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By Washington's teeth! U.S. presidential history, now on Google Arts & Culture
Did you know that the Bush Family has a favorite taco recipe, which First Lady Barbara Bush described as "loved by all who love Mexican food"? Or that George Washington's dentures were not made of wood as is popularly thought, but actually from human and cow teeth as well as ivory? Or how about that, to celebrate his Inauguration, Theodore Roosevelt received a lock of president Lincoln's hair as a gift?
No, we're not presidential scholars; we're just excited for Presidents' Day! Today, as a follow-up to our American Democracy collection, Google Arts & Culture is partnering with more than 30 cultural institutions to bring you history from the United States presidency, available at g.co/americandemocracy.
With over 2,000 new artifacts, photos, pictures and more, and 63 new exhibits (for 158 exhibits, total) this collection invites you to remember and celebrate the history, lives and legacies of the 44 U.S. presidents. Take an immersive tour of presidents' iconic homes and get a sneak peek into their private lives—from childhood and family life, to favorite pastimes and chefs—in addition to their public accomplishments.
Explore the weird world of the presidential pets—other than dogs, there have been raccoons, sheep, horses, badgers, and even a pygmy hippopotamus and elephants.
You can view 25 presidential portraits captured using Google's Art Camera. These gigapixel quality images allow you to zoom in and explore details of these portraits more thoroughly than you could with the naked eye.
Dwight D. Eisenhower, 35th president of the United States.
We're making available 17 new 360-degree virtual tours that transport you to places full of presidential history. Using the Google Arts & Culture App (available on iOS and Android) and Google Cardboard, take a virtual tour of places like the home of Franklin D. Roosevelt and the Ulysses S. Grant National Historic Site. And, in addition, educators can use Google Expeditions to take students on a guided tour of the White House, right from their desks! There are 14 Google Expeditions relating to the Office of the President, including Presidential Museums and work by the First Ladies, all great trips for students across grades and subjects.
Take a virtual reality tour of the White House, right from wherever you are.
Ever wonder what it's like to travel like POTUS? Take a look at Ronald Reagan's Air Force One (now housed in his Presidential Library) and other ways presidents have traveled in safety and style.
Our Presidents' Day collection covers the vast political and personal histories of our U.S. heads of state, full of intriguing and surprising stories that allow for anyone with an internet connection to turn into a presidential historian. We hope you enjoy!
Source: The Official Google Blog
This entry was posted in Official Google Blog and tagged Arts & Culture, Public Policy on February 15, 2017 by Lucy Schwartz.
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This 2017-2018 school year, AASB will award fifteen graduating seniors each with a $1,500 scholarship to apply toward their post-secondary education. The scholarship may be applied toward the student's choice of a business, trade, or a college institution. Click here for application information.


Spinal manipulation for the early management of persistent non-specific low back pain--a critique of the recent NICE guidelines.
Ernst, E. "Spinal Manipulation for the Early Management of Persistent Non-specific Low Back Pain--a Critique of the Recent NICE Guidelines." International Journal of Clinical Practice, vol. 63, no. 10, 2009, pp. 1419-20.
Ernst E. Spinal manipulation for the early management of persistent non-specific low back pain--a critique of the recent NICE guidelines. Int J Clin Pract. 2009;63(10):1419-20.
Ernst, E. (2009). Spinal manipulation for the early management of persistent non-specific low back pain--a critique of the recent NICE guidelines. International Journal of Clinical Practice, 63(10), pp. 1419-20. doi:10.1111/j.1742-1241.2009.02163.x.
Ernst E. Spinal Manipulation for the Early Management of Persistent Non-specific Low Back Pain--a Critique of the Recent NICE Guidelines. Int J Clin Pract. 2009;63(10):1419-20. PubMed PMID: 19689486.
Guidelines in the evaluation and management of low back pain.
Spinal manipulation compared with back school and with individually delivered physiotherapy for the treatment of chronic low back pain: a randomized trial with one-year follow-up.
Early management of persistent non-specific low back pain: summary of NICE guidance.
Is spinal manipulation an effective treatment for low back pain? Yes: spinal manipulation is a useful adjunct therapy.
Comparison of general exercise, motor control exercise and spinal manipulative therapy for chronic low back pain: A randomized trial.
A clinical prediction rule for classifying patients with low back pain who demonstrate short-term improvement with spinal manipulation.
Congruence between decisions to initiate chiropractic spinal manipulation for low back pain and appropriateness criteria in North America.
Chiropractic for low back pain. Efficacy of spinal manipulation for low back pain has not been reliably shown.
A perspective for considering the risks and benefits of spinal manipulation in patients with low back pain.
What are the guidelines for acute low back pain?


The Astria Suparak firing: Up or down for West Fayette?
I had a few thoughts as I watched hundreds of art lovers Thursday night moving from The Delavan to The Red House to The Warehouse, along West Fayette Street.
I thought of how this would have been unimaginable 20 years ago, when The Warehouse was a warehouse and The Redhouse was a battered old tavern.
I thought of how incredible it would be, in this budding gallery district, if those chipped and battered railroad bridges were handed over to teams of artists as enormous examples of public art.
I thought of how much this district might achieve if this $56 million investment on the Near West Side comes to be.
And I thought of the irony of how this great burst of energy on a downtown street was potentially a form of a goodbye, touched off by the firing of a person hired with the mission of helping to generate that energy.
Astria Suparak, curator of The Warehouse Gallery, has been dismissed by the university. Jeffrey Hoone, the university official who dismissed her, said he is constrained from speaking publicly of the reasons, constraints that go with any personnel decision. In that void, angry members of the arts community contend that Suparak was let go for reasons from censorship to institutional jealousy at her success.
You can read many of those arguments in this blog, much of it assembled by Joanna Spitzner, an assistant professor at SU and one of those organizing support for Suparak.
I lead off the reader response with this note from Lonnie Chu, an instructor at SU and a community activist in Eastwood.
And I look forward to hearing your thoughts, one way or another, regarding West Fayette - and what happens after Suparak.
(http://syracuse-warehouse.blogspot.com/) They come from not only Syracuse, but from other cities, other states, even other countries.
Astria is a treasure that the city and the university need to support and nurture, for she is a one-woman Connective Corridor, making art fun, accessible, perhaps troubling, but always exciting.
There have been a number of themes expressed by those writing in. I'd like to summarize some of them here.
Syracuse Arts Community Unwilling to Let Dismissed Curator Go.
- Astria is a magnet that has attracted people to live in the city and artists to do shows here. In fact, "The Yes Men" agreed to come here because of Astria and have thus canceled their November exhibit because of her dismissal. It is not the Warehouse, it is Astria in the Warehouse that is attracting these artists. Without her, they don't want to be here. We can look forward to an exodus of those who came because of the excitement her work has generated and we can go back to getting used to being known as a "loser city."
- Astria has the skill and takes the time to build relationships with community members, organizations and businesses. She's been active in the downtown TNT and she and her visiting artists use the services and products of local vendors.
- The Warehouse Gallery does not just happen. It takes vision and connections like Astria's to make all of this happen. If she is made to leave, all of her work stops. Relationships will have been severed, trust will have been betrayed. Once that happens, it's very hard to start it up again. To paraphrase one writer, if you dynamite this span in the bridge that is the Connective Corridor, the whole thing can collapse. Astria's work has inspired hope in city residents. Take that away, and much of the progress we've been making can reverse itself.
- The timing of this dismissal does not bode well. Just when the most recent show, "COME ON: Desire Under The Female Gaze" is opening, and just when The Yes Men are about to come to town, Astria is suddenly fired. As one writer pointed out, it is very ironic that we learn of her dismissal on the same day as the Newhouse III dedication. Emblazoned on the outside wall of that building are the words of the First Amendment of the Constitution that protect freedom of speech. Yet Astria was made to change the original title for "COME ON" - apparently a "feminist gaze" was not acceptable. Now the appearance is that the entire theme is unacceptable. This may not be the case, but the impression it gives the outside world is seriously detrimental to Syracuse.
"Since opening in 2006, through Astria, the Warehouse Gallery has helped become a cornerstone of the area art scene going well beyond just the visual arts. In a very short period of time, it has gained a solid following from the students, artists and the community at large. She has been able to put together exciting, interesting events and exhibitions that most people would never suspect would/could happen in Syracuse. Through her work, she has helped enrich, rejuvenate and energize the people and community around her."
It is not too late to reinstate her. Syracuse does not have to be the loser any more.


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Delta posts $5.4 billion loss in another brutal quarter, warns recovery could take 2 years or more
Delta Air Lines passenger planes are seen parked due to flight reductions made to slow the spread of coronavirus disease (COVID-19), at Birmingham-Shuttlesworth International Airport in Birmingham, Alabama, U.S.
Elijah Nouvelage | Reuters
Delta Air Lines booked a $5.4 billion net loss for the third quarter after the coronavirus pandemic roiled what is usually the peak travel period of the year.
Delta's revenue dropped by more than 75% from $12.56 billion a year ago to $3.06 billion in the three months ended Sept. 30, slightly under analysts' forecasts for $3.1 billion, it said Tuesday, kicking off third-quarter reporting for the beleaguered sector. Delta's president warned revenues may not normalize for "two years or more."
Delta's shares were down 2% in premarket trading.
The airline also posted a wider-than-expected loss during the quarter.
Large airlines like Delta have been particularly challenged in the pandemic because they previously relied heavily on business travel and sprawling international networks, two areas that have been hard-hit in recent months.
Delta has spent recent months retiring dozens of aircraft and reducing its footprint to cut costs. About 18,000 Delta employees, about a fifth of its pre-pandemic workforce, accepted buyouts and early retirement packages, prompting a $3.1 billion restructuring charge.
Airlines received portions of $25 billion in federal payroll support Congress passed this spring but talks between the Trump administration and Congress for additional airline aid have repeatedly derailed.
Here's how Delta performed compared with what Wall Street expected, based on average estimates compiled by Refinitiv:
Adjusted EPS: a loss of $3.30 versus an expected loss of $3
Revenue: $3.06 billion versus $3.11 billion, expected
More airlines are competing for price-sensitive leisure travelers within the United States but demand has struggled, despite an uptick from multi-decade lows hit in April. The Transportation Security Administration screened nearly 64 million people at U.S. airports in the third quarter, up 150% from the three months ended June 30 but still down from the 221 million people TSA screened in the year-earlier period.
Delta and its competitors have scrambled to introduce enhanced cleaning procedures and other policies, to calm travelers nervous about flying during the pandemic. Delta, for example, leaves middle seats open on flights.
Delta was able to cut its daily cash burn by more than 44% from roughly $43 million during the second quarter to an average of $24 million a day. Delta got down to $18 million a day in September, an improvement but still far off its goal of breaking even by the end of the year.
"While our September quarter results demonstrate the magnitude of the pandemic on our business, we have been encouraged as more customers travel and we are seeing a path of progressive improvement in our revenues, financial results and daily cash burn," Delta's CEO Ed Bastian said in an earnings release.
Since the pandemic began, Delta has posted more than $11 billion in losses.
On an adjusted basis, Delta lost $3.30 per share, more than the $3 per share loss analysts polled by Refinitiv were expecting.
Delta has already retired dozens of planes to cut costs. In the last quarter, it added to the list, deciding to retire its Boeing 767-300 ERs and 717-200s by 2025 and CRJ-200s by 2023.
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redpajama_set_name: RedPajamaCommonCrawl


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Baron Annesley ist ein erblicher britischer Adelstitel, der bislang dreimal, je einmal in der Peerage of England, der Peerage of Ireland und der Peerage of the United Kingdom verliehen wurde.

Verleihungen und weitere Titel
In erster Verleihung wurde am 20. April 1661 der Titel Baron Annesley, of Newport Pagnel in the County of Buckingham, in der Peerage of England für Arthur Annesley, 2. Viscount Valentia, neu geschaffen, und zwar als nachgeordneter Titel zum gleichzeitig verliehenen Titel Earl of Anglesey. 1660 hatte er bereits von seinem Vater den Titel Viscount Valentia, geerbt, der 1622 in the Peerage of Ireland für diesen geschaffen worden war. Earldom und Baronie erloschen am 14. Februar 1761 beim Tod des 6. Earls, die Viscountcy Valentia fiel an seinen Verwandten Arthur Annesley als 8. Viscount.

