decidable.
\end{prop}

\begin{conjA}[\cite{Diu}]
$\RRR$-reduction is semi-convergent for every Artin--Tits monoid.
\end{conjA}

Conjecture~$\ConjA$ is true for every Artin--Tits monoid of FC~type, but other cases remain open. Every Artin--Tits monoid satisfies the finiteness assumptions of Proposition~\ref{P:WP} and, therefore, the word problem of the enveloping group of every Artin--Tits monoid satisfying Conjecture~$\ConjA$ is decidable.

\subsection{Padded semi-convergence}\label{SS:PaddedSC}

We now introduce a weakening of semi-conver\-gence, in which trivial initial entries may be added to multifractions. For $\pp$ a nonnegative integer and $\aav$ a multifraction, we write $\One{2\pp} / \aav$ for the multifraction obtained from~$\aav$ by adding $2\pp$~trivial entries on the left of~$\aav$. Note that the number of~$1$s that are added must be even in order to preserve the image of the multifraction in the group.

\begin{defi}\label{D:InflSC}
If $\ff$ is a map from~$\FR\MM$ to~$\NNNN$, we say that $\RRR$-reduction is semi-convergent for~$\MM$ \emph{up to $\ff$-padding} if $\One{2\ff(\aav)} / \aav \rds\nobreak \one$ holds for every
multifraction~$\aav$ that represents~$1$ in~$\EG\MM$.
\end{defi}

By definition, semi-convergence is semi-convergence up to $0$-padding. Because $\aav \rds\nobreak \bbv$ trivially implies $\One{2\pp} /\aav \rds \One{2\pp} / \bbv$ for every~$\pp$, semi-convergence of $\RRR$-reduction up to $\ff$-padding implies its semi-convergence up to $\ff'$-padding for~$\ff' \ge \ff$ (meaning $\forall \aav {\in} \FR\MM\,(\ff'(\aav) \ge \ff(\aav))$): appending trivial initial entries preserves the existing reductions and possibly adds new ones.

\begin{prop}\label{P:PaddedSC2WP}
If $\MM$ is a strongly noetherian gcd-monoid with finitely many basic elements for which $\RRR$-reduction is semi-convergent up to $\ff$-padding for some Turing-computable map~$\ff$ on~$\FR\MM$, then the word problem of~$\EG\MM$ is decidable.
\end{prop}

Speaking of a Turing-computable map defined on~$\FR\MM$ makes sense because, under the assumptions of Proposition~\ref{P:PaddedSC2WP}, $\MM$ admits a finite presentation in terms of the finite atom family~$\SS$ of~$\MM$, and every element of the monoid~$\MM$ is represented by finitely many words in~$\SS$ only, which implies the decidability of the word problem of~$\MM$ with respect to~$\SS$. Then a Turing-computable map on~$\MM$ means one that is induced by a Turing-computable map on words in~$\SS$.

If $(\SS, \RR)$ is a monoid presentation, that is, $\RR$ is a family of pairs of words in~$\SS$, we denote by~$\equivp_\RR$, or simply~$\equivp$, the congruence on~$\SS^*$ generated by~$\RR$, implying $\MON\SS\RR = \SS^*{/}{\equivp}$. For~$\ww$ a word in~$\SS$, we denote by~$\clp\ww$ the element of~$\MON\SS\RR$ represented by~$\ww$, that is, the $\equivp$-class of~$\ww$. Let~$\SSb$ be a disjoint copy of~$\SS$ that consists of a new element~$\INV\ss$ for each element~$\ss$ of~$\SS$. For~$\ww$ a word in $\SS \cup \SSb$, we denote by~$\INV\ww$ the word obtained from~$\ww$ by exchanging~$\ss$ and~$\INV\ss$ everywhere and reversing the order of letters. In this context, the element of~$\SS$ (\resp $\SSb$) are called positive (\resp negative), and we denote by~$\cl\ww$ the element of the group~$\GR\SS\RR$ represented by~$\ww$ when $\INV\ss$ represents~$\ss\inv$.

\begin{proof}[Proof of Proposition~\ref{P:PaddedSC2WP}]
The argument is the same as the one in~\cite{Diu} for Proposition~\ref{P:WP}. Let $\SS$ be the atom set of~$\MM$, and let $\ww$ be a word in $\SS \cup \SSb$. To decide whether $\ww$ represents~$1$ in~$\EG\MM$, we first express it as $\ww_1 \ww_2\inv \ww_3 \pdots$ with $\ww_1, \ww_2$, ... in~$\SS^*$ (no negative letters). Then, writing $\underline\ww = (\ww_1, \ww_2, ...)$ and $\clp{\underline\ww}$ for $\clp{\ww_1} / \clp{\ww_2} / \pdots$, we append~2$\ff(\clp{\underline\ww})$ trivial entries on the left, and exhaustively enumerate all sequences~$\underline{\ww}'$ satisfying $\One{2\ff(\clp{\underline\ww})} / \clp{\underline{\ww}} \rds \clp{\underline{\ww}'}$: this is possible because, under the assumptions of the statement, the existence of common multiples and, from there, the relation~$\rds$, are decidable. Then $\ww$ represents~$1$ in~$\EG\MM$ if and only if a trivial sequence $\ew \sdots \ew$ appears in the list, where $\ew$ is the empty word.
\end{proof}

We are thus led to the new conjecture, Conjecture~$\ConjAp$, stated in the introduction. Conjecture~$\ConjAp$ is \textit{a priori} weaker than Conjecture~$\ConjA$ but, by Proposition~\ref{P:PaddedSC2WP}, it nevertheless implies the decidability of the word problem of every Artin--Tits group for which it is true.

\subsection{Split reduction}\label{SSSplitRed}

We now connect the padded version of semi-conver\-gence of $\RRR$-reduction with the ordinary semi-convergence of a new rewrite system, a variant of $\RRR$-reduction that can naturally be called \emph{split reduction}.

In the definition of multifraction reduction used in \cite{Dit,Diu,Div} and repeated in Section~\ref{SS:PaddedSC}, in order for $\aav \act \Red{\ii, \xx}$ to be defined, $\xx$ and~$\aa_\ii$ have to admit a common multiple, a left- or right-multiple depending on the parity of~$\ii$. In order to define a variant of this reduction, we now relax the assumption and only require that $\xx$ and some divisor of~$\aa_\ii$ admit a common multiple; then, we again remove~$\xx$ from~$\aa_{\ii + 1}$ and push it to the left or right side using the lcm operation, this time through the divisor of~$\aa_\ii$ that is involved and not necessarily through the whole of~$\aa_\ii$.

\begin{defi}
(See Figure~\ref{F:Sharp} right.) Assume that $\MM$ is a gcd-monoid. For $\aav, \bbv$ in~$\FR\MM$, and for $\ii \ge 1$ even (\resp odd) and $\xx, \yy$ in~$\MM$ such that $\xx \lcm \yy$ (\resp $\xx \lcmt \yy$) exists, we write $\aav \act \Redbrk{\ii, \xx, \yy} = \bbv$ if we have $\dh\bbv = \dh\aav + 2$, $\bb_\kk = \aa_\kk$ for $\kk < \ii$, $\bb_\kk = \aa_{\kk - 2}$ for $\kk > \ii + 3$, and
$$\begin{array}{lccc}
\text{for $\ii$ even:}
&\yy \bb_\ii= \aa_\ii,
&\yy \bb_{\ii + 1} = \xx \bb_{\ii + 2} = \xx \lcm \yy,
&\xx \bb_{\ii+3} = \aa_{\ii+1},\\
\text{for $\ii$ odd:\qquad}
&\bb_\ii \yy= \aa_\ii,
&\bb_{\ii + 1} \yy = \bb_{\ii + 2} \xx = \xx \lcmt \yy,
&\bb_{\ii+3} \xx = \aa_{\ii+1}.
\end{array}$$
In addition we define ``trimming'' rules $\Redtrim\ii$, such that $\aav \act \Redtrim\ii = \bbv$ holds if we have $\aa_{\ii + 1} = 1$, $\dh\bbv = \dh\aav - 2$, and $\bb_\kk = \aa_\kk$ for $\kk < \ii$, $\bb_\kk = \aa_{\kk + 2}$ for $\kk > \ii + 1$, and $\bb_\ii = \aa_\ii \aa_{\ii + 2}$ (\resp $\aa_{\ii + 2} \aa_\ii$) if $\ii$ is odd (\resp even).
We write $\aav \rdsp \bbv$ if either $\aav \act \Redbrk{\ii,\xx,\yy} = \bbv$ holds for some~$\ii, \xx, \yy$ with $\xx\yy \not= 1$ or $\aav \act \Redtrim\ii = \bbv$ holds for some $\ii$, and denote by $\rdsp^*$ the reflexive--transitive closure of~$\rdsp$. The rewrite system~$\SD\MM$ so obtained is called \emph{split reduction} on~$\MM$.
\end{defi}

\begin{figure}[htb]
\begin{center}
\begin{picture}(50,29)(0,1)
\psset{nodesep=0.7mm}
\psline[style=back,linecolor=color2]{c-c}(0,0.5)(9.5,0.5)(9.5,20)(39.5,20)(39.5,30.5)(50,30.5)
\psline[style=back,linecolor=color1]{c-c}(0,0)(10,0)(10,10)(40,10)(40,30)(50,30)
\put(-5,0){...}
\pcline{->}(0,0)(10,0)
\pcline{<-}(10,0)(10,10)\trput{$\aa_{\ii - 1}$}
\pcline{<-}(10,10)(10,20)\trput{$\xx'$}
\psline[style=thin](9,1)(8,1)(8,20)(9,20)\put(1,10){$\bb_{\ii - 1}$}
\pcline{->}(10,10)(40,10)\tbput{$\aa_\ii$}
\pcline{->}(10,20)(40,20)\taput{$\bb_\ii$}
\psarc[style=thin](10,20){3}{270}{360}
\pcline[linewidth=1.5pt,linecolor=color3,arrowsize=1.5mm]{<-}(40,10)(40,20)\tlput{$\xx$}
\pcline{<-}(40,20)(40,30)\tlput{$\bb_{\ii + 1}$}
\pcline{->}(40,30)(50,30)
\put(52,30){...}
\psline[style=thin](41,10)(42,10)(42,29)(41,29)\put(43,20){$\aa_{\ii + 1}$}
\put(23,14){$\Leftarrow$}
\end{picture}
\hspace{5mm}
\begin{picture}(50,29)(0,1)
\psset{nodesep=0.7mm}
\psline[style=back,linecolor=color2]{c-c}(0,0.5)(9.5,0.5)(9.5,10.5)(25,10.5)(25,20)(39.5,20)(39.5,30.5)(50,30.5)
\psline[style=back,linecolor=color1]{c-c}(0,0)(10,0)(10,10)(40,10)(40,30)(50,30)
\put(-5,0){...}
\pcline{->}(0,0)(10,0)
\pcline{<-}(10,0)(10,10)\trput{$\aa_{\ii - 1}$}\tlput{$\bb_{\ii - 1}$}
\pcline{<-}(25,10)(25,20)\put(18,15.5){$\bb_{\ii + 1}$}
\pcline{->}(10,10)(25,10)\put(15,12){$\bb_\ii$}
\pcline[linewidth=1.5pt,linecolor=color3,arrowsize=1.5mm]{->}(25,10)(40,10)\put(31.5,11.5){$\yy$}
\psline[style=thin](11,9)(11,8)(40,8)(40,9)\put(23,5.5){$\aa_\ii$}
\pcline{->}(25,20)(40,20)\taput{$\bb_{\ii+2}$}
\psarc[style=thin](25,20){3}{270}{360}
\pcline[linewidth=1.5pt,linecolor=color3,arrowsize=1.5mm]{<-}(40,10)(40,20)\tlput{$\xx$}
\pcline{<-}(40,20)(40,30)\put(33,25.5){$\bb_{\ii + 3}$}
\pcline{->}(40,30)(50,30)
\put(52,30){...}
\psline[style=thin](41,10)(42,10)(42,29)(41,29)\put(43,20){$\aa_{\ii + 1}$}
\put(31,15){$\Leftarrow$}
\put(28,14){$\mathrm{\scriptscriptstyle part}$}
\end{picture}
\end{center}
\caption{\small Comparing the reduction~$\Red{\ii, \xx}$ (left) and the split reduction~$\Redbrk{\ii, \xx, \yy}$ (right), here for $\ii$ odd: in both cases, we divide~$\aa_{\ii + 1}$ by~$\xx$ and push~$\xx$ to the left using an lcm, but, for~$\Red{\ii, \xx}$, we demand that $\xx$ crosses the whole of~$\aa_\ii$, whereas, for~$\Redbrk{\ii, \xx, \yy}$, we only require that $\xx$ crosses the possibly proper divisor~$\yy$ of~$\aa_\ii$; as a consequence, we cannot gather the elements at level~$\ii - 1$ and the depth increases by~$2$.}
\label{F:Sharp}
\end{figure}

