The Sikat Pinoy National Arts & Crafts Fair is one of the key programs of the Department of Trade and Industry-Bureau of Domestic Trade Promotion (DTI-BDTP) to provide marketing assistance to MSMEs engaged in the creative industries.
Carefully screened to comply with stringent criteria in line with the Sikat Pinoy theme of Piling-Piling Produktong Pilipino, about 280 exhibitors from the country's 16 regions will be participating in the upcoming 2017 Sikat Pinoy National Arts & Crafts Fair, to be held from October 26 to 29, at Megatrade Halls 1-3, SM Megamall, Mandaluyong City. Admission is free.
Each region's unique products will be represented like fine jewelry from Bulacan, pearls from Mindanao, Filipino heritage cuisine, and a coffee pavilion with different coffee varieties. Millennial shoppers will be pleased to find that, although many of the products are made using traditional methods, they will have designs that are timeless and surprisingly trendy. For example, indigenous textiles and beads are used in place mats, table runners, shawls, bags, fashion accessories, home décor, and gift items.
From Luzon, there will be abel blankets from Ilocos Norte, crocheted products from La Union, ikat textiles from Ifugao, tingguian textiles and beads from Abra, as well as piña cloth from Camarines Norte. From the Visayas, women's groups will bring their hablon fabrics and embroidered clothing from Iloilo, piña textiles and bariw bags from Antique, and basey mats and bags from Samar. From Mindanao, there will be Yakan woven products from Zamboanga City, malong from Maguindanao, and hinabol textiles from Bukidnon.
To further educate Metro Manila mall goers, there will be a series of live demonstrations on indigenous and traditional weaving, embroidery, and embellishment techniques from the different regions.
Complementing the wealth of heritage and artisanal products on display, the fair will also be a "festival of the senses" with a daily program of onsite artistic performances, skills demonstrations, and entrepreneurship learning sessions.
On Friday, October 27, the live programs will focus on food, as follows: 11:00AM - Story of EchoStore (Pacita "Chit" Juan); 12:30PM - T'Boli Food Preparation (Lyn Lambago); 2:00PM - Philippine Culinary Heritage (Clang Garcia); 3:00PM - Story of G-Stuff (Jen Santos); 4:00PM – Tepo Mat, Tutup Food Cover (Tawi-Tawi Island Souvenir); 4:30PM - Muslim Mindanao Cuisine and Halal Food Preparation (Chef Datu Shariff Pendatun III); 6:30pm - Philippine Coffee Appreciation and Bartending (DTI-CAR); 8:00PM – Talk by weavers from different regions.
On Saturday, October 28. the sessions will center on crafts and finance for MSMEs, with the following topics: 10:30AM - MNL Perfumes (Mica Javier); 11:30AM - Pursuing Business in Design (Rachelle Leones); 1:00PM - Fair Trade and Handicrafts (Voltaire Alferez); 2:00PM - Yabang Pinoy (Maricris Sarino-Joson); 3:00PM - Finance Management for Business Entrepreneurs (Edwin Suson); 5:00PM – Talk by weavers from different regions; 7:00PM - NACF Fashion Show.
The culminating activities on Sunday, October 29, will be a music and arts festival consisting of: 11:00AM – Watercolor on Canvass (Art Nebula); 1:30PM - Painting (Fernando Sena); 3:30AM - Youth Empowerment through the Performing Arts (Shirley Kimmayong); 4:30PM - Lemuhen Cultural Dance Troupe (Lyn Lambago); 5:30PM – Talk by weavers from different regions; 7:00PM – Homeworkz mini concert featuring R&B Prince JayR, Kris Lawrence, Alcelle Santos, Mica Javier, Q-York, Bendeatha of Salbakuta, Francis Lim, and Deejay Poblete & Noah.
All the sessions are open to the public for free.
For more information, please call DTI-BDTP at 751-3223 or email bdtp@dti.gov.ph.


Hallo, ich heiße Danielle. My pronouns are she/her or they/them. I live in Berlin, Germany where I enjoy riding bikes, spinning Poi, and photography.
I sometimes write about computers and things on my blog.
I work on the Nomad cluster orchestrator at HashiCorp.
CircleCI: I worked on many aspects of build infrastructure, including writing an ESXi cluster orchestrator. A talk about some K8s + Nomad work can be found here. The monthly build count drastically increased since that talk.
CocoaPods: The dependency manager for mac and iOS developers. I worked on bug fixes, new features, release management, technical design, and web operations.
Fastlane: Deployment tools for iOS and Android developers. I primarily maintained CI and provided feedback on new design work.
I enjoy speaking about topics related to building reliable systems, developer tools, and sustainable teams. Due to Brexit, I'm taking 2019 off from speaking at international conferences, however, I would be interested in speaking at Conferences and Meetups in Germany.
If you'd like me to speak, email dani@builds.terrible.systems.
© Danielle Lancashire. CC BY-SA 4.0.


text:
MinorsOn Campus
Hiring Guidelines
Minors on Campus
The Office of Human Resources
Gulick Hall, 1st Floor
Monday-Friday 8:30 a.m.-5 p.m.
Hobart and William Smith Colleges is committed to a student-centered environment, is globally focused and grounded in the values of equity and service, and develops citizens who will lead in the 21st century. Education takes place not only in the classroom, but also outside the classroom through campus programs and service projects. The Colleges view civic responsibility, community engagement, and international education as integral components of a liberal arts education. In carrying out this charge the Hobart and William Smith community serves and partners with a broad array of internal and external constituents of all ages from a variety of communities, near and far, in an intentionally open environment.
Inherent to fulfilling this mission is a human obligation to provide safety for persons of all ages who participate in the Colleges' programs. All members of the Colleges' community are stewards of these relationships and must recognize differences of power that often exist. As members of this community, we all have a shared responsibility to protect those who are in a position of vulnerability. This is fulfilled when all individuals share in accountability for this obligation.
In light of these principles, the Colleges has implemented a Minors On Campus policy and associated safe guards for youth programs or events. This Policy applies to any program, activity or service involving Minors that occurs on the Colleges' property and to any HWS sponsored program, activity or service involving Minors, wherever it occurs. This Policy does not apply to general public events in the Colleges' facilities where parents or guardians are required to attend and appropriately supervise their child. Any Minor visiting the Colleges and not directly participating in a program, activity or service must be supervised by a parent or guardian at all times. Unsupervised minors who are not participating in a program, activity or service may be asked to leave the Colleges' property by Campus Safety.
Reporting RequirementsHWS Policy on MinorsTrainingBackground ChecksFAQ
Youth Program Questionnaire
Related Sites and Contacts:
HWS Department of Campus Safety, 315-781-3656
HWS Title IX Office, 315-781-3922
NYS Office of Children and Family Services, 800-342-3720
New York Statewide Central Register of Child Abuse and Maltreatment, 800-342-3720

redpajama_set_name: RedPajamaCommonCrawl


THE PRESIDENT: Thank you. Please sit down, ladies and gentlemen. I want to welcome all of you here and especially say a word of welcome and thanks to Senator Glenn and Mrs. Glenn and their daughter. Senator Glenn made this occasion possible today.
I want to welcome a group of extraordinary student athletes, the Marauders and Lady Marauders of Central State University, winners of the NAIA Championships in football, as well as men and women's indoor and outdoor track and field. I want to welcome the Central State President Dr. Author Thomas.
These teams have been remarkably successful. First of all, Central State's football team captured the 1992 NAIA Division One National Championship with a come-from-behind victory over -- what school? (Laughter.) This was no fluke. For Coach Billy Joe, named Division One Coach of the Year, it was the second time that he's won a national title in three years. Coach Joe has guided Central State to the playoffs for the past six seasons, and to the finals for the past three. His winning formula: the three Ds he preaches to his players -- drive, desire, and determination.
Coach Joe, you've earned a fourth D for the Marauders -- dynasty. That's what you've put together. And I must say, I've carried a special interest in this team because you had to run over the University of Central Arkansas a couple of times in playing for these championships. So we followed it very interestingly.
This was the first national title in both indoor and outdoor track. At the indoor championship, they captured the title by winning the mile relay in the final event. They also swept the 600-yard run behind the winning pace of team member Neil DeSilva. This young man went on to clock winning times in both the 200 and 400-meter dash, to help them win the outdoor championship.
The Lady Marauders took their indoor title and also their first, winning six out of 16 events with record-setting performance and double wins by both Carolyn Sterling and Sherdon Smith. Outdoors the Lady Marauders claimed their third consecutive NAIA championship -- a "threepeat." Dionne Hemming set a world record for the 400-meter hurdles on her way to earning the title of "Most Outstanding Female Performer."
Jumping hurdles can also be a useful skill in this city. But I understand Dionne could not be with us here today because she's in Spain.
On behalf of our nation, let me salute all of you for your fine performances. You are teams with truly a proven track record. As student athletes at an historically African American institution, you can be proud of your many achievements. Your drive and your desire and your determination are an example for all Americans.


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Purple Hair Dye Over Henna In spite of all the area of interest hair salons that have now been showing up, from braid pubs to blowout salons to interrogate studios, specialist hair dye jobs still induce the beauty industry. But we are rethinking the procedure after being ruined by this elegance terror narrative.
So we asked cosmetic chemist Ni'Kita Wilson, chief executive officer of Catalyst aesthetic Development and Elizabeth Cunnane Phillips, Philip Kingsley's trichological expert, to spell out what hair dye is doing for your own hairthinning.
Hair dye contains a couple of barriers it has to overcome before it deposits the shade upon your own hair. "It needs to be able to enter the hair shaft, and also obstacle among your cuticle, that's your hair of hair protection," explains Wilson. "In case you presume how tree bark includes that outside covering, then that's how exactly our cuticle acts. Therefore for [dye] to permeate throughout the cuticle, it needs to be lifted" Spread peppermint. This compound elevates the pH of the entire scalp, also in doing so, the cuticle lifts and calms. "The problem with this really is as soon as you've bothered the cuticle, the damage has begun because the cuticle is not supposed to be lifted," states Wilson.
Allergic reactions to hair dye are pretty frightful, but they are also scarce. "It is projected that 4 in one million folks are sensitive to hair dyes. Nevertheless they can happen, and so you ought to be careful," states Cunnane Phillips. "The little proportion of danger tends to place people off out of doing a patch test, however, we can not stress enough the value of performing a test for secure colouring."
Lots of Men and Women in the United States and Europe utilize hair dyes. It's projected more than one-third of women over age 18 and roughly 10% of adult males above age 40 utilize some hair dye.
Modern day hair dyes are classified as lasting (or amino ), semi-permanent , and temporary. Permanent hair dyes, which produce up roughly 80% of now marketed goods, include of colourless dye"intermediates" (compounds known as aromatic amines) and dye"couplers." |} At the presence of hydrogen peroxide, the intermediates and also couplers respond with the other person to form pigment molecules. Darker colours are created by using high doses of intermediates. Semi-permanent and short-term hair dyes are nonoxidative and include coloured compounds that blot hair right.
More than 5,000 unique compounds are applied in hair dye solutions, a few which are reported to be carcinogenic (cancer-causing) in animals (2--4). Because numerous people utilize hair dyes, boffins have attempted to decide whether contact with the compounds in hair colouring services and products is related to an increased risk of cancer in people.
Purple Hair Dye Over Henna Historical hair dye formulations contained compounds, like aromatic amines that were found to induce cancer in animals. From the mid century - the too late 1970s, however, suppliers shifted the components in dye services and products to eliminate some of these chemical substances (2, 3, 5). It's not understood if a number of these compounds still used in hair dyes can lead to cancer. Given the wide spread usage of hair dye merchandise, a tiny increase in risk may possess a considerable public wellness impact.
A number of reports have researched the association between your personal usage of hair dyes as well as the risk of NHL, together with contradictory outcomes. Because the tiny quantity of some studies may have restricted their capacity to detect an association between hair dye use and cancer, a pooled (joint ) analysis of 4 case-control studies have been carried out (8). Each studies had obtained detailed details on hair dye usage, including dates and duration of usage, and NHL sub type. The pooled analysis comprised 4,461 women with NHL along with 5,799 girls who didn't possess NHL. The results of the analysis showed that females who started applying hair dye before 1980 had a slightly (30 percent ) improved risk of NHL when compared with all girls who hadn't ever applied hair dye, while no sudden increase in risk was seen for women who started applying hair dye after 1980.
Once the investigators analyzed the hazards of various specific NHL subtypes, they found that hair dye end users had increased possibilities of both follicular lymphoma and chronic lymphocytic leukaemia/small lymphocytic lymphoma. For the most part, the gains were confined by women who started applying hair dye before 1980, but an higher risk of follicular lymphoma was detected among females who started employing dark-coloured dyes after 1980. Even though these outcomes are in agreement with the idea that previous hair dyes are somewhat more carcinogenic, it is also feasible the lack of increased hazards for hair dye people who began applying dyes after 1980 reveals lower cumulative vulnerability ranges or inadequate period because first vulnerability for virtually any increase in hazard to become clear.
Research of this association between hair dye usage as well as the chance of leukaemia have experienced contradictory outcomes. As an example, one particular case-control analysis analyzed hair dye usage one of 769 individuals with adult acute leukaemia and also 623 individuals with no leukaemia at the united states of america and Canada (9). It found that the dangers of severe leukaemia were high than users of formulas of both permanent and nonpermanent (i.e., semi permanent and temporary) dyes than one of those who had not used dyes, although the gains weren't statistically significant. No risk gains were noticed among users of recent dye formulations. The risk was highest among those who had consumed everlasting dyes for longer durations (15 or more years).
But a case-control research in Italy observed no association between usage of permanent hair dye overall and risk of leukaemia, but consumers of black everlasting stains, but not one of additional colour dyes, did possess an increased hazard. In this analysis, however, didn't collect information on the frequency or timing of hair dye usage Purple Hair Dye Over Henna.
Related Post "Purple Hair Dye Over Henna"


