text:
Youssef Megahed--The Question is Why?
by Dan Vara
https://www.investigativeproject.org/1393/youssef-megahed-the-question-is-why
Youssef Megahed, a permanent resident alien of the United States, was in immigration court in Miami last week. According to published reports, Megahed faced deportation based on terrorism charges stemming from his much publicized arrest in South Carolina in 2007 along with his friend Ahmed Mohamed.
After a five-day hearing before Immigration Judge Kenneth Hurewitz, however, the deportation case against Mr. Megahed was terminated because Immigration and Customs Enforcement (ICE) failed to prove its case.
The question the American public should be asking is "Why?"
I don't know if Mr. Megahed is a really a terrorist. I don't know if, given the chance, he would have or will engage in conduct to harm the United States. But what I do know as a former federal immigration prosecutor is that even the publicly available information evidences that this case appears to have involved what could be characterized as a textbook case of terrorists in the making.
In fact, and on many levels, this case reminds me very much of the Imran Mandhai case in Miami that I was involved in a few years ago.
Both involved a group of young men who, on the surface, seemed to be law abiding residents of this country. Both, however, also involved young men who for a variety of reasons but for the specific reason that they disagree with the policies of this country in relation to the situation in the Middle East, made a decision to proceed into activities that, left unchecked, could have resulted in a lot of Americans being very unpleasantly affected.
Some of the similarities are eerie. Each of the cases involved the respective group going to gun ranges in search of firearms and firearms training. Both involved trips to various locations on what appeared to be innocuous trips but were later determined by authorities to be something far more nefarious. Both involved incriminating physical evidence that members in the group at first disavowed but later admitted were items related to planned terrorist activities. And, both initially involved adamant claims of innocence, outcries of injustice, and later admissions of guilt by all or some of the involved individuals.
The biggest difference in the two cases, however, was in how and when the groups were taken down by authorities.
In the Mandhai case, the FBI learned of the group's activities, was able to proactively employ electronic surveillance in investigating the group, was able to put a covert operative in play, and was ultimately able to provide me with hours of recordings that demonstrated many overt acts; to include a chilling statement from the then 19-year old Mandhai to co-conspirators that bombing a hospital "full of poor people" was absolutely within the bounds of Jihad and that also included and equally chilling excerpt of Mandhai trying to buy automatic weapons, shoulder fired rockets, and explosives on the streets of Miami.
Though they never did manage to buy any weapons, he and his friends never got beyond the preliminary planning stages, and the group never did manage to carry out even one attack against this country, Mandhai and his friend are in now in prison. And, when he gets out, he will be deported.
Imagine, however, what we would have been faced with in the Mandhai case if, rather than having his own words about their terrorist target scouting mission, he and his friends had been stopped by local authorities for suspicious behavior and all that was found was the weapons shopping list that they had prepared. Imagine also what would have been the case if rather than have Mandhai gleefully laughing on videotape as he touches and describes the weapons, he had been stopped next to the van containing these items, arrested on suspicion of a conspiracy to violate federal firearms laws, and claimed ignorance of the contents of the van. Most of all, imagine if the FBI had been left to go back and seek to prove that what the group was doing in going to gun ranges, in trying to buy weapons at such gun ranges, and in meeting together as a group, was in fact activity in preparation for a planned terrorist attack.
In the Megahed case, the U.S. government was lucky, and then again, not so lucky. As published reports seem to establish, the FBI learned of Youssef Megahed and Ahmed Mohamed only after they were stopped and arrested by local police authorities in South Carolina. At the time of the arrest, Megahed and Mohamed were in a car which contained explosives materials, were in possession of a computer that contained a bomb-making training video intended for Jihadists and posted on You Tube just days before. After their arrest and while seated in a police car, Megahed and Mohamed engaged in a conversation about the materials in the trunk that, to me, evidences that they were worried about what authorities would make of the materials.
Missing in the Megahed case, however, was the ability of the FBI to covertly investigate the group. In short, what they got on that fateful day in 2006 was literally it. On that day, the group was on specific notice that the authorities were on to them. No more activities, no more videos, no more gun ranges, no more overt acts of any kind. Not when you have been "outed." And certainly not when you are faced with the major impediment of being behind bars.
While the FBI did its best to go back in time and seek to explore the details of the group's activities, it was left to discover only that which, without the actors own words to tell us what they intended, can be and was characterized as something less than precursor activity for a terrorist act.
But was it? Like the Mandhai group, the trips to the gun ranges could, if the FBI and DHS had the conversations of the group on tape, have been clear evidence of overt acts related to terrorism. The preparation and publication of the bomb-making video tape, if electronic surveillance had been available of Mr. Megahed and his friends discussing it, viewing it and taking specific actions to distribute it, could also have been evidence of Megahed's involvement in terrorism. And, perhaps if the U.S. government had been able to make audio tapes of Megahed and Mohamed during their road trip from Florida north to S. Carolina and maybe, if they had not been arrested, points further north on I-95, we would all have heard a discussion of objectives, issues, plans, and even potential targets.
But, the federal government didn't and doesn't have any of that. All they had, at least from what they were allowed to use, was what was introduced at the criminal trial and immigration hearing.
What they had, however, was pretty good evidence of what was going on. Beyond all the physical evidence that they had as noted above, and beyond what the supplemental investigation revealed about Megahed's other associations, they had one major thing. Mr. Megahed was in a car which contained explosives materials, a computer that showed a bomb-making video for Jihadists, and Ahmed Mohamed who, by his own later confession, was involved in terrorism.
This brings to mind another case that I handled. One involving solely circumstantial evidence. One involving a terrorist "sleeper." The U.S. government had no usable evidence whatsoever of any overt act that showed that he was a terrorist. What we did have, though, was evidence of a series of business social encounters, business relationships, and business transactions, with one person after another, whom we absolutely knew were involved in terrorism; including some persons who have been publicly acknowledged as terrorists.
In cross-examination of this Respondent, I confronted him with each and every one of these matters. His answers? "I did not know." "I wasn't aware." "That is a surprise to me." And, indirectly, "I'm shocked."
Drawing an objection, of course, I asked this person a question very much like this one, "Sir, given that most of your social and business encounters since you arrived in this country were with persons involved in terrorism, can you please tell us, are you a terrorist or are you the most unlucky person in America?"
The same question might apply to Mr. Megahed.
Yet, many more questions should also be asked of ICE.
At the outset, I must disclose that I personally know most of the reported federal legal actors in this latest ICE drama. Judge Hurewitz, before he became a judge, worked for me. So did the ICE prosecutors, Gina Garrett-Jackson and Loren Coy.
The first question, if anybody at ICE would talk, should be why they did not call Yousef Megahed to testify on his own behalf? While I was not there and thus don't really know if Mr. Megahed testified, and while we won't know for sure until we get the judge's written decision, I have to presume that Mr. Megahed was not called to testify. I make that presumption from a very simple unanswered analytical inquiry—if he had been called, how would he have been able to explain away his statements in the police car in South Carolina, the assertions set forth in the FBI (search warrant) affidavit, and most importantly, his intimate friendship with confessed terrorist supporter Ahmed Mohamed?
I trained most if not all of the new federal immigration attorneys hired between 1992 and 2005 in my role as an INS and ICE National New Attorney Training program instructor on cross-examination techniques.
I also specifically trained all INS Miami SCLU attorneys in my role as the INS District Counsel in Miami.
As all of these attorneys were told, the first thing to be done in a case like this one is to call the Respondent, what aliens in deportation /removal proceedings are called, as a witness.
That would be especially so here since Mr. Megahed, having been acquitted of criminal charges related to the matter before the immigration court, had no 5th amendment privilege against self-incrimination. Mr. Megahed had no protection against self-incrimination because in the correct application of the Constitutional concept of "double-jeopardy," having once been acquitted by a jury of his peers, his answers, as long as they were truthful, could not serve as a basis for the government to ever criminally prosecute him for the same conduct again.
Had he been called to testify, Mr. Megahed, unless he is the best witness ever to take the stand, would have had a very hard time. Especially at the hands of somebody as skilled and experienced as Ms. Garrett-Jackson.
Imagine the impact of Mr. Megahed having to answer even this one question in relation to whether he knew about what was in the trunk of the car, "Isn't true that, when you were arrested in South Carolina, you spoke to Ahmed Mohamed in the police car and asked him 'Did you tell them about the benzene?'
If he said yes to this question, what impact would it have had on the Judge Hurewitz? How could Judge Hurewitz, who was once one of INS Miami District's most accomplished criminal alien prosecutors, or any other judge, have found that Mr. Megahed was not involved in terrorism with Ahmed Mohamed, the now confessed and convicted terrorist supporter?
If Megahed answered "no," and Ms. Garrett-Jackson later introduced the audio tapes of the conversation and a government certified translator to testify as to what was said, could any judge have found that Mr. Megahed was not perjuring himself and seeking to conceal his involvement in terrorism just like Mohamed initially did?
So, why wasn't Youssef Megahed called to testify?