In zweiter Verleihung wurde am 20. September 1758 der Titel Baron Annesley, of Castlewellan in the County of Down, in der Peerage of Ireland an den Unterhausabgeordneten William Annesley verliehen. Dieser war väterlicherseits ein Enkel des 1. Viscount Valentia. Am 14. November 1766 wurde er in der Peerage of Ireland auch zum Viscount Glerawly, in the County of Fermanagh, erhoben. Sein Sohn, der 2. Viscount, wurde am 17. August 1789 in der Peerage of Ireland auch zum Earl Annesley, of Castlewellan in the County of Down, erhoben. Das Earldom wurde mit dem besonderen Zusatz verliehen, dass es in Ermangelung eigener legitimer männlicher Nachkommen auch an seinen jüngeren Bruder Richard Annesley und dessen männliche legitime Nachkommen vererbbar sei. Da der 1. Earl mehrere uneheliche, aber keine legitimen Kinder hatte, beerbte ihn demnach bei seinem Tod 1686 sein Bruder als 2. Earl. Heutiger Inhaber der Titel ist Michael Annesley als 12. Earl.

In dritter Verleihung wurde am 7. Mai 1917 der Titel Baron Annesley, of Bletchington in the County of Oxford, in der Peerage of the United Kingdom für Arthur Annesley, 11. Viscount Valentia, neu geschaffen. Dieser hatte 1863 bereits den Titel Viscount Valentia geerbt. Diese Baronie Annesley erlosch beim Tod seines Sohnes, des 12. Viscounts am 6. Oktober 1949. Die Viscountcy Valentia fiel an seinen Verwandten William Annesley als 13. Viscount.

Liste der Barone Annesley

Barone Annesley, erste Verleihung (1661)
 Arthur Annesley, 1. Earl of Anglesey, 1. Baron Annesley (1614–1686)
 James Annesley, 2. Earl of Anglesey, 2. Baron Annesley (1645–1690)
 James Annesley, 3. Earl of Anglesey, 3. Baron Annesley (1670–1702)
 John Annesley, 4. Earl of Anglesey, 4. Baron Annesley († 1710)
 Arthur Annesley, 5. Earl of Anglesey, 5. Baron Annesley (1678–1737)
 Richard Annesley, 6. Earl of Anglesey, 6. Baron Annesley (1690–1761)

Barone Annesley, zweite Verleihung (1758)
 William Annesley, 1. Viscount Glerawly, 1. Baron Annesley (um 1710–1770)
 Francis Annesley, 1. Earl Annesley, 2. Baron Annesley (1740–1802)
 Richard Annesley, 2. Earl Annesley, 3. Baron Annesley (1745–1824)
 William Annesley, 3. Earl Annesley, 4. Baron Annesley (1772–1838)
 William Annesley, 4. Earl Annesley, 5. Baron Annesley (1830–1874)
 Hugh Annesley, 5. Earl Annesley, 6. Baron Annesley (1831–1908)
 Francis Annesley, 6. Earl Annesley, 7. Baron Annesley (1884–1914)
 Walter Annesley, 7. Earl Annesley, 8. Baron Annesley (1861–1934)
 Beresford Annesley, 8. Earl Annesley, 9. Baron Annesley (1894–1957)
 Robert Annesley, 9. Earl Annesley, 10. Baron Annesley (1900–1979)
 Patrick Annesley, 10. Earl Annesley, 11. Baron Annesley (1924–2001)
 Philip Annesley, 11. Earl Annesley, 12. Baron Annesley (1927–2011)
 Michael Annesley, 12. Earl Annesley, 13. Baron Annesley (* 1933)

Voraussichtlicher Titelerbe (Heir apparent) ist der Sohn des aktuellen Titelinhabers Michael Annesley, Viscount Glerawly (* 1957).

Barone Annesley, dritte Verleihung (1917)
 Arthur Annesley, 11. Viscount Valentia, 1. Baron Annesley (1843–1927)
 Caryl Annesley, 12. Viscount Valentia, 2. Baron Annesley (1883–1949)

Literatur und Weblinks
 Charles Kidd: Debrett's Peerage and Baronetage. Debrett's Ltd, London 2014, ISBN 0-9929348-2-6.
 Leigh Rayment's Peerage Page

Annesley
Annesley
Annesley

Erstverleihung 1661

redpajama_set_name: RedPajamaWikipedia


Fabric: Mod 1960s thick wool. We did a burn test on this piece and it appears to be 100% pure wool, although it's more of a blanket feel rather than a soft apparel feel.
Purchase size: This remnant measures 70cm wide x 30cm long.


Search the Trident Library and professional Web sites to find sources from which you can develop your educational program on a topic related to stress management.
The sources should be sources of facts, data, and theoretical material relevant to your topic area. At least one or two sources must address the planning, development, and presentation of health-related educational programs.
Please prepare an annotated bibliography (not an essay) of 5 scholarly articles and/or professional resources. Include a summary of what each resource will provide that will be relevant and helpful for formulation and development of your program.
1) Your ability to prepare an annotated bibliography and use scolarly and/or professional sources.
2) Your ability to apply knowledge about the effects of stress and the benefits of good stress management that you have gained in this course to an educational program for a target audience in the health care professionals. Explain why the sources youve selected will help you achieve your learning and behavioral objectives for this audience.
ASSIGNMENT EXPECTATIONS: Please read before completing assignments.
Assignment should be approximately 2 pages in length (double-spaced).
Please use major sections corresponding to the major points of the assignment, and where appropriate use sub-sections (with headings).
Remember to write in an academic manner (avoid using the first person except when describing a relevant personal experience if asked in the assignment).
The focus of these assignments is on independent thinking and critical analysis. Use your own words and build on the ideas of others. Cite references for all facts and ideas that are not common knowledge.
When material is copied verbatim from external sources, it MUST be properly cited. This means that material copied verbatim must be enclosed in quotes and the reference should be cited either within the text or with a footnote. Quotations should be very rarely used. I cannot evaluate what you have learned if you copy someone elses words.
Use of peer-reviewed articles is required. Websites must be professional sources, not consumer sources (i.e. find government or university sites, or nonprofit organizations sites).
Be sure to cite your references within the text of all papers and on the reference list at the end. For examples, look at the way the references are listed within the modules and on the background reading list.
Proofread your paper to be sure grammar and punctuation are correct, and that each part of the assignment has been answered clearly.
BACKGROUND INFORMATION WILL BE UPLOADED.


mediums could include, acrylics, watercolour, pastels, paper clay, collage, artist trading cards.
I am approaching you, the parents of PAC's, as I have not only been a PAC member as a parent, and a PAC participant as a teacher, but, lately I have had the pleasure of being hired by parents to teach after school art classes and/or create murals in their schools… I am pleased to offer a special discount rate to get my small business known in my new community… and, please know that you can use me for fund raising… because… any of these classes would be a wonderful opportunity to any family at twice or even three times the listed price, therefore, leaving PAC with the potential of fund raising, if you so choose. As a parent myself, I would have welcomed this as a fund raising event… imagine getting quality art instruction, while knowing that you are supporting the school's fundraising efforts.
I would be so pleased to work with you in any capacity … do not hesitate to write or call with any questions or thoughts.


Quite a dramatic crop. Not that the rest if the original frame was uninteresting; in fact, it worked quite well, however, what I was really taken with was the ripples on the water reflecting the branches above. The ripples being caused by the drips from those overhanging branches (now out of shot). I love the way they move and interact, I love how something that seems so simple is in fact so complex.


What are the impacts of recycled fibres?
The European Outdoor Group is pioneering a new project seeking to compare the impacts of recycled and virgin polyester and nylon on synthetic textile microfibre pollution; as well as a separate project that examines textile coloration.
The two new textiles projects will be trialled with partner mills in Taiwan, and will be managed by UK-based consultancy biov8tion on behalf of the Outdoor Industry Microfibre Consortium. The latter is a part of the European Outdoor Group.
The results of both projects are expected to be revealed at the ISPO sportswear event in Munich in January 2019.
One must bear in mind that an estimated 190 000 tons of fibres from polyester, acrylic and nylon enter the world's marine ecosystem through washing machine wastewater each year, argues a report from Eunomia Research & Consulting.
Meanwhile, US retailer Target has pledged to invest US$ 1 million in post-consumer textile recycling technologies by 2020. Also, US fashion firm VF Corporation has revealed plans to train all of its product designers based in Europe in circular economy principles in a bid to boost the amount of second-hand garments and footwear collected at European owned and operated Timberland and The North Facestores by 10% in the next two years.
The dawn of 'slow fashion'?
Indeed, fashion does not need to have such an extremely short shelf life, notes Jane Milburn, social activist and author of the new book "Slow clothing – Finding meaning in what we wear". Born in Australia, she points out that her fellow countrymen purchase no less than 27 kg worth of new clothes every year, while sending some 23 kg to landfill.
To counter this consumption trend, the author advocates exploring vintage shops, mending existing items, supporting local small business owners, crafting personal styles, and to customise pieces after they have lost their appeal.


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Васи́лий Ива́нович Роздяло́вский (12 апреля 1914, Гродненская губерния, Российская империя — 23 июня 2007) — советский учёный в области теоретической и строительной механики и сопротивления материалов. Участник Второй мировой войны.

Биография
Василий Роздяловский родился 12 апреля 1914 года в Гродненской губернии Российской империи (ныне Брестская область Белоруссии) в семье студента лесного института Ивана Васильевича Роздяловского и учительницы Марии Степановны. Во время Первой мировой войны в 1915 году семья эвакуировалась в Екатеринослав. В 1916 году отец Василия окончил Харьковский сельскохозяйственный институт, и семья, переехав в Смоленскую губернию, в 1923 году окончательно осела в Рославле.

После окончания в Рославле средней школы Василий Роздяловский работал рабочим связи на строительстве железной дороги. В 1931 году поступил на третий курс рославльского вечернего рабфака, после окончания которого в 1933 году поступил в Московский институт инженеров железнодорожного транспорта. В 1939 году окончил институт и был направлен по распределению в московскую контору по проектированию Байкало-Амурской магистрали. Через короткое время был призван в Красную армию в железнодорожные войска Забайкальского военного округа. Начав рядовым, окончил службу инженер-капитаном. Участвовал в военных действиях Второй мировой войны на Карельском фронте в 1943—1944 годах. Как военный инженер занимался восстановлением железнодорожных путей на Украине, завершив службу в Днепропетровске в 1946 году. За время военной службы был награждён двумя орденами и двенадцатью медалями.

В 1946—1953 годах работал в проектном институте «Приднепровский Промстройпроект» в Днепропетровске. В 1953 году возглавил кафедру строительных материалов и работ инженерного факультета Днепропетровского сельскохозяйственного института; в 1955 году было присвоено звание доцента. В 1955 году факультет был ликвидирован, и Роздяловский перешёл в Днепропетровский химико-технологический институт на должность доцента кафедры теоретической механики и сопротивления материалов. С 1963 года до выхода на пенсию в 1979 году работал в Днепропетровском горном институте на должности доцента кафедры теоретической и строительной механики, где преподавал теоретическую механику, сопротивление материалов, строительную механику и занимался научно-исследовательской работой. В 1965—1967 годах преподавал в Камбодже; командор Королевского ордена Камбоджи.

Умер 23 июня 2007 года в возрасте 93 лет.