UIS) today announced that Australia and New Zealand healthcare provider Healthscope – a new client for Unisys – has awarded the company a three-year contract to provide a package of Unisys security solutions, including a Security Information and Event Management (SIEM) system and associated services, to protect critical IT systems used to deliver quality healthcare to patients.
Under the engagement, signed in the second quarter of 2018, the SIEM solution, which will include technology from Unisys partner LogRhythm, a leader in security intelligence and analytics, will enable Healthscope to secure its IT environment by quickly and efficiently detecting and neutralizing advanced cyber threats before they can disrupt operations and the delivery of healthcare services. A Unisys Security Operations Center will manage the SIEM system and provide 24/7 detection and response services to Healthscope.
Healthscope is a leading private healthcare provider with 43 hospitals in Australia and market-leading international pathology operations in New Zealand. The company's more than 18,000 employees provide always-on services in an environment where safety and quality are paramount.
"Protecting patient information and critical medical systems is a top priority for Healthscope," said Lachlan McGill, information security manager at Healthscope. "Preventing attacks is just one piece of the puzzle. Timely detection and response is critical to minimizing the impact on business operations and, most importantly, patient care."
Healthscope is planning a further rollout to New Zealand this year.
Rick Mayhew, vice president and general manager, Unisys Asia Pacific, said: "It's really all about minimizing the impact and disruption to the organization, enabling clinicians to focus on the patients' health and recovery, not worrying about technology or security. Our collective knowledge of healthcare operations as well as legal and compliance requirements, will enable Healthscope to accelerate their objective to provide a secure, care-focused experience for its patients."
Unisys security solutions combine experienced consulting, advanced products and managed services for the entire security lifecycle from prediction and prevention to detection and remediation of risks and advanced threats. For more information on Unisys security solutions, click here.
Unisys leverages deep domain expertise across vertical markets, including life sciences and healthcare. Recently, Unisys also announced that its award-winning Stealth™ security software suite featured upgrades to extend protection to purpose-built Internet of Things (IoT) devices and other connected equipment, including medical devices like health monitors and IV pumps. Unisys supports more than 1.5 million life science incidents and service requests per year through its global service desk offerings.
For more information about how Unisys can help you manage your life sciences and healthcare needs across the globe, please click here.
About Unisys Unisys is a global information technology company that builds high-performance, security-centric solutions for the most digitally demanding businesses and governments on Earth. Unisys offerings include security software and services; digital transformation and workplace services; industry applications and services; and innovative software operating environments for high-intensity enterprise computing. For more information on how Unisys builds better outcomes securely for its clients across the Government, Financial Services and Commercial markets, visit www.unisys.com.


Since the house is a little bit chaotic as a result of the work, I've been taking the kids out and about. We've been to the National Capital Radio and Television Museum when my husband's mom was in town, the Hagerstown City Park just the other day (which has oodles of ducks and geese as well as a swan with lovely waterways), and various parks and play dates. Oh, and we've been around looking at houses, since we're hoping to sell soon. Today, we looked at a lovely 1930s brick home smack in the middle of cropland. The views were gorgeous and tranquil which is really appealing right about now in contrast to the craziness that is house renovation. haha!
So, while I don't have any ration cooking to show for this month's menus (hopefully soon!), I thought I'd post a few highlights from our trip to the awesome Radio & Television Museum! It's ridiculous how many pictures I took. It's so hard to narrow it down!
I love the Art Deco details on this radio!
A seriously cool hands-on activity!
The photo on the right is the table with all the fantastic hands-on stuff they have.
My kids spent a long time at this table!
It's a good thing we had the museum all to ourselves on a cold, rainy day.
We had a really great time, and I learned a lot! Their collection is fantastic and though the museum is crammed into a small house, it's jam packed with drool-worthy artifacts. I also love that they host radio repair workshops where you can learn to take apart your old radio and fix it!
The only downside is that it's not stroller friendly with only stairs going up to the 2nd floor. It's also not really a good place for little kids under the age of 6, as was evidenced by my stir-crazy 16-month old stuck in the stroller or my arms the whole time, the poor little guy. I'd still go visit this museum again. I'm not much of a sign reader, but I'd like to go back and do some more reading about all that awesome technology!
If you're ever in the Maryland/D.C. area, definitely look this museum up along with its sister museum The National Electronics Museum. It's off the beaten path compared to all the fancy Smithsonian Museums, but I think it's worth the trip.


It is made to be admired, worn and made your own. Crafted in rose gold or steel, it flatters every wrist in its two sizes: small and medium. With its two back and front dials and a double set of hands driven by a single movement, it radiates an enduringly seductive elegance. On the back, the deep black dial features an Art Deco style enhanced by two rows of diamonds that flirt with the night hours.


With 2016 under his belt, it appears Anderson .Paak is gathering a whole new roster for his next project. And the selections speak for themselves. In an Instagram post scored by NxWorries' "Khadijah," .Paak can be seen putting in work with tour mate, Bruno Mars, as well as the living legend, Nile Rodgers, and Britain's own wunder-brothers, Disclosure.
The caption reads, "Caught a wave in London," and the clip arrives one week after footage surfaced of The Internet joining .Paak onstage at a one-off headlining show in London. All of which is to say, if you happen to be in the UK today or Thursday evening, you should probably consider grabbing a ticket for one of the hottest shows to grace a stage in some time.
This summer, .Paak and Knxwledge will release a remix project of their debut album Yes Lawd!, while .Paak's band, The Free Nationals, is also readying its own debut. Watch as Anderson .Paak hits the studio with some heavyweights in the clip below and hold tight for the fruits of their labor.


text:
MainTags listJoe Lieberman
Joe Lieberman: Trump 'has every right to take this case to court'
Arutz Sheva Staff , 11/11/2020, 2:52 AM (Yoni Kempinski)
Joe Lieberman: I hope Trump and Biden will put the country first
i24NEWS ,11/3/2020, 9:03 PM
NAACP accuses Joe Lieberman's son, a Senate candidate, of using 'racist tropes' in book
Ron Kampeas, JTA ,8/9/2020, 3:39 PM
The Shabbat of Senator Joe Liberman
Former US Senator speaks to Shabbat UNPLUGGED founder about the importance of Shabbat to the Jewish people.Arutz Sheva Staff, 6/9/2020, 8:35 PM
Joe Lieberman: Oppression and persecution cannot exterminate us
Former Senator: 'No nation on earth has endured such struggles and sufferings as we have. Jew-baiting merely winnowed out our weaklings.'Arutz Sheva Staff, 10/31/2019, 11:56 AM
Joe Lieberman: Israel wrong to ban Omar and Tlaib
Former Senator Joe Lieberman says Israel made a "serious mistake" by denying entry to Reps. Rashida Tlaib and Ilhan Omar.Elad Benari, Canada, 8/17/2019, 1:44 AM
'Democratic Party not anti-Jewish but some members are'
Former US Senator comments on controversy surrounding Rep. Ilhan Omar and her anti-Semitic comments.Elad Benari, 3/18/2019, 1:06 AM
Former Senator urges Democrats to act on anti-Semitism
Joe Lieberman calls on Democrats not to delay resolution to condemn anti-Semitism that was drafted following Ilhan Omar's comments.Ben Ariel, 3/7/2019, 2:34 AM
Lieberman: Don't rush to judgment on Gov. Northam's racist photo
Ex-Senator and 2000 vice presidential candidate Joe Lieberman warns against 'rush to judge' Virginia governor for racist yearbook photo.Marcy Oster, JTA, 2/6/2019, 11:36 AM
Joe Lieberman lauds McCain's accommodation of Jewish observance
In eulogy for John McCain, Joe Lieberman recalls McCain's accommodation of Lieberman's Jewish observance and his love of Jerusalem.JTA, 9/2/2018, 10:24 AM
Joe Lieberman: Proud my daughter is making aliyah
Joe and Hadassah Lieberman part from daughter Hana Lowenstein as she immigrates to Israel with her family.Orly Harari, 7/26/2018, 9:45 AM
Joe Lieberman: Trump had the courage past presidents lacked
2000 VP candidate tells Arutz Sheva why past presidents declined to move embassy, and why Trump was right to ignore their precedent.Yoni Kempinski, Jerusalem, 5/14/2018, 10:41 AM
McCain regrets not picking Joe Lieberman as 2008 running mate
Arizona Senator battling cancer says he regrets not choosing Joe Lieberman, Democrat turned independent, for VP in 2008 race.JTA, 5/8/2018, 7:26 AM
Joe Lieberman: Trump did the right thing on Iran deal
2000 Democratic Vice Presidential candidate and Iran nuke deal opponent Joe Lieberman praises Trump's refusal to certify Iranian compliance.David Rosenberg, 10/15/2017, 10:56 PM
Joe Lieberman withdraws from consideration for FBI chief
Former Senator writes to US President to withdraw his name for consideration to replace fired FBI director James Comey.JTA, 5/25/2017, 9:22 PM
Report: Joe Lieberman no longer in running to head FBI
Former Senator and front-runner to replace James Comey as the head of the FBI reportedly removed from consideration.JTA, 5/24/2017, 7:40 PM
Lieberman emerges as front-runner for FBI Director
Former Democratic vice-presidential nominee the front-runner to be named FBI director, say White House officials.Ben Ariel, 5/19/2017, 12:02 AM
Jewish former senator a candidate for FBI Director
Trump to interview four candidates for FBI Director, including Joe Lieberman.Arutz Sheva Staff, 5/18/2017, 1:08 AM
Joe Lieberman says he will vote for Clinton
Former Connecticut senator says he would vote for Hillary Clinton days after leaving open the possibility he would vote for Donald Trump.JTA, 8/12/2016, 5:24 AM
Lieberman: Media exaggerates Bibi-Obama hostility
Former leading Democrat says final 14 months of Obama Admin will see improved ties; launches fierce criticism of White House policy.Ari Soffer, 11/24/2015, 9:44 PM
'Politics of Haggadah' at YU
Rabbi Lord Jonathan Sacks and former US Senator Joe Lieberman look at the Pesach text in a new light - as work of political Jewish thought.Arutz Sheva Staff, 3/23/2015, 7:03 PM
Joe Lieberman to Congress: "Hear Netanyahu Out!"
Former vice presidential candidate urges Democrats to attend Netanyahu's upcoming Congressional address.Hillel Fendel, 2/23/2015, 9:33 PM
Joe Lieberman 'Becomes A Mensch'
Former Senator and VP candidate joins group aiming to increase unity in Israeli society; Arutz Sheva speaks with founder of organization.Yoni Kempinski, Ari Yashar, 11/14/2014, 7:52 AM
Lieberman Reflects on Career
Former Senator Joseph Lieberman addressed Yeshiva University students, faculty and staff in a lecture titled "Judaism and Public Service."Arutz Sheva Staff, 10/30/2014, 5:45 AM
Sntr. to Discuss "Judaism and Public Policy" at YU
Senator Joseph Lieberman to begin lectures at the end of October; public invited to attend.Arutz Sheva Staff, 10/6/2014, 6:34 PM
Senator Joe Lieberman to Chair Public Policy at YU
US Senator to give undergraduate courses, give public lectures at YU and Stern.Arutz Sheva Staff, 7/25/2014, 11:53 AM
Lieberman, Chambliss Support Operation in Gaza
Senators Saxby Chambliss and Joe Lieberman both express support for Israel's Operation Pillar of Defense in Gaza.Elad Benari, 11/19/2012, 3:12 AM
McCain, Lieberman Urge Arms for Syrian Rebels
Senators McCain, Lieberman and Graham called for direct military aid for Syrian rebels and warned Obama is letting Al Qaeda move in.Tzvi Ben Gedalyahu, 8/6/2012, 10:01 AM
US Lawmakers to Probe Underwear Bomb Leaks
Lawmakers from both US parties say leaks about a dramatic CIA operation thwarting an Al-Queda plot demand an investigation.Gabe Kahn, 5/10/2012, 3:31 PM
Netanyahu: Talks with Iran Gave it a 'Freebie'
PM Netanyahu meets U.S. Senator Joe Lieberman, warns against Iran using talks to further advance its nuclear program.Elad Benari, 4/16/2012, 2:16 AM
Sen. McCain Visiting PA
Senators McCain, Lieberman and Graham are visiting PA leaders in Ramallah on a "fact-finding tour." They will meet with Barak on Wednesday.Tzvi Ben Gedalyahu, 7/6/2010, 4:04 PM
Senators Defy Mitchell's Threat
McCain and Lieberman knocked down Mitchell's threat to slice aid to Israel if more concessions are not made to the PA. "Congress won't back cuts."IsraelNN Staff, 1/10/2010, 7:17 PM