Adding the trimming rules~$\Redtrim\ii$ is a technical artefact that will make statements simpler. It directly follows from the definition that, if $\aav \rdsp^* \bbv$ is true, then $\aav$ and~$\bbv$ represent the same element in the group~$\EG\MM$. A major difference between $\RRR$- and $\SSS$-reductions is that, even if the monoid~$\MM$ is noetherian, no termination can be expected for the latter in general.

\begin{exam}
Let $\MM$ be the Artin--Tits monoid of type~$\Att$, that is,
$$\MON{\tta, \ttb, \ttc}{\tta\ttb\tta = \ttb\tta\ttb, \ttb\ttc\ttb = \ttc\ttb\ttc, \ttc\tta\ttc = \tta\ttc\tta},$$
and let $\aav = \tta\ttb / \ttc$. Then we find $\aav \act \Redbrk{1, \ttc, \ttb} \Redbrk{1, \ttc, \tta} = 1 / \tta\ttc / \ttc\tta / \ttb / \ttc\ttb / 1$, whence
$$\aav \act \Redbrk{1, \ttc, \ttb} \Redbrk{1, \ttc, \tta} \Redbrk{3, \ttb, \tta} \Redbrk{3, \ttb, \ttc} \Redbrk{5, \tta, \ttc} \Redbrk{5, \tta, \ttb} = 1 / \tta\ttc / 1 / \ttc\ttb / 1 / \ttb\tta / \tta\ttb / \ttc/ \tta\ttc / 1 / \ttb\tta / 1 / \ttc\ttb / 1,$$
that is, with obvious notation, $\aav \rdsp^6 1 / \tta\ttc / 1 / \ttc\ttb / 1 / \ttb\tta / \aav/ \tta\ttc / 1 / \ttb\tta / 1 / \ttc\ttb / 1$, leading to an infinite sequence of split reductions from~$\aav$.
\end{exam}

The following result directly follows from the definition; we state it both and as a useful example for practice and for future reference.

\begin{lemm} \label{lem:move_to_left}
If $\MM$ is a gcd-monoid, and $\aav = \pdots/ \aa / 1 / \bb \cc / \pdots $ is a multifraction on~$\MM$ with elements $\aa$, $1$, and~$\bb \cc $ in positions $i-1$, $i$, and $i+1$, then we have
$$\begin{array}{cccl}
\aav \act \Red{\ii , \bb} &=\pdots / \aa / 1 / \bb\cc / \pdots\act \Red{\ii , \bb} &=\pdots/\aa \bb /1/\cc /\pdots &\text{for $i$ even,}\\
\aav \act \Red{\ii , \cc} &=\pdots / \aa / 1 / \bb\cc / \pdots\act \Red{\ii , \cc} &=\pdots /\cc \aa /1/\bb /\pdots &\text{for $i$ odd}.
\end{array}$$
In particular, for $\aa =1$, we find
$$\begin{array}{cccl}
\aav \act \Red{\ii , \bb} &=\pdots / 1 / 1 / \bb\cc / \pdots\act \Red{\ii , \bb} &=\pdots/ \bb /1/\cc /\pdots &\text{for $i$ even,}\\
\aav \act \Red{\ii , \cc} &=\pdots / 1 / 1 / \bb\cc / \pdots\act \Red{\ii , \cc} &=\pdots / \cc /1/\bb /\pdots &\text{for $i$ odd}.
\end{array}$$
\end{lemm}

\begin{proof}
The diagrams provide a proof.
\begin{center}
\begin{picture}(45,7)(0,7)
\psset{nodesep=0.7mm}
\pcline{->}(0,6)(15,0)\tbput{$\aa\bb $}
\pcline{<-}(15,0)(30,0)\tbput{$1$}
\pcline{->}(30,0)(45,6)\tbput{$\cc $}
\pcline{->}(0,6)(15,12)\taput{$\aa$}
\pcline{<-}(15,12)(30,12)\taput{$1$}
\pcline{->}(30,12)(45,6)\taput{$\bb \cc $}
\pcline{->}(15,12)(15,0)\tlput{$\bb $}
\pcline{->}(30,12)(30,0)\trput{$\bb $}
\psarc[style=thin](15,0){3}{0}{90}
\end{picture}
\HS{10}
\begin{picture}(45,7)(0,7)
\psset{nodesep=0.7mm}
\pcline{<-}(0,6)(15,0)\tbput{$\cc\aa$}
\pcline{->}(15,0)(30,0)\tbput{$1$}
\pcline{<-}(30,0)(45,6)\tbput{$\bb $}
\pcline{<-}(0,6)(15,12)\taput{$\aa$}
\pcline{->}(15,12)(30,12)\taput{$1$}
\pcline{<-}(30,12)(45,6)\taput{$\bb \cc $}
\pcline{<-}(15,12)(15,0)\tlput{$\cc $}
\pcline{<-}(30,12)(30,0)\trput{$\cc $}
\psarc[style=thin](15,0){3}{0}{90}
\end{picture}
\vspace{4mm}
\end{center}
\end{proof}

\begin{lemm}\label{L:Part}
Assume that $\MM$ is a gcd-monoid and $\aav, \bbv$ belong to~$\FR\MM$. \\
\null\HS{3}\ITEM1 If $\aav \rds \bbv$ holds, then so does $\aav \rdsp^* \bbv$.\\
\null\HS{3}\ITEM2 If $\aav \rdsp^* \bbv$ holds, then $\One{2\pp} / \aav \rds \bbv / \One{2\qq}$ holds for some~$\pp$ and~$\qq$.
\end{lemm}

\begin{proof}
\ITEM1 For an induction it is sufficient to prove that $\aav \rd \bbv$ implies $\aav \rdsp^* \bbv$. So assume that $\aav \act \Red{\ii, \xx} = \bbv$ with, say, $\ii$~even. Writing $\xx \lcm \aa_\ii = \xx \bb_\ii = \aa_\ii \xx'$ and applying the definition of~$\Redbrk{\ii, \xx, \yy}$, we find
$$\aav \act \Redbrk{\ii, \xx, \aa_\ii} = \aa_1 \sdots \aa_{\ii - 1} /1 / \xx' / \bb_\ii \sdots \bb_\nn = \bb_1 \sdots \bb_{\ii - 2} / \aa_{\ii - 1} /1 / \xx' / \bb_\ii \sdots \bb_\nn.$$
The definition of~$\Red{\ii, \xx}$ gives $\bb_{\ii - 1} = \aa_{\ii - 1}\xx'$, so we deduce
$$\aav \act \Redbrk{\ii, \xx, \aa_\ii} \Redtrim\ii = \bb_1 \sdots \bb_{\ii - 2} / \aa_{\ii - 1}\xx' / \bb_\ii \sdots \bb_\nn = \bbv,$$
whence $\aav \rdsp^2 \bbv$. The argument is similar for $\ii$~odd.

\ITEM2 We prove using induction on~$\mm$ that $\aav \rdsp^\mm \bbv$ implies the existence of~$\pp$ and~$\qq$ satisfying $\One{2\pp} / \aav \rds \bbv / \One{2\qq}$. Put $\nn := \dh\aav$.
The result is straightforward for~$\mm = 0$ with $\pp = \qq = 0$.
Suppose that $\mm = 1$. Suppose first $\aav \act \Redbrk{\ii, \xx, \yy} = \bbv$ with, say, $\ii$~even.
By hypothesis, $\yy$ left-divides~$\aa_\ii$, and we have $\aa_\ii = \yy \bb_\ii$.
Starting from $\One2 /\aav$, we can first use $\ii$~reduction steps to split~$\aa_\ii$ into~$\bb_\ii$ and~$\yy$ by pushing $\aa_1 \wdots \aa_{\ii - 1}$, and~$\bb_\ii$ to the left using Lemma~\ref{lem:move_to_left} repeatedly: precisely, we find
$$\One2 / \aav \act \Red{2, \aa_1}\Red{3, \aa_2} \pdots \Red{\ii, \aa_{\ii - 1}} \Red{\ii + 1, \bb_\ii}= \aa_1 \sdots \aa_{\ii - 1} / \bb_\ii / 1 / \yy / \aa_{\ii + 1} \sdots \aa_\nn,$$
whence $\One2 /\aav \rd^\ii \bb_1 \sdots \bb_\ii / 1 / \yy / \aa_{\ii + 1} / \bb_{\ii + 4} \sdots \bb_{\nn + 2}$. Let $\aav'$ be the latter multifraction. By hypothesis, $\xx$ divides~$\aa_{\ii + 1}$, which is~$\aa'_{\ii + 3}$, and the lcm of~$\xx$ and~$\yy$, that is, of~$\xx$ and~$\aa_{\ii + 2}$, exists, so $\aav'$ is eligible for~$\Red{\ii + 2, \xx}$. Then we obtain
$$\aav' \act \Red{\ii + 2, \xx} = \bb_1 \sdots \bb_\ii / \bb_{\ii + 1} / \bb_{\ii + 2} / \bb_{\ii + 3} / \bb_{\ii + 4} \sdots \bb_{\nn + 2}\bbv,$$
whence $\One2 /\aav \rds \aav' \rd \bbv$, the expected result with $\pp = 1$ and~$\qq = 0$.

Suppose now $\aav \act \Redtrim\ii = \bbv$, still with $\ii$~even. By hypothesis, we have $\aa_{\ii + 1} = 1$, and we find
$$\aav \act \Red{\ii + 1, \aa_{\ii + 2}} = \aa_1 \sdots \aa_{\ii - 1} / \aa_\ii\aa_{\ii + 2} /1 / 1 / \aa_{\ii + 3} \sdots \aa_\nn$$
and from there, by the same argument as above with $\One2 / \aav$, we can push $\aa_{\ii + 3} \wdots \aa_\nn$ to the left, obtaining
$$\aav \act \Red{\ii + 1, \aa_{\ii + 2}} \Red{\ii + 2, \aa_{\ii + 3}} \sdots \Red{\nn - 1, \aa_\nn} = \aa_1 \sdots \aa_{\ii - 1} / \aa_\ii\aa_{\ii + 2} /\aa_{\ii + 3} \sdots \aa_\nn / 1 / 1,$$
which implies $\aav \rds \bbv/ \One2$, the expected result with $\pp = 0$ and~$\qq = 1$. The verifications for $\ii$~odd are similar.