A second question to be asked would be, why, as reported, was ICE limited to presenting only two witnesses? As my experience would tell me, and even if the inquiry were limited to only what has here-to-date been released publicly, there were many more officers and federal agents who participated in the government investigation of Mr. Megahed and Mr. Mohamed. Agents such as Daniel J. McTavish, who wrote the search warrant affidavit, agents from the Tampa JTTF who handled all of the background and extended investigation facets of the case, and the FBI Headquarters personnel who oversaw and directed the efforts of the involved FBI field offices.
So, why only two? And, of those two, why only one actual investigator and one forensics analyst?
I don't know the answers to these questions and I don't think that ICE Headquarters is likely to publicly provide them anytime soon, if ever. But what is clear, at least to me, is that somebody, somewhere, made the decision to limit what the Garrett-Jackson team could present.
This has to be the case because it would be almost impossible to believe that Judge Hurwitz, had additional factual witnesses been presented, would have prevented ICE from putting them on the stand or disallowed their testimony. And it would be absolutely impossible for me to believe that Gina Garrett-Jackson, having been allowed the opportunity, would have simply chosen not to present everything she had.
It is therefore a question that the public, in my opinion, should ask.
A third question that should be posed is, what did all the other intelligence agencies say about Mr. Megahed and Mr. Mohamed, and was Ms. Garrett-Jackson given the ability to see whatever they may have had and to ask whether it could be declassified for use in the hearing? While this would be a very sensitive question that would likely get absolutely no response at all, unless it was asked by a Member of Congress, standard operating procedures would tell me that such evidence exists and that Ms. Garrett-Jackson, left with discretion to use her talents, would have used the hard-won professional relationships and credibility that she has with such agencies to get their cooperation in the deportation/removal effort.
Yet the same experience that tells me that such would have been the case suggests to me that Ms. Garrett-Jackson was not allowed to proceed with such an effort. For if she had been, and knowing the level of respect she enjoys in the community, I strongly believe that the case presented to Judge Hurewitz would have been much stronger and would have likely included, at the very least, a few more witnesses on behalf of the US government.
A fourth, and probably the most vexing, set of related questions, are why did ICE agree to conditions of release and how did they do so as quickly as they did?
The decision to release Yousef Megahed and the manner in which it was determined is so far out of the realm of standard operating procedure for ICE that it has be considered as determinative evidence of the fact that something went awry in the management of the prosecution effort in the case.
While those who aren't familiar with the inner-workings of ICE and the litigation of national security matters might not understand much less agree with this, the U.S. government would not have initiated this case unless it was truly believed that the evidence that they had against Mr. Megahed proved that he was involved in terrorism.
That is the way in all national security cases. Especially since those in the government well know that these types of cases will get the most scrutiny, the most media coverage, and, the most public attention.
The belief that evidence proves that someone has engaged in terrorism is not based on whim. It is based on vetting, after vetting, after vetting of the evidence, of the law, of the procedure and of the proposed litigation strategy at many levels of the federal bureaucracy, across many agencies, and before many very senior officials.
Where the disconnect occurs is when one of three things happen. The first is when operational developments interfere. The second is when politics kick in. And the last is when incompetence and fear take control.
Operational developments. They interfere, in general, when the analysis of and proposed effort in a pending action is materially affected by a related action elsewhere. One example is where developments in the case in litigation result in a determination that certain information that could come out in the litigation might compromise other national security efforts elsewhere.
Politics. Anyone involved in or who follows the subject of immigration in this country knows that politics are always part and parcel of the debate regarding the immigration issue of the moment.
Anyone involved in the prosecution of immigration national security matters knows that politics are at the forefront of the determinations made by those with the authority to make them.
What most people don't realize, however, is that rather than encourage national security prosecutions, politics impede such efforts.
The simple reason for this is that those at the top of the Executive Branch are, for the most part, politicians. Some are elected and some are appointed. But most, if not all, of them look first, if they have to be involved, at what one of these efforts will mean for their political agenda, political party and political futures.
When the political factor is allowed to control, all bets are off. And, in most cases, you can rest assured that what could have been accomplished by taking the path with the shortest distance between point A and point B will now likely be a treacherous trek through an unknown and ever-changing universe.
Fear and incompetence. The most dangerous of the three. The most likely to be fatal.
Fear always surfaces when any one of the other two factors apply; when operational developments or politics jump in and change the game. And, fear always controls when incompetence is involved.
In the Megahed case, I will take a guess and suggest that operational factors alone did not likely materially change the course of the prosecution.
I will also guess that politics were not the only issue.
Instead, I will assert, as a guess based on my experience in such matters, that fear and incompetence resulting from some possible operational developments and absolutely resulting from political considerations, took hold.
How can I be so sure?
The Department of Homeland Security placed Megahed in removal proceedings because they believed he was involved in terrorism. When they made the decision to do that, they asserted that he was an alien described in the terrorism sections of the Immigration and Nationality Act (INA). Even when they lost before the Immigration Judge, DHS/ICE had the legal ability to keep Mr. Megahed in custody pending an appeal. In fact, and as standard operating procedure should dictate, IF they really believed that he was a terrorist, the DHS/ICE plan should have been to keep Mr. Megahed in custody until and if the courts ordered otherwise. Instead, ICE agreed, on their own volition, to release him.
ICE did not wait for a judge to tell them that they had to. ICE did not wait until the Board of Immigration Appeals (BIA), on appeal, agreed with the judge. Instead, ICE simply made the decision to release Mr. Megahed on their own volition.
ICE did so in record time, and as reported, to the surprise of even Mr. Megahed's lawyer.
What happened was pre-planned. In the world of federal immigration bureaucracy, such decisions are never made quickly. In fact, they are mostly never even made anywhere within the time that they should have been.
And being that the decision was what it was, I guarantee you that it was not made by Ms. Garrett-Jackson, by her boss, or even by his boss. It was made at a much higher level. In Washington, D.C. By a politician, for political reasons, and based on the agreement of lesser senior federal officials, by way of an exercise akin to the "Cowardly Lion before the Great Wizard of Oz," whose continued federal careers are, if nowhere else but in their minds, in jeopardy if they even remotely think to protest.
Rather than a sudden revelation that Mr. Megahed was in fact not involved in terrorism, the pressure of public opinion, the call of those who cannot or will not understand or agree that a deportation action is a separate animal from a criminal prosecution, the whims of those in political power, and the fear and incompetence of those who might be able to change the result, in my opinion, led to the decision to release Mr. Megahed.
For these reasons, and because the American people really deserved to find out if Mr. Megahed, as previously asserted by ICE, really poses a danger to our community, I suggest that all of us should be asking: If he really is a terrorist, how could the federal government have presented its case without requiring Youssef Megahed to tell us all what he was doing with now convicted terrorist Ahmed Mohammed in South Carolina in a car with explosives and a computer that contained a bomb-making video for terrorists? How, could DHS prosecute its case against Youssef Megahed with the testimony of only one investigative witness and one forensics witness? How could ICE have failed to obtain and present whatever information could be released for use as evidence by the other agencies with information on Youssef Megahed? And, how could the federal government say one minute that it had evidence to arrest, charge and deport Youssef Megahed on terrorist grounds, and in the next say, on their own volition, that he could be released from custody?
Most importantly, however, we should be asking--Why?
The bottom line is that ICE lost this case at the trial level.
One of the reasons may have been because of the good luck that Megahed and Mohamed were arrested before anything bad could happen, if that is what they intended to do, and the bad luck of the FBI in not, because of the situation, having being afforded the opportunity to pro-actively and specifically investigate their activities.
Yet, from the looks of it, albeit from the outside looking in only at what is publicly available, they did so once again because of the very same factors that have caused them to lose similar cases.
So, and for the moment, the final question in this case should be, what is ICE going to do now to fix this?
The final question for the war on terrorism as it relates to immigration enforcement is, however, what should be done with ICE?
Dan Vara was the INS District Counsel in Miami, Florida from 1990 until 2003. He was also the Chief Counsel, ICE, in Orlando , Florida from 2003 until 2006. As the highest ranking federal immigration attorney in the State of Florida , he was at the forefront of many significant immigration enforcement matters involving counterterrorism and counterintelligence. He was also an instructor on such matters at INS, ICE and FBI conferences. He is now in private practice in South Florida .
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Working with your family can be challenging but at Il Piave, a leading Italian restaurant in Sydney's trendy inner west, they manage the restaurant like it's their home.
Il Piave is owned and run by three siblings: Andrew, Vanessa and Robert Martin.
Vanessa is the young passionate head chef, Robert and Andrew runs the restaurant and the business. Even Mum and Dad are involved, with Mrs Martin assisting Vanessa in making their homemade pasta and gelato.
Venture inside and see why their skills, passion and expertise have the customers coming back.