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On November 7th, The Carver Project welcomed Kristina Arriaga of the U.S. Commission on International Religious Freedom. At a morning breakfast, John Inazu interviewed Commissioner Arriaga about her role in negotiations for the release of Andrew Brunson from imprisonment in Turkey. In the afternoon, Inazu and Arriaga joined Judge Lech Garlicki, formerly a member of the European Court of Human Rights, for a discussion on international religious freedom at Washington University's School of Law. The event was co-hosted by The Carver Project, Christian Legal Fellowship, and the Jewish Law Society.


Todays card is a quick and simple one for the new challenge over at Cardz 4 Guyz. This week we would like to see your 'Speed' themed, male cards.
I have been very short of time this week, so my card was a quick and simple decoupaged effort, but it did fit the theme really well.
Love this image and your card, such a great masculine card, something I never seem to have enough of. Thanks for joining us at TTCRD this week, hope you'll join us for more of our weekly challenges.
An elegant and classy card.


Gluten-Free Cupcake Recipe (I like using my gluten-free pumpkin cupcake batter colored yellow to resemble the candy corn base).
Make cupcakes in mini muffin pan, baking 8-10 minutes, or until tops spring back when pressed lightly. You don't want these to turn brown.
While cupcakes cool, make frosting (color orange, if you like, to resemble center of candy corn).
Make whipped topping. Keep chilled until ready to use.
Assemble cupcakes by frosting cooled cupcakes with orange frosting, then topping with whipped topping.
Garnish with candy corn or your favorite white, yellow or orange candies.
Store in refrigerator to keep whipped topping cool.
In this recipe, you'll find links to my gluten-free pumpkin cupcake recipe, my no butter creamy frosting, and dairy-free whipped topping. If you eat dairy products, feel free to use regular gluten-free frosting and whipped topping. Use the products that suit your own special diet, and enjoy creating something fun based on this bit of festive inspiration!
Quick to fix, versatile and delish, I know you'll love these Mini Gluten Free Candy Corn Cupcakes as much as my friends and family do!


text:
Roborock unveiling some goodies at CES today
I don't generally do hype for companies on unreviewed products, but after reviewing a lot of robot vacuum cleaners and pretty much deciding Rhonda, our beloved Roborock S7 is about the best for my needs, figured I'd keep up with what's new in their world.
Today at CES Roborock was slated to release details on their new S7 MaxV Ultra. I'm not entirely sure if they have yet, but the embargo lifted at 8:30am ET so we're good to go. Hopefully everything went off without a hitch at CES.
I've been waiting to pounce
S7 MaxV Ultra
What the S7 MaxV Ultra does differently from the S7 appears to be, at first glance mostly in the docking functions.
Water is injected, waste is emptied, and the mop is washed between uses. Basically the S7+Dock and then you don't have to clean the mop or refill for longer periods of time.
Under the hood it's got a new obstacle avoidance system which uses a camera, 3D structured light, a neural processing unit, and a certification for cybersecurity standards.
You can see a very CGI demo of the S& MaxV Ultra here
The S7 MaxV Ultra will still have the vibrating lift/lower mop pad that means you're not dragging a dirty mop pad over carpet, and you don't have to constantly put in and remove a mop attachment.
Basically they've turned the robot vacuum process into something you have to pay attention to only once every few weeks, or when / if it gets trapped.
The suction power has been upped to 5100pa according to the specs, which is beyond anything I've played with so far. That level of suction would be the max setting, so you'd probably only see that on areas you defined, or if it's getting ticked at trying to pick up some dirt.
The S7 MaxV Ultra will be coming out sometime in the second quarter. In the meantime get pricing and expected availability on the Roborock website. Really check that out, you can see how the mop is cleaned and the waste water removed.
What's really interesting about this unit is not listed in the press release, but listed in the video above is video calling. The ability to scream at your kids to knock off destroying stuff while you're in a hammock is shown about 1:30 into the above YouTube video. This means you're probably able to power it up and take a stroll around your house if you're paranoid that someone's there while you're at work.
Also means you need to be paranoid enough to have a real password for your Roborock account as a unit with a camera and audio capabilities being compromised could be… compromising.
"We know you're not a bad dog and undeserving of love." Counseling sessions with the vacuum are not listed as a feature yet, but one can hope.
The Q series
Not a lot of information out on this except it's their new mid-range line that starts with auto-empty technology.
The Roborock Dyad
It's a wetmop/vacuum combo. Not quite in my wheelhouse as I don't know whether this has been done before (I went down a very long rabbit hole for my still unreleased H7 review). Looks sort of like a carpet cleaner… you know what, I'm just going to cut and paste the PR I got and you decide if it's interesting.
Cut and paste PR on the Roborock Dyad:
Las Vegas, Jan. 4, 2022: Roborock, creator of ultra-intelligent home robotics made to simplify life, today unveiled the all-new Dyad, its first wet and dry vacuum cleaner designed to make floor cleaning easier and more effective. The new all-in-one solution vacuums, washes, and dries floors to get rid of dirt and messes from edge-to-edge.
"We set out to tackle some unresolved pain points in the wet-dry vacuum space like poor edge cleaning, weak pick-up and maneuverability issues, and we achieved that and more with Roborock Dyad," said Richard Chang, CEO of Roborock. "Featuring our proprietary technology, DyadPower™, Dyad is the only dual roller motor, multi-roller wet and dry vacuum that cleans edge-to-edge to eliminate messes in one sweep. This multi-use product has the capability to remove everything from dirty water to encrusted dirt seamlessly in a single cleaning session."
Adaptive and Effective Cleaning with DyadPower™
DyadPower™ Vigorously Cleans With Every Stroke: Two motors work separately to power three rollers, one in the front and two in the rear, enabling the cleaning system to rotate in opposite directions. This combination, unique to the market, allows Dyad to efficiently lift both wet and dry messes in one sweep.
Spotless Cleaning Along the Edges: Edge-to-edge cleaning is no longer a problem. Featuring two rear rollers spanning the width of the head, Dyad cleans thoroughly along edges and flush with the wall to get hard-to-reach dust and debris.
Adaptive Cleaning Intelligence: Dyad detects tough stains and automatically increases water flow and suction power for better cleaning results.
Not Just for Cleaning Dust
Switch to Drying Mode: Dyad suctions wet messes into its 620 ml dirty water tank while simultaneously releasing fresh water onto the floor and working its rollers to actively remove dirt. Its multi-use functions make it easy to clean up spilled coffee or lift milk and cereal at the same time.
Make Floors Tea Tree-Scented: Created by Unilever under its OMO brand, this unique cleaning solution formulated exclusively for Roborock helps to remove stains from hard floors and ceramic tiles, while leaving floors with an extra touch of freshness.
Effortless to Clean, Maintain and More
Self-Cleans With The Push of a Button: The rollers clean themselves automatically for an added level of convenience, so no dirty hands in the process.
Clean Effortlessly: Thanks to its multi-roller system that rotates in opposite directions, Dyad glides smoothly across the floor. Its agile cleaning head twists 180 degrees to navigate easily around furniture.
Power to Clean Large Spaces: A 5000 mAh high-capacity battery can run nonstop for up to 35 minutes while cleaning up to 3056 sqft (284 sqm).
Intelligent and Convenient: At a glance, its LED screen indicates cleaning mode, battery life and even notifies customers to floor areas with a build up of dirt. To add even more convenience, a voice alert system announces when it's time to refill the water tank or charge the battery.
← Instant XLR just add USB Pyle XLR to USB Interface
I watched The Matrix Resurrections in VR goggles →


text:
LIVE Adult Education: Shared Experience of the Orthodox Christian Faith
In this installment of Our Life in Christ in America, Bill Hinkle explores the true meaning of Philanthropy. He explains how this word, of Greek origin, describes the love Christians ought to have for others, as mirroring the Philanthropy of God to man.
almsgiving10
Politics in America
Culture War or the Spiritual Life?
Bill Hinkle
Bill (Basil) Hinkle, Former Minority Whip of the Washington State House of Representatives was a firefighter paramedic for 17 years before entering the world of politics. Bill has been an elected official since 1996, serving first as Kittitas County Commissioner from 1996 to 2006, then serving five terms in the House of Representatives, where he served on the leadership team as the Minority Whip. He was the former ranking minority member of the House Healthcare Committee and has served on numerous commissions and tasks forces over the years with a particular emphasis on social welfare and healthcare policy issues, as well as water law and environmental issues. Bill is currently the Executive Director for the Rental Housing Association of Washington, a statewide association of landlords and their associates, headquartered in Seattle, WA, where he is still active in government affairs at the state and local level. Additionally he is serves on the Board of Trustees of Yakima Regional Medical Center and is the Founder and past Dean of the Jennifer Dunn Leadership Institute. Bill and his wife Debra are committed to Christ and serving His Holy Church. They have two children at home. They are founding members of The Prophet Elijah Antiochian Orthodox Christian Mission, in Ellensburg, WA, where Bill serves as a Subdeacon with the blessing of his bishop. As a graduate of the Antiochian House of Studies, St. Stephen's Theology Course, finishing a Master's Degree in Applied Theology, he is the first President of the Antiochian House of Studies Alumni Association. Bill attended Tacoma Community College, South Puget Sound Community College, and the University of Washington's Paramedic Training Program. Bill is also a lifetime member of the National Rifle Association, and a member of Rotary International, the Order of St. Ignatius, Washington Cattlemen Association, and the Rocky Mountain Elk Foundation.