Suppose finally $\mm \ge 2$. Write $\aav \rdsp^{\mm - 1} \ccv \rdsp \bbv$. The induction hypothesis implies the existence of~$\pp$ and~$\qq$ satisfying $\One{2\pp} / \aav \rds \ccv / \One{2\qq}$,
and that of~$\pp'$ and~$\qq'$ satisfying $\One{2\pp'} / \ccv \rds \bbv / \One{2\qq'}$.
As $\RRR$-reduction is compatible with left and right multiplication, $\One{2\pp} / \aav \rds \ccv$ implies $\One{2\pp + 2\pp'} / \aav \rds \One{2\pp'} / \ccv / \One{2\qq}$, and $\One{2\pp'} / \ccv \rds \bbv / \One{2\qq'}$ implies $\One{2\pp'} / \ccv / \One{2\qq} \rds \bbv / \One{2\qq + 2\qq'}$. By transitivity of~$\rds$, we deduce $\One{2\pp + 2\pp'} / \aav \rds \bbv / \One{2\qq + 2\qq'}$ and this is the expected result.
\end{proof}

We deduce a simple connection between $\RRR$- and $\SSS$-reductions to~$\one$:

\begin{prop}\label{P:Part}
If $\MM$ is a gcd-monoid then, for every multifraction~$\aav$ on~$\MM$,
\begin{equation}
\aav \rdsp^* \one \text{\quad is equivalent to \quad} \exists\pp\,(\One{2\pp}/ \aav \rds \one).
\end{equation}
\end{prop}

\begin{proof}
Assume $\aav \rdsp^* \one$, say $\aav \rdsp^* \One\rr$. By Lemma~\ref{L:Part}, there exist~$\pp, \qq$ satisfying $\One{2\pp} / \aav \rds \One{\rr + 2\qq}$, whence $\One{2\pp}/\aav \rds \one$.

Conversely, assume $\One{2\pp}/ \aav \rds \one$. Suppose first $\dh\aav \le 1$. Then $\One{2\pp}/ \aav \rds \one$ is possible only for $\aav = 1$, in which case $\aav \rdsp^* \one$ is trivial. Now suppose $\dh\aav \ge 2$ with $\aav$ not trivial, that is, at least one entry of~$\aav$ is not~$1$. Let $\ii$ be minimal satisfying $\aa_\ii \not= 1$. Then $\aav$ is eligible for~$\Redbrk{\ii, 1, \aa_1}$, and we find $\aav \act \Redbrk{\ii, 1, \aa_\ii} = \One2 / \aav$, whence $\aav \rdsp^* \One{2\pp} / \aav$ in $\pp$~steps. Next, by Lemma~\ref{L:Part} again, $\One{2\pp}/ \aav \rds \one$ implies $\One{2\pp} / \aav \rdsp^* \one$. We deduce $\aav \rdsp^* \one$ by transitivity of~$\rdsp^*$.
\end{proof}

We naturally say that $\SSS$-reduction is semi-convergent for~$\MM$ if $\aav \rdsp^*\nobreak \one$ holds for every multifraction~$\aav$ that represents~$1$ in~$\EG\MM$. Proposition~\ref{P:Part} implies:

\begin{coro}\label{C:Part}
If $\MM$ is a gcd-monoid, then $\SSS$-reduction is semi-convergent for~$\MM$ if and only if $\RRR$-reduction is semi-convergent for~$\MM$ up to padding.
\end{coro}

\subsection{Connection with subword reversing and Property~$\PropH$}\label{SSPropH}

In the absence of a
a solution of the word problem for general Artin--Tits groups, Property~$\PropH$ was proposed~\cite{Dia} as a weaker statement in the direction of such a solution along the line of Dehn's algorithm for hyperbolic groups, and it was proved that Property~$\PropH$ is indeed satisfied for the standard presentation of all Artin--Tits groups of type~FC~\cite{Dib} and of sufficiently large type~\cite{GR}. We now establish a simple connection between Property~$\PropH$ and multifraction reduction:

\begin{prop}\label{P:H}
If $\MM$ is a gcd-monoid and $(\SS, \RR)$ is an lcm-presentation for~$\MM$ (on both sides), then the following are equivalent: \vspace{0.5mm}\\
\null\HS3\ITEM1 $\RRR$-reduction is semi-convergent for~$\MM$ up to padding;\\
\null\HS3\ITEM2 $\SSS$-reduction is semi-convergent for~$\MM$;\\
\null\HS3\ITEM3 Property~$\PropH$ is true for~$(\SS, \RR)$.
\end{prop}

By Corollary~\ref{C:Part}, \ITEM1 and~\ITEM2 in Proposition~\ref{P:H} are equivalent, and we are left with establishing the equivalence of~\ITEM2 and~\ITEM3, which will occupy the rest of the section.

We first recall a few definitions. If $(\SS, \RR)$ is a monoid presentation, and $\ww, \ww'$ are words in $\SS \cup \SSb$, then we say that $\ww'$ is obtained from~$\ww$ by one step of \emph{right $\RR$-reversing} if $\ww'$ is obtained by deleting some length~$2$ factor $\INV\ss \ss$ of~$\ww$ or replacing some length~$2$ factor~$\INV\ss \tt$ of~$\ww$ with~$\vv \INV\uu$ such that $\ss\vv = \tt\uu$ is a relation of~$\RR$. Symmetrically, we say that $\ww'$ is obtained from~$\ww$ by one step of \emph{left $\RR$-reversing} if $\ww'$ is obtained by deleting some length~$2$ factor $\ss \INV\ss$ of~$\ww$ or replacing some length~$2$ factor~$\ss \INV\tt$ of~$\ww$ with~$\INV\vv \uu$ such that $\vv\ss = \uu\tt$ is a relation of~$\RR$.

\begin{defi}\label{D:PropH}
Suppose that $(\SS, \RR)$ is a monoid presentation. If $\ww, \ww'$ are words in $\SS \cup \SSb$, we say that $\ww \spec \ww'$ is true if one can go from~$\ww$ to~$\ww'$ using finitely many \emph{special} transformations of the following types: \ITEM1 replacing a positive factor~$\uu$ of~$\ww$ by~$\uu'$ with $\uu' \equivp \uu$ (``positive equivalence''), \ITEM2 replacing a negative factor~$\INV\uu$ of~$\ww$ by~$\INV{\uu'}$ with $\uu' \equivp \uu$ (``negative equivalence''), \ITEM3 right $\RR$-reversing, \ITEM4 left $\RR$-reversing. We say that $(\SS, \RR)$ satisfies \emph{Property~$\PropH$} if $\cl\ww = 1$ implies $\ww \spec \ew$, where $\ew$ is the empty word.
\end{defi}

All special transformations map a word in $\SS \cup \SSb$ to one that represents the same element in~$\EG\MM$, so $\ww \spec \ew$ always implies $\cl\ww = 1$: Property~$\PropH$ says that this implication is an equivalence. Note that special transformations add no new trivial factor~$\ss\INV\ss$ or~$\INV\ss\ss$, a situation reminiscent of Dehn's algorithm for hyperbolic groups~\cite[Sec.\,1.2]{Dib}.

Let us say that a presentation~$(\SS, \RR)$ of a monoid~$\MM$ is a \emph{right-lcm presentation} if $\RR$ contains one relation for each pair~$(\ss, \tt)$ in~$\SS \times \SS$ such that $\ss$ and~$\tt$ admit a common right-multiple and this relation has the form $\ss\vv = \tt\uu$ where both $\ss\vv$ and $\tt\vv$ represent the right-lcm $\ss \lcm \tt$. A left-lcm presentation is defined symmetrically. By~\cite[Thm.~4.1]{Dfx}, every strongly noetherian gcd-monoid admits a right-lcm presentation and a left-lcm presentation. For instance, the standard presentation of an Artin-Tits monoid is an lcm presentation on both sides. The point here is that, if $(\SS, \RR)$ is a right-lcm presentation, then right-reversing computes the right-lcm in~$\MON\SS\RR$ in the following sense:

\begin{lemm}[{\cite[Prop.~3.6]{Dia}}]\label{L:Lcm}
Assume that $\MM$ is a strongly noetherian gcd-monoid and $(\SS, \RR)$ is a right-lcm presentation for~$\MM$. Then, for all words~$\uu, \vv$  in~$\SS^*$, the elements~$\clp\uu$ and~$\clp\vv$ of~$\MM$ admit a common right-multiple if and only if the right $\RR$-reversing of~$\INV\uu \vv$ leads to  word~$\vv' \INV{\uu'}$ with~$\uu', \vv'$ in~$\SS^*$ and, in this case, one has $\clp{\uu \vv'} = \clp{\vv \uu'} = \clp\uu \lcm \clp\vv$.
\end{lemm}

Of course, if $(\SS, \RR)$ is a left-lcm presentation of~$\MM$, left $\RR$-reversing computes the left-lcm in~$\MM$ in a similar sense.

Our aim from now on is to connect the relation~$\spec$ on words in $\SS \cup \SSb$ with partial reduction of multifractions. The convenient bridge between words and multifractions is as follows:

\begin{defi}\label{def:represents}
Suppose that $\MM = \MON\SS\RR$. A word~$\ww$ in $\SS \cup \SSb$ is said to \emph{represent} a depth~$\nn$ multifraction~$\aav$ on~$\MM$ if there exist words $\ww_1 \wdots \ww_\nn$ in~$\SS^*$ satisfying $\ww = \ww_1 \INV{\ww_2} \ww_3 \INV{\ww_4} \pdots$ and $\clp{\ww_\ii} = \aa_\ii$ for every~$\ii$. We say that $\ww$ \emph{sharply} represents~$\aav$ if, in addition, $\ww_\ii$ is nonempty for $1 < \ii < \nn$.
\end{defi}

A word~$\ww$ in $\SS \cup \SSb$ may represent several multifractions, since the expression of~$\ww$ as $\ww_1 \INV{\ww_2} \ww_3 \INV{\ww_4} \pdots$ need not be unique if no maximality condition is required, but it sharply represents a unique multifraction.

\begin{lemm}\label{L:Connection1}
Assume that $\MM$ is a gcd-monoid with lcm presentation~$(\SS, \RR)$ and $\ww$ is a word in $\SS \cup \SSb$ representing a multifraction~$\aav$. Then $\aav \rdsp \bbv$ implies $\ww \spec \ww'$ for some word~$\ww'$ in $\SS \cup \SSb$ representing~$\bbv$.
\end{lemm}

\begin{proof}
Put $\nn = \dh\aav$, and let $\ww_1 \wdots \ww_\nn$ satisfy $\ww = \ww_1 \INV{\ww_2} \ww_3 \INV{\ww_4} \pdots$ and $\clp{\ww_\ii} = \aa_\ii$ for every~$\ii$. Suppose first $\aav \act \Redtrim{\ii} = \bbv$ with, say, $\ii$~odd. By hypothesis, we have $\aa_\ii = 1$, which requires $\ww_\ii = \ew$, because there is no nontrivial invertible element in~$\MM$, so the empty word must be the only element of its $\equivp$-class. Then we have $\bb_\ii = \aa_\ii \aa_{\ii + 2}$, and $\ww$ also represents~$\bbv$, as witnessed by the decomposition $\ww_1 \wdots \ww_{\ii - 1}, \ww_\ii\ww_{\ii + 2}, \ww_{\ii + 3} \wdots \ww_\nn$. The argument is similar for $\ii$ even, replacing $\ww_\ii\ww_{\ii + 2}$ with $\ww_{\ii + 2}\ww_\ii$.