Награды
 Медаль «За оборону Советского Заполярья» (1945)
 Медаль «За победу над Германией в Великой Отечественной войне 1941—1945 гг.»
 Медаль «За восстановление предприятий чёрной металлургии юга»
 Командор Королевского ордена Камбоджи

Библиография

Примечания

Ссылки

Выпускники МИИТа
Военные железнодорожники Великой Отечественной войны
Преподаватели Днепропетровского государственного аграрно-экономического университета
Преподаватели Днепровской политехники


Since the age of six, Trevor has undergone 10 surgeries to correct his scoliosis. As he grew, rods placed on his spine to correct its shape had to be adjusted through surgery. Then his doctors at the Herbert J. Louis Center for Pediatric Orthopedics at Phoenix Children's decided to try something new. They implanted a device on his spine called the Magec Rod. It allows the rods that correct his spine to be magnetically adjusted in a short clinic visit, reducing the amount of surgeries. Today, Trevor is spending less time in the operating room and doing more of what he loves.


text:
For you roadies... Basso and Ullrich OUT of the TDF!
DIRT BOY
Weight Weenie Shop Owner
Ivan Basso and Francisco Mancebo joined Jan Ullrich on the sidelines of the Tour de France after organisers dumped two of the race's top riders. Ullrich had already been suspended by his T-Mobile team Friday morning when Tour director Christian Prudhomme announced the exclusion of Giro victor Basso.
"The enemy is not cycling, the enemy is doping," Prudhomme declared less than 24 hours before the start of the sport's greatest race.
Prudhomme praised T-Mobile's earlier decision to suspend Ullrich, sporting director Rudy Pevenage, and rider Oscar Sevilla, while at the same time promising fortitude on the part of the Tour's governing body.
"We have been in contact with T-Mobile since the start of the week, and since yesterday we've been waiting for the official document, finally we received it, from the Spanish Civil Guard and the Interior ministry."
Prudhomme said that after a unanimous decision by the association of sporting directors, the riders who have been banned will not be replaced.
"This morning there was a meeting of the AGCP and they decided to respect the ethical code 100 percent, so it means that the riders who will be asked to leave the Tour de France will not be replaced by other ones."
During a grim press conference Prudhomme sat alongside longtime Tour boss Jean Marie LeBlanc and AGCP chief and Quick Step sporting director Patrick Lefevere.
"I'm not very proud to be here today," Lefevere declared, adding that the decision not to replace the banned riders was a unanimous one taken "to send a strong signal."
"The sporting directors of each team are informing the riders concerned right now," Prudhomme said.
Basso, the Tour's runner-up in 2005 and this year's favourite, had already been warned by CSC sporting director Bjarne Riis the Frenchman added.
"I am cool, I am waiting for someone to bring me evidence of my guilt," Basso had said earlier on Friday when leaving his hotel for a training session.
Prudhomme said that Vincent Lavenu, the AG2R team boss, was also talking to Mancebo, who finished fourth overall in last year's Tour.
Astana up in the air
As far as Alexandre Vinokourov's Astana-Wurth team, which has been at the centre of the affair since majority owner Manalo Saiz's arrest, Prudhomme indicated the situation was still unresolved.
"For the team Astana-Wurth it's a bit more complex, because so many riders are on the official list that it seems to be a team doping system."
While Vino's name did not appaear on the list of riders implicated, nine of his team-mates did.
Before the Tour de France received the official report, the Court of Arbitration for Sport ruled on Thursday against the race organisers' request to exclude Astana-Wurth (formerly Liberty Seguros) from the Grande Boucle.
"I think this is normal. I think it is a brave move and a good example sent to the other sport directors. Today, we can do only one thing: be brave," Leblanc said in an interview with French daily Le Monde.
Both LeBlanc and Prudhomme said five or six riders would ultimately be suspended.
New favourite Valverde reacts
The suspensions now leave the race to start with no clear front-runner, though one of the top contenders who was not implicated, Alejandro Valverde, was upset by the news.
"It is a big blow for the race. You could see something like this might happen," Valverde told Spanish daily Marca's website on Friday.
"Oscar [Sevilla] is a great friend of mine and Ullrich is one of the favourites. I am calm about it all because I have nothing to fear from something which affects lots of other people.
"I just want the race to start as soon as possible and hope it is clean and that everything goes well."
Maybe a Frenchman can fianlly win the TDF . So sad for the sort of Cycling we all love!
I really thought Basso was clean. years rooting for this guy have just gone down the toilet? Or should we just shrug this off as another athlete trying to get an edge. I have said it for 10 yrs SPORTS on ALL LEVELS is drug ladden!
Light-Bikes.com
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Axis II
SUBLIM8er
WTF, what are these guys thinking! This was going to be Ulrich's race! Whatever. Why couldn't he just get Lance's doc on board?
Every man has inside himself a parasitic being who is acting not at all to his advantage.
well I guess it's time to burn my CSC and Fasso Bortolo jerseys as one of my favorite riders is doping.
Maybe I can keep my Fasso one, because as on now Pettachi is clean, for now.
I guess with these thoughts, I should sart burning my NFL jerseys too....
blatido
MTB'er
Pro Cycling (and any "pro" sport) sucks...
One thing for sure that everybody knows about but few admit as a fact is doping to achieve better performance in sports. Now the attention is on road cycling, but fact is, ANY sport is affected by that fact. MTB, soccer, basketball, all kind of olympic sports, american football, baseball, etc., etc., etc...
Fact is, along as the "pro" competitors receive big amounts of money for winning a race and even bigger amounts of money from sponsors, they will do ANYTHING to keep being at the podium and to keep sponsorship. Sport doping WILL NOT BE STOPPED by tests, and laws, and athletes suspensions. There is too much money and interests involved not to be tempted to take "something" that can give me that extra boost to win.
I think the solution is to allow doping. At least that way athletes would certainly know what they're taking, they personally would take responsability for what they're doing and risks would be reduced.
"I will not relent"
ShortTravelMag
Getting too old for this...
Allowing doping would solve nothing. Everyone would take the allowable drugs, because they could, and the next guy is doing it...then it would start with a few big guys taking or doing something that isn't on the list of allowable drugs, because they still need that edge over the next guy...then someone drops dead...you just can't open the floodgates in my opinion. I'd rather see drastic events like this happen - the top names and teams go down in flames. Keep throwing these guys out, every year, let the clean ones rise to the top. Maybe some of the clean pros would be the top guys if they could race against other clean guys...give them their chance.
ShortTravelMagazine.com - For the XC Mtb Enthusiast.
bbarend
Now Floyd and Levi will be even more motivated. Perhaps we will see two Americans on the podium.
The SS Boz
Want to vent?
Why is this in save some weight forum??? Unless you are refering to the "weight" that was lost from the TDF???
-Boz
"I'm supa-fly TNT, I'm tha guns of the navarone!"
artnshel
I'm disapointed that Basso and Ullrich won't be in the race. I'm less excited to watch it now.
However It's interesting that their own teams and not the tour pulled them.
That Vino was on the same list but may get to race and possibly win is bogus.
Whoever wins the race will have a cloud over their status as leader.
"El Whatever"
This is so sad....
Man, I lost my interest on road riding after reading this.
And, of course, it takes my respect for champions of old like Lance, Indurain and others, to a new level.
psinsyd
Faller - Expert Class
As a fan of Basso's also, that's horrible news. It will be an interesting TdF to say the least!!
Ah well, let's just hope F1 can race a full field at Indianapolis on Sunday!:thumbsup:
BMC FourStroke
azonicrider188a said:
well just get your asses some juicy 7s and huck down a mountain. thoose v brakes will snap like my butt
Ultra Magnus
Thread derail, F1 in USA?
Back to the thread... Seams like since Lance is gone who was the #1 focus for doping allogations (I can say it but not sure how to spell it), it's an accusation feeding frenzy.
"I've come to believe that common sense is not that common" - Matt Timmerman
BipedZed
recovering racer
Warp said:
Ummm...you honestly think World Cup MTB racing is clean?
mtb_biker
-> SickLines.com <-
the only thing i question is their timing... how long have they known for? Anyone have details? I wonder if they waited until this time so that they could not compete.
SickLines
http://www.sicklines.com
Well first of this is a CYCYLING site.
This is a CYCYLING site
Lot's of roadies here too and people who care about cycling in general
I thought others who did not know would appricate this, as I vist this froum more than any other.
This is probaly the biggest news any of us have seen in cycling in our life time! The whole Operación Puerto thing.
Onl a lighetr note, with the doping is SPAIN, maybe this is the answwer to alot of Malol psots and thoughts.... :idea: ut: :crazy:
DIRT BOY said:
LOL.....doh!!!!
:skep:
AZ-X
It's a sad thing to see. They need to do some housecleaning on the VTT/MTB side as well, I'd wager. It would be nice to see a completely clean sport (or as clean as it can possibly be) emerge from all of this.
I hate seeing people robbing others of victory as riders like Filip Meirhaeghe and others have done in the past. The only thing that's going to stop it is ever increasing enforcement and vigilance. Punishments have to be elevated well above the occasional "slap on the wrist" stuff as well. I really think that an absolute Zero Tolerance policy really has to be instated at this point.
As far as I'm concerned, Filip shouldn't have been allowed back for at least 2 years, of which he should have had a steady barrage of tests during the entire period. That is more of a real probationary sentence. Other than just full banishment from the sport for life, I see no other way.
As for waiting this long to pull the riders...
I feel sorry for the event itself and for the towns and French (as well as other) people who are helping to host the event, but I sincerely hope that the shock value of this entire episode sinks in and helps to convince the sponsors, as well as the riders, to take all of this extremely seriously.
Ride to Live...
Niner MCR 9 Framest and XO Cranks w/ Ceramic BB for sale in MTBR Classifieds!
BipedZed said:
Philip Meirhaege proved that this would be utopic... but then, I don't follow the WC's either. Much less defend them.
At least they're not pulling out the top three contenders for the World Championship as they did in the TdF.
Maybe that's why I've only raced once and don't plan on racing either... nothing as riding for the pleasure of riding.
eurorider
Always good laughs here:thumbsup:
About the doping though, I don't know why people are surprised.
Anyways, go Rasmussen!:thumbsup:
aussie_yeti
eurorider said:
Or Cadel (Evans), a MTBer and Aussie on the podium of the Tour - now's as good a time as any.
Sad to see the big names go - the winners status will be marred by the fact that the field wasn't at it's strongest. But atleast a hard line stance is finally being taken. Hopefully they get to the bottom of it quickly and accurately, the cheats are booted and we can once again admire the champions for their natural talent and dedication to the sport.
Mr. Scary said:
Why is there always one in every forum that asks this stupid question? I'd say the thread heading was pretty clear from the onset as to the subject of this post. If you aren't interested, don't read it. Instead one person always feels compelled to go in, read, and then post why is this subject being discussed here? Beavis and Butthead for sure.
Now that matter is dealt with, this news is very disappointing. It makes for a race with alot of what-if's...
Maybe next year the doping will be gone and people will race clean. Yeah right...
You read all the other posts and figured this was the best one to reply to?....flame it up bro........Geez...do you understand that wink icon at all???
DaFireMedic Jul 4, 2006


Is someone able to show me where I'm wrong with that script please ?
It work very well except it don't move to bottom when we read the web page.
Can you show me where I've done the mistake please ?
Try re-implementing the script. It appears none of the dynamic features are working.
I have had the same problem as the person above.
When I comment out this tag the Menu works fine. It moves down and up when you scroll. But when you put back in that tag the menu won't work.
cr3ative can you explain this as to why this DOCType tag is stopping the menu from working. Or anyone else who might know why this is happening. Because all my HTML files has the DOCTYPE tag in them.
Ah yes, this has happened before.
Is it necessary to have a DOCType in your HTML files?
It should allow the script to function properly, but still indicates what type of HTML/XHTML you are using.
It may not be quite correct, but until more of the "kinks" are worked out, using HTML 4.01 DOC tags may be best. Especially seeing as the DOC tag referenced by sm30 is listing it as a "transistional" document.
Thanks, I will try this. But will any code or tags act or display different beacuse of the Doctype change?


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text:
\section{Introduction}
\label{sec:intro}
Let $\AA = (A, +)$ be, unless otherwise specified, an additively written semigroup, viz. an ordered pair consisting of a set and a binary associative operation on it; note that, in this paper, ``additive'' does not imply ``commutative''.
We address the reader to \cite[\S{} 1.1]{How} for basic aspects of semigroup theory.

If $X_1, \ldots, X_n \subseteq A$, we let $X_1 + \cdots + X_n$ denote, as usual, the sumset, relative to $\mathbb A$, of the $n$-tuple $(X_1, \ldots, X_n)$, namely
$$
X_1 + \cdots + X_n := \{x_1 + \cdots + x_n: x_1 \in X_1, \ldots, x_n \in X_n\};
$$
we replace $X_i$ with $x_i$ in this notation if $X_i = \{x_i\}$ for some $i$, provided it does not cause confusion, and we use $n X_1$ for $X_1 + \cdots + X_n$ if $X_1 = \cdots = X_n$.

We write $\AA^\times$ for the set of units (or invertible elements) of $\AA$, and for $X \subseteq A$ we set $X^\times := X \cap \mathbb A^\times$ when there is no danger of ambiguity.

In particular, $A^\times = A$ if and only if $\mathbb A$ is a group or $A$ is empty, and $A^\times \ne \emptyset$ if and only if $\AA$ is a monoid, i.e. there exists a (provably unique) element $0_\mathbb{A} \in A$, labeled as the identity of $\mathbb A$, such that $x + 0_\mathbb{A} = 0_\mathbb{A} + x = x$ for all $x \in A$.