Ethiopian authorities are releasing their preliminary report on last month's fatal airline crash. Also, former Vice President Joe Biden has spoken out about his past interactions with women.
Members of Robert Mueller's team reportedly say Attorney General William Barr sanitized the special counsel's report. Also, Joe Biden responded to allegations he has inappropriately touched women.
After several women have come forward to say close, physical interactions with former Vice President Joe Biden have made them uncomfortable, he is speaking out for the first time about the issue.
The Vermont senator's campaign says it raised $18.2 million from more than 500,000 donors. Former Texas Rep. Beto O'Rourke has raised $9.4 million since entering the race three weeks ago.
A divided court ruled against a Missouri man who said that because of a rare medical condition, death by lethal injection would constitute "cruel and unusual punishment."
NPR's Morning Edition has begun speaking to 2020 Democratic presidential candidates in a series of conversations to explore their core campaign messages. We listen back to some of those exchanges.
Lee Boyd Malvo was convicted in the 2002 killings and is serving a life sentence without the possibility of parole. But he was a minor, and the Supreme Court says that's not allowed in most cases now.
Former Texas congressman Beto O'Rourke announced that he's running for president.
Sen. Bernie Sanders, I-Vt., signed a loyalty pledge required by the Democratic Party to run for president and serve as a Democrat. The requirement grew out of his own 2016 campaign.
The billionaire businessman — who has been a Republican, a Democrat and an independent — is not running for president in a field growing more crowded by the day.


text:
Todd Keeler-Wolf
Todd Keeler-Wolf is an ecologist specializing in vegetation classification and mapping. A true Californian, Todd was born in Oakland and received his B.S. and Ph.D. from the University of California Santa Cruz. He was the Senior Vegetation Ecologist at the California Department of Fish and Wildlife for 25 years, where he lead the Vegetation Classification and Mapping Program. During this period, he studied the vegetation and flora of virtually every terrestrial ecosystem in the state. Todd is the author of numerous books and publications. He co-developed the "Sawyer and Keeler-Wolf classification system" and along with John Sawyer and Julie Evens, the Second Edition of A Manual of California Vegetation, with its accompanying online database. This guide features over 485 descriptions, 352 vegetation maps, fire/flood data, restoration and management considerations, and more. Todd also has traveled widely working outside of the State in The Caribbean, Australasia, South America, and East Africa, and has been a long-time active member of the Ecological Society of America's Vegetation Classification Panel.
As an "experienced" ecologist having studied the interdependence of plants, animals, and their environments for several decades, Todd offers a California-based perspective on biodiversity. What is unique about the state's biodiversity? How does adversity shape the kind of biodiversity we have here? What similar patterns of diversity exist here and elsewhere? What are the most vulnerable habitats to loss of biodiversity? What do we have to know about species to assess their vulnerability? And what kind of hope is there for restoration, protection, and recovery of the loss of biodiversity?


OVERLOOK POINTE Condos For Sale in PORTLAND | Portland Condo Mania!
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Ronda Rousey is an entertainer of enormous talent who without a doubt knows how to put on a good show both inside and outside the Octagon. And she proved as much with a bravado performance at UFC 193 media day at Etihad Stadium in Australia that was alternately provocative to subtlety — and inadvertently — profane.
Rousey found herself completely flustered after uttering something that came out entirely wrong when discussing her upcoming bout against Holly Holm at UFC 193 on Nov. 15. After noting that Holm is a strong counter-puncher, Rousey unleashed the following easily misinterpreted commentary.
"If I came right in her face right away … wow, that sounded terrible," Rousey said. "Oh my God. You know what I mean. How do I even say this right now?
The second way she articulated her thoughts were much clearer. Not as funny, but clearer nonetheless.
On a more serious note, the UFC women's bantamweight champion also took on the controversial issue of performance-enhancing drugs — specifically how marijuana is treated as such — when she criticized how fellow UFC fighter Nick Diaz was suspended for five years by the Nevada State Athletic Commission after testing positive for a third time for pot.
Rousey certainly isn't the first high-profile athlete to rail against the notion that marijuana shouldn't be treated as a performance-enhancing drug. And odds are good she won't be the last.
But as far as her presser is concerned, watching Rousey get all flustered is far more entertaining.


All projects by Jay Developers are approved by regularised housing financial banks and institution. We also help our customers by submitting the master file to get the project approval with third-party financial institutions. We make sure to meet the most stringent conditions put forward by any financial institution.


text:
Q: Simple json array structure i know the next question must seem stupid but i'll do it anyway :)
I've got this structure:
{
      "name": "Erjet Malaj",
      "first_name": "Erjet",
      "work": [
        {
          "employer": {
            "id": "110108059015688",
            "name": "Art of Living Foundation"
          },
          "position": {
            "id": "137081429661437",
            "name": "Volunteer"
          },
          "start_date": "0000-00",
          "end_date": "0000-00"
        },
        {

I retrieve response.data[i].first_name like this, how am i supposed to retrieve work id?

A: work is an array. You need to loop through it
var work = response.data[i].work,
    worklen = work.length,
    j, workdetails;

for(j = 0; j < worklen; j++){

    //simplify for readability
    workdetails = work[j];

    //where prop is "employer" or "position"
    workdetails[prop].id
    workdetails[prop].name
}

A: did you try response.data[i].work.position.id ?
on second thought, work is an array of objects, so it's accessed like :
response.data[i].work[0].position.id

A: Assuming you mean employer id of a given work
This should get the employer id
response.data[i].work[j].employer.id

You could also do this (depending on preference)
response.data[i].work[j]['employer'].id

In this case the work position is j (used as an array indexer), replace j with 0 to tget the first, with 1 to get the second etc.

A: There is an id for the employer and position of each record in the work array.  If you want to loop over a person's work entries and collect those IDs, use:
var person = response.data[i];
var work = person.work;

for(var j = 0; j < work.length; ++j) {
    var work_entry = work[j];
    console.log(work_entry.employer.id);
    console.log(work_entry.position.id);
}

The entries in the work array are accessed numerically from 0.  The entries themselves are objects, and you can access their properties using obj_name.property_name or obj_name["property_name"]

redpajama_set_name: RedPajamaStackExchange


At Community MD, a family clinic in Brampton, Ontario, the number one priority is their patients. By incorporating an online patient portal into their practice, they have taken action toward improved overall patient safety and satisfaction.
Recognizing the importance of embracing new healthcare technologies, Community MD implemented the miDASH solution, which assists them in improving the patient experience at their clinic. This technology allows patients to send secure online messages to their health care providers to request prescription renewals and refills, where before they would need to attend an in-clinic appointment to make these requests. This solution has improved the patient experience by noticeably reducing unnecessary appointments and allowing patients to make speedy and efficient requests with a click of a button, and at their convenience.


Two respected trainers and personal coaches in the Western performance horse world are offering a free online training series to help riders prepare for this year's show season.
Be Unstoppable, hosted by Sandy Collier and Barbra Schulte, kicks off Monday, Jan. 28, at 6:30 p.m. Central with the first of a four-class series. Classes are scheduled over the next four days.
Each class in the Be Unstoppable series is a one-hour session with a live question-and-answer section to evaluate attendees' skills. The program is designed for riders of all experience levels in the cutting, reined cow horse, ranch versatility and stock horse disciplines. The class themes are: competitive mental skills; boxing/down the fence; cutting/herd work; and reining.
"This is a time of the year that can be so frustrating for people," said Schulte, who has $984,325 in lifetime earnings. "There is so much down time when they're anxious to get going.
Barbra Schulte and Sandy Collier, two respected trainers and personal coaches in the Western performance horse world are offering a free online training series. • Courtesy photo.
Collier, of Santa Maria, California, has earned $833,000 since posting her first check to Equi-Stat in 1985. She is the only woman to win the National Reined Cow Horse Association (NRCHA) Snaffle Bit Futurity Open. With multiple NRCHA and American Quarter Horse Association (AQHA) World championships under her belt, Collier also serves as an international clinician and judge. The author of "Reining Essentials," and educational DVDs is an honoree in both the National Cowgirl Hall of Fame and the NRCHA Hall of Fame, as well as an internationally certified life coach.
Schulte, who was the first woman to win the National Cutting Horse Association (NCHA) Summer Spectacular Derby and Classic, has a master's degree in speech pathology and audiology, and certification as a personal performance coach for riders from the Human Performance Institute in Orlando, Florida.
Named National Female Equestrian of the Year Award by the AQHA, Schulte's an honoree in the National Cowgirl Hall of Fame and in the NCHA Members Hall of Fame. She is a personal performance coach, a cutting horse trainer, publisher of online programs, clinician, author and speaker.
To sign up for Collier and Schulte's online sessions or for more information, visit beunstoppable.lpages.co/rss-save-my-spot.


"The Corfu Mare is a new and friendly boutique hotel. The staff were great and very polite.
The rooms were clean and cleaned every day. The only downside was the bathroom which, was very small and the shower leaked across the bathroom floor.
The food menu for the cafe/restaurant was very small which, is not the be all and end all (as there are many local restaurants a stone's throw away) the breakfast is what you come to expect from Europe. Eggs, bacon, cereals, cold hams and cheeses etc.
The location is not the best, as Corfu Town is over a km away and to walk it could be quite dangerous as there are no well built pedestrian paths or sufficient lighting when it is dark (walking the main road). Walking during the day could be pleasant if you walked the back way to Corfu town (better and more sufficient paths) but it takes at least 30 - 45Mins. The big blue buses travel to Corfu town on a regular basis and are roughly 2 Euros. Getting a taxi back from Corfu town is relatively cheap; we never paid over 8 Euros.
Remember to take your insect repellent and bite cream, you will be very lucky if you do not get nibbled on at some point during your holiday"
"We were domiciled here by Easyjet during the volcanic ash cloud situation. Stayed here from 16-23 April.
The best feature of this hotel is the staff! The three ladies on the front desk, whose names I don't know, were lovely. We were in a stressful situation, having to relocate and not knowing when we would be going home. These ladies were very kind and supportive of us in this situation. Thank you for your kindness ladies.
The hotel itself is very smart, furnished in a boutique style. We were in a double room with a sofa bed for our son. All very clean, lovely duvets and pillows. Only complaint was the underneath mattress protector, which was plastic backed and generated an awful lot of heat. Once this was removed, everything perfect. The fan system in the room was very impressive and super-effective. The hotel had wi-fi, but were unable to offer internet access if you didn't have a lap top to use. This was frustrating, given our situation being stranded there.
Breakfast- the usual Greek style breakfast of cereals, bacon, eggs, beans, cold meat, cheese, yogurt and honey and fruit etc. The orange juice wasn't very nice - it was more like undiluted orange squash, but OK when diluted with water.
The hotel is situated about half an hour from Corfu Town - quite a pleasant walk at the back of the hotel into town, until you hit the main road, but then you are nearly there. The main road on the front side of the hotel, is not easy to walk on and very busy with traffic. The blue bus ran on this busy main road, if you could work out the times. A very nice Mexican restaurant on the main road, Poco Loco, served lovely food in a lovely atmosphere."
"The hotel is just perfect .
The room was a very good size and very nice decorated . One of the big advantage is the cleanliness of the room and all the rest of the hotel .
Near the hotel you can find a wide variety of restaurant and cafeterias that their prices are resonable .
I would definetly visit this hotel in the future .
Thanking all the staff members and especially to Yulli that make our short stay there very pleasant and make us feel that we are in our home .
See you soon all ."
"When arrived, 1 hour waiting for a room to be ready. No spacious parking. Modern decoration at room but very small even for 2 people. The bathroom was a joke. No bathtub and the room easy get flooded.
The worst for a 120 euro per night hotel-room was the breakfast. Unhealthy cheap stuff without basic things like orange juice etc. The view from rooms was not great and the pool was behind a 70s building. Definitely was not worth it!"
"Corfu Mare Hotel is recently innovated and has beautiful rooms with modern decoration. Its location is superb because lies close to town and close to high way so you can easily access beaches and villages. Free parking and internet access are available. We stayed in a double room and even though the room was not so spacious we loved it and the bathroom was also really nice! The hotel staff was kind and obliging. Two things we did not like: breakfast ( poor) and hotel clothing condition (especially the towels were nearly rugs)."