Suppose now $\aav \act \Redbrk{\ii, \xx, \yy} = \bbv$ with, say, $\ii$~even. The assumption that $\aav \act \Redbrk{\ii, \xx, \yy}$ is defined implies that $\yy$ left-divides~$\aa_\ii$, and $\xx$ left-divides~$\aa_{\ii + 1}$. It follows that there exist words~$\uu, \vv, \uu'', \vv''$ in~$\SS$ satisfying
$$\ww_\ii \equivp \uu\uu'' \text{\quad and \quad} \ww_{\ii + 1} \equivp \vv\vv'' \text{\quad with \quad} \clp\uu = \yy \text{\quad and \quad}\clp\vv = \xx.$$
By Lemma~\ref{L:Lcm}, the assumption that $\xx \lcm \yy$ exists implies that the right-$\RR$-reversing of~$\INV\uu \vv$ leads to some -word~$\vv' \INV{\uu'}$ with~$\uu', \vv'$ in~$\SS^*$ such that $\uu\vv'$ and $\vv\uu'$ both represent~$\xx \lcm \yy$. Put $\ww'_\kk := \ww_\kk$ for $\kk < \ii$, $\ww'_\ii := \uu''$, $\ww'_{\ii + 1} := \vv'$, $\ww'_{\ii + 2} := \uu'$, $\ww'_{\ii + 3} := \vv''$, and $\ww'_{\kk + 2} := \ww_\kk$ for $\kk > \ii + 1$, and $\ww':= \ww'_1 \INV{\ww'_2} \ww_3 \INV{\ww_4} \pdots$. By construction, $\ww'$ is obtained from~$\ww$ using special transformations (namely positive and negative equivalences and right-reversing) and, by the definition of~$\Redbrk{\ii, \xx, \yy}$, it represents~$\bbv$. The argument is similar for $\ii$ odd, exchanging right- and left-reversings.
\end{proof}

Some care is needed for the other direction, because positive and negative equivalences might cause trouble. This is where sharp representations are useful.

\begin{lemm}\label{L:Connection2}
Assume that $\MM$ is a gcd-monoid with lcm presentation~$(\SS, \RR)$ and $\ww$ is a word in $\SS \cup \SSb$ sharply representing a multifraction~$\aav$. Then $\ww \spec \ww'$ implies $\aav \rdsp^* \bbv$ for some multifraction~$\bbv$ sharply represented by~$\ww'$.
\end{lemm}

\begin{proof}
Put $\nn = \dh\aav$, and let $\ww_1 \wdots \ww_\nn$ satisfy $\ww = \ww_1 \INV{\ww_2} \ww_3 \INV{\ww_4} \pdots$ and $\clp{\ww_\ii} = \aa_\ii$ for every~$\ii$.
Suppose first that $\ww'$ is obtained from~$\ww$ by positive equivalence. Because the representation is sharp, the transformation involves one word~$\ww_\ii$, with odd~$\ii$ only. Then we have $\ww'_\ii \equivp \ww_\ii$, and $\ww'_\kk = \ww_\kk$ for~$\kk \not= \ii$, so $\ww'$ sharply represents~$\aav$, and the result is true with $\bbv = \aav$. The case when $\ww'$ is obtained from~$\ww$ by negative equivalence is similar.

Suppose now that $\ww'$ is obtained from~$\ww$ by one step of right-reversing. This means that there exists an even integer~$\ii$, generators~$\ss, \tt$ in~$\SS$, words~$\uu', \vv'$ in~$\SS^*$, and a relation $\ss\vv = \tt \uu$ in~$\RR$ satisfying $\ww_\ii = \tt \vv'$, $\ww_{\ii + 1} = \ss \uu'$, and such that $\ww'$ is obtained from~$\ww$ by replacing the factor $\INV{\ww_\ii} \ww_{\ii + 1}$, which is $\INV{\vv'} \, \INV\tt \ss \uu'$, with $\INV{\vv'} \uu \INV{\vv} \uu'$. Now, if we define $\ww'_\kk := \ww_\kk$ for $\kk < \ii$, $\ww'_\ii := \vv'$, $\ww'_{\ii + 1} := \uu$, $\ww'_{\ii + 2} := \vv$, $\ww'_{\ii + 3} := \uu'$, and $\ww'_{\kk + 2} := \ww_\kk$ for $\kk > \ii + 1$, then we have $\ww' = \ww'_1 \INV{\ww'_2} \ww'_3 \INV{\ww'_4} \pdots$, whereas, by very definition, the words $\ww'_1 \wdots \ww'_{\nn + 2}$ witness that $\ww'$ represents the multifraction $\aav \act \Redbrk{\ii, \ss, \tt}$. If all words~$\uu, \vv, \uu', \vv'$ are nonempty, then the representation is sharp, and taking $\bbv := \aav \act \Redbrk{\ii, \ss, \tt}$ gives the result. Otherwise, we restore sharpness by inserting trimming steps. Assume for instance that $\uu$ is empty, but $\uu'$ and $\vv\vv'$ are not. Then $\aav \act \Redbrk{\ii, \ss, \tt}$ is eligible for~$\Redtrim\ii$, and $\ww'$ sharply represents $\bbv := \aav \act \Redbrk{\ii, \ss, \tt}\Redtrim\ii$. The other cases are treated similarly, possibly using several trimmings. Note that free reduction, which corresponds to~$\ss = \tt$ and $\uu = \vv = \ew$, is not special. Finally, the argument for left-reversing is similar, with $\ii$ now necessarily odd.
\end{proof}

We can now complete the argument.

\begin{proof}[Proof of Proposition~\ref{P:H}]
As said above, it only remains to establish the equivalence of~\ITEM2 and~\ITEM3. Suppose first that $\SD\MM$ is semi-convergent, and let $\ww$ be a word in $\SS \cup \SSb$ representing~$1$ in~$\EG\MM$. Let $\ww_1 \wdots \ww_\nn$ be words in~$\SS^*$ satisfying $\ww = \ww_1 \INV{\ww_2} \ww_3 \pdots$, and let $\aav:= \clp{\ww_1} \sdots \clp{\ww_\nn}$. Then we find
\begin{equation}\label{E:Conn}
\can(\aav) = \can(\clp{\ww_1}) \can(\clp{\ww_2})\inv \can(\clp{\ww_3}) \pdots = \cl{\ww_1} \cl{\ww_2}\inv \cl{\ww_3} \pdots = \cl{\ww},
\end{equation}
so $\aav$ represents~$1$ in~$\EG\MM$. By hypothesis, this implies $\aav \rdsp \one$. By Lemma~\ref{L:Connection1}, we must have $\ww \spec \ww'$ for some word~$\ww'$ in $\SS \cup \SSb$ representing~$\one$. The word~$\ww'$ must be the empty word, so we have $\ww \spec \ew$. Thus \ITEM2 implies~\ITEM3.

Suppose now that Property~$\PropH$ is satisfied for~$(\SS, \RR)$, and let $\aav$ be a multifraction satisfying $\can(\aav) = 1$. First, by applying as many trim steps~$\Redtrim\ii$ as possible to~$\aav$, we obtain $\aav \rdsp^* \aav'$ for some multifraction~$\aav'$ satisfying $\aa'_\ii \not= 1$ for $1 < \ii < \dh{\aav'}$. Now let $\ww$ be a word in $\SS \cup \SSb$ representing~$\aav'$. The conditions $\aa'_\ii \not= 1$ imply that $\ww$ sharply represents~$\aav'$. By construction, $\can(\aav') = 1$ holds in~$\EG\MM$, and, therefore, by the computation of~\eqref{E:Conn}, $\ww$ represents~$1$ in~$\EG\MM$. By Property~$\PropH$, we must have $\ww \spec \ew$. By Lemma~\ref{L:Connection2}, there exists a multifraction~$\bbv$ sharply represented by~$\ew$. The latter is necessarily of the form~$\one$, implying $\aav' \rdsp \bbv$, whence $\aav \rdsp \one$. Thus \ITEM3 implies~\ITEM2.
\end{proof}

It was noted in~\cite{Dia} that Property~$\PropH$ need not imply the decidability of the word problem in general; Proposition~\ref{P:H} precisely measures the gap between the two properties: by Proposition~\ref{P:PaddedSC2WP}, the decidability of the word problem follows from the semi-convergence of $\RRR$-reduction up to a Turing-computable padding, whereas Property~$\PropH$ entails no effective upper bound on the padding that is needed.

\begin{rema}
As stated above, Property~$\PropH$ makes sense only for those gcd-monoids that admit a presentation that is both a right-lcm and a left-lcm presenta\-tion---as is the case for the standard presentation of Artin--Tits groups. If $\MM$ is a strongly noetherian gcd-monoid with atom set~$\SS$, then $\MM$ admits a right-lcm presentation~$(\SS, \RR)$ and a left-lcm presentation~$(\SS, \RRt)$ which need not coincide. Then Property~$\PropH$ can be stated for the double presentation~$(\SS, \RR, \RRt)$, using~$\RR$ for right-reversing and~$\RRt$ for left-reversing, and all results stated above extend without change to this general version.
\end{rema}

\section{Artin--Tits groups of sufficiently large type}
\label{sec:sufflarge}

We now investigate a special family of gcd-monoids, namely Artin--Tits monoids of sufficiently large type, for which we establish the padded semi-convergence result stated as Theorem~1.

A presentation~$(\SS, \RR)$ of a monoid~$\MM$ is an \emph{Artin--Tits presentation} if, for each pair~$(\ss, \tt)$ in~$\SS \times \SS$ with $\ss \neq \tt$, the set~$\RR$ contains at most one relation, and that relation has the form $\ss\tt\pdots = \tt\ss\pdots$, relating two alternating products of length $\mst$, for some integer $\mst \geq 2$. If $\mst=2$, then $(\ss,\tt)$ is called a \emph{commuting pair}, and if there is no such relation between $\ss$ and $\tt$, then $(\ss,\tt)$ is called  a \emph{free pair}, and we define $\mst=\infty$. The associated \emph{Coxeter diagram} is a complete graph with vertex set  $\SS$, and a label $\mst$ on the edge joining $\ss$ and $\tt$. A monoid $\MM$ is called an Artin--Tits monoid if it possesses an  Artin--Tits presentation $(\SS,\RR)$. We say that $\MM$ and its enveloping group $\EG\MM$ are of {\em sufficiently large type} if, in any triangle in the associated Coxeter diagram, either no edge has label $2$, or all three edges have label $2$, or at least one edge has label $\infty$.

Since all relations in $\RR$ have the form $\uu = \vv$, where $\uu$ and~$\vv$ are positive words of the same length, we see that the word-length map~$\wl$ induces a well-defined map on the presented monoid, verifying strong noetherianity. We use $\wl$ also to denote the induced map on the monoid and, for $\aa \in \MM$, we call $\wl(\aa)$ the \emph{word-length} of~$\aa$. We extend~$\wl$ to multifractions by~$\wl(\aav) := \sum_\ii \wl(\aa_\ii)$, and call~$\wl(\aav)$ the \emph{word-length} of~$\aav$.

\subsection{The result}\label{sec:result}

The main result that we shall prove is:

\begin{prop}\label{P:Main}
If $\MM$ is an Artin--Tits monoid of sufficiently large type and $\aav$ is a multifraction of word-length~$\ell$ that represents~$1$ in~$\EG\MM$, then we have $\One{2\pp} /\aav \rds \one$ with $\pp = 3 \ell (\ell+2)/4$.
\end{prop}

Note that, since all of the defining relators of the Artin-Tits groups have even length, a word in $\SS \cup \SSb$ that represents the identity must have even length, so $\ell$ must be even in the above proposition.

So, $\RRR$-reduction is semi-convergent for every Artin--Tits monoid of sufficiently large type up to a quadratic padding. As the square of word-length is obviously Turing-computable, Proposition~\ref{P:Main} verifies that $\MM$ satisfies Conjecture~$\ConjAp$, and therefore that the word problem of the associated Artin--Tits group is decidable (as was known already). By Proposition~\ref{P:H}, we also obtain an alternative proof to the main result of~\cite{GR}, namely that the standard presentation of Artin-Tits groups of sufficiently large type satisfies Property~$\PropH$.