For $X, Y \subseteq A$ we let $X - Y := \{z \in A: X \cap (z + Y) \ne \emptyset\}$ and ${-Y}+X := \{z \in A: X \cap (Y+z) \ne \emptyset\}$, which extends the notion of difference set from groups to semigroups, at the cost that $X-Y$ or ${-Y}+X$ may be empty even if $X$ and $Y$ are not. We use, respectively, $X - y$ and ${-y} + X$ in place of $X - Y$ and ${-Y} + X$ if $Y = \{y\}$ and no confusion can arise; note that $X - y = \{x + (-y): x \in X\}$ if $y \in A^\times$, and similarly for ${-y} + X$.

Given $X \subseteq A$, we denote by $\langle X \rangle$ the smallest subsemigroup of $\mathbb A$ containing $X$, and we set $\langle\fixed[-0.6]{ \text{ }}\langle X \rangle\fixed[-0.6]{ \text{ }}\rangle := \langle X \cup \{-x: x \in X^\times\} \rangle$ and $\ord(X) := |\langle X \rangle|$. If $X = \{x\}$ and there is no risk of misunderstanding, we just write $\ord(x)$ in place of $\ord(X)$, and we use $\langle x \rangle$ in an analogous way.

Lastly, we say that an element $z \in A$ is cancellable (in $\mathbb A$) if both the functions $A \to A: x \mapsto x + z$ and $A \to A: x \mapsto z + x$ are injective, and we refer to $\mathbb A$ as a cancellative semigroup if every $z \in A$ is cancellable.
\section{The Cauchy-Davenport constant of an \texorpdfstring{$n$-tuple}{n-tuple}}
\label{sec:CD-constants}
With the above in mind, we now introduce a quantity that happens to capture, as discussed below, interesting features of the ``combinatorial structure'' of $\mathbb A$.

To start with, we set, for every $X \subseteq A$, $\gamma(X) := |X|$ if $|X| \le 1$, otherwise
\begin{equation}
\label{equ:CD_constant}
\gamma(X) := \sup_{x_0 \in X^\times} \inf_{x_0 \ne x \in X} {\rm ord}(x - x_0),
\end{equation}
where $\sup(\emptyset) := 0$ and $\inf(\emptyset) := \infty$;
note that \eqref{equ:CD_constant} can be slightly simplified if $\mathbb A$ is a group, by replacing $X^\times$ with $X$.

Throughout, we will see how to use $\gamma(X)$ to obtain, for $X, Y \subseteq A$, non-trivial lower bounds on $|X+Y|$. But first, for all $X_1, \ldots, X_n \subseteq A$ let
\begin{equation*}
\label{equ:def_of_large_CD}
\gamma(X_1, \ldots, X_n) :=
\left\{
\begin{array}{ll}
\!\! 0 & \text{if }X_i = \emptyset\text{ for some }i \\
\!\! \max_{1 \le i \le n\fixed[0.2]{ \text{ }}} \gamma(X_i) & \text{otherwise}
\end{array}
\right.\!\!\!;
\end{equation*}
we refer to $\gamma(X_1, \ldots, X_n)$ as the \textit{Cauchy-Davenport constant}, relative to $\mathbb A$, of the $n$-tuple $(X_1, \ldots, X_n)$. Occasionally, we may add a subscript `$\mathbb A$' to the right of the letter $\gamma$ in the above definitions if we need, for any reason, to be explicit about the semigroup on which they do actually depend.

The Cauchy-Davenport constant of a tuple was first introduced in \cite{Tring12}, though in a different notation and in a somewhat different form, and further investigated in \cite{Tring13}, as part of a broader program aimed at the extension of aspects of the theory on Cauchy-Davenport type inequalities from groups to more abstract settings, where certain properties (of groups) are no longer necessarily true.

In particular, the author proposed in \cite{Tring13} to prove (or disprove) the following:
\begin{conjecture}
\label{conj:1}
If $\AA$ is a cancellative semigroup and $X_1, \ldots, X_n \subseteq A$, then
\begin{equation*}
|X_1+\cdots+X_n| \ge \min(\gamma(X_1, \ldots, X_n), |X_1| + \cdots + |X_n| + 1 - n).
\end{equation*}
\end{conjecture}
This is plainly true if $n = 1$ or $X_1^\times = \cdots = X_n^\times = \emptyset$, is straightforward when $|X_i| \ge \gamma(X_1, \ldots, X_n)$, and especially $|X_i| = \infty$, for some $i = 1, \ldots, n$ (see Lemma \ref{lem:trivialieties}\ref{item:trivialieties_(ii)} below), and has been so far confirmed in a couple more of cases:
\begin{enumerate}[label={\rm (\roman{*})}]
\item\label{item:partial_results} if $n = 2$ and each of $X_1$ and $X_2$ generates a commutative subsemigroup of $\mathbb A$, see \cite[Theorem 8 and Corollary 10]{Tring12};
\item if $\gamma(X_1) = \cdots = \gamma(X_n)$ and $\mathbb A$ is commutative (in addition to being cancellative), see the note added in proof at the end of \cite[\S{} 6]{Tring13}.
\end{enumerate}
The conjecture was first motivated by point \ref{item:partial_results} above and the following theorem, see \cite[Theorem 7]{Tring13}:
if $\mathbb A$ is a cancellative semigroup and $X, Y \subseteq A$, then
\begin{equation}
\label{equ:weaker_conjecture}
|X+Y| \ge \min(\gamma(X+Y), |X| + |Y| - 1).
\end{equation}
In fact, \eqref{equ:weaker_conjecture} is weaker than the bound provided by Conjecture \ref{conj:1} in the case of two summands,
as $\gamma(X, Y) \ge \gamma(X+Y)$ in general, and actually $\gamma(X, Y) \gg \gamma(X + Y)$ in many common situations, see \cite[Lemma 3 and Example 4]{Tring13}. (The
symbol `$\gg$' means that, for a suitable choice of the semigroup $\mathbb A$ and the sets $X$ and $Y$, the left-hand side can be made larger than the right-hand
side by an arbitrary factor.)

With all this said, here comes the main contribution of the present work, which is, in the first place, a strengthening of the Cauchy-Davenport theorem \cite{Cauchy1813, Daven35, Daven47}.
\begin{theorem}
\label{th:mainA062}
Assume $\mathbb A$ is a cancellative semigroup, and let $X, Y \subseteq A$ such that $\langle Y \rangle$ is com\-mu\-ta\-tive and $Y \ne \emptyset$. Then at least one of the following holds:
\begin{enumerate}[label={\rm (\roman{*})}]
\item\label{cor:main_condition_(i)} $|X + Y| \ge |X| + \min(\gamma(Y), |Y| - 1)$;
\item\label{cor:main_condition_(ii)} $X + 2Y = X + Y + \bar{y}$ for some $\bar{y} \in Y^\times$.
\end{enumerate}
\end{theorem}
Loosely speaking, the theorem says that, for all $X, Y \subseteq A$, and in the presence of cancellativity, $|X+Y|$ cannot be ``too small'', unless $X+Y$ has ``structure'', which
is made more precise by the next statement, whose proof we defer to the end of \S{} \ref{sec:preliminaries}.
\begin{proposition}
\label{prop:equivalent_form_with_generated_subgroup}
Assume $\mathbb A$ is a cancellative semigroup, and let $X, Y \subseteq A$ be finite sets such that $\langle Y \rangle$ is com\-mu\-ta\-tive and $Y^\times \ne \emptyset$. Then the following are equivalent:
\begin{enumerate}[label={\rm (\roman{*})}]
\item\label{cor:equ_condition_(i)} $X + 2Y = X + Y + \bar{y}$ for some $\bar{y} \in Y^\times$;
\item\label{cor:equ_condition_(ii)} $X + 2Y = X + Y + y$ for all $y \in Y$;
\item\label{cor:equ_condition_(iii)} $X + \langle\fixed[-0.6]{ \text{ }}\langle Y - \bar y \rangle\fixed[-0.6]{ \text{ }}\rangle = X + \langle Y - \bar y \rangle = X + Y - \bar y$ for every $\bar{y} \in Y^\times$.
\end{enumerate}
\end{proposition}
We use Theorem \ref{th:mainA062} and Proposition \ref{prop:equivalent_form_with_generated_subgroup} to prove a couple of corollaries: the first has essentially the same content of \cite[Theorem 8]{Tring12}, and deriving it from Theorem \ref{th:mainA062} shows, in the end, that the results of this paper subsume and strengthen \textit{all} those obtained in \cite{Tring12}; the second is reminiscent of an addition theorem of Y.~O.~Hamidoune \cite[p.~249]{Hami} we refer to as the Hamidoune-Shatrowsky theorem, as it is a generalization of an earlier (and weaker) result of L.~Shatrowsky \cite{Shat}.
\begin{corollary}
\label{cor:UDT_improved}
Assume $\mathbb A$ is a cancellative semigroup, and let $X, Y \subseteq A$ such that $X \ne \emptyset$ and $\langle Y \rangle$ is com\-mu\-ta\-tive. Then
$
|X + Y| \ge \min(\gamma(Y), |X| + |Y| - 1)$.
\end{corollary}
\begin{corollary}
\label{cor:HS}
Let $\mathbb A$ be a cancellative monoid with identity $0_\mathbb{A}$, and let $X, Y \subseteq A$ such that
$\langle Y \rangle$ is commutative and $X \cup (X + Y) \ne X + \langle\fixed[-0.6]{ \text{ }}\langle Y \rangle\fixed[-0.6]{ \text{ }}\rangle$. Then
\begin{equation}
\label{equ:stronger_than_HS}
|X \cup (X + Y)| \ge |X| + \min(\gamma(Y \cup \{0_\mathbb{A}\}), |Y| - \mathbf 1_Y(0_\mathbb{A})),
\end{equation}
where $\mathbf 1_Y(0_\mathbb{A}) := 1$ if $0_\mathbb{A} \in Y$, otherwise $\mathbf 1_Y(0_\mathbb{A}) := 0$.
\end{corollary}
As long as $\langle Y \rangle$ is commutative, Corollary \ref{cor:HS} is indeed stronger, and can be \textit{much} stronger, than the Hamidoune-Shatrowsky theorem, according to which we would rather have that if $\mathbb A$ is a group, $Y \ne \emptyset$, and $X \cup (X + Y) \ne X + \langle\fixed[-0.6]{ \text{ }}\langle Y \rangle\fixed[-0.6]{ \text{ }}\rangle$, then
\begin{equation}
\label{equ:hamidoune-shatrowsky_bound}
|X \cup (X + Y)| \ge |X| + \min(v(Y), |Y|),
\end{equation}
where $v(Y)$ is the minimal order of the elements of $Y$. In fact, there are two cases:
\begin{enumerate}[label={\rm (\roman{*})}]
\item $0_\mathbb{A} \in Y$. Then $X \subseteq X + Y$ and $v(Y) = 1$, hence \eqref{equ:hamidoune-shatrowsky_bound} simplifies to $|X + Y| \ge |X| + 1$. If $|Y| \ge 2$, this is (strictly) weaker than \eqref{equ:stronger_than_HS}, and actually \textit{much} weaker than \eqref{equ:stronger_than_HS} for (comparatively) large values of both $\gamma(Y \cup \{0_\mathbb{A}\})$ and $|Y|$ (which can be easily attained). Otherwise, $Y = \{0_\mathbb{A}\}$ and $X + \langle\fixed[-0.6]{ \text{ }}\langle Y \rangle\fixed[-0.6]{ \text{ }}\rangle = X + Y = X$, which, however, would be a contradiction.
\item $0_\mathbb{A} \notin Y$. Then $|Y| - \mathbf 1_Y(0_\mathbb{A}) = |Y|$, and of course $\gamma(Y \cup \{0_\mathbb{A}\}) \ge v(Y)$.
\end{enumerate}
On the other hand, the Hamidoune-Shatrowsky theorem holds, provided $\mathbb A$ is a group, without the additional assumption on $\langle Y \rangle$ made in Corollary \ref{cor:HS}, which leads us to believe that a more general version of Theorem \ref{th:mainA062} should be true.

Incidentally, let us mention here that, while the original proof of the Hamidoune-Shatrowsky theorem relies on Hamidoune's theory of atoms, our proof of Theorem \ref{th:mainA062}, and hence of Corollary \ref{cor:HS}, is essentially based on a non-commutative variant of the Davenport transform first considered, to our knowledge, in \cite[\S{} 4]{Tring12}.