12-IN-ONE AMAZING LEAVE IN VOLUMIZING Addresses 12 common hair care concerns in one bottle. Maximizes volume, swells hair strands, provides anti-aging protection. Notable ingredients: Raw Avocado, Keratin Proteins, Amber Extracts, Whole Wheat derived proteins. Non-greasy.


Hello everyone! Today is the first day that All You Need Is Unicorns is featuring original work by other writers. We don't own any of these works, we just want to share these works with you with their permission. We're going to be doing this every Friday, so it will be called #featurefriday!
We love the idea of sharing ideas and thoughts and art. We love the inspiration that others' words give to our own creativity.
If you'd like your poetry featured on our site, feel free to send us some poetry, prose, haikus, lyrics, art, whatever floats your fancy.
So, it is with great pleasure that we showcase this first piece. We think it is fantastic. Let us know what you think!


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Adr Legal B&T Slp can be considered one of the first professional Societies of lawyers, experts in lobbying & public affairs, in the world to have contributed to individuate and naming with the terms "Cryptoeconomics" and "Tokenomics" some social-economic behaviours spread on internet from 2009 until now, but never classified.
The terms "Cryptocurrency economics" or "Cryptonomics" refers to cryptocurrencies and currency-based ledgers economy.
Cryptonomics concentrates on the creation of cryptocurrencies, involved in directing the behaviour of customers, sellers, service providers and some other parts inside the network.
In addition to this, it tries to determinate the users' behaviour, the coin price, the transactions' volume and the provided services.
Drawing on economy and the theory of games, it helps the developers to choose the best "cryptocurrency system parameters" to generate the desired results.
Microtokenomics concentrates on forces from a single ledger system.
Macrotokenomics analyses the entire ecosystem of a DLT network, that depends on third-party aggregated actions, like exchanges, regulators, governance groups and commercial partners.
Like Bergs, Davidson and Potts at RMIT identified, on the internet, a wider category called "Institutional Cryptoeconomics", that comprehends ledgers systems, their rules and the way in which other companies and societies interact between them, on the same way, Adr Legal B&T Slp introduced the concept of "Institutional Tokenomics" with which we mean the wider system of economic DLT interactions between all the token-based ledgers, organisations and societies and that differs from the traditional economy in the way that it is not based on predictive objectives, but on design goals.
For all these reasons, ADR LEGAL B&T Slp, with the help of its "Tokencomics lawyers", provide to governments, to ONG, to companies, all the technical and legal support to produce a good or a DLT technology-based (in particular on blockchain) and that can be distributed through a determined token and acquire force inside the Cryptocurrency market, that in 2018 achieved a 350 billion dollars capitalisation.


Long ago, Aristotle observed, "We are what we repeatedly do. Excellence, then, is not an act but a habit." After spending another January helping to pave the road to hell, I realize that my habits have again defeated my New Year's resolutions.
Years ago, after numerous struggles and frustrations, Arnold tried something different: "I assigned myself a small but meaningful behavioral change — a microresolution — and I succeeded in changing myself immediately. Yet it was only after succeeding at several microresolutions modeled on the first that I realized I had stumbled onto a method for making targeted mini-commitments that succeeded virtually every time." She had established a new pattern of behavior, a new habit.
1. We make the wrong resolutions whereas microresolutions focus on doing, not being. Being different follows, rather than precedes, deliberate action.
2. We depend solely on willpower to succeed whereas a microresolution is designed to reform a precise autopilot activity and requires little willpower to succeed.
3. We're too impatient whereas, when completing microresolutions, the key to lasting transformation is not speed or force but nurture.
4. We underestimate our mental and emotional resistance to change whereas microresolutions foster self-awareness and expose the hidden attitudes that thwart success.
5. We expect to fail whereas microresolutions are easy to keep.
Just as it is so easy to make a list of resolutions, it is just as easy to make a list of reasons and key points. It is important to note that, in her Introduction and first few chapters, Arnold identifies the WHAT and then devotes most of the material that follows to explaining HOW to use microresolutions to transform a life permanently.
I hope you are now resolved to read Small Move, Big Change: Using Microresolutions to Transform Your Life Permanently.
To learn more about Caroline and her work, please click here.
To watch her presentation at Microsoft headquarters, please click here.


text:
Simon Says, January 3, 2020
Thought Provoking Insights by Rabbi Simon Rosenbach
Last week, Pharaoh says to Joseph, "I hear that you are able to interpret dreams," and Joseph answers, "God will answer your questions." In the Joseph story, people need to put their trust in God, but between humans, there is a lot of mistrust going around.
As this Parashah, Vayigash, opens, Judah is explaining why his aged father (Jacob) will die when Benjamin does not come home. Many times, Judah alludes to Jacob's being alive: "we have an aged father," "and so, should I come to your servant, my father, and the lad [Benjamin] be not with us, for his life is bound to the lad's, when he saw the lad was not with us, he would die," and, "for how shall I go up to my father, if the lad be not with us? Let me see not the evil that would find out my father."
Joseph can not take it anymore, and he reveals himself to his brothers: "I am Joseph." And he adds, notwithstanding Judah's references to his living father, "Is my father still alive?"
All of the medieval commentators who are included in the traditional commentary don't comment on this. Arbarbanel (Isaac ben Judah Arbarbanel), who lived 300 years after Rashi, asks, "Why did Joseph ask, 'Is my father still alive,' when Judah made clear two or three times that Jacob was still alive?" Robert Alter suggests that Joseph does not trust Judah's description of his father, but I see something different.
I have said this many times, but Joseph does not ask, "Is our father alive?" Even though Joseph forgives the brothers in the following verses, he harbors lingering resentment that his brothers sold into slavery. The pronoun "my" signifies that Joseph has special relationship with Jacob.
Additionally, Joseph wants to know whether Jacob still holds a special place in his heart for Joseph, after all these years. "My father" signifies that special relationship, whereas "our father" would equate all of the brothers' relationship to Jacob.
The lesson to be learned is that a parent is obligated to develop a unique relationship with all of his or her children. The Torah treats favoritism as a given: Sarah loved Isaac more than she loved Ishmael, Rebekah loved Jacob more than she loved Esau, and favoritism moves the story along. Isaac was more deserving to perpetuate the Abrahamic tradition, and Jacob as more deserving to perpetuate the Abrahamic tradition, as well. But in the modern world, we should love each of children equally, but we can love them differently.