Proposition~\ref{P:Main} follows from directly the following technical result.
We say that a multifraction~$\aav$ is \emph{geodesic} if its word-length is minimal over all multifractions that represent the same group element.

\begin{prop}\label{P:MainProp}
If $\MM$ is an Artin--Tits monoid of sufficiently large type and $\aav$ is a multifraction of word-length~$\ell$ that is not geodesic, then $\One{6\ell}/\underline{a}$ is $\RRR$-reducible to a multifraction of word-length $\ell -2$.
\end{prop}

The proof makes use of the existence of an effective rewrite system for  Artin--Tits groups of sufficiently large type that is defined and explored in \cite{HR1,HR2}, most of whose rules involve two generators only.  We show that  appropriate sequences of rules from that system can be simulated within multifraction reduction to reduce  $\One{6\ell}/\underline{a}$ as required.

The following subsection, Subsection~\ref{sec:strategy}, explains our strategy. Further technical results are established in Subsection~\ref{sec:technical}, and then the proof given in Subsection~\ref{sec:proof}.

\subsection{Strategy for the proof of Proposition~\ref{P:MainProp}}\label{sec:strategy}

For the rest of this section, we assume that $\MM$ is an Artin--Tits monoid of sufficiently large type, and
that $(\SS,\RR)$ is the associated presentation.

We shall refer to the elements of $\SS$ as
{\em generators} and to those of $\SS \cup \SSb$ as {\em letters}.
For $\ss \in \SS$, we call $\ss$ the {\em name} of both of
the letters $\ss$ and $\INV\ss$. We use $\f{\ww}$ and $\l{\ww}$ to denote the
first and last letter of a word $\ww$ in $\SS \cup \SSb$.
Furthermore, in situations where an element~$\aa$ of~$\MM$ is represented
by a unique word $\ww$ in $\SS$, we use $\f{\aa}$ and $\l{\aa}$ also to
denote the first and last letters of $\ww$.

The rewrite system described in \cite{HR1,HR2} is based on two kinds of length
preserving rules relating words, $\tau$-moves and $\kappa$-moves, together with
free reduction. By definition, a $\tau$-move rewrites a certain type of 2-generator subword
$w$ on non-commuting non-free generators to another word $\tau(w)$ on the same
two generators, where $w$ and $\tau(w)$ have the same length and represent
the same group element. A $\kappa$-move applies commutation relations to
subwords on two or more generators. In this article, in order to simplify the
exposition, we choose to break $\kappa$-moves down as sequences
of $\kappa$-moves on 2-generator subwords, and we introduce the term
$\tau^+$-move to mean either a $\tau$-move or
a $\kappa$-move applied to a 2-generator subword.

So $\tau^+$-moves can be applied only to certain geodesic 2-generator words,
which we call {\em critical} words. When $v$ is critical, then $\tau^+(v)$ is
also critical, on the same set of generators.
In all cases, $\f{v}$ and $\f{\tau^+(v)}$ have distinct
names, as do $\l{v}$ and $\l{\tau^+(v)}$. Furthermore, if $v$ is neither a
positive nor a negative word, then exactly one of $\f{v}$ and $\f{\tau^+(v)}$
lies in $\SS$, and the same is true for $\l{v}$ and $\l{\tau^+(v)}$.

Suppose that $w$ is a freely reduced non-geodesic word in~$\SS \cup \SSb$. Then, by \cite[Proposition 3.2]{HR2}, $w$ admits a so-called {\em leftward length reducing critical sequence}. This consists of a certain type of factorisation (see \cite{HR2} for a precise definition) $w = \alpha w_1 w_2 \pdots w_k \beta$ for some $k \ge 1$, where $w_1 \wdots w_k$ are geodesic $2$-generator subwords of $w$, and $w_k$ is critical. We can locate such a decomposition, if it exists, by a finite search.

To perform the length reduction, as described in \cite{HR2}, we first replace $w_k$ by $\tau^+(w_k)$. Then, if $k>1$, we replace $w'_{k-1}$ by $\tau^+(w'_{k-1})$, where $w'_{k-1} := w_{k-1}\f{\tau^+(w_k)}$. We continue in this manner, finally replacing $w'_1$ by $\tau^+(w'_1)$, where $w'_1 := w_1\f{\tau^+(w_2)}$. At that stage, we have $\l{\alpha} = \INV{\f{\tau^+(w'_1)}}$, and we can freely cancel these two adjacent letters.

To mimic the length reduction summarised above, we need to simulate the replacement of $w_k$ by $\tau^+(w_k)$. If $w_k$ is a positive or a negative word then, by definition, $\tau^+(w_k)$ is another such word representing the same group element, with $\f{\tau^+(w_k)} \neq \f{w_k}$. In that case, the only multifraction operations that we may need to carry out are moves that ensure that the part of the multifraction that is represented by $w_k$ occupies a single component.

When $w_k$ is neither a positive nor a negative word, the simulation of the replacement of $w_k$ by $\tau^+(w_k)$ is generally carried out in several stages. Initially, a certain part of the multifraction, which may be spread across several components, is represented by the word $w_k$.  We use $\RRR$-reduction (if necessary) to adjust that part of the multifraction so that it occupies exactly two adjacent components, numbered $i$ and $i+1$ for some $i$, and represents a word $v_k$ with $\cl{w}=\cl{v_k}$. Furthermore, if $\f{w_k} \in \SS$ (i.e. $w_k$ starts with a positive letter), then $i$ is odd and $v_k$ is a concatenation of nonempty positive and negative words, whereas if $\f{w_k} \in \SSb$, then $i$ is even and $v_k$ is a concatenation of nonempty negative and positive words. We then carry out a single $\RRR$-reduction, after which the part of the multifraction in question occupies components $i-1$ and $i-2$, and represents a word $u'_k$ with $\cl{u'_k}=\cl{v_k}$, which is a concatenation of negative and positive words in the first case, and of positive and negative words in the second.

Let $u_k$ be the maximal proper suffix of $u_k'$. We shall show later in Lemma~\ref{lem:sameletter} that $\f{u_k'}= \f{\tau^+(w_k)}$, so $u_k$ represents the same group element as $\f{\tau^+(w_k)}^{-1}\tau^+(w_k)$. Then, for $k>1$, we adjoin $\f{u_k'}$ to $w_{k-1}$ to give the word $w_{k-1}'$ defined above.

We repeat the above procedure on $w_{k-1}'$ to define words $u'_{k-1}$, $u_{k-1}$ and then, provided that $k>2$, to define $w'_{k-2}$, $u'_{k-2}$, $u_{k-2}$, ..., $w'_1$, $u'_1$, $u_1$. Then the first letter $\f{u'_1}$ of $u'_1$ freely reduces with the final letter $\l{\alpha}$ of $\alpha$, and this cancellation can be effected by a single $\RRR$-reduction.

So the word has effectively been transformed to $\alpha'u_1 \pdots u_k \beta$, where $\alpha'$ is the maximal prefix of~$\alpha$. Now we use $\RRR$-reduction to simulate the replacement of each $u_i$ by a $2$-generator geodesic $v_i$. The resulting word $v = \alpha' v_1 \pdots v_k \beta$, where $|\alpha'| = |\alpha| - 1$, $|v_k| = |w_k|-1$ and, for $1 \le i < k$, $|v_i| = |w_i|$, is the same as the word resulting from the application of sequence of $\tau^+$-moves to $w$ followed by the free cancellation. So we have successfully used $\RRR$-reduction to simulate the reduction resulting from the leftward critical sequence, and reduce the length of the word by~2.

\subsection{Technical results for two-generator Artin--Tits monoids}\label{sec:technical}

In order to simulate the $\tau^+$ transformations, we need a few preliminary results. By definition, the elements involved in the transformations belong to a non-free two-generator submonoid of the ambient monoid, and we establish here various technical statements involving the elements of such submonoids and their enveloping groups.

The first point is that every element in the group admits distinguished fractional decompositions.

\begin{prop}\label{prop:length2}
If $\MMt$ is a non-free two-generator Artin--Tits monoid, then every element of~$\EG\MMt$ admits a unique expression $\aa \bb\inv$ with $\aa, \bb$ in~$\MMt$ and $\aa \gcdt \bb = 1$, and a unique expression $\cc\inv \dd$ with $\cc, \dd$ in~$\MMt$ and $\cc \gcd \dd = 1$.
\end{prop}

\begin{proof}
The lcm~$\Delta$ of the two generators is a Garside element in~$\MMt$ and,
as a consequence, any two elements of~$\MMt$ admit a common right-multiple
and a common left-multiple. By the classical Ore's theorem, this implies that
every element of~$\EG\MMt$ can be expressed as a right fraction~$\aa\bb\inv$
and as a left fraction~$\cc\inv \dd$. As any two elements of~$\MMt$ admit a
left gcd and a right gcd, dividing the numerator and the denominator of a
right (\resp left) fraction by their right gcd (\resp left gcd) yields
irreducible fractions. Uniqueness follows from the fact that, if the right
gcd of~$\aa$ and~$\bb$ is trivial, then $\cc\aa$ is the right lcm
of~$\cc$ and~$\dd$ whenever we have $\cc \aa = \dd \bb$
(see for instance \cite[Lemma~9.3.5]{ECHLPT}).
\end{proof}

In the above situation, we shall call $\aa\bb^{-1}$ and $\cc^{-1}\dd$ the
\emph{right} and \emph{left} {\em fractional normal forms} of the associated
element.

The next point is that, under mild assumptions, the numerator and the denominator of these fractional normal forms are represented by unique words in the generators:

\begin{lemm}\label{cor:uniquewd}
If $\MMt$ is a non-free two-generator Artin--Tits monoid, and $\aa\bb\inv$ and $\cc\inv\dd$ are the fractional normal forms of an element~$\gg$ of~$\EG\MMt$ and, moreover, $\aa$ and $\bb$ (or $\cc$ and $\dd$) are both $\not= 1$, then none of $\aa$, $\bb$, $\cc$ or $\dd$ is divisible by the Garside element $\Delta$, and there are unique positive words in $\MMt$ representing each of them.
\end{lemm}

\begin{proof}
If, for example, $\aa$ were divisible by $\Delta$ then it would have a representative word with the same last letter as a representative word for~$\bb$, which would violate the condition that $\aa$ and~$\bb$ have no common right divisor. The uniqueness of the word representing $\aa$ is proved in~\cite[Proposition~4.3]{MM}.
\end{proof}

In this situation, we shall unambiguously write $\f{\aa}$ and $\l{\aa}$ for the first and last letters of the unique 2-generator word that represents $\aa$, and similarly for $\bb$, $\cc$, and~$\dd$.