text:
George Lincoln Rockwell (1918. március 19. – 1967. augusztus 25.), amerikai nemzetiszocialista politikus.

Élete

Született 1918. március 19-én az Amerikai Egyesült Államokban, az Illinois állam beli Bloomington nevű szénbányász/gazdálkodó településen. Édesapja humorista, míg édesanyja táncos volt. Szülei hatéves korában elváltak.

Gyermekkorát Maine, Rhode Island és New Jersey államokban töltötte. Apja egy mainei kisvárosban élt, így a fiatal Rockwell ott töltötte nyarait, amiket rendszerint túrázással, kirándulással és vitorlázással töltött el. Az apjával eltöltött nyári vitorlázások miatt egész életében a vitorlázás kedvelője lett.
Kapcsolata a szüleivel igen szeretetteljes volt, ami a család felbomlása után is megmaradt. Egészséges és átlagos gyermekkora volt.

1938 őszén a Brown Egyetemen kezdett tanulni, ahol filozófiát tanult, bár érdeklődött a tudományok iránt is.

A második világháború kitörése után – mint több millió amerikai hazafi – ő is jelentkezett a hadseregbe.

1944-ben a Guam elleni támadásban a légi támogatást irányította, amiért előléptették '45-ben. A háború után visszatért a civil élethez, ahol művészként kezdett új életet.

Összeházasodott egy lánnyal, akit a Brown Egyetemen ismert meg. A házasság nem volt tökéletes, bár így is 10 évig tartott.

Politikai életútja

A hadsereg elhagyása utáni első 5 évben New York és Maine államokban művészeteket tanult, reklámfotózott, festett, írt. 1950-ben visszatért a seregbe, a Koreai háború miatt, Kaliforniában képzett ki vadászpilótákat. A politikai pályafutása is ekkor kezdődött.
Egy ismerőse ekkor adott neki kölcsön kommunista-ellenes könyveket. A politikába való elmerülés után a zsidóság és a kommunizmus összefonódását állapította meg és ekkor kezdett a hitleri eszmerendszer után érdeklődni. Ekkor döbbent rá, hogy a rossz oldalon harcolt a háborúban.

Mein Kampf példány keresett a könyvtárakban, és nekiült elolvasni. A nemzetiszocializmus eszméje akkor érintette meg.

1953 után kezdett bele a kampányába. Előbb egy csendes szerveződés vezetője lett, de a terve már megvolt egy keménymagú nemzetiszocialista szervezet felépítésére. Határozott elképzelései voltak mind a gazdaságról, mind a faji és szociális kérdésekről.

1957-1958 nehéz két év volt számára. New York, New Jersey és Pennsylvania között utazgatott, a gondolatait vetette papírra.

1958-ban szervezte meg az első tiltakozást a Fehér Ház előtt, de több államban is szerveződtek tüntetések, Atlantában (Georgia) és Louisvilleben (Kentucky). Ezek a megmozdulások még nem nyílt nemzetiszocialista demonstrációk voltak. 10 héttel később Atlantában egy zsinagóga titokzatos körülmények között felrobbant, ekkor a média hadjáratot indított Rockwell és támogatói ellen.

Ekkor már a nemzetiszocializmust látta az egyetlen kiútnak. Nem érdekelte, hogy mit terjeszt a média, a saját újtját járva "náciként" terjesztette a szélsőjobboldali propagandát. Házára egy hatalmas horogkeresztet tett és megkezdődött a valódi szerveződés. Rohamosztagosok csatlakoztak hozzá, létszámuk rohamosan növekedett. A rendőrség folyamatos házkutatásokkal nehezítette ő és emberei helyzetét, a média "Amerikai Führer"-nek hívta. Megalapította az "Amerikai Párt"-ot, amit később "Amerikai Náci Párt"-nak nevezett át.

Mivel akcióik folyamatosan a média kereszttüzében voltak, így terjeszthette az eszmét. Pártja leginkább a zsidóság befolyása, a kommunizmus, a marxizmus, a bolsevizmus és a keményvonalas kapitalizmus ellen emelte fel szavát.

1960-ban adták ki a "Stormtrooper" nevű magazinukat. A közszereplését egyre több rádió és televízió kezdte letiltani, ámbár a Playboynak is adott interjút. 5 éven keresztül elég erős nyomás alá került.

1965-ben Virginia állam kormányzóválasztásán indult. Erős kampányba kezdett, 5730 szavazatot kapott. (Virgina lakossága  fő)

1966. január 1-jén az Amerikai Náci Párt nevét Nemzetiszocialista Fehéremberek Pártjára változtatta.

1967-ben fejezte be a White Power (fehér hatalom) című könyvét.

1967. augusztus 25-én, egy pénteki napon az arlingtoni főhadiszállásuk közelében egy golyó leterítette. A gyilkosa egy néhány hónappal azelőtt a pártból kirúgott férfi volt, aki marxista ideológiát csempészett a párt publikációi közé.
Miután az Amerikai Egyesült Államok kormánya megvonta Rockwell jogát, hogy temetőben temethessék el, elhamvasztása után a párttársai egy nemzetiszocialista búcsúszertartás keretében emlékeztek meg róla Arlingtonban.

Publikációi
 A White Power könyv
 A kacsák és a csirkék története
 This time the world

Interjúk
 A Playboy vele készített interjúja
 A Playboy vele készített interjújának magyar fordítása

Jegyzetek

Fordítás

Amerikai politikusok
1918-ban született személyek
1967-ben elhunyt személyek
Nemzetiszocializmus
Meggyilkolt amerikai politikusok

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction}
A \emph{proper total $k$-colouring} of a graph $G=(V,E)$ is an assignment $c$ of colours from the set $\{1,2,\ldots,k\}$ to the edges and the vertices of $G$ such that
adjacent edges and vertices are coloured differently and the colour of every edge is distinct from those assigned to its end-vertices.
A \emph{total neighbour sum distinguishing $k$-colouring} of $G$, or \emph{tnsd $k$-colouring} for short,
is its proper total $k$-colouring $c$ such that for every edge $uv\in E$, there is no \emph{conflict} between $u$ and $v$,
{\it i.e.}, $s(u)\neq s(v)$, where $s(w)$ is the sum of colours taken on the edges incident with $w$ and the colour of the vertex $w$ for $w\in V$.
In other words, for every vertex $w\in V$, $\displaystyle s(w) = \sum_{e \in E_w} c(e)+c(w)$, where $E_w$ is the set of edges incident with $w$ in $G$.
We denote by $\chi''_{\Sigma}(G)$ the \emph{total neighbour sum distinguishing index} of $G$, which is the least integer $k$ such that a tnsd  $k$-colouring of $G$ exists.
The roots of this branch of graph theory date back to the '80s, and the papers~\cite{ChartrandErdosOellermann,Chartrand}
on degree irregularities in graphs (and multigraphs) and the parameter \emph{irregularity strength} of a graph. For more details concerning a motivation for investigating integer graph colourings and a few crucial results on the irregularity strength see {\it e.g.}~\cite{Aigner
Lazebnik
Frieze,KalKarPf,Lehel,MajerskiPrzybylo2,Nierhoff}.

By definition and the requirement of properness of colourings investigated, the total neighbour sum distinguishing index of every graph $G$ is not smaller than $\Delta(G)+1$.
The following conjecture was proposed by Pil\'{s}niak and Wo\'{z}niak in~\cite{PW},
where it was also verified for a few classical graph families, including, {\it e.g.},
complete graphs, bipartite graphs and
graphs with maximum degree at most three.

\begin{conjecture}[\cite{PW}]\label{MoMa}
For every graph $G$, $\chi''_{\Sigma}(G) \leq \Delta(G) + 3$.
\end{conjecture}
Note that such postulated upper bound exceeds just by one the bound implied by the famous Total Colouring Conjecture
posed by Vizing~\cite{Vizing2} in 1968 and independently by Behzad~\cite{Behzad} in 1965, and
concerning proper total colourings (without our additional requirement on sum distinction between adjacent vertices).
The both conjectures seem to be very challenging and are in general widely open. The best general result concerning the latter one~\cite{MolloyReedTotal} confirms however the Total Colouring Conjecture up to a (large) additive constant.

The best general upper bound concerning Conjecture~\ref{MoMa} implies that $\chi''_{\Sigma}(G)\leq (1+o(1))\Delta$ for every graph $G$ with maximum degree $\Delta$, see~\cite{Przybylo_asym_optim_total} and~\cite{Przybylo_asymptotic_note}. See also~\cite{total_sum_planar,LiLiuWang_sum_total_K_4,PW,Przybylo_CN_3} for partial results on this conjecture.
In particular Ding et al. first confirmed Conjecture~\ref{MoMa} for planar graphs with sufficiently large maximum degree:
\begin{theorem} [\cite{total_sum_planar}]
Any planar graph $G$ with $\Delta(G) \geq 13$ satisfies $\chi''_\Sigma(G) \leq \Delta(G)+3$.
\end{theorem}
This was then improved by Yang et al. in the following form.
\begin{theorem} [\cite{total_sum_planar2}]
Any planar graph $G$ satisfies $\chi''_\Sigma(G) \leq \max\{\Delta(G)+2,13\}$.
\end{theorem}

Let ${\rm mad}(G)=\max\left\{\frac{2|E(H)|}{|V(H)|},\;H \subseteq G\right\}$ be the \emph{maximum average degree} of a graph $G$, where $V(H)$ and $E(H)$ are the sets of vertices and edges of $H$, respectively.
This is a conventional measure of sparseness of an arbitrary graph (not necessary planar). For more details on this invariant see {\it e.g.} \cite{Coh10,Toft}.
Dong and Wang first made the link between the maximum average degree and the total neighbour sum distinguishing index.
They proved the following result.

\begin{theorem}[\cite{DongWang_mad}] \label{th:summadCite1}
Any graph $G$ with $\Delta(G) \geq 5$ and ${\rm mad}(G)<3$ satisfies $\chi''_\Sigma(G) \leq \Delta(G)+2$.
\end{theorem}