Our last result is a special case of Theorem \ref{th:mainA062} and a strengthening of \cite[Corollary 15]{Tring12}, which is in turn a generalization of the Chowla-Pillai theorem (on sumsets in finite cyclic groups), see \cite[Theorem 1]{Chow} and \cite[Theorems 1--3]{Pil}.
\begin{corollary}
\label{cor:pillai&chowla_improved}
Fix $n \in \mathbf N^+$. Denote by $(\ZZb_n, +)$ the additive group of the integers modulo $n$ and by $\pi$ the canonical projection $\ZZb \to \ZZb_n$. Let $X, Y \subseteq \ZZb_n$ be non-empty sets such that $X+2Y \ne X + Y + \bar y$ for some $\bar y \in Y$. Then
\begin{displaymath}
|X+Y| \ge |X| + \min(\delta_Y^{-1} n, |X|+|Y|-1),
\end{displaymath}
where $\delta_Y := 1$ if $|Y| = 1$, otherwise
$$
\delta_Y := \min_{y_0 \in \pi^{-1}(Y)} \fixed[0.05]{ \text{ }} \max_{y \in \pi^{-1}(Y),\fixed[0.2]{ \text{ }} \pi(y) \fixed[0.1]{ \text{ }} \ne \fixed[0.1]{ \text{ }} \pi(y_0)}  \gcd(n, y - y_0).
$$
\end{corollary}
We will prove Theorem \ref{th:mainA062} and its corollaries in \S{} \ref{sec:proofA062}, but first we need to gather together a few facts that play a role in the proofs: this is done in the next section.
\section{Preparations}
\label{sec:preliminaries}
We start with basic properties of semigroups that are readily adapted from the case of groups and used repeatedly in the sequel (with or without a comment).
\begin{lemma}
\label{lem:trivialieties}
The following hold:
\begin{enumerate}[label={\rm (\roman{*})}]
\item\label{item:trivialieties_(i)} If $z \in A$ is cancellable and $X \subseteq A$, then $|z+X|=|X+z|=|X|$.
\item\label{item:trivialieties_(ii)} If $X_1, \ldots, X_n \subseteq A$ and $X_i$ contains at least one cancellable element for each $i$, then
$\left|X_1 + \cdots + X_n\right| \ge \max_{1 \le i \le n} |X_i|$.
\item\label{item:trivialieties_(iii)} Let $z \in A$, and assume $z$ is cancellable and $n := \ord(z) < \infty$. Then $\mathbb A$ is a monoid and $nz$ is the identity of $\mathbb A$.
\item\label{item:trivialities_(iv)} Let $\mathbb A$ be a monoid and $X \subseteq A$, and let
$
z \in \mathcal C(X) \cap A^\times$, where
$$
\mathcal C(X) := \{z \in A: x + z = z + x \text{ for all }x \in X\}
$$
is the center of $X$ (in $\mathbb A$).
Then $-z \in \mathcal C(X)$, and $\langle X - z \rangle$ and $\langle -z+X\rangle$ are both commutative subsemigroups if $\langle X \rangle$ is.
\end{enumerate}
\end{lemma}
\begin{proof}
\ref{item:trivialieties_(i)}--\ref{item:trivialieties_(ii)} Units are cancellable elements, and for a cancellable $z \in A$ both the functions $A \to A: x \mapsto x + z$ and $A \to A: x \mapsto z + x$ are injective.

\ref{item:trivialieties_(iii)} By hypothesis, $(n+1)z = kz$ for some $k = 1, \ldots, n$. We claim that $k = 1$. Indeed, if $k \ge 2$ then $z$ being cancellable (in $\mathbb A$) implies $(n+2-k)z = z$, which is impossible, since $2 \le n+2-k \le n$ and $z, \ldots, nz$ are pairwise distinct.

So, for all $x \in A$ we have $x+z = (x + nz) + z$ and $z + (nz + x) = z + x$, which, by using again that $z$ is cancellable, yields $x + nz = nz + x = x$, and ultimately means that $\mathbb A$ is a monoid with identity $nz$.

\ref{item:trivialities_(iv)} Let $z \in \mathcal C(X)$ and $x \in X$, and for ease of notation denote by $\tilde z$ the inverse of $z$. By the cancellativity of $\mathbb A$, it is immediate that $x + \tilde z = \tilde z + x$ if and only if $x = (x + \tilde z) + z = \tilde z + x + z$, which is
true, as $\tilde z + x + z = \tilde z + z + x = x$ by the assumptions on $x$ and $z$. It follows that $\tilde z \in \mathcal C(X)$.

With this in hand, suppose $\langle X
\rangle$ is a commutative subsemigroup of $\mathbb A$ and pick $v, w \in \langle X - z
\rangle$. Then, there exist $k,\ell \in \mathbf N^+$ and $x_1, \ldots, x_k, y_1, \ldots, y_\ell \in X$ such that $v = \sum_{i=1}^k (x_i + \tilde z)$ and $w = \sum_{i=1}^\ell (y_i + \tilde z)$, with the result that $v + w = w + v$ by induction on $k + \ell$ and the observation that, for all $u_1, u_2 \in X$, it holds
$$
(u_1 + \tilde z) + (u_2 + \tilde z) = u_1 + u_2 +
2\tilde z = u_2 + u_1 + 2\tilde z = (u_2 +
\tilde z) + (u_1 + \tilde z),
$$
where we have used, in particular, that $\tilde
z \in \mathcal C(X)$, as proved in the above. Hence $\langle X -
z\rangle$ also is commutative, and the case of $\langle {-z} + X\rangle$ is analogous.
\end{proof}
We omit the proof of the next elementary lemma, but the interested reader can refer to \cite[Lemma 12 and Remark 13]{Tring13} for details.
\begin{lemma}\label{lem:old_units_distribution}
If $X_1, \ldots, X_n \subseteq A$, then $X_1^\times + \cdots + X_n^\times \subseteq (X_1 + \cdots + X_n)^\times$, and the inclusion is actually an equality provided that $\mathbb A$ is cancellative.

Moreover, if $\mathbb A$ is a monoid and $x_1 \in X_1^\times, \ldots, x_n \in X_n^\times$, then $x := x_1 + \cdots + x_n$ is too an invertible element and ${-x} = (-x_n) + \cdots + (-x_1)$.
\end{lemma}
The following result reveals a certain invariance of the Cauchy-Davenport constant; we address the reader to \cite[Proposition 14]{Tring13} for a proof.
\begin{lemma}\label{lem:unital_shifts_vs_gamma_equality}
Assume $\mathbb A$ is a monoid, and let $X \subseteq A$ and $z \in A^\times$. Then
$$
\gamma(X) = \gamma(X - z) = \gamma({-z} + X).
$$
\end{lemma}
Given $X, Y \subseteq A$, we say that a pair $(X_0, Y_0)$ of subsets of $A$ is an inv\-ar\-i\-ant transform, relative to $\mathbb A$, of $(X, Y)$ if:
\begin{enumerate}[label={\rm (\textsc{s}\arabic{*})}]
\item $|X + Y| = |X_0 + Y_0|$;
\item $|X| = |X_0|$ and $|Y| = |Y_0|$;
\item $\gamma(X) = \gamma(X_0)$ and $\gamma(Y) = \gamma(Y_0)$.
\end{enumerate}
This notion is motivated by the next lemma, cf. \cite[Corollary 15]{Tring13}.
\begin{lemma}
\label{lem:units_give_invariant_n_transform}
Let $\mathbb A$ be a cancellative monoid with identity $0_\mathbb{A}$, and pick $X, Y \subseteq A$. Assume that
$Y^\times \ne \emptyset$ and $|Y| \ge 2$, and let $\kappa$ be an integer $\le \gamma(Y)$.
Then, there exists an in\-var\-i\-ant transform $(X_0, Y_0)$ of $(X, Y)$ such that:
\begin{enumerate}[label={\rm (\roman{*})}]
\item\label{lem:subinvariance(i)} $0_\mathbb{A} \in Y_0$ and $\gamma(Y_0) \ge \ord(y) \ge \kappa$ for every $y \in Y_0 \setminus \{0_\mathbb{A}\}$;
\item\label{lem:subinvariance(ii)} if $\langle Y \rangle$ is commutative, then so is $\langle Y_0 \rangle$;
\item\label{lem:subinvariance(iii)} if $\langle Y \rangle$ is commutative and $X + 2Y \neq X + Y + \bar{y}$ for every $\bar{y} \in Y^\times$, then $X_0 + 2Y_0 \ne X_0 + Y_0 + \bar{y}_0$ for all $\bar{y}_0 \in Y_0^\times$.
\end{enumerate}
\end{lemma}
\begin{proof}
Fix an integer $\kappa \le \gamma(Y)$, and using that $Y^\times \ne \emptyset$ and $|Y| \ge 2$, let $y_0 \in Y^\times$ such that
$
\gamma(Y) \ge \inf_{y_0 \fixed[0.1]{ \text{ }} \ne \fixed[0.2]{ \text{ }} y \fixed[0.15]{ \text{ }} \in Y} \ord(y - y_0) \ge \kappa$.
Then, set
$$
X_0 := X + y_0\ \ \text{and}\ \ Y_0 := -y_0 + Y.
$$
Clearly, $0_\mathbb{A}$ is in $Y_0$, and a straightforward computation gives that
\begin{equation}
\label{equ:a_simple_identity}
X + Y = (X + y_0) + ({-y_0} + Y) = X_0 + Y_0.
\end{equation}
In addition, since $y_0 \in A^\times$ and units are cancellable, we have from Lemma \ref{lem:unital_shifts_vs_gamma_equality} that $\gamma(X) = \gamma(X_0)$ and $\gamma(Y) = \gamma(Y_0)$, and from Lemma \ref{lem:trivialieties}\ref{item:trivialieties_(i)} that $|X| = |X_0|$ and $|Y| = |Y_0|$. Lastly, using that $0_\mathbb{A} \in Y_0$ and, on the other hand, $\upsilon \in Y_0$ if and only if $\upsilon = y - y_0$ for some $y \in Y$, we find
\begin{displaymath}
\gamma(Y_0) \ge \inf_{0_\mathbb{A} \fixed[0.1]{ \text{ }} \ne \fixed[0.2]{ \text{ }} \upsilon \fixed[0.15]{ \text{ }} \in Y_0} \ord(\mathfrak \upsilon) = \inf_{y_0 \fixed[0.1]{ \text{ }} \ne \fixed[0.2]{ \text{ }} y \fixed[0.15]{ \text{ }} \in Y} \ord(y - y_0) \ge \kappa.
\end{displaymath}
Putting it all together, this shows that $(X_0, Y_0)$ is an invariant transform of $(X, Y)$. So point \ref{lem:subinvariance(i)} is proved, and  \ref{lem:subinvariance(ii)} follows from Lemma \ref{lem:trivialieties}\ref{item:trivialities_(iv)}.

As for \ref{lem:subinvariance(iii)}, suppose that $\langle Y \rangle$ is commutative and $X + 2Y \ne X + Y + \bar{y}$ for all $\bar{y} \in \langle Y \rangle$, yet $X_0 + 2Y_0 = X_0 + Y_0 + \bar{\upsilon}$ for some $\bar{\upsilon} \in Y_0^\times$. Accordingly, note that $Y_0^\times = {-y_0} + Y^\times$ by Lemma \ref{lem:old_units_distribution}, and let $\bar{y} \in Y^\times$ such that $\bar{\upsilon} = {-y_0} + \bar{y}$. Then, we get from \eqref{equ:a_simple_identity} and point \ref{lem:subinvariance(ii)} above that
$$
X + Y + \bar{y} - y_0 = X_0 + 2Y_0 = X + 2Y - y_0,
$$
which yields $X + 2Y = X + Y + \bar{y}$ (again, because $-y_0$ is a unit, and hence we can cancel it out). This, however, is absurd and leads to the desired conclusion.
\end{proof}
Last but not least, we will need the following proposition, which is essentially a revised version of \cite[Proposition 23]{Tring12}.
\begin{proposition}\label{prop:properties_of_the_modified_Davenport_transform}
Assume $\mathbb A$ is a cancellative semigroup, and let $X, Y \subseteq A$ be such that $X + 2Y \not\subseteq X + Y$ and $\langle Y \rangle$ is a commutative subsemigroup of $\mathbb A$. Accordingly, fix $z \in (X + 2Y) \setminus (X + Y) \ne \emptyset$, and define
\begin{equation}\label{equ:defining_Davenport_transformed_pair}
\tilde Y_z := \{y \in Y: z \in X + Y + y\}
\ \ \text{and}\ \ %
Y_z := Y \setminus \tilde Y_z.
\end{equation}
Then the following hold:
\begin{enumerate}[label={\rm(\roman{*})}]
\item\label{item:D_transform_ii} $(X + Y_z) \cup (z - \tilde Y_z) \subseteq X + Y$;
\item\label{item:D_transform_iii} $(X + Y_z) \cap (z - \tilde Y_z) = \emptyset$;
\item\label{item:D_transform_iv} $|z - \tilde Y_z| \ge |\tilde Y_z|$;
\item\label{item:D_transform_v}  $|X + Y| + |Y_z| \ge |X + Y_z| + |Y|$.
\end{enumerate}
\end{proposition}
\begin{proof}
\ref{item:D_transform_ii} Let $w \in z - \tilde Y_z$. Then, there exists $y \in \tilde Y_z$ such that $z = w + y$. But $y \in \tilde Y_z$ if and only if $z = \tilde w + y$ for some $\tilde w \in X+Y$, so $w = \tilde w$ by
cancellativity, and hence $w \in X+Y$. This shows that $z - \tilde Y_z \subseteq X + Y$, and then we are done, as it is clear, on the other hand, that $X+Y_z \subseteq
X+Y$.