Next, the generators occurring at the ends of the numerators and denominators obey some constraints:

\begin{lemm} \label{lem:letterchange}
If $\MMt$ is a non-free two-generator Artin--Tits monoid, and $\aa\bb\inv$ and $\cc\inv\dd$ are the fractional normal forms of an element~$\gg$ of~$\EG\MMt$ and, moreover, $\aa$ and $\bb$ (or $\cc$ and $\dd$) are both $\not= 1$, then we have $\f{a} \ne \l{c}$ and $\f{b} \ne \l{d}$.
\end{lemm}

\begin{proof}
We proved in Lemma~\ref{cor:uniquewd} that none of $a,b,c,d$ is divisible by the Garside element $\Delta$ of $\MMt$. Now, from the equality $ca = db$ and the fact that $c$ and $d$ have no nontrivial common left divisor, we have $\f{c} \ne \f{d}$, and so the element $ca$ of~$\MMt$ is represented by more than one word in the generators. By~\cite[Proposition~4.3]{MM} again, any such word must admit $\Delta$ as a subword. Since neither $c$ nor $a$ is divisible by $\Delta$, we have $\Delta = c'a'$, where $c'$ and $a'$ are nontrivial left and right divisors of $c$ and $a$. But the two words representing $\Delta$ consist of alternating letters, and so we have $\f{a} \ne \l{c}$ and similarly $\f{b} \ne \l{d}$.
\end{proof}

Using the previous result, we obtain some control about the first (or the last) generator occurring in the fractional normal form of an arbitrary geodesic word.

\begin{lemm} \label{lem:sameletter}
Assume that $\MMt$ is a non-free two-generator Artin--Tits monoid, and $\aa\bb\inv$ and $\cc\inv\dd$ are the fractional normal forms of an element~$\gg$ of~$\EG\MMt$ lying neither in~$\MMt$ nor in~$\MMt\inv$. Then, if $w$ is a geodesic word representing~$\gg$, we have $\f{a} = \f\ww$ if $\f\ww$ is positive, and $\l\cc = \INV{\f\ww}$ if $\f\ww$ is negative.
\end{lemm}

\begin{proof}
The proof is by induction on the length $n$ of $w$. In this proof, for a generator $x \in S$, we shall write $x^{\pm}$ to mean
either $x$ or $\INV\xx$. Suppose that $w= x_{i_1}^{\pm} \pdots x_{i_n}^{\pm}.$ There is nothing to prove
for $n \le 1$. Let us assume that that $\f{w} =x_{i_1}$, that is, $\f\ww$ is
positive; the other case, when $\f{w}$ is negative, is similar.
Let $v$ be the maximal prefix
$x_{i_1}x_{i_2}^{\pm} \pdots x_{i_{n-1}}^{\pm}$ of $w$.
If $v$ is a positive word, then $ab^{-1}$ with $a = \cl{v}$ and
$b=\cl{x_{i_n}}$ is a fractional normal form of $\cl{w}$, so $\f{a} = x_{i_1}$.
Otherwise, let $a'b'^{-1}$ be the right fractional normal form of $\cl{v}$.
Then by induction we have $\f{a'} = \f{v}=\f{w}$.

Suppose first $\l{w} = \INV\xx_{i_n}$. Now $\cl{a'}$ and $\cl{x_{i_n}b'}$
may have a nontrivial greatest common right divisor $\aa''$ but,
since $w$ is not a negative element, this cannot be the whole of~$a'$.
By Lemma~\ref{cor:uniquewd}, $a'$ is represented by a unique positive word.
Hence $a'=a \aa''$ and so $\f{a}=\f{a'}=\f{w}$.

Otherwise we have $\l{w} = x_{i_n}$. If $x_{i_n}=\f{b'}$
then $x_{i_n}$ simply cancels
with $\l{b'^{-1}}$ and we are done.
Otherwise the left fractional normal form of $\cl{b'^{-1}x_{i_n}}$ is
$\cl{b'}^{-1}\cl{x_{i_n}}$ and, since $\l{a'} \ne \l{b'}$, it follows from Lemma
\ref{lem:letterchange} that the right fractional normal form $a''b''^{-1}$ of
$\cl{b'^{-1}x_{i_n}}$ has $\f{a''} \ne \l{b'}$ and hence $\f{a''} = \l{a'}$.
So $a'a''$ is not divisible by $\Delta$.
Hence $a'a''$ has a unique geodesic representative and so, if $a'a''$ and
$b''$ had a nontrivial common right divisor, then so would $a''$ and $b''$,
which is not the case.
So $a'a''b''^{-1}$ is the right fractional normal form of $\cl{w}$,
with $\f{a'}=\f{w}$, and we are done.
\end{proof}

We now establish a series of results involving reduction of multifractions, still in the non-free two-generator case. Although very easy, the following result will be crucial.

\begin{lemm}\label{lem:swap_forms}
If $\MMt$ is a non-free two-generator Artin--Tits monoid, and $\aa\bb\inv$ and $\cc\inv\dd$ are the fractional normal forms of an element of~$\EG\MMt$, then we have
$$\begin{array}{cccc}
\pdots /\ss /c/d\tt /\pdots & \rd^{R_{i,d}} &\pdots/\ss a/b/\tt /\pdots &
\text{when $i$ is even}\\
\pdots /\ss /a/\tt b/\pdots & \rd^{R_{i,b}} & \pdots /c\ss /d/\tt /\pdots &
\text{when $i$ is odd},
\end{array}$$
where the multifractions are of depth at least 3, and the entries shown are in
positions $i-1,i,i+1$.In particular, for $\ss =1$, we have
$$\begin{array}{cccc}
\pdots /1/c/d\tt /\pdots & \rd^{R_{i,d}} &\pdots/a/b/\tt /\pdots &
\text{when $i$ is even}\\
\pdots /1/a/\tt b/\pdots & \rd^{R_{i,b}} & \pdots /c/d/\tt /\pdots &
\text{when $i$ is odd},
\end{array}$$
\end{lemm}

\begin{proof}
The diagrams provide a proof.
\begin{center}
\begin{picture}(45,7)(0,7)
\psset{nodesep=0.7mm}
\pcline{->}(0,6)(15,0)\tbput{$\ss a$}
\pcline{<-}(15,0)(30,0)\tbput{$b$}
\pcline{->}(30,0)(45,6)\tbput{$\uu $}
\pcline{->}(0,6)(15,12)\taput{$\ss $}
\pcline{<-}(15,12)(30,12)\taput{$c$}
\pcline{->}(30,12)(45,6)\taput{$d\uu $}
\pcline{->}(15,12)(15,0)\tlput{$a$}
\pcline{->}(30,12)(30,0)\trput{$d$}
\psarc[style=thin](15,0){3}{0}{90}
\end{picture}
\HS{10}
\begin{picture}(45,7)(0,7)
\psset{nodesep=0.7mm}
\pcline{<-}(0,6)(15,0)\tbput{$c\ss $}
\pcline{->}(15,0)(30,0)\tbput{$d$}
\pcline{<-}(30,0)(45,6)\tbput{$\uu $}
\pcline{<-}(0,6)(15,12)\taput{$\ss $}
\pcline{->}(15,12)(30,12)\taput{$a$}
\pcline{<-}(30,12)(45,6)\taput{$\uu b$}
\pcline{<-}(15,12)(15,0)\tlput{$c$}
\pcline{<-}(30,12)(30,0)\trput{$b$}
\psarc[style=thin](15,0){3}{0}{90}
\end{picture}
\end{center}
\vspace{4mm}
\end{proof}

Using Lemma~\ref{lem:swap_forms} repeatedly, we deduce that $\RRR$-reduction can be used to go from a fractional normal form to a geodesic expression.

In the sequel, $\One\mm$ stands for $1 \sdots 1$, $\mm$~terms. Note that, for~$\aa, \bb$ in~$\MM$, the multifraction~$\One{\mm}/a/b$ represents the element~$ab^{-1}$ of~$\EG\MM$ when $\mm$ is even, and the element~$a^{-1} b$ when $\mm$ is odd.

\begin{lemm}\label{lem:to_geo}
Assume that $\MMt$ is a non-free two-generator Artin--Tits monoid, and $\aa\bb\inv$ and $\cc\inv\dd$ are the fractional normal forms of an element~$\gg$ of~$\EG\MMt$ lying neither in~$\MMt$ nor in~$\MMt\inv$. Let $\vv$ be a geodesic word representing~$\gg$. Then, using $\RRR$-reduction, we can transform each of the multifractions $\One{4|v|}/a/b$ and $\One{4|v|-1}/c/d$ to a multifraction that is represented by $v$.
\end{lemm}

\begin{proof}
The proof is by induction on $|v|$. Since we are assuming that $v$ is not
positive or negative, there is nothing to prove when $|v| \le 1$.

Let $x=\f{v}$ and let $v'$ be the maximal proper suffix of $v$ (of geodesic  length $|v|-1$). By Lemma~\ref{lem:sameletter}, $x$ is equal to either $\f{a}$ or to $\f{c^{-1}}$.  If $x=\f{a}$ then we use Lemma \ref{lem:swap_forms} to transform $\One{4|v|-1}/c/d$ to $\One{4|v|-2}/a/b/1$, and if $x = \f{c^{-1}}$, then we transform $\One{4|v|}/a/b$ to $\One{4|v|-1}/c/d/1$.  In either case, from the fractional normal form for $g$ whose representative word does not start with $x$, we derive a multifraction represented by a word beginning with $x$, and this is equal to the multifraction representing the other fractional normal form of $g$ (or that minus two  initial entries containing $1$). We apply Lemma~\ref{lem:move_to_left} with $t=1$ to move the letter $x$ left to the first or second entry of the multifraction.

In the first case we have identified reductions
\[ \One{4|v|}/a/b \rds \cl{x}/\One{4|v|-1}/a'/b, \quad
 \One{4|v|-1}/c/d \rds \cl{x}/\One{4|v|-3}/a'/b/1, \]
and in the second case we have identified reductions
\[\One{4|v|}/a/b \rds 1/\cl{x}/\One{4|v|-3}/c'/d/1,
\quad
 \One{4|v|-1}/c/d \rds 1/\cl{x}/\One{4|v|-3}/c'/d.
\]

If $a'$ or $c'$ is equal to the identity, then $b$ or $d$ is a positive or
negative element of $\GG'$ and hence by Lemma~\ref{cor:uniquewd}
is represented by the unique geodesic word $v'$, and we are done.
Otherwise, by induction, we can apply sequences of $\RRR$-reductions to transform each of $\One{4|v|-4}/a'/b$ and $\One{4|v|-5}/c'/d$ to
multifractions represented by~$v'$. Hence there are sequences of $\RRR$-reductions that transform each of $\One{4|v|} / a/b$ and $\One{4|v| - 1}/c/d$ to multifractions represented by $v$.
\end{proof}

\subsection{Proof of Proposition~\ref{P:MainProp}}\label{sec:proof}

Using the results of Section~\ref{sec:technical}, we now describe how the steps outlined in Subsection~\ref{sec:strategy} can be executed using $\RRR$-reduction. As in Subsection~\ref{sec:strategy} and for the rest of this section, $\MM$ is a fixed Artin--Tits monoid of sufficiently large type.

\renewcommand\theta{\yy}
\renewcommand\eta{\zz}
\renewcommand\xi{\xx}

Let $w$ be a non-geodesic word representing the multifraction
$\underline{a} = a_1/a_2/\pdots /a_n$ on~$\MM$.
Then $w$ has a leftward length reducing critical factorisation
$w=\alpha w_1 \pdots w_k\beta$. Suppose that the subword $w_i$ starts within
the entry $a_{j_i}$ of the multifraction $\underline{a}$, that is, that we have
\[ a_{j_i}=
 \left \{ \begin{array}{ll} \cl{\theta_i \eta_i} & \text{for $j_i$ odd}\\
 \cl{\eta_i \theta_i} & \text{for $j_i$ even}
\end{array} \right .
\]
for positive words $\eta_i,\theta_i$, where
$w_i$ shares a prefix with $\eta_i$ or~$\INV\eta_i$.
Specifically, unless $w_i$ is a positive or negative word, one of
$\eta_i$ or~$\INV\eta_i$ is a proper prefix of
$w_i$, but if $w_i$ is positive or negative, then $w_i$ might instead be a
prefix of $\eta_i$ or~$\INV\eta_i$.

It follows from Lemma \ref{lem:move_to_left} applied with $t=1$ that, via an application of the sequence of moves
$R_{2k,a_1}R_{2k+1,a_2}\pdots R_{2k+j_1-1,a_{j_1-1}}$, we have
$$\One{2k}/a_1/\pdots /a_n\rds \One{2k-2}/a_1/a_2/\pdots /a_{j_1-1}/1 / 1 /a_{j_1}/\pdots /a_n$$
and then application of $R_{2k+j_1,\cl{\eta_i}}$  transforms this to
$$\One{2k-2}/a_1/a_2/\pdots /a_{j_1-1}/\cl{\theta_1}/1/\cl{\eta_1}/a_{j_1+1}/ \pdots/ a_n.$$
Hence we see that a sequence of such reductions transforms the multifraction  $\One{2k}/a_1/\pdots /a_n$ to a multifraction
$$\underline{a}' := a_1/a_2/\pdots /a_{j_1-1}/\cl{\theta_1}/1/\cl{\eta_1}/a_{j_1+1}/\pdots
/1/\cl{\eta_2}/\pdots\pdots/1/\cl{\eta_k}/\pdots/a_n$$
for which $\cl{\eta_i}$ is an entry of~$\underline{a}'$ for each of the subwords $\eta_i$. Let us redefine $j_i$ to denote the index of the entry of $\underline{a}'$ in which $\cl{\eta_i}$ now lies.