redpajama_set_name: RedPajamaCommonCrawl


Young children who live near a major roadway are at greater risk of developmental delays because of traffic-related pollutants, concluded a new study published in the journal Environmental Research. They were twice as likely to score lower on tests of communications skills, compared to those who live farther away from a major roadway.
An additional finding from the study was that the offspring of women who were exposed to high levels of ultra-fine airborne particles and ozone (traffic-related pollutants) during pregnancy were also at risk of developmental delays during infancy and early childhood.
Data from the Upstate KIDS Study was analyzed. The addresses of 5,825 study participants to a roadway data set were matched, calculating the distance of each address to the nearest major roadway. The home address, mother's work address during pregnancy and address of the child's day care location were matched to an Environmental Protection Agency data set to estimate the levels of air pollution.
Fine motor skills, large motor skills, communication, personal social functioning and problem-solving ability were evaluated for each child.
Compared to children living more than half a mile from a major roadway, children living from roughly 164 feet to .3 miles from a major roadway were twice as likely to have failed at least one screen of the communications domain.
Prenatal exposure to raised PM2.5 was also associated with a 1.6 to 2.7% higher risk of failing any developmental domain, while higher ozone exposure resulted in a .7 to 1.7% higher risk of failing a developmental domain.
Higher postnatal exposure to ozone was linked to a 3.3% higher risk of failing most domains of the developmental screen at 8 months, a 17.7% higher risk of overall screening failure at 24 months, and a 7.6% higher risk of overall screening failure at 30 months.
Exposure to pollution during pregnancy has been shown to be associated with low birthweight, preterm birth and stillbirth in earlier studies.
What this study has shown is that exposure to air pollutants during early childhood is associated with a higher risk for developmental delay, compared to similar exposures in the womb.
While the study has only demonstrated an association and not a causal relationship, these findings do bring out one more aspect of the harmful effects of air pollution and highlight the need to control the rising levels of air pollution.


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text:
\section{Introduction and concepts}

Labeled graphs are becoming an increasingly useful family of
mathematical models for a broad range of applications, such as
\emph{time tabling and scheduling}, \emph{frequency assignment},
\emph{register allocation}, \emph{computer security} and so on. In
\cite{Burris-Schelp1997}, Burris and Schelp introduced that a proper
edge $k$-coloring of a simple graph $G$ is called a \emph{vertex
distinguishing edge $k$-coloring} ($k$-vdec, or vdec for short) if for any two distinct
vertices $u$ and $v$ of $G$, the set of the colors assigned to the
edges incident to $u$ differs from the set of the colors assigned to
the edges incident to $v$. The minimum number of colors required for
a vertex distinguishing coloring of $G$ is denoted by $\chi
\,'_s(G)$. The maximum and minimum degrees of $G$ are denoted by $\Delta (G)$ and $\delta(G)$, respectively. Let $n_d(G)$ denote the number of vertices of degree $d$ with respect to $\delta(G)\leq d\leq
\Delta(G)$ in $G$, or write $n_d=n_d(G)$ if there is no confusion.
Burris and Schelp \cite{Burris-Schelp1997} presented the following
conjecture:

\begin{conj} \label{conj:c4-Burris-Schelp-conjecture}
Let $G$ be a simple graph having no isolated edges and at most one isolated vertex, and let $k$ be the smallest integer such that $(^k_d)\geq n_d$ for all $d$ with respect to $\delta(G)\leq d\leq
\Delta(G)$. Then $k\leq \chi \,'_s(G)\leq k+1$.
\end{conj}

It seems very difficult to settle down Conjecture \ref{conj:c4-Burris-Schelp-conjecture}, since no more results to verify it (cf. \cite{Balister-Bollobas-Schelp}, \cite{Burris-Schelp1997}, \cite{Cerny-Hornak-Sotak1996}). We show some results on trees to confirm positively Conjecture \ref{conj:c4-Burris-Schelp-conjecture}, and try to approximate $\chi\,'_s(G)$ of a graph $G$ by $\chi\,'_s(T)$ of some tree $T$ generated from $G$. Graphs mentioned here are finite, simple and undirected. We use standard terminology and notation of graph theory, and write $[m,n]=\{m,m+1,\dots ,n\}$ for integers $m,n$ with respect to $0\leq m<n$. Let $\pi$ be a $k$-$vdec$ of a graph $G$, and $S_i=\{uv:\pi(uv)=i,uv\in
E(G)\}$ for $i\in [1, k]$. We call $\pi$ an \emph{equitable}
$k$-\emph{vdec} if $S_i$ and $S_j$ differ in size at most one for
distinct $i,j\in [1, k]$. The smallest value of $k$ such that $G$ has
an equitable $k$-\emph{vdec} is denoted by $\chi\,'_{es}(G)$. Clearly, $\chi\,'_{s}(G)\leq \chi\,'_{es}(G)$. The set of neighbors of a vertex $u$ of $G$ is denoted as $N_G(u)$, or $N(u)$ if no confusion. The diameter of $G$ is written as $D(G)$. A \emph{leaf} is a vertex of degree one, and a \emph{$k$-degree vertex} is one of degree $k\geq 2$. $P_n$ is a path of length $n-1$.

A tree $Q$ of diameter four has its own vertex set $V(Q)=W_r\cup X_m\cup \big (\bigcup ^n_{i=1}Y_i\big )$, where $W_r=\{w_0,w_i:i\in [1,r]\}$ if $r\geq 1$, $X_m=\{s_i,s\, '_i:i\in [1,m]\}$ if $m\geq 1$, $Y_i=\{t_i,t\, '_{i,j}:j\in [1,r_i]\}$ for $r_i\geq 2$ and $i\in [1,n]$ if $n\geq 1$; and $Q$ has its own edge set $E(Q)=E(W_r)\cup E(X_m)\cup \big (\bigcup ^n_{i=1}E(Y_i)\big )$, where $E(W_r)=\{w_0w_i:i\in [1,r]\}$, $E(X_m)=\{w_0s_i,s_is\, '_i:i\in [1,m]\}$, $E(Y_i)=\{w_0t_i,t_it\, '_{i,j}:j\in [1,r_i]\}$ for $r_i\geq 2$ and $i\in [1,n]$. Thereby, we can write this tree $Q$ having diameter $D(Q)=4$ as $Q=Q(r,m,n)$ with the center $w_0$ hereafter. Especially, $Q(0,2,0)=P_5$ such that  $\chi\,'_s(Q(0,2,0))=n_1(Q(0,2,0))+2$.  let $U=\{Q(0,2,0)$, $Q(r,2,0)$, $Q(0,1,1)\}$. We will show the following results in this note.

\begin{thm} \label{them:main-theorem}
Let $T$ be a tree with $n_2(T)\leq n_1(T)$ and at least three vertices. Then $\chi\,'_s(T)=n_1(T)+1$ if $D(T)=3$ and $T\in U\setminus \{Q(0,2,0)\}$; $\chi\,'_s(T)=n_1(T)+2$ if $T=Q(0,2,0)$; and $\chi\,'_s(T)=n_1(T)$ if $D(T)\geq 4$ and $T\not \in U$.
\end{thm}

\begin{thm} \label{thm:GLT2-theorem-222}
Let $T$ be a tree with $n_2(T)\leq n_1(T)$ and at least three vertices. If $|E(T)|\leq 2[n_1(T)+1]$, then $\chi\,'_{es}(T)=\chi\,'_{s}(T)$.
\end{thm}