\ref{item:D_transform_iii} Suppose for a contradiction that $W := (X + Y_z) \cap (z - \tilde Y_z)$ is non-empty, and let $w \in W$. Then $w = x + y_1$ and $z = w + y_2$ for some $x \in X$, $y_1 \in Y_z$, and $y_2 \in \tilde Y_z$. Since
$\langle Y \rangle$ is commutative, it follows that
$$
z = x + y_1 + y_2 = x + y_2 + y_1,
$$
which implies by \eqref{equ:defining_Davenport_transformed_pair} that $y_1 \in \tilde Y_z$, because
$Y_z, \tilde Y_z \subseteq Y$. This is, however, absurd, as
$Y_z$ and $\tilde Y_z$ are obviously disjoint.

\ref{item:D_transform_iv} We have from
\eqref{equ:defining_Davenport_transformed_pair} that for each $y \in \tilde Y_z$
there exists $w \in X+Y$ such that $z = w + y$, and hence $w
\in z - \tilde Y_z$. On the other hand, $\mathbb A$ being cancellative yields that $w + y_1 \ne w + y_2$ for all $w \in
A$ and distinct $y_1, y_2 \in \tilde Y_z$. Thus, we see that there is an injection $\tilde Y_z \to z - \tilde Y_z$, with the result that $|z - \tilde Y_z| \ge |\tilde Y_z|$.

\ref{item:D_transform_v} Note first that $X$ and $Y$ are non-empty, because otherwise we would have
$(X + 2Y) \setminus (X + Y) = \emptyset$, in contrast to our assumptions.

Using that $\mathbb A$ is can\-cel\-la\-tive, it follows from Lemma
\ref{lem:trivialieties}\ref{item:trivialieties_(ii)} that $|X+Y| \ge |Y|$. This implies the claim if $|Y| = \infty$, so suppose from now on that $Y$ is a finite set.

Then, the inclusion-exclusion
principle and the above points \ref{item:D_transform_ii}-\ref{item:D_transform_iv} give
\begin{equation*}
 |X+Y| \ge |X+Y_z| + |z - \tilde Y_z| \ge |X+Y_z| + |\tilde Y_z|.
\end{equation*}
But $\tilde Y_z = Y \setminus Y_z$ and $Y_z \subseteq Y$, so in the end
$
|X+Y| \ge |X+Y_z| + |Y| - |Y_z|$,
and the proof is thus complete.
\end{proof}
We conclude this section with the following:
\begin{proof}[Proof of Proposition \ref{prop:equivalent_form_with_generated_subgroup}]
\ref{cor:equ_condition_(i)} $\Rightarrow$ \ref{cor:equ_condition_(ii)}. Assume that $X + 2Y = X + Y + \bar{y}$ for some $\bar{y} \in Y$, and let $y \in Y$. Then $X + Y + y \subseteq X + Y + \bar{y}$, and on the other hand, we have from Lemma \ref{lem:trivialieties}\ref{item:trivialieties_(i)} that $|X + Y + y| = |X + Y + \bar{y}|$. But since $X$ and $Y$ are finite, this is possible only if $X + Y + y = X + Y + \bar{y}$, and we are done.

\ref{cor:equ_condition_(ii)} $\Rightarrow$ \ref{cor:equ_condition_(iii)}. Pick $\bar y \in Y^\times$. By hypothesis, we have $X + 2Y = X + Y + \bar y$, and using that $\langle Y \rangle$ is com\-mutative, this is equivalent to $X + 2(Y - \bar y) = X + Y - \bar y$.

It follows (by induction) that $X + n(Y - \bar y) = X + Y - \bar y$ for all $n \in \NNb^+$, and since $\langle W \rangle = \bigcup_{n \ge 1} nW$ for every $W \subseteq A$, we obtain
\begin{equation}
\label{equ:identity_with_subsgrp}
X + \langle Y - \bar y \rangle = X + Y - \bar y.
\end{equation}
Now, the conclusion is trivial if $X$ is empty.
Otherwise, we get by points \ref{item:trivialieties_(i)} and \ref{item:trivialieties_(ii)} of Lemma \ref{lem:trivialieties}, equation \eqref{equ:identity_with_subsgrp}, and the assumption that $X$ and $Y$ are finite that
$$
\ord(Y - \bar y) \le |X + \langle Y - \bar y \rangle| = |X + Y - \bar y| = |X + Y| \le |X| \cdot |Y| < \infty,
$$
Thus, $\ord(y - \bar y) < \infty$ for all $y \in Y$, which, together with Lemma \ref{lem:trivialieties}\ref{item:trivialieties_(iii)}, implies $\langle Y - \bar y \rangle = \langle\fixed[-0.6]{ \text{ }}\langle Y - \bar y \rangle\fixed[-0.6]{ \text{ }}\rangle$. This leads to the desired conclusion.

\ref{cor:equ_condition_(iii)} $\Rightarrow$ \ref{cor:equ_condition_(i)}. Let $\bar y \in Y^\times$. By hypothesis, we have $X + \langle Y - \bar y \rangle = X + Y - \bar y$. Together with the commutativity of $Y$, this implies
$$
X + Y - \bar y \supseteq X + 2(Y - \bar y) = X + 2Y - 2\bar y,
$$
and hence $X + 2Y \subseteq X + Y + \bar y$, which is enough to conclude the proof (since, of course, $X + Y + \bar y \subseteq X + 2Y$).
\end{proof}
\section{Proofs}
\label{sec:proofA062}
We start with Theorem \ref{th:mainA062}, whose proof is actually a ``transformation proof'', extending to a non-commutative setting ideas first used by H.~Davenport in \cite{Daven35}.

In fact, the reasoning follows the same broad scheme of the proof of \cite[Theorem 8]{Tring12}, but differs from the latter in significant details.
\begin{proof}[Proof of Theorem \ref{th:mainA062}]
Set $\kappa := |X+Y|$ for brevity's sake, and suppose that $X + 2Y \ne X + Y + \bar{y}$ for all $\bar{y} \in Y^\times$. We have to prove that
\begin{equation}
\label{equ:reduction_to_the_main_case}
\kappa \ge |X| + \min(\gamma(Y), |Y| - 1).
\end{equation}
This is obvious if $Y^\times = \emptyset$ or $|Y| = 1$, since in that case the right-hand side of \eqref{equ:reduction_to_the_main_case} equals $|X|$, and $\kappa \ge |X|$ by Lemma \ref{lem:trivialieties}\ref{item:trivialieties_(ii)}. So we assume for the sequel that $Y^\times$ is non-empty and $|Y| \ge 2$.

Then, also $X$ is non-empty, otherwise $X + 2Y = X + Y + \bar{y} = \emptyset$ for every unit $\bar{y} \in Y^\times$, in contrast to our hypotheses (as $Y^\times \ne \emptyset$). Hence, we are done if $X$ or $Y$ is infinite, since $\kappa \ge \max(|X|,|Y|)$, again by Lemma \ref{lem:trivialieties}\ref{item:trivialieties_(ii)}.

Putting it all together, we are thus reduced to the case where
\begin{equation}
\label{equ:first_summary}
1 \le |X| < \infty, \ \ 2 \le |Y| < \infty,\ \ \text{and}\ \ Y^\times \ne \emptyset,
\end{equation}
which means, among other things, that $\mathbb A$ is (necessarily) a monoid; as usual, we will denote the identity of $\mathbb A$ by $0_\mathbb{A}$.

Building on these premises, we now suppose, towards a contradiction, that
\begin{equation}
\label{equ:for_the_sake_of_contradiction}
\kappa < |X| + \min(\gamma(Y), |Y| - 1).
\end{equation}
More precisely, we assume that $(X, Y)$ is a minimal counterexample to \eqref{equ:reduction_to_the_main_case}, in the sense that if $(\bar X, \bar Y)$ is another pair of non-empty subsets of $A$ such that $\langle \bar Y \rangle$ is commutative, $|\bar Y| \ge 2$ and $\bar X + 2\bar Y \ne \bar X + \bar Y + \bar{\bar{y}}$ for every $\bar{\bar{y}} \in \bar{Y}^\times$, and
$$
|\bar X + \bar Y| < |\bar X| + \min(\gamma(\bar Y), |\bar Y| - 1),
$$
then $|Y| \le |\bar Y|$; of course, this is always possible and involves no loss of generality.
Lastly, we may further assume, as we do, that
\begin{equation}
\label{equ:special_form_of_gamma}
0_\mathbb{A} \in Y \ \ \text{and}\ \ \gamma(Y) \ge \inf_{0_\mathbb{A} \ne y \in Y} \ord(y) \ge \kappa - |X| + 1,
\end{equation}
for we get by \eqref{equ:first_summary}, \eqref{equ:for_the_sake_of_contradiction}, and Lemma \ref{lem:units_give_invariant_n_transform} that this, again, does not affect the generality of the reasoning.
Accordingly, we have that
\begin{equation*}
X + 2Y \not\subseteq X + Y.
\end{equation*}
In fact, $0_\mathbb{A} \in Y$ yields that $X + Y \subseteq X + 2Y$; therefore, $X+2Y \subseteq X+Y$ would imply $X + 2Y = X + Y + 0_\mathbb{A}$, which is, however, impossible, as we are supposing $X + 2Y \ne X + Y + \bar{y}$ for every $\bar{y} \in Y^\times$.

So, let $z$ be some element in the non-empty set $(X + 2Y) \setminus (X + Y)$, and define
$$
\tilde Y_z := \{y \in Y: z \in X + Y + y\}
\ \ \text{and}\ \ %
Y_z := Y \setminus \tilde Y_z.
$$
Clearly, $\tilde Y_z \ne \emptyset$ and $0_\mathbb{A} \notin \tilde Y_z$, so we have by \eqref{equ:special_form_of_gamma} that
$0_\mathbb{A} \in Y_z$ and $
1 \le |Y_z| < |Y|$.
Then, exploiting that $\langle Y \rangle$ is commutative and $|Y| < \infty$, we obtain by Proposition \ref{prop:properties_of_the_modified_Davenport_transform}\ref{item:D_transform_v} and equation \eqref{equ:for_the_sake_of_contradiction} that
\begin{equation}
\label{equ:transference}
|X + Y_z| \le |X+Y| + |Y_z|  - |Y| < |X| + |Y_z| - 1.
\end{equation}
It follows that $|Y_z| \ge 2$, as otherwise we would have from Lemma \ref{lem:trivialieties}\ref{item:trivialieties_(i)}
and \eqref{equ:transference} that $|X| = |X+Y_z| < |X|$, which is absurd. To summarize, we have found that
\begin{equation}
\label{equ:props_of_the_Davenport_transform}
0_\mathbb{A} \in Y_z \subsetneq Y,
\ \ 2 \le |Y_z| < |Y|,
\ \ \text{and}\ \
|X+Y_z| < |X| + |Y_z| - 1,
\end{equation}
which, along with \eqref{equ:for_the_sake_of_contradiction} and \eqref{equ:special_form_of_gamma}, gives
\begin{equation}
\label{eq:towards_a_contradiction}
|X+Y_z| \le \kappa < |X| + \inf_{0_\mathbb{A} \ne y \in Y} \ord(y) \le |X| + \inf_{0_\mathbb{A} \ne y \in Y_z} \ord(y) \le |X| + \gamma(Y_z),
\end{equation}
where we have used, in particular, that $|Y_z| \ge 2$ and $Y_z^\times \ne \emptyset$ by \eqref{equ:props_of_the_Davenport_transform}, and that $\inf(C) \le \inf(B)$ provided $\emptyset \ne B \subseteq C \subseteq \mathbf N \cup \{\infty\}$.