Let us assume that $j_k$ is odd, and hence that $w_k$ has a positive prefix -
the other case is similar.

Suppose first that $w_k$ is a positive word.
Then $w_k$ might be distributed over more than one entry of $\underline{a}'$ as a
prefix of a multifraction $\underline{b} :=
\cl{\eta_k}/1/a_{j_k+2}/1/\pdots/a_{j_k+2s}$.
But then a sequence of reductions
$R_{j_k+2s-1,a_{j_k+2s}}\pdots R_{j_k+1,a{j_k+2}\pdots a_{j_k+2s}}$ would
transform $\underline{b}$ to $\underline{b}':=\cl{\eta_k} a_{j_k+2}\pdots a_{j_k+2s}/\One{2s}$.
So we can assume that $w_k$ lies within a single multifraction entry, as a
prefix of $\eta_k$. The required $\tau^+$-move on $w_k$ can then be executed
just by choosing a different representative word for $w_k$, which does not
change~$\underline{b}'$.

Otherwise, $w_k$ is distributed over more than one entry of $\underline{a}'$, and has a positive prefix. Suppose that $w_k$ occupies $s>2$ such entries. So it has a suffix $\xi_k$ (or~$\INV\xi_k$) which defines a left (resp. right) divisor of the $(j_k+s-1)$-th entry of $\underline{a}'$. Then, by Lemma \ref{lem:swap_forms}, we can apply a reduction $R_{j+k+s-2,\cl{\xi_k}}$, which effectively replaces $w_k$ by a word representing the same group element that occupies occupies at most $s-1$ multifraction entries. A sequence of at most two $s-2$ such reductions transforms $w_k$ into a word occupying two entries, which represents the fractional normal form $ab^{-1}$ of $\cl{w_k}$, with $a$ the $j_k$-th entry and $b$ a right divisor of the $(j_k+1)$-th entry of the new multifraction $\underline{a}''$. (Since $w_k$ is not a positive or negative element of the group, it could not be transformed to a word that occupies fewer than two multifraction entries.) By Lemma~\ref{lem:sameletter} we have $\f{a}=\f{\eta_k}$. Now we apply Lemma~\ref{lem:swap_forms}, with $i = j_k$ and $t=1$, to transform $\underline{a}''$ into a multifraction in which $ab^{-1}$ is represented by its other fractional normal form, $c^{-1}d$, where now $c$ is the $(j_k-1)$-th entry and $d$ the $j_k$-th entry.

Let $u_k'$ be a product of geodesic words representing $c^{-1}$ and $d$. Then, from Lemma~\ref{lem:letterchange}, we have $\l{c} \ne \f{a} = \f{w_k}$, and so $c^{-1}$ and hence also $u_k'$ begins with the first letter that we wish to append to $w_{k-1}$ to give $w_{k-1}'$.

We can repeat this process on $w_{k-1}',w_{k-2}' \wdots w_1'$ in turn. Then we can carry out the free reduction between the first letter of $u'_1$ and the final letter of $\alpha$ using a single $\RRR$-reduction. Denote the multifraction after the free reduction by $\underline{b}$. Then a word representing $\underline{b}$ contains each of the words $u_1 \wdots u_k$ as a subword, where each $u_i$ is either a positive or negative word within a single entry of $\underline{b}$, or else it is a word in fractional normal form $ab^{-1}$ or $c^{-1}d$, where the entries of $\underline{b}$ in position $j_i-1$ and $j_i$ are $a$ and $b$, or $c$ and $d$, respectively.

We complete the proof by describing $\RRR$-reductions that transform the multifraction $\One{N}/\underline{b}$ into one that is represented by the word $\alpha'v_1\pdots v_k \beta$, for an appropriately chosen integer $N$. In the case where $u_i$ is a positive or a negative word, $u_i$ and  $v_i$ are geodesic words with $\clp{u_i} = \clp{v_i}$, so no change is needed to the multifraction.

We can apply Lemma~\ref{lem:to_geo} to deal with those $u_i$ that are neither positive nor negative words. We deduce that we can find a sequence of $\RRR$-reductions  that transform $\One{N+2k}/\underline{a}$ via $\One{N}/\underline{b}$ to a multifraction that is represented by the word $\alpha'v_1\pdots v_k \beta$ provided that $N \ge 4 \sum_{i=1}^k |v_i|$. Since clearly both $k$ and $\sum_{i=1}^k |v_i|$ are bounded above by $|w|$, the proof of the Proposition~\ref{P:MainProp} is completed using Lemma~\ref{lem:to_geo}. This completes also the proof of Proposition~\ref{P:Main}. \hfill$\square$

\medskip The results of the current paper leaves the question of whether Artin--Tits monoids of sufficiently large type satisfy Conjecture~$\ConjA$ open, but, at least, we know now that $\RRR$-reduction is relevant for them. In view of this result, and those of~\cite{Dit}, which settle the case of type~FC, the ``first'' case for which nothing is known in terms of reduction (nor of the word problem) is the Artin--Tits monoid with exponents $3, 3, 3, 3, 3, 2$, that is, the monoid\\
\parbox{\linewidth}{\medskip\quad$\MON{\mathtt{a,b,c,d}}{\mathtt{aba=bab, aca=cac, bcb=cbc, ada=dad, bdb=dbd, cd=dc}}$.\medskip}\\
We hope that further progress will arise soon

redpajama_set_name: RedPajamaArXiv


archi.exe - archi, Ftrap Virus, Virus Notice!
archi.exe – Here is the scoop on Ftrap Virus as it pertains to computer network security. The big question: what is archi.exe and is it spyware, a trojan and if so, how do I get rid of Ftrap Virus?
Finding a program with the name archi.exe running on your computer means that your computer could be infected with a form of the ftrap virus.
archi.exe is considered to be a security risk, not only because antivirus programs flag Ftrap Virus as a virus, but also because a number of users have complained about its performance.
Ftrap Virus is likely a virus and as such, presents a serious vulnerability which should be fixed immediately! Delaying further investigation of archi.exe may cause serious harm to your system and will likely cause a number of problems, such as slow performance, loss of data or leaking private information to websites.
Whenever you have a concern about a file like archi.exe, feel free to leave a comment (you'll find the comment section at the bottom of this page); we'll research your process or dll file and post the results to the site.
We do our best to update process information as often as possible but inaccuracies may still exist; a prime example would be a virus that is named after a legitimate file and perhaps Ftrap Virus is such a case. If you find an inaccuracy or have information that will help others, please, don't hesitate to leave a comment!
Every attempt has been made to provide you with the correct information for archi.exe or FTRAP VIRUS. Many spyware / malware programs use filenames of usual, non-malware programs. If we have included information about archi.exe that is inaccurate, we would greatly appreciate your help by leaving a comment with the correct information below and we'll do our best to correct it.
You should verify the accuracy of information we provided about archi.exe.


Some time ago I mentioned a project I was working on to move an application from Heroku to a virtual server running on Rackspace. One of the challenges I faced in that project was setting up some background jobs.
Given that we were using Ubuntu we decided to use Upstart to manage the jobs.
Create a config file to make upstart manage each job. These files should be placed under /etc/init/. These files should include some generic information for upstart (like the user that will run the service) and also some specific information related to job (like how to start it). Here you can find and example to configure Clockwork (a cron replacement app built on Ruby).
A client component that takes care of collecting log information and sending it to centralized storage.
One company that offers this kind of service is Logentries. I used it for a couple of Ruby applications, some of them hosted on Heroku and others hosted on my own server. In the case of Heroku apps, the integration with LogEntries is transparent (you don't need to modify your app code), you just have to add logentries plugin to your Heroku app , configure it and that's all, all log information is sent to LogEntries. For applications running outside Heroku, you can use logentries gem to your app, and just make your app to use LogEntries logger.
Other service I have used is Papertrail. I used it for Ruby and .Net applications hosted on my own servers and in each case I have used a different integration strategy. For .Net apps I like to use Log4Net library which natively integrates with Papertrail (details can be found here). In the case of Ruby apps, I have used a different strategy. I install on my server a client application that takes care of reading local log files and push them to the Papertrial website. This strategy is totally transparent for the application and is application agnostic: no matter what technology your app is built on.


text:
Stoptechy
HomeEntertainment4 Pinay Girl Video Goes Viral On Internet
4 Pinay Girl Video Goes Viral On Internet
Jules Ari video
People on TikTok, Twitter, and Reddit have seen updates on the 4 Pinay Girl video, and they want to see that part of the video. The 4 Pinay Girl video quickly went viral after it was posted on Twitter and Reddit. The news then got around to other social media sites. At that point, there were already a few more videos online about his account.
Watch the "Jabol TV Girl" video that is going around on Twitter and Reddit.
People on the internet want to watch the video, but they can't find it on social media without doing very specific searches. Unlike other movies, you can't find anything on social media sites about this one. Customers can also go to websites to find explicit recordings. They have nothing else to do. They are stuck and can't move.
One of the "Jabol TV Girl Viral Video" videos is becoming more and more popular and is being shared on a lot of different networks. because you can find it online. Even though it has been proven beyond a reasonable doubt that the video has s*xual content, new questions are being asked.
The full version of "Jabol TV Girl Video" is shared on Reddit.
Many websites say they can get people to the video, but you can't always count on them to do so. There aren't many sites on the Internet that can do the same thing. Since the video just started going viral on social media, the process should only take a few days. So, it should take a few days to finish everything. This is true whether or not people who buy things online want to know the movie's story. People who use the Internet are just as interested as people who don't use the Internet in the company's past and present CEOs.
pic.twitter.com/K4jtPgPsMy
— A$ET NEGARA (@clubasetnegara) January 4, 2023
When there isn't much information about the company's owner or the services they offer, it's hard to decide what to do. Everyone in the world is crazy about the movie. Please follow these rules if people find the video. They will do their investigation in secret because it is probably safe to do so. People shouldn't do it in front of them.
4 Pinay Girl Video
Jabol TV Girl Video
Watch the Leaked Eatpraydong Video
Watch Georgina Njenga video go viral on Twitter and Reddit
Adminhttps://stoptechy.com
The Marissa Lamb video has gone viral on Twitter and Reddit
Ragazza disco Treviso video get so popular
Angela Aguilar Viral Video On Twitter And Internet
Watch Skimaskgirl video Get Out viral
Watch Thinjen video viral over the social media
What reason of Marshall Tucker death? Marshall Tucker Band dies at the age of 99
Viral News465
Stoptechy is a top leading news, technology and entertainment Blog site where you read all latest news like trending news, World News on a one portal.
Contact us: stoptechy@gmail.com
© 2022 stoptechy.com | All Right Reserved