This subject was intensively studied afterwards, and the following improvement has been announced recently (but no proof of such a supposed fact
was published thus far).
\begin{theorem}[\cite{Qiu_Wang_Liu_Xu}] \label{th:summadCite2}
Any graph $G$ with $\Delta(G) \geq 8$ and ${\rm mad}(G)< \frac92$ satisfies $\chi''_\Sigma(G) \leq \Delta(G)+3$.
\end{theorem}

In this paper, we prove a stronger statement than
the above:

\begin{theorem}\label{th:summad}
Any graph $G$ with $\Delta(G) \geq 8$ and ${\rm mad}(G)<\frac{14}{3}$ satisfies $\chi''_\Sigma(G) \leq \Delta(G)+3$.
\end{theorem}

Recall that the girth of a graph is the length of a shortest cycle in it.
As every planar graph with girth $g$ satisfies ${\rm mad}(G) < \frac{2g}{g-2}$, the following corollary can be easily derived from Theorem~\ref{th:summad}:

\begin{corollary}
Any triangle-free planar graph $G$ with $\Delta(G) \geq 8$ satisfies $\chi''_\Sigma(G) \leq \Delta(G)+3$.
\end{corollary}

\section{Proof of Theorem~\ref{th:summad}}

\subsection{Preliminaries}

Fix an integer $k\geq 8$.
In the following, $n_i(G)$ denotes the number of vertices of degree $i$ in a graph $G$ (and similarly for $n_{i^+}(G)$ with ``at least $i$'' and for $n_{i^-}(G)$ with ``at most $i$'').
We say a graph $G$ is \emph{smaller} than a graph $H$, $G \prec H$ if $|E(G)|+|V(G)|<|E(H)|+|V(H)|$.
 We say a graph is \emph{minimal} for a property when no smaller graph verifies it.
We shall also call any vertex of degree $d$ ($\geq d$, $\leq d$) in a given graph a \emph{$d$-vertex} (\emph{$d^+$-vertex}, \emph{$d^-$-vertex}, resp.) of this graph.
The same nomenclature shall be used for neighbours as well.\\

\subsection{Structural properties of H}

Suppose that $H$ is a minimal graph with maximum degree $\Delta\leq k$,
${\rm mad}(H)<\frac{14}{3}$ and $\chi''_\Sigma(H)>k+3$ (hence $H$ is connected and $\delta(H)\geq 1$).
In the remaining part of the paper we argument that in fact $H$ cannot exist, {\it i.e.} that there exists a tnsd  $(k+3)$-colouring of $H$,  and thus prove Theorem~\ref{th:summad}.

In this subsection we exhibit some structural properties of $H$.

The following lemma shall be very useful to this end.
Its proof was inspired by the research from~\cite{BP14}.
The same result but in the case of lists of integers can also be derived from~\cite{Alon}.

\begin{lemma}[\cite{HocquardPrzybylo1}]\label{MartheLemma}
For any finite sets $L_1,\ldots,L_t$ of real numbers with $|L_i|\geq t$ for $i=1,\ldots,t$,
the set $\{x_1+\ldots+x_t:x_1\in L_1,\ldots,x_t\in L_t; x_i\neq x_j~{\rm for}~i\neq j\}$ contains
at least $\sum_{i=1}^t|L_i|-t^2+1$ distinct elements.
\end{lemma}

\begin{observation}\label{obs3v}
Every $3^-$-vertex $v$ in $H$ can be recoloured (or coloured if it has no colour assigned) so that it has a different colour than its adjacent vertices and incident edges, and so that $v$ is not in conflict with any of its neighbours.
\end{observation}

\begin{proof}
This follows directly by the fact we have $k+3\geq 11$ colours available, while at most $9$ of them might be blocked by the requirements from the thesis.
\end{proof}

\begin{lemma}\label{lemma4v}
For every vertex $v \in V(H)$,
$n_{4^+}(v) \ge n_{2^-}(v)+1+ n_{3^-}(v)\times(k - d(v))$.
\end{lemma}

\begin{proof}
Suppose on the contrary that $v\in V(H)$ with $d(v)=d\geq 1$ is adjacent to $\alpha$ $2^-$-vertices $u_1,\dots,u_\alpha$, $\beta$ $3$-vertices $w_1,\dots,w_\beta$ and to $\gamma$ $4^+$-vertices where $\gamma < \alpha+1+(\alpha+\beta)(k-d)$, hence $\alpha+\beta\geq 1$.
Colour $H'=H-\{vu_1,\ldots,vu_\alpha,vw_1,\ldots,vw_\beta\}$ by minimality ({\it i.e.} fix any tnsd  $(k+3)$-colouring of $H'$, which must exist due to the fact that $H$ is our minimal counterexample, while $H' \prec H$, $\Delta(H')\leq k$ and ${\rm mad}(H')<\frac{14}{3}$) and uncolour $u_i$ and $w_j$ for all $i \in \{1,\ldots,\alpha\}$ and $j \in \{1,\ldots,\beta\}$. Let $L_i$ and $L'_j$, with $i \in \{1,\ldots,\alpha\}$ and $j \in \{1,\ldots,\beta\}$ be the sets of available colours respectively for the edges $vu_i$ and $vw_j$ ({\it i.e.} those colours in $\{1,\ldots,k+3\}$ not used by their adjacent edges in $H$ or $v$).
Note that
$|L_1|,\dots,|L_\alpha| \ge \alpha + \beta + 1 + k -d$ and $|L'_1|,\dots,|L'_\beta| \ge \alpha + \beta+ k -d$.
By Lemma~\ref{MartheLemma}, we may extend this colouring to a (partial) proper colouring of $H$ in different ways, obtaining at least $\alpha(\alpha+\beta)+\alpha+\alpha(k-d)+\beta(\alpha+\beta)+\beta(k-d)-(\alpha+\beta)^2+1=\alpha+1+(\alpha+\beta)(k-d)>\gamma$ distinct sums for~$v$.
Thus we can do it in such a way that $v$ is no in conflict with any of its $4^+$-neighbours.
By Observation~\ref{obs3v} we therefore obtain a contradiction. $\blacksquare$\\

\end{proof}

\begin{corollary}\label{obs2v}
For every vertex $v \in V(H)$ with $d(v) \ge 7$, $n_{2^-}(v) \le d(v)-5$.
\end{corollary}

\begin{proof}
Suppose to the contrary that $n_{2^-}(v) \ge d(v)-4$ for some $v \in V(H)$ with $d(v) \ge 7$. By Lemma~\ref{lemma4v} we have: $n_{4^+}(v) \ge n_{2^-}(v)+1+ n_{3^-}(v)\times(k - d(v))\ge d(v)-4+1+n_{3^-}(v)\times(k - d(v)) \ge d(v)-3$.

Consequently, we have:
$d(v) \ge n_{2^-}(v)+n_{4^+}(v) \ge d(v)-4+d(v)-3=2d(v)-7$.
Hence, $d(v) \le 7$, {\it i.e.} $d(v)=7$. As $k \ge 8$, then by Lemma~\ref{lemma4v} we thus obtain:
$n_{4^+}(v) \ge n_{2^-}(v)+1+ n_{3^-}(v)\times(k - d(v))\ge 4 + n_{3^-}(v)\times(k - d(v)) \ge d(v)$, a contradiction with the fact that $n_{2^-}(v) \ge d(v)-4$. $\blacksquare$
\end{proof}

Within the proof of the remaining structural properties of $H$, aggregated in Claim~\ref{claimstructure} below, we shall apply several times the following algebraic tool due to Alon~\cite{Alon}.

\begin{theorem}[Combinatorial
Nullstellensatz]\label{Comb_Nul} Let $\mathbb{F}$ be an arbitrary
field, and let $P=P(x_1,\ldots,x_n)$ be a polynomial in
$\mathbb{F}[x_1,\ldots,x_n]$.
Suppose the coefficient of a monomial $x_1^{k_1}\ldots
x_n^{k_n}$, where each $k_i$ is a non-negative
integer, is non-zero in $P$ and the degree ${\rm deg}(P)$ of
$P$ equals $\sum_{i=1}^n k_i$.
If moreover $S_1,\ldots,S_n$ are any
subsets of $\mathbb{F}$ with $|S_i|>k_i$ for $i=1,\ldots,n$,
then there are $s_1\in
S_1,\ldots,s_n\in S_n$ so that $P(s_1,\ldots,s_n)\neq 0$.
\end{theorem}

\begin{claim}\label{claimstructure}
The graph $H$ does not contain any of:

\begin{enumerate}
\item[(C1)] \label{c1} a $2^-$-vertex $v$ adjacent to a $(\frac{k}{2}+1)^-$-vertex $u$;
\item[(C2)] \label{c2} a $4^-$-vertex $v$ adjacent to a $4^-$-vertex $u$;
\item[(C3)] \label{c3} a $3^-$-vertex $v$ adjacent to a $5^-$-vertex $u$;
\item[(C4)] \label{c4} a $5$-vertex $v$ adjacent to three $4$-vertices $v_1,v_2,v_3$;
\item[(C5)] \label{c5} a $6$-vertex $v$ adjacent to a $3^-$-vertex $u$ and to a $4^-$-vertex $w$;
\item[(C6)] \label{c12} a $7$-vertex $v$ adjacent to a $2^-$-vertex $u$, to a $3^-$-vertex $w$ and to a $4^-$-vertex $y$.
\item[(C7)] \label{c6} a vertex $v$ of degree $d\geq 8$ adjacent to $(d-7)$ $2^-$-vertices $v_1,\ldots,v_{d-7}$, to two $3^-$-vertices $u_1,u_2$ and to a $4^-$-vertex $w$;
\item[(C8)] \label{c11} a vertex $v$ of degree $d=\Delta\geq 3$ adjacent to $(d-2)$ $3^-$-vertices $v_1,\ldots,v_{d-2}$ and to one $4^-$-vertex $u$.
\end{enumerate}
\end{claim}
\medskip

\begin{figure}[!h]
\begin{center}
\centering
\begin{tikzpicture}[scale=0.75,auto]
	\tikzstyle{w}=[draw,circle,fill=white,minimum size=5pt,inner sep=0pt]
	\tikzstyle{b}=[draw,circle,fill=black,minimum size=5pt,inner sep=0pt]
	\tikzstyle{t}=[rectangle,minimum size=5pt,inner sep=0pt]

	 \tikzstyle{whitenode}=[draw,ellipse,fill=white,minimum size=9pt,inner sep=0pt]
	 \tikzstyle{blacknode}=[draw,circle,fill=black,minimum size=9pt,inner sep=0pt]
	 \tikzstyle{texte}=[minimum size=9pt,inner sep=0pt]

\draw (0,0) node[whitenode] (u) [label=above:$u$] {$\left(\frac{k}{2}+1\right)^-$}
--++ (0:4cm) node[whitenode] (b) [label=above:$v$] {$2^-$};

\draw (2.2,-1.5) node[t] (t1) {$(C_1)$};

\draw (6,0) node[whitenode] (b) [label=above:$u$] {$4^-$}
--++ (0:2.5cm) node[whitenode] (b) [label=above:$v$] {$4^-$};

\draw (7.2,-1.5) node[t] (t1) {$(C_2)$};

\draw (11,0) node[whitenode] (b) [label=above:$u$] {$5^-$}
--++ (0:2.5cm) node[whitenode] (b) [label=above:$v$] {$3^-$};

\draw (12.3,-1.5) node[t] (t1) {$(C_3)$};

\draw (17.3,0) node[whitenode] (v) [label=below:$v$] {$5$}
--++ (0:-2.2cm) node[w] (u)  [label=below:$v_4$] {}
--++ (0:4.4cm) node[w] (w) [label=below:$v_5$] {};