\section{Lemmas and proofs}

\begin{lem} \label{them:lemma-11}
Let $T$ be a tree with $n_2(T)\geq n_1(T)$. Then there exists a $2$-degree vertex such that one of its neighbors is either a leaf or a vertex of degree $2$.
\end{lem}
\begin{proof} Write $n_d=n_d(T)$ for $d\in [\delta(T),\Delta(T)]$, here $n_d=n_d(T)=0$ if $T$ has no vertices of degree $d$. Let $n_{\geq k}=|V(T)|-n_1-n_2-\cdots -n_{k-1}$ for $k\geq 2$. By contradiction. Assume that each $2$-degree vertex $x$ has its neighborhood $N(x)=\{x_1,x_2\}$ such that degree $\ud_T(x_i)\geq 3$ for $i=1,2$. So, we have $n_2\leq n_{\geq 3}$ by the assumption. Applying the formula $n_1=2+\sum _{3\leq d\leq \Delta(T)}(d-2)n_d$ shown in the article \cite{Yao-Zhang-Wang2010}, we have $n_1\geq 2+n_d$ for $d\geq 3$, and $n_1\geq 2+n_3+2n_{\geq 4}=2+2n_{\geq 3}-n_3$, which shows that $n_{\geq 3}\leq \frac{1}{2}(n_1+n_3)-1$. Thereby, the following inequalities
$$n_2\leq n_{\geq 3}\leq \frac{1}{2}(n_1+n_3)-1\leq n_1-2$$
show a contradiction.
\end{proof}

\begin{lem} \label{them:lemma-aa}
Let $T$ be a tree with diameter three. Then $\chi\,'_{s}(T)=n_1(T)+1$, and $\chi\,'_{es}(T)=\chi\,'_{s}(T)$.
\end{lem}
\begin{proof} Let two complete graphs $K_{1,m}$ and $K_{1,n}$ have their own vertex and edge sets as follows: $V(K_{1,m})=\{s,s_1,s_2,\dots,s_m\}$ and $E(K_{1,m})=\{ss_1,ss_2,\dots,ss_m\}$, $V(K_{1,n})=\{t,t_1,t_2,\dots,t_n\}$ and $E(K_{1,n})=\{tt_1,tt_2,\dots,tt_n\}$, where $s,t$ are the centers of $K_{1,m}$ and $K_{1,n}$. Each tree of diameter 3, denoted as $S_{m+1,n+1}$, can be obtained by joining two centers of $K_{1,m}$ and $K_{1,n}$ with an edge. So,  $S_{m+1,n+1}$ has its own vertex set $V(S_{m+1,n+1})=V(K_{1,m})\cup V(K_{1,n})$ and edge set $E(S_{m+1,n+1})=E(K_{1,m})\cup E(K_{1,n})\cup \{st\}$. Suppose that $f$ is a $k$-vdec of $S_{m+1,n+1}$ such that $k=\chi\,'_{s}(S_{m+1,n+1})$. Let $C(u,f)$ be the set of the colors assigned to the edges incident to $u$ of $S_{m+1,n+1}$. Since $C(s_i,f)\neq C(s_j,f)$, $C(s_i,f)\neq C(t_l,f)$, $C(t_i,f)\neq C(t_j,f)$, and $C(s,f)\neq C(t,f)$, so we know $k\geq m+n+1=n_1(S_{m+1,n+1})+1$. This $k$-vdec $f$ can be exactly defined as: $f(ss_i)=i$ for $i\in [1,m]$; $f(st)=m+1$; $f(tt_j)=m+1+j$ for $j\in [1,n]$. Thereby, $k=m+n+1$. By the definition of the $k$-vdec $f$, we are not hard to see $\chi\,'_{es}(T)=\chi\,'_{s}(T)$.
\end{proof}

\begin{lem} \label{them:lemma-bb}
Let $U=\{Q(0,2,0)$, $Q(r,2,0)$, $Q(0,1,1)\}$. For all trees $Q=Q(r,m,n)$ of diameter four we have $\chi\,'_s(Q)=n_1(Q)+1$ if $Q\in U\setminus \{Q(0,2,0)\}$; $\chi\,'_s(Q)=n_1(Q)+2$ if $Q=Q(0,2,0)$; and $\chi\,'_s(Q)=n_1(Q)$ for $Q\not \in U$. Furthermore,  $\chi\,'_{es}(Q)=\chi\,'_{s}(Q)$.
\end{lem}
\begin{proof} Using the description of a tree $Q=Q(r,m,n)$ of diameter four in Section 1. Let $n_1=n_1(Q)$, and $f$ be a $k$-vdec of $Q$ such that $k=\chi\,'_s(Q)$. Note that $D(Q)=4$.

\emph{Case A.} $(r,m,n)=(r,m,0)$ with $m\geq 2$ and $r\geq 0$. Clearly, $n_1=r+m$.

\emph{Case A1.} $(r,m,0)=(r,2,0)$, $n_1=r+2$. We can easily see $\chi\,'_s(Q)=n_1+1$ for $r\geq 0$, since we have to color one of two edges $w_0s_1$ and $w_0s_2$ with one color that is not in $\{f(w_0w_s),f(s_is\, '_i):s\in [1,r], i\in [1,2]\}$. As $r=0$, $Q$ is a path with $4$ vertices, so $\chi\,'_s(Q)=n_1+2=4$.

\emph{Case A2.} $(r,m,0)$ with $m\geq 3$, $n_1=r+m$. We show $f$ in the following: $f(s_is\, '_i)=i$ for $i\in [1,m]$; $f(w_0s_i)=i+1$ for $i\in [1,m-1]$, and $f(w_0s_m)=1$; $f(w_0w_j)=m+j$ for $j\in [1,r]$ if $r\neq 0$. Clearly, $C(u,f)\neq C(v,f)$ for any two vertices $u,v\in V(Q)$, which means  $\chi\,'_s(Q)=n_1$.

\emph{Case B.} $(r,0,n)$ with $n\geq 2$, $n_1=r+\sum^n_{i=1}r_i$. It is easy to show $\chi\,'_s(Q)=n_1$ in this case, since $n_2\leq 1$.

\emph{Case C.} $(0,m,n)$ with $m\geq 1$ and $n\geq 1$, $n_1=m+\sum^n_{i=1}r_i$.

\emph{Case C1.} $(0,m,n)=(0,1,1)$, $n_1=1+r_1$. Since $f(s_1s\, '_1)=f(w_0t_1)$ if $\chi\,'_s(Q)=n_1$ in this case, we can see $C(w_0,f)=C(s_1,f)$; a contradiction. So $\chi\,'_s(Q(0,1,1))=2+r_1=n_1+1$.

\emph{Case C1.} $(0,m,n)=(0,1,n)$ with $n\geq 2$, $n_1=1+\sum^n_{i=1}r_i$. To show $\chi\,'_s(Q)=n_1$, we have $f$ defined as: $f(s_1s\, '_1)=1$; $f(t_1t\, '_{1,j})=1+j$ for $j\in [1,r_1]$; $f(t_it\, '_{i,j})=f(t_{i-1}t\, '_{i-1,j})+j$ for $j\in [1,r_i]$ and $i\in [2,n]$; $f(w_0s_1)=f(t_{n}t\, '_{n,1})$; $f(w_0t_j)=f(t_{j+1}t\, '_{j+1,1})$ for $j\in [1,n-1]$; $f(w_0t_n)=f(s_1s\, '_1)$.