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Covariant vectors and change of basis matrix I'm studying the basics of special relativity and I've been having some difficulties with the concept of contravariant and covariant vectors. Now, at least in the basic way in which we introduce them in this beginners course, I think to have understood enough about them. Still, I'm a little bit confused about the following question:
Given a finite dimensional vector space and two different basis, let's say $A$ and $B$, it's always possible to find the change of basis matrix $M$. Given a vector $v$ and it's representation in $A$ given by $(V_{a1}, V_{a2}, ..., V_{an})$, the matrix $M$ can be used to transform it and obtain the representation in $B$ $(V_{b1}, V_{b2}, ..., V_{bn})$.
Now, the Minkowski space is a vector space. And applying a Lorentz Transformation, for example the usual $x$-boost, should be mathematically equivalent to a change of base. Therefore, how come that some vectors transform according to a matrix (the matrix of the lorentz transformation) and some accorting to it's inverse matrix? Isn't this in contrast with the fact that the change of basis matrix is unique?
I suppose the answer must involve the metric tensor $g$ and the fact it isn't the identical matrix but has $(1,-1,-1,-1)$ as the diagonal, however it's not very clear to me. Can somebody help?

redpajama_set_name: RedPajamaStackExchange


Tiny Tina can be used with a Mini Travel Bag. When purchasing Tiny Tina with Mini Travel Bag an additional doll is included. This set is a perfect car / church toy and fits nicely in our Mini Travel Bag.
Please Note: The Mini Travel Bag is NOT included with your Tiny Tina Set but can be purchased separately. Mini Travel Bags are listed under Felt Board page.
Tiny Tina fits nicely into a pencil box. Your child will have something to do when you need a quiet activity. We include a blue piece of felt with this set. Glue the blue felt piece into your pencil box as shown in photo.
The Tiny Tina Mini Set, along with Dinosaur Play Set, Dazzling Dancers, and All Star Alex make Perfect Car Toys. Makes traveling much more fun!


text:
Company Expansion & Consolidation
Houston Wire & Cable to Sell Southwest Wire, Completes Southern Wire Sale
HWCC is netting $25 million from the divestments, which it is using to reduce debt.
Houston Wire & Cable
HOUSTON — Houston Wire & Cable Company (HWCC) announced Wednesday that it has entered into a definitive asset purchase agreement for the sale of its Southwest Wire Rope business, a provider of lifting slings and wire rope and rigging products, for a purchase price of $5 million, subject to a working capital adjustment. The sale excludes approximately $2.6 million of trade accounts receivable, which HWCC will retain. HWCC expects to use the net sales proceeds and accounts receivable collections to further reduce Revolver debt. HWCC has already reduced Revolver debt to approximately $21.4 million as of January 13, from a peak of $95.2 million in the first quarter of 2020.
James L. Pokluda III, HWCC President & CEO commented, "Southwest Wire Rope is a company that for over fifty years has been a leading supplier of quality lifting slings, wire rope and rigging supplies to the lifting industry. We wish them much success in the future. Proceeds from the sales of Southern Wire and Southwest Wire Rope allow us to reduce debt, which lessens financial risk in these uncertain times."
The transaction is subject to customary closing conditions and is expected to close in the first quarter.
Southern Wire Sale
HWCC announced Jan. 6 that it completed the sale of its Southern Wire business, a wholesale distributor of wire rope and rigging products, to Southern Rigging Companies. The sale closed on December 31, 2020 for a purchase price of $20 million, subject to a working capital adjustment. HWCC used the net proceeds of approximately $18.1 million to reduce debt.
The Company made great progress reducing debt during the year as Revolver debt decreased $66.4 million from its peak of $95.2 million in the first quarter to $22.6 million at year-end. The Revolver debt reflects proceeds from the sale of the Southern Wire division at December 31, 2020 of $18.1 million. Excluding the Southern Wire proceeds, Revolver debt at year-end would have been $40.7 million.
As previously communicated, the Company's year-end goal was to lower Revolver debt to between $40 to $45 million, excluding a divestiture. The Company achieved its Revolver debt goal and believes the substantial debt reduction of $66.4 million lessens financial risk without any deterioration of its ability to provide excellent customer service.
The Company also has a Paycheck Protection Program ("PPP") loan of $6.2 million which was received on May 4, 2020 and funded under the Coronavirus Aid, Relief, and Economic Security Act. We anticipate approximately $5.8 million of this loan will be forgiven in 2021. Total debt at December 31, 2020, which includes the Revolver and PPP debt, was $28.8 million.
James L. Pokluda III, HWCC President & CEO commented, "We are excited we were able to complete the Southern Wire sale in 2020 and I am pleased to report that the sale of this division, together with our prudent management of working capital and expense reduction initiatives, has allowed us to reduce our debt to $28.8 million at year end, down 70% from the peak in 2020. In addition, we were able to achieve this debt reduction while maintaining our outstanding customer service."
With 45 years' experience in the industry, Houston Wire & Cable Company, an industrial distributor, is a large provider of products in the U.S. market. Headquartered in Houston, Texas, the Company has sales and distribution facilities strategically located throughout the United States.
Standard stock items available for immediate delivery include continuous and interlocked armor cable; instrumentation cable; medium voltage cable; high temperature wire; portable cord; power cable; primary and secondary aluminum distribution cable; private branded products, including LifeGuard™, a low-smoke, zero-halogen cable; mechanical wire and cable and related hardware, including wire rope, lifting products and synthetic rope and slings; corrosion resistant fasteners, hose clamps, and rivets.
Latest in Company Expansion & Consolidation
Optimas Solutions Achieves Global IATF Creditation
Fortna Rebrands With New Website, Logo
Hillenbrand Slims Down by Selling Abel Pumps, Red Valve Businesses
More in Company Expansion & Consolidation
Global Industrial Debuts New Branding
The company says a "We Can Supply That" tagline conveys deep customer insights and ability to deliver today's and tomorrow's solutions.
Motion Industries Breaks Ground on $11M Shop Near Birmingham
When completed, the 104,000 square-foot building will house Motion's area fluid power shop, hose and rubber shop and engineering department.
Optimas Completes Refinancing Aimed for Long-Term Growth
The fastener distributor and manufacturer said the move will deliver against its Manufacturing Solutions business strategy announced in August.
Würth Action Bolt and Tool Joins WINA's New Construction Services Unit
The new division is now comprised of Weinstock Bros., Würth House of Threads, and Würth Action Bolt and Tool.
Master Distributor AgoNow Launches Welding Subsidiary, WeldNow
Digital wholesale partner WeldNow gives distributors access to more than 170,000 SKUs from hundreds of top welding, cutting, filler metal, consumable and accessory brands.
Timken Invests $75M in Wind, Solar Energy Markets
Timken is going all-in on the renewable energy market, stemming from its 2018 acquisition of Cone Drive.
Houston Wire & Cable Selling Southern Wire Unit for $20M
HWCC expects to use the net sales proceeds to reduce debt.
MRO Distributor & Integrator 3B Supply Launches New Website
The site enables easy navigation between 3B's integrated supply services, technology & vending, brands & suppliers and an online catalog.
3M to Cut 2,900 Jobs in Restructuring
It follows layoffs of 1,500 employees this past January.
MSC to Move, Consolidate its Long Island Customer Support Center
The metalworking and MRO supplies distributor is relocating it to a much smaller facility near its headquarters in Melville, NY.
Sandvik Coromant Partners With Autodesk on Metal-Cutting
The long-term collaboration aims to help customers stay competitive with new technologies and provide digital design and manufacturing solutions.


Challenge Roth: it's a distant but good memory. I am not sure this counts as a race report as it comes so long after the race. Since then I have holidayed with Brett and my family in Garmisch, returned home to Boulder, spectated at the inauguaral Ironman Boulder and well, I am now inhabiting a hazy world familiar to me as that feeling that accompanies Kona prep. Anyway, without further ado here are a few reflections on my Challenge Roth experience.
Welcome:If I told you I arrived on a Sunday evening in Munich and was met by Marcus, part of the Challenge Team, you would think "So what?". However, it wasn't any Sunday evening. It was the Sunday evening of the World Cup final, featuring Germany. That's a faux pas right there! Despite my ill timed arrival I still got a friendly pick up and that about sums up how well I was looked after all week. We returned to the same homestay family we had in 2012. Moni and Fribo made Brett and I feel as welcome as ever. Moni even remembered my pre-race meal and had the chicken and rice ready for me on Saturday evening.
Rookie Mistakes: Vaseline on the inside of your goggles. Yes I did this. After spending the first few hundred metres berating myself I got on with it in a rather stop start kind of way. Advice: Don't do this.
The Joyce Clan: I felt like I was seeing my peeps all around the course. My family was out in force: my parents, my sister Katie, Pete and Vicki my brother and, sister-in-law and Walt my nephew. The Union Jack flags were out in force on the long hill at the southern most point of the bike course. When I saw them again in the final few kms of the run they had me cracking up (laughing!) with their loud and enthusiastic cheers. It meant a lot to share the day with them.
Hotness: Dang! Bavaria put on a stinking hot day for us. I felt it hotting up during the bike and the first part of the run was seriously warm.
The Walchshoefers: The entire family give this event the personal touch. I am sure Felix cheered every person he saw out on the race course and he was out there doing it himself. I had a high five from Kathrin out on the run course. Their passion drives this race and makes it a "must do" event.
Sponsors and Support Crew: I say it everytime but it is true. I couldn't do this without the fantastic sponsors I have in my corner.
THANK YOU Lesley and Jason for all your help during race week. THANK YOU to my coach Dave for getting me ready for action and dealing with my hiccups on the way. THANK YOU to Brett for supporting me everyday…especially for enduring me during the current haze days! Six weeks till Kona!


1- What is GifCam ?
GifCam is a free animation software with simple manipulations. It operates like a camera on top of all windows to help you take a picture of any desired area and save them as animated images. This program provides a variety of high-quality animation methods.
GifCam captures an area or full screen and saves it as an animated image. The created product is saved in a GIF (or AVI) file with small size.
The operating principle of the GifCam is to continually capture the area which a user wants. Each photo taken is a frame. The GifCam automatically compares the difference between previous and following frames, and renders only different pixels and increases the delay for a frame if the following frame doesn't change. Based on this principle the created product is a GIF file with small size but its quality is very good.
The GifCam also allows you to edit the content of animated image to be created, such as adding text to the animated image, changing the delay of a frame, etc.
Capture screen areas or full screen and save as animated image.
Support with many animation methods.
The latest version of the GifCam at this time (11-2017) is 5.0.
The GifCam program consists of only one gifcam.exe file. The GifCam window opens, you can adjust the area you want to record to create an animated image.
Use "Full Screen" option, if you want to record full screen and use "Capture Cursor" if you want to record the image of mouse when it moves within the recording area of the GifCam.
After finishing alignment of areas to be recorded, press "Rec" button to start recording.
After finishing the recording, you can save the result in a file. At this time, you have more other options for the quality of output, and the format of the file created.
Quantize: Create an image with the best quality.
GrayScale: Animate with the 2 colors such as white and black..
Preview: Preview results before saving in a file.
Export To AVI: Select output format as AVI (a video format), if this option is not selected, the file created will have GIF format.
The following are some results based on quality options.
On "Edit" window, you can see frames. Frame 0 is the first image. The following frames have green areas which are positions where fixels don't change from its previous frame. It will be considered as transparent area.
Right click on the F rame that you want to edit.
Delete the frames from the first frame to the selected frame.
Delete frames from the selected frame to the final frame.
Delete frames with even order number.
Edit the size of all frames.
Delete frames from N1 position to N2 position.
Add text to current frame.
Change the size of all frames.
Crop all frames, and retains only the areas in which you are interested.
Add reverse frames to the end. Example: your animated image has 10 frames (F0, F1, ... F9). After using this function, the frames (F9, F8,... F0) will be added to the end, thus, your animated image will have 20 frames.


JOIN LEGION M FOR FREE!
PLEASE NOTE: I am not an employee of Legion M, nor am I a paid spokesperson. I am simply a VERY SATISFIED member, and I want to grow the Legion to 1,000,000+, like anybody else!
I was lucky enough to be employed and #woke when Legion M opened up their first round to individual investors like me, thanks to the JOBS ACT, on Wefunder in 2016. I've been having a blast as a member of the Legion ever since!