This is, however, absurd, as \eqref{equ:props_of_the_Davenport_transform} and \eqref{eq:towards_a_contradiction} together contradict the minimality of the pair $(X, Y)$, and it concludes the proof.
\end{proof}
Now we can proceed to prove the corollaries of Theorem \ref{th:mainA062}.
\begin{proof}[Proof of Corollary \ref{cor:UDT_improved}]
The claim is trivial if $
|X + Y| \ge |X| + \min(\gamma(Y), |Y| - 1)$, or $|Y| \le 1$, or $Y^\times$ is empty.
Otherwise, we have from Theorem \ref{th:mainA062} and Proposition \ref{prop:equivalent_form_with_generated_subgroup}\ref{cor:equ_condition_(iii)} that $
X + Y - \bar y = X + \langle\fixed[-0.6]{ \text{ }}\langle Y - \bar y \rangle\fixed[-0.6]{ \text{ }}\rangle$ for some $\bar y \in Y^\times$. Consequently, points \ref{item:trivialieties_(i)} and \ref{item:trivialieties_(ii)} of Lemma \ref{lem:trivialieties}, along with Lemma \ref{lem:old_units_distribution}, give that
\begin{equation*}
\begin{split}
|X + Y| & = |X + Y - \bar y| \ge |\langle\fixed[-0.6]{ \text{ }}\langle Y - \bar y \rangle\fixed[-0.6]{ \text{ }}\rangle| \ge \ord((y - \bar y) - (y_0 - \bar y)) = \ord(y - y_0)
\end{split}
\end{equation*}
for all $y \in Y$ and $y_0 \in Y^\times$, and this yields $|X+Y| \ge \gamma(Y)$.
\end{proof}
\begin{proof}[Proof of Corollary \ref{cor:HS}]
Of course, $X$ is non-empty, otherwise $X \cup (X + Y) = X + \langle\fixed[-0.6]{ \text{ }}\langle Y \rangle\fixed[-0.6]{ \text{ }}\rangle$. Consequently, the claim is trivial if $X$ and $Y$ is infinite, since in that case $|X \cup (X + Y)| = \infty$ by Lemma \ref{lem:trivialieties}\ref{item:trivialieties_(ii)}, and it is still trivial if $Y$ is empty, since then either side of equation \eqref{equ:hamidoune-shatrowsky_bound} is equal to $|X|$.

So, we assume for the sequel that $X$ and $Y$ are both finite and non-empty, in such a way that $X \cup (X + Y)$ is finite too, and set $Y_0 := Y \cup \{0_\mathbb{A}\}$.

We claim that $X + Y_0 = X \cup (X + Y) \ne X + \langle\fixed[-0.6]{ \text{ }}\langle Y_0 \rangle\fixed[-0.6]{ \text{ }}\rangle$. If $\langle\fixed[-0.6]{ \text{ }}\langle Y \rangle\fixed[-0.6]{ \text{ }}\rangle$ is infinite, this is clear from the above; otherwise,
$\langle\fixed[-0.6]{ \text{ }}\langle Y_0 \rangle\fixed[-0.6]{ \text{ }}\rangle = \langle\fixed[-0.6]{ \text{ }}\langle Y \rangle\fixed[-0.6]{ \text{ }}\rangle$ by Lemma \ref{lem:trivialieties}\ref{item:trivialieties_(iii)}, and we have (by hypothesis) $X \cup (X + Y) \ne X + \langle\fixed[-0.6]{ \text{ }}\langle Y \rangle\fixed[-0.6]{ \text{ }}\rangle$.

Therefore, we get from Theorem \ref{th:mainA062} and Proposition \ref{prop:equivalent_form_with_generated_subgroup}\ref{cor:equ_condition_(iii)} that
$$
|X \cup (X + Y)| = |X+Y_0| \ge |X| + \min(\gamma(Y_0), |Y_0| - 1),
$$
which concludes the proof, because $|Y_0| - 1 = |Y| - \mathbf 1_Y(0_\mathbb{A})$.
\end{proof}
\begin{proof}[Proof of Corollary \ref{cor:pillai&chowla_improved}]
To begin, let $\tilde{w}$ denote, for every $w \in \ZZb_n$, the smallest non-negative integer in $w$.
The claim is trivial if $Y$ is a singleton. Otherwise, since
$
\ord(w-w_0) = n/\gcd(n, \tilde{w}-\tilde{w}_0)
$
for all $w,w_0 \in \mathbf Z_n$, we have
\begin{equation*}
\begin{split}
\gamma(Y) & = \max_{y_0 \in Y} \min_{y_0 \ne y \in Y} \ord(y-y_0)  = \frac{n}{\min_{y_0 \in Y} \max_{y_0 \ne y \in Y} \gcd(n,\tilde{y}-\tilde{y}_0)}.
\end{split}
\end{equation*}
It follows that $\gamma(Y) = \delta_Y^{-1} n$, because $\gcd(n, \tilde y) = \gcd(n, \xi)$ for every $y \in \mathbf Z_n$ and $\xi \in \ZZb$ with $\xi \equiv \tilde y \bmod n$. So we are done by Theorem \ref{th:mainA062} and Proposition \ref{prop:equivalent_form_with_generated_subgroup}\ref{cor:equ_condition_(ii)}.
\end{proof}
\section{Closing remarks}
\label{sec:remarks}
The bound provided by Theorem \ref{th:mainA062}\ref{cor:main_condition_(i)} is meaningful only if $\gamma(X) > 0$,
insofar as $\mathbb A$ being a cancellative semigroup implies, by Lemma \ref{lem:trivialieties}\ref{item:trivialieties_(ii)}, that $|X + Y| \ge |X|$. This means, in particular, that the theorem is not very useful unless $\mathbb A$ is a monoid, and raises the challenge of further generalizing the result (and its corollaries) so as to replace $X^\times$ in \eqref{equ:CD_constant} with a subset of $A$ that is significant also when $A^\times = \emptyset$.

On a similar note, every \textit{commutative} cancellative semigroup can be embedded into a group.
It was, however, proved by A.~Malcev in \cite{Malcev37} that there are finitely generated cancellative semigroups that do \textit{not} embed into a group, which serves as a ``precondition'' for some aspects of the present work and its prequels \cite{Tring12, Tring13}, as
it shows that the study of sumsets in cancellative semigroups cannot be systematically
reduced,
in the absence of commutativity, to the case of groups (at least, not in an
obvious way).
\section{Acknowledgments}
This research was supported by the Austrian FWF Project M1900-N39, and partly by the French ANR Project ANR-12-BS01-0011. The author is grateful to Paolo Leonetti (Universit\`a Bocconi, Italy) for some useful comments.

redpajama_set_name: RedPajamaArXiv


text:
Monthly Archives: July, 2022
VELVEETA® IS BOLDY BACK WITH STENHOUSE JR. AND THE NO. 47 CAMARO TEAM AT ROAD AMERICA
Official Release - July 1, 2022
JTG Daugherty Racing is pleased to have VELVEETA® return to the No. 47 Camaro ZL1 with Pick 'n Save on the hood for the Kwik Trip 250 presented by JOCKEY Made in America at Road America this weekend.
RCR Event Preview – Road America
Although Road America has been in operation since the 1950's and has challenged some of the world's best racers for over 60 years, the first modern day-era NASCAR Cup Series race at the 4-mile, 14-turn road course was in 2021.
Extreme E returns for summer showdown in Sardinia
Extreme E returns to Sardinia for the series' first-ever double-header event, co-organised by the Automobile Club d'Italia and Regione Sardegna (6-7 and 9-10 July).
Aidan Fassnacht and Sword Performance Back In Action At Road America As MX-5 Race Winner
Aidan Fassnacht earned his first career Idemitsu Mazda MX-5 Cup win last weekend at Watkins Glen, but has limited time to celebrate as he and the No. 15 Sword Performance Mazda MX-5 are back in action this week at Road America, July 1-3.
Paul Menard Back with Showtime Motorsports as Trans Am Summer Heats Up
The Franklin Road Apparel Showtime Motorsports team will head for the second stint of the midsummer doubleheader this weekend coming from Mid Ohio to Road America, July 1-3.
Buying your first quad. Everything you need to know
SM - July 1, 2022
A quad is traditionally used for off-road sports and activities. They are becoming very popular as both a hobby and in agriculture.
Kenny Murillo captures pole for Michelin Pilot Challenge at Canadian Tire Motorsport Park 120
Angela Campbell - July 1, 2022
Kenny Murillo was fastest during qualifying Friday afternoon, topping the leaderboard in his No. 72 Murillo Racing Mercedes-AMG GT4 with a track record lap at 106.997 mph and will start on the pole for Saturday's Michelin Pilot Challenge Canadian Tire Motorsport Park 120 for the sixth round of the Michelin Pilot Challenge season.


Even supermodels get lice! And then they get ribbed about their hygeine.
Pop quiz: What's the one four-letter word that can strike fear into the heart of any mom? No, we're not talking about puke (although that is pretty bad). It's lice, the tiny critters that mean life as you know it in your house is about to change, and not for the better.
Many a mom has detected the pesky pests on the heads of their babes, or worse, had a school nurse deliver the bomb that Junior's not welcome back at school until you get rid of his tresses' new tenants. Even supermodels have to deal with infestations, Heidi Klum revealed on Wednesday's episode of The Ellen DeGeneres Show.
"My daughter was itching and then she had to go to the nurse and then they looked and sure enough they found a little egg," the gorgeous mom-of-four told DeGeneres.
And it wasn't just Klum's daughter who ended up scratching her head; the lice spread to the former Victoria's Secret model herself! Talk about no fun.
To make matters worse, DeGeneres decided to rub salt in Klum's figurative wound, asking whether she bathes.
"Yeah, of course we do!" Klum replied indignantly.
Hey, don't worry Heidi. You aren't the first mom to find yourself at the receiving end of false assumptions about lice (even if it was a joke!) There's definitely that stereotype that if your child gets lice, he or she is dirty. But even sharing a hat with another kid at school can spread the infestation.
The best you can do is try to educate your children about how to avoid getting lice. And if the worst happens, and your kiddo's head is taken hostage by those evil little bugs, know you aren't alone!


text:
Electric Public Transport Market - Global Industry Analysis, Size, Share, Growth, Trends, and Forecast 2018 - 2026
Transparency Market Research
Electric public transport is a public transport system with electric propulsion, which includes electric vehicles and locomotives. Electric public transport comprises electric and hybrid taxis, public transportation buses, and locomotives such as metros, subways, monorails, and trams. Electric public transport reduces emission from transportation at a rapid rate and thereby protects the environment.
Surge in demand for green vehicle and necessity to curb emissions are major factors that are expected to augment the electric public transport market during the forecast period. Electric public transport is a major trend growing toward the rising transition of the public transportation. Furthermore, incentives and tax credit offerings on electric vehicles by regulatory bodies across the globe are significant factors that are likely to boost the electric public transport market during the forecast period. According to a report published by Zero Emission Urban Bus System Project, 19 public transport authorities and operators have a circulated e-bus strategy for 2020 covering 25 cities in Europe. By the end of 2017, there were more than 2,500 electric buses representing around 6% of the total fleet of 40,000 public buses operating in these cities. Additionally, more than 13 public transport authorities and operators in a further 18 more cities in Europe are projected to account more than 6,100 electric buses in service, representing nearly 43% of their total fleet of 14,000 public buses. Increase in rate of adoption of electric public transportation in China, especially by Shenzhen bus group, is another major adoption of electric public transport. This is projected to offer lucrative opportunities to manufacturers and is likely to lead to government ties regarding electric vehicle and locomotives. However, higher cost of electric vehicles and limited availability of charging stations are estimated to hamper the electric public transport market during the forecast period.
The global electric public transport market can be segmented based on transportation vehicle, propulsion, and region. Based on transportation vehicle, the electric public transport market can be bifurcated into buses, taxi, locomotives, and others. Among transportation vehicle, buses is a rapidly expanding segment. This is due to mean rise in electrification of public transport by government bodies to curb emission.
In terms of propulsion, the electric public transport market can be divided into fully electric and hybrid. The fully electric segment accounts for a prominent share of the electric public transport market, as compared to the hybrid segment. Rapid transition of public transport and surge in inclination to opt for electric vehicles are key factors that are likely to boost the fully electric segment of the electric public transport market during the forecast period.
In terms of geography, the global electric public transport market can be segregated into Asia Pacific, Europe, North America, Middle East & Africa, and Latin America. Among regions, Asia Pacific accounts for a key share of the global electric public transport market. Higher adoption of electric buses and coaches and rise in penetration of electric vehicles are major factors fuelling the number of electric buses being introduced in the region. Countries such as China, India, Japan, and South Korea are planning to replace their public transport system with electric transport system. This is a major factor that is anticipated to boost the market in the region.
Prominent players operating in the global electric public transport market include General Motors, Ford Motor Company, AB Volvo, BMW AG, BYD Company Limited, Daimler AG, BYD Company Limited., Shenzhen Wuzhoulong Motors Co., Ltd, Yutong, Solaris Bus & Coach S.A. Nippon Seiki Co., Ltd., Proterra, Ashok Leyland, Tata Motors, Foton Motor Inc., Zhongtong Bus Holding Co., Ltd., FAW Group, IVECO, Dongfeng Automobile Co., Ltd., Bombardier Inc., Siemen AG, Mitsubishi Electric Corporation, CRRC Corporation Limited, and Hitachi Ltd.
Automotive Block Chain Market - Global Industry Analysis, Size, Share, Growth, Trends, and Forecast 2018 - 2026
Automotive Cabin Air Quality Sensor Market - Global Industry Analysis, Size, Share, Growth, Trends, and Forecast 2018 - 2026
Automotive Intelligent Rear-view Mirror Market - Global Industry Analysis, Size, Share, Growth, Trends, and Forecast 2018 - 2026
Automotive MRO Market - Global Industry Analysis, Size, Share, Growth, Trends, and Forecast 2018 - 2026
Automotive Power Distribution Block Market - Global Industry Analysis, Size, Share, Growth, Trends, and Forecast 2018 - 2026