\emph{Case C2.} $(0,m,n)=(0,m,1)$ with $m\geq 2$, $n_1=m+r_1$. For showing $\chi\,'_s(Q)=n_1$ we define: $f(s_is\, '_i)=i$ for $i\in [1,m]$; $f(w_0s_i)=i+1$ for $i\in [1,m-1]$, and $f(w_0s_m)=f(t_1t\, '_{1,1})$; $f(t_1t\, '_{1,j})=m+j$ for $j\in [1,r_1]$ and $f(w_0t_1)=m$.

\emph{Case C3.} $(0,m,n)=(0,m,n)$ with $m\geq 2$ and $n\geq 1$, $n_1=m+\sum^n_{i=1}r_i$. We have $\chi\,'_s(Q)=n_1$ by defining $f$ appropriately by the methods showing in Case C1 and Case C2.

\emph{Case D.} $(r,m,n)$ with $r\geq 1$, $m\geq 1$ and $n\geq 1$, $n_1=r+m+\sum^n_{i=1}r_i$. By the techniques use in the above cases, we can define $f$ to show $\chi\,'_s(Q)=n_1$.

Through the above Cases A, B, C and D, we conclude $\chi\,'_{es}(Q)=\chi\,'_{s}(Q)$, since one color is used at most twice under the vdec $f$. The lemma is covered.
\end{proof}

\textbf{The proof of Theorem \ref{them:main-theorem}.} Let $n_i=n_i(T)$ for $i=1,2$, and neighborhoods $N(w)=N_T(w)$ for $w\in V(T)$. Because $T=K_{1,n-1}$ when $D(T)=2$, so $\chi\,'_{s}(T)=n_1=n-1$. For $D(T)=3,4$, by Lemmas \ref{them:lemma-aa} and \ref{them:lemma-bb}, the theorem holds true. So, we show $\chi\,'_{s}(T)=n_1$ by induction on vertex numbers of trees $T$ with diameter $D(T)\geq 5$. We will use the description of trees having diameter four in the following discussion.

\emph{Case 1.} There exists a leaf $v$ having a neighbor $u$ with degree $\ud_T(u)\geq 4$. Let $T\,'=T-v$, so $n_1(T\,')=n_1-1$. Clearly, $D(T\,')\geq 5$.

\emph{Case 1.1.} $n_2(T\,')\leq n_1(T\,')$. Then by induction hypothesis, there is an edge coloring $\xi_{T\,'}:E(T\,')\rightarrow \{1,2,\dots,b\,'\}$ such that $\chi\,'_s(T\,')=b\,'=n_1(T\,')=n_1-1$. It is straightforward to define a vdec $\xi_{T}$: $E(T)\rightarrow \{1,2,\dots,b\,', b\,'+1\}$ such that $\xi_{T}(e)=\xi_{T\,'}(e)$ for $e\in E(T)\setminus \{uv\}$, and $\xi_{T}(uv)=b\,'+1$.

\emph{Case 1.2.} $n_1(T\,')=n_2(T\,')-1$. By Lemma \ref{them:lemma-11}, there exists a $2$-degree vertex $x\in V(T\,')$ with $N_{T\,'}(x)=\{x_1,x_2\}$ such that $\ud_{T}(x_1)=\ud_{T\,'}(x_1)\leq 2$. Notice that $\ud_{T\,'}(x)=\ud_{T}(x)$ and $\ud_{T\,'}(x_1)=\ud_{T}(x_1)$. Let $T_1$ be the tree obtained from $T\,'$ by suppressing the vertex $x$, that is, $T_1=T\,'-x+x_1x_2$. Clearly, $D(T_1)\geq 4$, $n_1(T_1)=n_1(T\,')$ and $n_2(T_1)=n_2(T\,')-1$.

If $D(T_1)=4$, we have $T_1=Q(0,m,0)$ with $m\geq 3$, since $n_1(T_1)=n_2(T_1)$ and  $\ud_T(u)\geq 4$. Thus, $u$ is the center of $T_1=Q(0,m,0)$, the leaf $v$ is the leaf $w_1$ of $Q(1,m,0)$, and $T$ is the tree obtained by subdivision of the edge $s_1s\, '_1$. It is not difficult to see $\chi\,'_{s}(Q(1,m,0))=n_1(Q(1,m,0))$, so $\chi\,'_{s}(T)=n_1$.

If $D(T_1)\geq 5$, again by induction hypothesis, there is a vdec $\xi_{T_1}:~E(T_1)\rightarrow \{1,2,\dots,b_1\}$ such that $\chi\,'_s(T_1)=b_1=n_1(T_1)=n_1-1$. Hence, $T$ has a proper edge coloring $\xi_{T}$ defined as: $\xi_{T}(e)=\xi_{T_1}(e)$ for $e\in E(T)\setminus \{uv, xx_1,xx_2\}$, and $\xi_{T}(uv)=b_1+1$, $\xi_{T}(xx_1)=b_1+1$ and $\xi_{T}(xx_2)=\xi_{T_1}(x_1x_2)$. It is easy to check $\xi_{T}$ to be a desired $n_1$-vdec of $T$, since $C(\xi_{T},u)\neq C(\xi_{T}, x_1)$.

\emph{Case 2.} There is a leaf $v$ having a 3-degree neighbor $u$, and Case 1 is false.

\emph{Case 2.1.} $v\,'$ is another leaf in the neighborhood $N(u)=\{v,v\,',u\,'\}$, and $T$ has a $2$-degree vertex $x$ having its neighborhood $\{x_1,x_2\}$ such that $x_1$ is a leaf of $T$.

We have a tree $T_1=T-\{v,v\,',x\}+x_1x_2$. Clearly, $D(T_1)\geq 3$ since $D(T)\geq 5$, $n_1(T_1)=n_1-1$ and $n_2(T_1)=n_2-1$.

If $D(T_1)=3$, $T_1$ is a $3$-diameter tree $S_{m+1,n+1}$ (we use the description shown in the proof of Lemma \ref{them:lemma-aa}) obtained by joining two centers of $K_{1,m}$ and $K_{1,n}$ with an edge. Since $D(T)\geq 5$, so $T$ can be obtained by subdividing the edge $ss_1$ ($=x_1x_2$) to form a path $sxs_1=x_1xx_2$ of $T$ and joining $t_1$ ($=u$) to two new vertices $v$ and $v\, '$ to $S_{m+1,n+1}$. It is not hard to make a desired $n_1$-vdec of $T$ based on the structure of $T$.

If $D(T_1)=4$, then $T_1=Q(r,m,n)$. If $T_1=Q(0,2,0)$, thus it goes to $T$ such that $n_1=3$ and $n_2=4$; a contradiction. As $T_1=Q(r,m,n)\neq Q(0,2,0)$, we can show $\chi\,'_{s}(T)=n_1$ by the edge colorings used in the proof of Lemma \ref{them:lemma-bb}.

For $D(T_1)\geq 5$, $T_1$, by induction hypothesis, has a vdec $\xi_{T_1}:~E(T_1)\rightarrow \{1,2,\dots,b_1\}$ such that $\chi\,'_s(T_1)=b_1=n_1(T_1)=n_1-1$. Notice that $u$ is a leaf of $T_1$. So, we can extend $\xi_{T_1}$ to a vdec $\xi_{T}$ of $T$ as follows: $\xi_{T}(e)=\xi_{T_1}(e)$ for $e\in E(T)\setminus \{uv, uv\,',xx_1,xx_2\}$; if $\xi_{T_1}(x_1x_2)\not \in C(\xi_{T_1},u\,')$, we set $\xi_{T}(uu\,')=b_1+1$, $\xi_{T}(uv\,')=\xi_{T_1}(uu\,')$, $\xi_{T}(uv)=\xi_{T_1}(x_1x_2)$, $\xi_{T}(xx_2)=\xi_{T_1}(x_1x_2)$ and $\xi_{T}(xx_1)=b_1+1$; if  $\xi_{T_1}(x_1x_2)\in C(\xi_{T_1},u\,')$, we define $\xi_{T}(uu\,')=\xi_{T_1}(uu\,')$, $\xi_{T}(uv)=\xi_{T_1}(x_1x_2)$, $\xi_{T}(uv\,')=b_1+1$, $\xi_{T}(xx_2)=b_1+1$  and $\xi_{T}(xx_1)=\xi_{T_1}(uu\,')$.