\draw (v)
--++ (150:2cm) node[w] [label=right:$v_1$] (v5) {$4$};

\draw (v)
--++ (90:1.5cm) node[w] [label=right:$v_2$] (v6)  {$4$};

\draw (v)
--++ (30:2cm) node[w] [label=right:$v_3$] (v7)  {$4$};

\draw (17.3,-1.5) node[t] (t1) {$(C_4)$};

\draw (2.2,-4.5) node[whitenode] (v) [label=below:$v$] {$6$}
--++ (0:-2.2cm) node[w] (u)  [label=below:$u$] {$3^-$}
--++ (0:4.4cm) node[w] (w) [label=below:$w$] {$4^-$};

\draw (v)
--++ (150:2cm) node[w] [label=right:$v_1$] (v5) {};

\draw (v)
--++ (110:1.5cm) node[w] [label=right:$v_2$] (v6)  {};

\draw (v)
--++ (70:1.5cm) node[w] [label=right:$v_3$] (v7)  {};

\draw (v)
--++ (30:2cm) node[w] [label=right:$v_4$] (v8)  {};

\draw (2.2,-7) node[t] (t1) {$(C_5)$};

\draw (10.3,-4.5) node[whitenode] (v) [label=below:$v$] {$7$}
--++ (0:-2.2cm) node[w] (u)  [label=below:$u$] {$2^-$}
--++ (0:4.4cm) node[w] (w) [label=below:$y$] {$4^-$};

\draw (v)
--++ (90:1.5cm) node[w] [label=right:$w$] (v6)  {$3^-$};

\draw (v)
--++ (-30:1.5cm) node[w] (v8) [label=below:$z_4$] {};

\draw (v)
--++ (-60:1.5cm) node[w] (v9) [label=below:$z_3$] {};

\draw (v)
--++ (-120:1.5cm) node[w] (v10) [label=below:$z_2$] {};

\draw (v)
--++ (-150:1.5cm) node[w] (v11) [label=below:$z_1$]  {};

\draw (10.3,-7) node[t] (t1) {$(C_6)$};

\draw (17.3,-4.5) node[whitenode] (v) [label=below:$v$] {$8^+$}
--++ (0:-2.2cm) node[w] (u)  [label=below:$u_1$] {$3^-$}
--++ (0:4.4cm) node[w] (w) [label=below:$u_2$] {$3^-$};

\draw (v)
--++ (150:2cm) node[w] [label=left:$v_1$] (v5) {$2^-$};

\draw (v)
--++ (90:1.5cm) node[w] [label=right:$v_{d-7}$] (v6)  {$2^-$};

\draw (v)
--++ (30:2cm) node[w] [label=right:$w$] (v7)  {$4^-$};

\draw (v)
--++ (-30:1.5cm) node[w] (v8) [label=below:$y_4$] {};

\draw (v)
--++ (-60:1.5cm) node[w] (v9) [label=below:$y_3$] {};

\draw (v)
--++ (-120:1.5cm) node[w] (v10) [label=below:$y_2$] {};

\draw (v)
--++ (-150:1.5cm) node[w] (v11) [label=below:$y_1$] {};

\draw (v5) edge [bend left,loosely dotted,thick] node {} (v6);

\draw (17.3,-7) node[t] (t1) {$(C_7)$};

\draw (2.2,-10) node[whitenode] (v) [label=below:$v$] {$d=\Delta$}
--++ (0:-2.2cm) node[w] (u)  [label=below:$v_1$] {$3^-$}
--++ (0:4.4cm) node[w] (w) [label=below:$u$] {$4^-$};

\draw (v)
--++ (50:2cm) node[w] [label=right:$v_{d-2}$] (v7)  {$3^-$};

\draw (u) edge [bend left,loosely dotted,thick] node {} (v7);

\draw (v)
--++ (-50:1.5cm) node[w] (v8) [label=below:$w$] {};

\draw (2.2,-12.5) node[t] (t1) {$(C_8)$};

\end{tikzpicture}
\caption{Forbidden configurations in $H
}\label{FCG_fig}
\end{center}
\end{figure}

\begin{proof}
We shall argument `reducibility' of each of these 8 configurations separately,
following a similar pattern of reasoning.
I.e., we shall first suppose by contradiction that a given configuration exists in $H$.
Then we shall consider a graph $H'$ smaller than $H$ with $\Delta(H')\leq k$ and ${\rm mad}(H')<\frac{14}{3}$
(usually guaranteing these properties by constructing $H'$ simply via deleting some edges or vertices from $H$),
and \emph{colour} it \emph{by minimality},
what shall mean from now on that we choose any tnsd $(k+3)$-colouring for $H'$.
Finally, in each case, we shall obtain a contradiction by extending the colouring chosen to a tnsd $(k+3)$-coloring of the entire $H$.
Whenever we analyze a partial colouring of a graph, the \emph{sum at a} given \emph{vertex}, $s(v)$ is defined as above, but every uncoloured edge and vertex contributes $0$ to this sum. We write that $u$ and $v$ are \emph{sum-distinguished}, if $s(u)\neq s(v)$.