text:
import Vue from 'vue'
import VueRouter from 'vue-router'

import Splash from '../views/splash.vue'
import WardLeaderList from '../views/ward-leader-list.vue'
import WardLeader from '../views/ward-leader.vue'
import CityMap from '../views/city-map.vue'
import ContentPage from '../views/content-page.vue'

Vue.use(VueRouter)

const routes = [
  {
    path: '/',
    name: 'splash',
    component: Splash
  },
  {
    path: '/leaders',
    redirect: '/leaders/democratic'
  },
  {
    path: '/leaders/:party',
    name: 'ward-leader-list',
    component: WardLeaderList,
    props: true
  },
  {
    path: '/leaders/:party/:ward/:slug',
    name: 'ward-leader',
    component: WardLeader,
    props: true
  },
  {
    path: '/map',
    name: 'city-map',
    component: CityMap
  },
  {
    path: '/:slug',
    name: 'content-page',
    component: ContentPage,
    props: true
  }
]

const router = new VueRouter({
  mode: 'history',
  routes,
  scrollBehavior
})

function scrollBehavior (to, from, savedPosition) {
  if (savedPosition) {
    return savedPosition
  } else {
    return { x: 0, y: 0 }
  }
}

export default router

redpajama_set_name: RedPajamaGithub


text:
NYU Alumni & Friends Connect February 2021
Nine NYU and NYU Alumni Podcasts to Listen to Right Now
The Latest on Spring 2021, NYU Commencement, and More
Enjoyed our last roundup of podcasts hosted and produced by NYU alumni, students, faculty, and staff? If you're looking to start a new career, meet women of color who are passionate outdoor advocates, or ready to be inspired by stories about and from the NYU community, check out our latest list of podcasts that keep us caught up with the world beyond NYU.
Quicklinks: Podcasts by NYU Alumni | Podcasts by the NYU Community
(Note: all links below open in new tabs)
Please note this is not an exhaustive list and only includes a small number of the many talented NYU alumni who work in the field of podcasts.
Podcasts by NYU Alumni
Amanda Kersey (CAS '11) is the producer of this Harvard Business Review podcast featuring conversations about the workplace and women's place in it. These conversations focus on supporting young women entering the workforce or in their first jobs.
Lemonada Media
Co-founded by Jessica Cordova Kramer (CAS '00) and Stephanie Wittels Wachs (TSOA '03), Lemonada Media launched 10 podcasts in 2020, including Last Day and In the Bubble with Andy Slavitt.
See Us Outside
Gabaccia Moreno (TSOA '13), first-generation Mexican-American multi-disciplinary creator, consultant, and passionate outdoor advocate, is the host and guide of this mini-series, which explores the unique relationships girls and young women of color have with the natural world and outdoor recreation.
Hosted by Jonathan Sonnenberg (GAL '20), Nucleus explores how students cultivate theory and practice in their diverse fields of research at NYU's Gallatin School of Individualized Study. Through interviews focused on the process of developing an individualized Rationale and Concentration, the podcast seeks to elucidate the natures and possibilities of interdisciplinary study.
Podcasts by the NYU Community
The NYU SPS Replay Podcast
Be inspired by the journeys and stories from our NYU School of Professional Studies alumni, faculty, students and administration.
NYU LabCast
Ro Reddick, who provides specialized services to students with career interests in creative industries at the Wasserman Center and the NYU Production Lab, brings you all new things at NYU, from the game center to helping you get on your feet in starting a new podcast.
The Wasserman Career Podcast features members of the NYU community doing interesting work and navigating the professional world. Listen now on Spotify, Apple Podcasts, or Anchor!
You Matter!
Brought to you by the Department of Public Safety, each episode informs the NYU community about resources that are available to victims of violence, assault or other criminal behaviors, both on and off campus, and to give first-hand experiences of victims and those in the community who assist them.
Ask a Political Scientist
Andrea Jones-Rooy, visiting Associate Professor and Director of Undergraduate Studies at the NYU Center for Data Science, standup comedian and trapeze artist, provides listeners with (mostly) non-partisan and science-forward discussions on American politics—what's going on, how we got here, and what to do about it.
NYU Alumni Podcast Roundup
Want more podcast recommendations? Check out our podcast roundup


All true teas (whether white, green, oolong or black) come from the same plant, called the Camellia sinensis. Any leaf, root, fruit or flower that comes from a plant other than the tea plant is considered an herbal tea. For example, chamomile flowers, peppermint leaves, yerba maté and rooibos are considered herbal teas because they do not come from the traditional tea plant.
The teas in this section are all made from herbal ingredients. Since ancient times, herbal teas have been revered for their bountiful health benefits and therapeutic properties. Herbal teas are often used to aid digestion, sooth allergies, relieve headaches and reduce stress. Continuing this tradition, Octavia's herbal teas are based on ancient homeopathic remedies, blended with quality, flavor and health benefits in mind.
With the exception of yerba maté (which comes from a plant with naturally occuring caffeine), all of our herbal teas are 100% caffeine-free, making them an excellent choice for evening or for people with caffeine-sensitivities.


text:
Home 49ers NFC Notes, 49ers, Rams, Jared Goff, Seahawks
NFC Notes, 49ers, Rams, Jared Goff, Seahawks
Nate Bouda
49ers GM John Lynch didn't completely rule out the idea of a highly-drafted quarterback to challenge Jimmy Garoppolo. However, as things stand now Lynch doesn't foresee San Francisco using their first-round pick at No. 12 overall on a passer. He added another defense of Garoppolo's standing in the organization.
"We certainly could [select a QB], I don't anticipate that right now, but we're so early in the process, we'll do what's best for this team," Lynch said via NBC Sports' Alex Didion. "I do wanna make it clear, and I think Kyle and I have been fairly transparent since we've been here, we have a lot of belief in Jimmy Garoppolo. Where does that come from? It comes from the sample that we have, and when he's been out there, we've been better. We've been successful.
"In addition to that, his teammates understand that. I've always believed leaders not only play at a high level, but they also make everyone else around them better. He's had that quality on our team, I think he's a good fit for what we do system-wise. Are there some areas he can improve on, sure, and we'll always challenge him to do that. But we have liked working with him, and we're excited to have him back."
Rams QB Jared Goff was healthy enough to be active against the Seahawks in the Wildcard Round on Saturday. But HC Sean McVay made the decision earlier in the week to go with backup QB John Wolford since he was healthy enough to practice the whole week. That plan went out the window when Wolford went down with an injury of his own and Goff came in for the rest of the game, but he acknowledged it was a tough experience not being the starter.
"Sean made it very clear to me, 'We're obviously not benching you. We need to have a plan moving forward and this week, with where you're at physically and what not…' I got it, to an extent. But it was tough," Goff said via NFL Media's Michael Silver.
"As a competitor, you want to be out there — but I prepared accordingly," Goff added. "We were doing almost all I could 'on air' throughout the week, just off to the side. I wasn't taking any reps against the defense, but I was just trying to get my thumb into a good place."
The performance wasn't Goff's best as he missed passes all day. But he did just enough for the win and to get the ball in the zip code of players like WR Cooper Kupp, whose 44-yard catch was one of a few huge plays in the game. Afterward, Goff said he's not going to use the thumb as an excuse despite being less than two weeks removed from surgery.
"I don't want to use (the injury) as an excuse," Goff said. "Like, I was out there playing; I'm capable. Yeah, there were limitations, but I'm not gonna sit here and say that I can't make that throw… or I can't make any of the throws on the field. I can make any throw I want to make. Well, that one in particular, I was trying to get into a rhythm. I was trying to figure it out a little bit with where I was at with my hand and get into a good place, and get into a good place, and Cooper made a great play for me and settled me down."
Rams WR Cooper Kupp hobbled off late in the game with a knee injury, which was scary given his history with an ACL injury. But he avoided a major injury and has just a knee contusion. (Mike Garafolo)
Rams DC Brandon Staley will likely receive a raise and a contract extension from Los Angeles if he doesn't take a head coaching job this cycle. (Garafolo)
The Rams are hopeful DT Aaron Donald can play next week after X-rays on his ribs came back negative. (Garafolo)
The Seahawks cooled off dramatically after a torrid start to the season as teams put a lid on their passing game, literally. After QB Russell Wilson and WR D.K. Metcalf shredded teams downfield, the gameplan from most opponents was to focus on limiting the deep shots and make Seattle work its way down the field. No one did it better than the Rams, who won the third matchup on Saturday 30-20 to knock Seattle out of the playoffs. Seahawks HC Pete Carroll lamented the team's inability to adjust that was finally exposed.
"That doesn't mean we throw the ball over their head all of the time and going for just bombs," Carroll said via ESPN's Brady Henderson. "But there's a lot of space we create in the play-passing game, and it seemed like during the course of the season, after the halfway point, we had hit so much early, we had been so effective that people found a way to stay back and just try to bleed us and make us have to throw the ball underneath, and we were maybe really going for it more than we needed to and didn't take advantage of switching gears a bit there as effectively as we would like."
There seemed to be an opportunity for Seattle to get back into the game when Rams DT Aaron Donald left the game with an abdomen injury. But even that wasn't enough for the Seahawks to get untracked on offense, as they continued to give up pressure, fail to get any offensive momentum and turned it over on a fumbled punt return as the final nail in the coffin.
"We needed to get going there," Wilson said via Adam Jude of the Seattle Times. "The game kind of felt stale for us in a way — we kind of flatlined. We needed to get going and make that happen. And the next thing you know, we didn't."
Brandon Staley
D.K. Metcalf
John Lynch
John Wolford
Eagles Request Interview With Rams DC Brandon Staley For HC Job
Seahawks Have Spoken To Doug Pederson About Offensive Coordinator Job
NFC Notes: 49ers, Rams, Seahawks
49ers Sign QB Josh Johnson & OT Corbin Kaufusi To Futures Deals
NFL Rescinds Josh Gordon's Conditional Reinstatement, Suspended Indefinitely


Jazz lyricist and vocalese singer Jon Hendricks passed away peacefully with family by his side in New York City on Wednesday, November 22 at the age of 96.
Known for bringing the style of vocalese to mainstream audiences, Hendricks rose to fame in the late 1950s as the principle lyricist for the vocal trio Lambert, Hendricks & Ross, earning a Grammy® Award for Best Performance by a Vocal Group for their 1961 album High Flying. His lasting legacy continues to stand the test of time with generations of world-renowned singers including Al Jarreau, Bobby McFerrin, and Kurt Elling influenced by Hendricks' work.
Jon Hendricks is survived by his daughters Michele and Aria Hendricks, son Jon Hendricks, Jr., and three grandchildren Colleen, Azaria and Daniel. His wife of 57 years, Judith, died two years ago, on November 18, 2015.