\emph{Case 2.2.} $\ud_T(v\,')=1$ for $v\,'\in N(u)=\{v,v\,',u\,'\}$, and $T$ has no a $2$-degree vertex that is adjacent to a leaf of $T$. Let $P=p_1p_2\cdots p_{D-1}p_D$ be a longest path of $T$, where $D$ is the diameter of $T$. Here, $\ud_T(p_2)=3=\ud_T(p_{D-1})$.

Without loss of generality, $p_2\neq u$, so $p_2$ is adjacent to two leaves $p_1,p\,'_1\in N(p_2)=\{p_1,p\,'_1,p_3\}$ with $\ud_T(p_3)\neq 1$. Suppose that $x,y$ are two $2$-degree vertices of $T$. Let $N(x)=\{x_1,x_2\}$ and $N(y)=\{y_1,y_2\}$, then we know that $\ud_T(x_i)\geq 2$ and $\ud_T(y_i)\geq 2$ for $i=1,2$ by the hypothesis of Case 2.2. Make a tree $T_2=T-\{v,v\,',p_1,p\,'_1,x,y\}+\{x_1x_2,y_1y_2\}$ with $n_1(T_2)=n_1-2$ and $n_2(T_2)=n_2-2$.

If $D(T_2)=2$, so $T_2=K_{1,m}$ with $V(K_{1,m})=\{s,s_1,s_2,\dots,s_m\}$ and $E(K_{1,m})=\{ss_1,ss_2,\dots,ss_m\}$ for $m\geq 2$. Without loss of generality, $u=s_1$, $p_2=s_2$, $ss_1=x_1x_2$ and $ss_j=y_1y_2$ for $j\neq 1$ since $D(T)\geq 5$. So, $D(T)=5$ or $6$, $n_1\geq 4$ and $n_2=2$. The structure of $T$ enables us to show $\chi\,'_{s}(T)=n_1$ by defining a desired $n_1$-vdec of $T$.

Consider the case $D(T_2)=3$. As $T_2=P_4$, $T$ is obtained by replacing by a path $w_0xx_1yt_1$ the edge $w_0t_1$ of $Q(0,0,2)$ with $r_1=r_2=2$ ($t_2=u$, $t_1=p_2$), so $n_1=n_2=4$. It is easy to define a desired $4$-vdec of $T$ by means of the techniques used in the proof of Lemma  \ref{them:lemma-bb}. When $T_2=S_{m+1,n+1}$ with $m+n\geq 3$, the structure of $T$ is clear such that we can show easily $\chi\,'_{s}(T)=n_1$.

For $D(T_2)=4$, thus $T_2=Q(r,0,n)$ with $0\leq r\leq 1$ and $n\geq 2$ according to $T$ has no a $2$-degree vertex that is adjacent to a leaf of $T$. We have two subcases $T_2=Q(1,0,2)$ and $T_2=Q(0,0,n)$ with $n\geq 3$. So, $x_1x_2=w_0t_1$ and $y_1y_2=w_0t_2$, and $u$ and $p_2$ both are the leaves of $T_2=Q(1,0,2)$ or $T_2=Q(0,0,n)$ with $n\geq 3$. We claim $\chi\,'_{s}(T)=n_1$ by the structure of $T$ in this situation.

When $D(T_2)\geq 5$, by induction hypothesis, we take a vdec $\xi_{T_2}:~E(T_2)\rightarrow \{1,2,\dots,b_2\}$ having $\chi\,'_s(T_2)=b_2=n_1(T_2)=n_1-2$, and extend $\xi_{T_2}$ to a proper edge coloring $\xi_{T}$ of $T$ by setting $\xi_{T}(e)=\xi_{T_2}(e)$ for $e\in E(T)\setminus \{uv, uv\,',uu\,',xx_1,xx_2, p_2p_1,p_2p\,'_1,p_2p_3,yy_1,yy_2\}$, and $\xi_{T}(uu\,')=b_2+2$, $\xi_{T}(uv\,')=\xi_{T_2}(uu\,')$, $\xi_{T}(uv)=b_2+1$, $\xi_{T}(xx_2)=\xi_{T_2}(x_1x_2)$, and $\xi_{T}(xx_1)=b_2+1$; $\xi_{T}(p_2p_3)=b_2+1$, $\xi_{T}(p_2p\,'_1)=\xi_{T_2}(p_2p_3)$, $\xi_{T}(p_2p_1)=b_2+2$, $\xi_{T}(yy_2)=\xi_{T_2}(y_1y_2)$, and $\xi_{T}(yy_1)=b_2+2$. Since $C(\xi_{T_2},p_3)\setminus \{\xi_{T_2}(p_2p_3)\}$ is not equal to one of $C(\xi_{T_2},x_i)\setminus \{\xi_{T_2}(x_1x_2)\}$ for $i=1,2$, and $C(\xi_{T_2},u)\setminus \{\xi_{T_2}(uu\,')\}$ is not equal to one of $C(\xi_{T_2},y_i)\setminus \{\xi_{T_2}(y_1y_2)\}$ for $i=1,2$, so $C(\xi_{T},p_3)\neq C(\xi_{T},x_1)$ and $C(\xi_{T},u\,')\neq C(\xi_{T},y_1)$, without loss of generality. Hence, it is not hard to verify that $\xi_{T}$ is a vdec of $T$ such that $\chi\,'_s(T)= b_2+2=n_1$.

\emph{Case 2.3.} The above Case 2.1 and Case 2.2 are false, simultaneously. So, any leaf of $T$ is either adjacent to a $2$-degree vertex or a 3-degree vertex having two neighbors of non-leaves. Thereby, $\ud_T(v)=1$, $\ud_T(v\,')\geq 2$ and $\ud_T(u\,')\geq 2$ in $N(u)=\{v,v\,',u\,'\}$. Let $P=x_1x_2\cdots x_{D-1}x_D$ be a longest path of $T$ with $\ud_T(x_2)=2$ or $\ud_T(x_{D-1})=2$, where $D=D(T)$. Without loss of generality, $x_2$ has its two neighbors $x_1$ and $x_3$ holding $\ud_T(x_1)=1$ and $\ud_T(x_3)\geq 2$.

We have a tree $T_3=T-\{v,u,x_2\}+\{x_1x_3,v\,'u\,'\}$. Clearly, $D(T_3)\geq 3$ since $D(T)\geq 5$, $n_1(T_3)=n_1-1$ and $n_2(T_3)=n_2-1$. If $D(T_3)=3$, the possible case is $T_3=P_4=x_1x_3x_4x_5=x_1u\, 'v\, 'x_5$, which implies $T=x_1x_2u\, 'uv\, 'x_5+uv$. Thereby, we have $\chi\,'_{s}(T)=n_1$. If $D(T_3)=4$, then  $T_3=Q(1,2,0)$ or $T_3=Q(0,m,0)$ with $m\geq 3$ ($T_3=Q(0,2,0)$ will induce $T$ has $n_1=3$ and $n_2=4$; a contradiction). Therefore, $s_1=u\,'$, $w_0=v\,'$, and $x_1$ is a leaf of $T_3=Q(1,2,0)$ or $T_3=Q(0,m,0)$ with $m\geq 3$. The structure of $T$ is clear when $D(T_3)=4$, so we can show a $n_1$-vdec  of $T$ by the methods used in the proof of Lemma  \ref{them:lemma-bb}.

For $D(T_3)\geq 5$, by induction hypothesis, $T_3$ admits a vdec $\xi_{T_3}:~E(T_3)\rightarrow \{1,2,\dots,b_3\}$ such that $\chi\,'_s(T_3)=b_3=n_1(T_3)=n_1-1$. It is straightforward to define a $n_1$-vdec $\xi_{T}$ of $T$ as follows: $\xi_{T}(e)=\xi_{T_3}(e)$ for $e\in E(T)\setminus \{uv, uv\,',uu\,',x_2x_1,x_2x_3\}$, and $\xi_{T}(uu\,')=b_3+1$ if $\xi_{T_3}(x_1x_3)\not \in C(\xi_{T_3},u\,')$, $\xi_{T}(uv\,')=\xi_{T_3}(u\,'v\,')$, $\xi_{T}(uv)=\xi_{T_3}(x_1x_3)$, $\xi_{T}(x_2x_3)=\xi_{T_3}(x_1x_3)$, and $\xi_{T}(x_2x_1)=b_3+1$.

\emph{Case 3.} The above Case 1 and Case 2 do not appear simultaneously, and every leaf is adjacent to a $2$-degree vertex in $T$, which means $n_2=n_1$.

\emph{Case 3.1.} Each $2$-degree vertex $x$ has its neighborhood $N(x)=\{x_1,x_2\}$ such that $\ud_T(x_1)=1$ and $\ud_T(x_2)\geq 3$, and no two $2$-degree vertices have a common neighbor. Let $R=p_1p_2\cdots p_l$ be a longest path in $T$, where $l=D(T)$. Thereby, $\ud_T(p_3)\geq 3$ and $\ud_T(p_{l-2})\geq 3$ since  $n_2=n_1$. Notice that $\ud_T(x)\leq 2$ for $x\in N(p_3)\setminus \{p_2,p_4\}$ (resp. $x\in N(p_{l-2})\setminus \{p_{l-3},p_{l-1}\}$) because $x\not \in V(R)$, and furthermore this vertex $x$ is neither a leaf (Case 2 has been assumed to disappear) nor a $2$-degree vertex (by the hypothesis of this subcase and $n_2=n_1$). We claim that this subcase does not exist.

\emph{Case 3.2.} Case 3.1 does not exist at all. We have a subgraph $H$ with $V(H)=\{w,u\}\cup V\,'$ with $m\geq 2$ and $V\,'=\{y_i,x_i:i\in [1,m]\}$ and $E(H)=\{wu, uy_i, y_ix_i: i\in [1,m]\}$, where $\ud_T(w)\geq 2$, every $y_i$ is a $2$-degree vertex and every $x_i$ is a leaf for $i\in [1,m]$. Then we have a tree $T_1=T-V\,'$ such that  $D(T_1)\geq 3$ because $D(T)\geq 5$.

If $D(T_1)=3$, the possible structure is $T_1=P_4$, but, which implies $n_2=n_1+1$; a contradiction.

For $D(T_1)=4$, thus $T_1=Q(1,m,0)$ with $m\geq 2$ under the restriction of Case 3.2, and $s\, '_1=u$. The clear structure of $T$ enables us to show $\chi\,'_{s}(T)=n_1$.

For $D(T_1)\geq 5$, $n_1(T_1)=n_1-m+1> n_2(T_1)=n_2-m$ from $n_1=n_2$, and $u$ is a leaf of $T_1$. By induction hypothesis, $T_1$ admits a vdec $\xi_{T_1}:~E(T_1)\rightarrow S=\{1,2,\dots,b_1\}$ having $\chi\,'_s(T_1)=b_1= n_1(T_1)=n_1-(m-1)$. We can define a $n_1$-vdec $\xi_{T}$ of $T$ as follows: $\xi_{T}(e)=\xi_{T_1}(e)$ for $e\in E(T)\setminus E(H)$, and $\xi_{T}(wu)=\xi_{T_1}(wu)$; $\xi_{T}(uy_i)=b_1+i$ for $i\in [1,m-1]$, and $\xi_{T}(uy_m)\in S\setminus \{\xi_{T_1}(wu)\}$; $\xi_{T}(y_1x_1)=\xi_{T_1}(wu)$, $\xi_{T}(y_jx_j)=b_1+j-1$ for $j\in [2,m]$.

Thereby, Theorem \ref{them:main-theorem} follows from the principle of induction.\qqed

\vskip 0.2cm

\textbf{Proof of Theorem \ref{thm:GLT2-theorem-222}.} It is not hard to let the number of a color be one or two in the desired vdecs $\xi_{T}$ of $T$ in all cases of the proof of Theorem \ref{them:main-theorem}, since $|E(T)|\leq 2(n_1+1)$. Thereby, Theorem \ref{thm:GLT2-theorem-222} follows from the principle of induction.\qqed

\begin{cor} \label{thm:GLT2-111}
Let $G$ be a connected graph having cycles, $p$ vertices and $q$ edges. If $n_2(G)\leq 2(q-p+1)+n_1(G)$, then $\chi\,'_{s}(G)\leq 2(q-p+1)+n_1(G)$.
\end{cor}
\begin{proof} Let $H$ be a spanning tree of $G$,
so $E\,'=E(G)\setminus E(H)$. Hence, we have another tree $T$
obtained by deleting every edge $uv\in E\,'$, and then adding two
new vertices $u\,',v\,'$ by joining $u\,'$ with $u$ and $v\,'$ with
$v$ simultaneously. Clearly, $n_1(T)=2(q-p+1)+n_1(G)$, and $n_2(T)=n_2(G)$ and
$\Delta(T)=\Delta(G)$. On the other hand, each $k$-vdec $\pi$
of $G$ with $k=\chi\,'_{s}(G)$ corresponds to an edge coloring $\pi\,'$ of $T$ such that
for any two distinct
non-leaf vertices $u$ and $v$ of $T$, the set of the colors assigned to the
edges incident to $u$ differs from the set of the colors assigned to
the edges incident to $v$.  Also, $\pi\,'$ is a
proper edge coloring of $T$ by setting
$\pi\,'(uu\,')=\pi\,'(vv\,')=\pi(uv)$ for $uu\,',vv\,'\in E(T)$
and $uv\in E\,'$; and $\pi\,'(e)=\pi(e)$ for $e\in E(H)\subset
E(T)$. Thereby, $k\leq
\chi\,'_{s}(T)$ since $k=\chi\,'_{s}(G)$. This corollary follows by Theorem \ref{them:main-theorem}.
\end{proof}

\begin{cor} \label{thm:GLT2-222}
Let $T$ be a spanning tree of a connected graph $G$, and $G[E\,']$ be an induced graph over the edge subset $E\,'=E(G)\setminus E(T)$. Then $\chi\,'_{s}(G)\leq \chi\,'_{s}(T)+\chi\,'(G[E\,'])$, where $\chi\,'(H)$ is the chromatic index of a graph $H$.
\end{cor}

As further work we present the following

\begin{conj} \label{conj:shan-yybb-chen}
Let $T$ be a tree with $2n_2(T)\leq (n_1(T)+k-1)^2$ for $k\geq 1$. Then $\chi \,'_s(T)\leq n_1(T)+k$.
\end{conj}

\vskip 0.4cm

\noindent \textbf{Acknowledgment.} The authors thank sincerely two referees' sharp opinions and helpful suggestions that improve greatly the article. B. Yao was supported by the National Natural Science
Foundation of China under Grant No. 61163054 and No. 61363060; X.-en Chen was supported by the National Natural Science Foundation of China under Grant No. 61163037.

{\footnotesize

redpajama_set_name: RedPajamaArXiv


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Leather long wallet | Monogrammed long wallet by nobrand.
A favorite of our artisans, an exquisite long wallet in Italian leather paired with suede for a sumptuous tactile finish.
Handmade Leather long wallet. Pictured in dark grey leather & cobalt blue suede and grey stitching.
Now choose how you want your perfect leather long wallet made.