\begin{enumerate}
\item[1.] Suppose there exists a $2^-$-vertex $v$ adjacent to a $(\frac{k}{2}+1)^-$-vertex $u$ in $H$. Colour $H'=H-\{uv\}$ by minimality and uncolour $v$. In order to colour $uv$ so that a (partial) $(k+3)$-colouring of $H$ obtained is proper we have to avoid at most $\frac{k}{2}+2$ colours,
     and possibly at most $\frac{k}2$ more colours to ensure the sum-distinction (of $u$ from its neighbours other than $v$). Hence, we have at least one colour left to extend the colouring, and thus obtain
     a tnsd $(k+3)$-colouring of $H$ via Observation~\ref{obs3v} applied to $v$, a contradiction.
\item[2;3.] Suppose there exists an edge $uv$ with $d(u),d(v)\leq 4$ or with $d(u)\leq 5$ and $d(v)\leq 3$  in $H$. Denote the neighbours of $u$ other than $v$ by $u_1,\ldots,u_q$, and denote the neighbours of $v$ other than $u$ by $v_1,\ldots,v_p$ (hence $d(u)=q+1$ and $d(v)=p+1$).
    By the minimality of $H$ there exists a tnsd $(k+3)$-colouring $c$ of $H'=H-\{uv\}$.
    Let us now undelete the edge $uv$ and remove the colours from $u$ and $v$.
    In order to extend the current partial colouring of $H$ to its proper total $(k+3)$-colouring we may use at least $11-6=5$ colours for $u$, at least $11-6=5$ colours for $uv$ and at least $11-6=5$ colours for $v$ if $d(u),d(v)\leq 4$, or otherwise: at least $3$ colours for $u$, at least $5$ colours for $uv$ and at least $7$ colours for $v$. Denote the respective lists of available colours by $L_u,L_{uv},L_v$, the sum at $u_i$
    by $s_i$ and the sum at $v_j$
    by $s'_j$ for $i=1,\ldots,q$, $j=1,\ldots,p$. Consider a polynomial with real variables:
    \begin{eqnarray}
    f(x_0,x_1,x_2) &=& (x_0-x_1)(x_0-x_2)(x_1-x_2)\left(x_0+\sum_{i=1}^q c(uu_i)-x_2-\sum_{i=1}^p c(vv_i)\right)\nonumber\\
    &\times&\prod_{i=1}^q\left(x_0+x_1+\sum_{j=1}^qc(uu_j)-s_i\right)\prod_{i=1}^p\left(x_2+x_1+\sum_{j=1}^pc(vv_j)-s'_i\right).\nonumber
    \end{eqnarray}
    Note that in order to extend the colouring $c$ to a tnsd $(k+3)$-colouring
    of $H$ it is now sufficient to find a non-zero (i.e. with non-zero value of $f$) substitution for $f$ such that $x_0\in L_u$, $x_1\in L_{uv}$ and $x_2\in L_v$. It is thus the more sufficient to find a non-zero substitution from these list for the polynomial $g$ defined as $g(x_0,x_1,x_2):=f(x_0,x_1,x_2)\cdot(x_0+x_1)^{3-q}(x_2+x_1)^{3-p}$ if $d(u),d(v)\leq 4$ or by $g(x_0,x_1,x_2):=f(x_0,x_1,x_2)\cdot(x_0+x_1)^{4-q}(x_2+x_1)^{2-p}$ otherwise.
    In the first of these cases however, the coefficient of the monomial $x_0^4x_1^3x_2^3$ in $g$ is the same as in $$h_1(x_0,x_1,x_2)=(x_0-x_1)(x_0-x_2)^2(x_1-x_2)(x_0+x_1)^{3}(x_2+x_1)^{3},$$ and equals\footnote{This and further computations
    were obtained
    by means of a computer program; one may verify these using {\it e.g.}
    \emph{Wolfram Mathematica}.}: $2$. Analogously, in the second case, the coefficient of the monomial $x_0^2x_1^3x_2^5$ in $g$ is the same as in $$h_2(x_0,x_1,x_2)=(x_0-x_1)(x_0-x_2)^2(x_1-x_2)(x_0+x_1)^{4}(x_2+x_1)^{2},$$ and equals also: $2$.
    In the both cases we thus obtain a contradiction by the Combinatorial Nullstellensatz.
\item[4.] Suppose there exists a $5$-vertex $v$ with $N(v)=\{v_1,\ldots,v_5\}$ such that $d(v_1)=d(v_2)=d(v_3)=4$ in $H$.
    By the minimality of $H$ there exists a tnsd $(k+3)$-colouring $c$ of $H'=H-\{vv_1,vv_2,vv_3\}$.
    Delete the colours of $v,v_1,v_2,v_3$. We associate variables $x_0,x_1,x_2,x_3,x_4,x_5,x_6$ with $v,vv_1,vv_2,vv_3,v_1,v_2,v_3$, respectively.
    For these we
    denote the lists of their available colours by $L_0,\ldots,L_6$ (obtained after excluding from $\{1,\ldots,k+3\}$ the colours already used on their respective adjacent or incident vertices and edges),
    respectively. Then $|L_0|\geq 7, |L_1|,|L_2|,|L_3|\geq 6, |L_4|,|L_5|,|L_6|\geq 5$.
    Let
    \begin{eqnarray}
    f(x_0,\ldots,x_6) &=& (x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)(x_0-x_5)(x_0-x_6)(x_1-x_2)(x_1-x_3)(x_2-x_3)\nonumber\\
                      &\times& (x_1-x_4)(x_2-x_5)(x_3-x_6)\prod_{i=1}^3(x_0+x_1+x_2+x_3+s(v)-x_i-x_{i+3}-s(v_i))\nonumber\\
                      &\times& \prod_{i=4}^5(x_0+x_1+x_2+x_3+s(v)-s(v_i)) \prod_{i=1}^3\prod_{u\in N(v_i)\smallsetminus\{v\}}(x_i+x_{i+3}+s(v_i)-s(u))\nonumber
    \end{eqnarray}
    (where $s(w)$ refers to the contemporary partial sum for every vertex $w$ in $H$). Note that the coefficient of the monomial $x_0^6x_1^5x_2x_3^5x_4^2x_5^3x_6^4$ in $f$ is the same as in the following polynomial:
    \begin{eqnarray}
    g(x_0,\ldots,x_6) &=& (x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)(x_0-x_5)(x_0-x_6)(x_1-x_2)(x_1-x_3)(x_2-x_3)\nonumber\\
                      &\times& (x_1-x_4)(x_2-x_5)(x_3-x_6)(x_0+x_2+x_3-x_4)(x_0+x_1+x_3-x_5)(x_0+x_1+x_2-x_6)\nonumber\\
                      &\times& (x_0+x_1+x_2+x_3)^2(x_1+x_4)^3(x_2+x_5)^3(x_3+x_6)^3,\nonumber
    \end{eqnarray}
    and equals $16$. By the Combinatorial Nullstellensatz we thus may extend our colouring to a tnsd $(k+3)$-colouring of $H$, a contradiction.
\item[5.] Suppose there exists a $6$-vertex $v$ adjacent to a $3^-$-vertex $u$, to a $4^-$-vertex $w$ and to vertices $v_1,v_2,v_3,v_4$ in $H$.
    By the minimality of $H$ there exists a tnsd $(k+3)$-colouring $c$ of $H'=H-\{vu,vw\}$.
    Delete the colours of $u,v,w$ and associate variables $x_0,x_1,x_2,x_3,x_4$ with $v,vu,vw,u,w$, respectively.
    Denote the lists of available colours for these by $L_0,\ldots,L_4$, resp., and note that $|L_0|\geq 3, |L_1|\geq 5, |L_2|\geq 4, |L_3|\geq 7, |L_4|\geq 5$.
    Consider a polynomial:
    \begin{eqnarray}
    f(x_0,\ldots,x_4) &=& (x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)(x_1-x_2)(x_1-x_3)(x_2-x_4)\nonumber\\
                      &\times& (x_0+x_2+s(v)-x_3-s(u))(x_0+x_1+s(v)-x_4-s(w))\nonumber\\
                      &\times& \prod_{i=1}^4(x_0+x_1+x_2+s(v)-s(v_i))\nonumber\\
                      &\times& \prod_{y\in N(u)\smallsetminus\{v\}}(x_1+x_3+s(u)-s(y))
                      \prod_{y\in N(w)\smallsetminus\{v\}}(x_2+x_4+s(w)-s(y))\nonumber
    \end{eqnarray}
    and set $g(x_0,\ldots,x_4)=f(x_0,\ldots,x_4)\cdot(x_1+x_3)^{3-d(u)}(x_2+x_4)^{4-d(w)}$.
    Note that the coefficient of the monomial $x_0^2x_1^4x_2^3x_3^5x_4^4$ in $g$ is the same as in:
    \begin{eqnarray}
    h(x_0,\ldots,x_4) &=& (x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)(x_1-x_2)(x_1-x_3)(x_2-x_4)\nonumber\\
                      &\times& (x_0+x_2-x_3)(x_0+x_1-x_4)(x_0+x_1+x_2)^4(x_1+x_3)^2(x_2+x_4)^3,\nonumber
    \end{eqnarray}
    and equals $-10$. By the Combinatorial Nullstellensatz there exists a non-zero substitution for $g$, hence the more for $f$, from the corresponding lists $L_0,\ldots,L_4$, and thus we may extend our partial colouring to a tnsd $(k+3)$-colouring of $H$, a contradiction.
\item[6.] Suppose there exists a $7$-vertex $v$ adjacent to a $2^-$-vertex $u$, to a $3^-$-vertex $w$ and to a $4^-$-vertex $y$ in $H$.
    Denote the remaining neighbours of $v$ by $z_1,z_2,z_3,z_4$.
    By the minimality of $H$ there exists a tnsd $(k+3)$-colouring $c$ of $H'=H-\{vu,vw,vy\}$.
    Delete the colours of $u,w,y$ and associate variables $x_1,x_2,x_3,x_4$ with $vu,vw,vy,y$, respectively.
    Denote the lists of available colours for these by $L_1,L_2,L_3,L_4$, resp., and note that $|L_1|\geq 5, |L_2|\geq 4, |L_3|\geq 3, |L_4|\geq 4$.
    Consider a polynomial (and note that by Observation~\ref{obs3v} we shall be able to colour properly vertices $u$ and $w$ at the end so that these are sum distinguished from their neighbours, thus we omit the corresponding requirements within the polynomial below):
    \begin{eqnarray}
    f(x_1,x_2,x_3,x_4) &=& (x_1-x_2)(x_1-x_3)(x_2-x_3)(x_3-x_4)\prod_{i=1}^4(x_1+x_2+x_3+s(v)-s(z_i))\nonumber\\
                      &\times& (x_1+x_2+s(v)-x_4-s(y))
                      \prod_{z\in N(y)\smallsetminus\{v\}}(x_3+x_4+s(y)-s(z)).\nonumber
    \end{eqnarray}
    Let $g(x_1,x_2,x_3,x_4)=f(x_1,x_2,x_3,x_4)\cdot (x_3+x_4)^{4-d(y)}$.
    Note that the coefficient of the monomial $x_1^4x_2^3x_3^2x_4^3$ in $g$ is the same as in:
    $$h(x_1,x_2,x_3,x_4) = (x_1-x_2)(x_1-x_3)(x_2-x_3)(x_3-x_4)(x_1+x_2+x_3)^4(x_1+x_2-x_4)(x_3+x_4)^3,$$
    and equals $-6$. Therefore, analogously as above we may extend our colouring, first to $vu,vw,vy,y$ by the Combinatorial Nullstellensatz, and then to $u$ and $w$ by Observation~\ref{obs3v}, to a tnsd $(k+3)$-colouring of $H$, a contradiction.
\item[7.] Suppose there exists
    a vertex $v$ of degree $d\geq 8$
    adjacent to $(d-7)$ $2^-$-vertices $v_1,\ldots,v_{d-7}$, to two $3^-$-vertices $u_1,u_2$ and to a $4^-$-vertex $w$ in $H$. The remaining neighbours of $v$ we denote by $y_1,y_2,y_3,y_4$.
    By the minimality of $H$ there exists a tnsd $(k+3)$-colouring $c$ of $H'=H-\{vv_1,vu_1,vu_2,vw\}$.
    Delete the colours of $v,v_1,\ldots,v_{d-7},u_1,u_2,w$ and associate variables $x_0,x_1,x_2,x_3,x_4,x_5$ with $v,vv_1,vu_1,vu_2,vw,w$, respectively.
    Denote the lists of available colours for these by $L_0,\ldots,L_5$, resp., and note that $|L_0|\geq 3, |L_1|\geq 6, |L_2|,|L_3|\geq 5, |L_4|\geq 4, |L_5|\geq 5$.
    Consider a polynomial
    (the vertices $v_1,\ldots,v_{d-7},u_1,u_2$ shall be coloured at the end via Observation~\ref{obs3v}):
    \begin{eqnarray}
    f(x_0,\ldots,x_5) &=& (x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)(x_0-x_5)(x_1-x_2)(x_1-x_3)(x_1-x_4)\nonumber\\
                      &\times& (x_2-x_3)(x_2-x_4)(x_3-x_4)(x_4-x_5)\prod_{i=1}^4(x_0+x_1+x_2+x_3+x_4+s(v)-s(y_i))\nonumber\\
                      &\times& (x_0+x_1+x_2+x_3+s(v)-x_5-s(w))\prod_{z\in N(w)\smallsetminus\{w\}}(x_4+x_5+s(w)-s(z)).\nonumber
    \end{eqnarray}
    Let $g(x_0,\ldots,x_5)=f(x_0,\ldots,x_5)\cdot(x_4+x_5)^{4-d(w)}$. Then the coefficient of the monomial $x_0x_1^5x_2^4x_3^3x_4^3x_5^4$ in $g$ is the same as in:
    \begin{eqnarray}
    h(x_0,\ldots,x_5) &=& (x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)(x_0-x_5)(x_1-x_2)(x_1-x_3)(x_1-x_4)\nonumber\\
                      &\times& (x_2-x_3)(x_2-x_4)(x_3-x_4)(x_4-x_5)(x_0+x_1+x_2+x_3+x_4)^4\nonumber\\
                      &\times& (x_0+x_1+x_2+x_3-x_5)(x_4+x_5)^3.\nonumber
    \end{eqnarray}
    and equals $5$. By the Combinatorial Nullstellensatz and Observation~\ref{obs3v} we may thus extend our partial colouring to a tnsd $(k+3)$-colouring of $H$, a contradiction.
\item[8.] Suppose $v$ is a vertex of degree $d=\Delta\geq 3$ adjacent to $(d-2)$ $3^-$-vertices $v_1,\ldots,v_{d-2}$ and to one $4^-$-vertex $u$ in $H$.
    Denote the remaining neighbour of $v$ by $w$.
    By the minimality of $H$ there exists a tnsd $(k+3)$-colouring $c$ of $H'=H-\{vv_1,\ldots,vv_{d-2},vu\}$.
    Delete the colours of $v,v_1,\ldots,v_{d-2},u$. First extend such a partial colouring of $H$ by choosing
    a colour (in $\{1,\ldots,k+3\}$) for $v$ in the following manner. If $d(v_1)=3$, denote the colours associated to the edges incident with $v_1$ and different from $vv_1$ by $a$ and $b$, and if $c(vw)\notin \{a,b\}$, choose for $v$ any colour in $\{a,b\}\smallsetminus\{c(w)\}$.
    In all other cases, choose for $v$ any colour distinct from $c(vw)$ and $c(w)$. Denote the colour of $v$ by $c(v)$.
    Then choose a colour $c(u)$ for $u$ as small as possible
    (and note that as our total colouring must be proper, this implies that either $u$ or some of its incident edges other than $uv$ has now colour at most $5$). Next we choose any colour $c(uv)$ so that the obtained (partial) colouring of $H$ is proper and $u$ is sum distinguished from its neighbours other than $v$ (this is possible, as $k+3>9$). Then we subsequently choose greedily colours for $vv_2,\ldots,vv_{d-2}$ so that the obtained partial total colouring of $H$ is proper. Finally we choose a colour $c(vv_1)$ for $vv_1$ distinct from the colours of its incident edges and the colour of $v$ (by our choice of $c(v)$, this blocks at most $\Delta+1$ choices) so that the sum at $v$ is distinct from the sum at $w$,
    and if $\Delta\leq k-1$, also distinct from the sum at $u$.
    We complete our colouring by choosing the colours for $v_1,\ldots,v_{d-2},u$ consistently with
    Observation~\ref{obs3v}. In order to see that the obtained colouring of $H$ is sum distinguishing it is sufficient to note that the sum at $v$ is distinct from the sum at $u$ when $\Delta=k$. Indeed,
    the sum of colours incident with $v$ except for the colour of $uv$ equals at least $1+\ldots+k=k(k+1)/2$, while by our choice of the colour for $u$, the sum of its incident colours except the one of $uv$ is at most $5+(k+3)+(k+2)+(k+1)=3k+11<k(k+1)/2$ (for $k\geq 8$). Thus we obtain a contradiction with the minimality of $H$. $\blacksquare$

\end{enumerate}

\end{proof}

\subsection{Discharging procedure}

In this subsection we use the discharging technique exploiting the vertices of the graph $H$.
For this aim we first define the \emph{weight
function} $\omega: V(H) \rightarrow \mathbb{R}$ by setting $\omega(x)=d(x)-\frac{14}{3}$ for every $x\in V(H)$.
Next we shall apply so called \emph{Ghost vertices method}, introduced earlier by Bonamy, Bousquet and Hocquard~\cite{BonamyEtAl},
and based on the following observation (where given any subsets $U,U'\subseteq V(H)$ and a vertex $v$, $d_U(v)$ denotes the number of neighbours of $v$ from $U$, while $E(U,U')$ is the set of edges joining $U$ and $U'$ in the graph $H$).

\begin{observation}\label{obs1v}
\noindent Let $V_1 \cup V_2$ be a partition of $V(H)$ where, say $V_1$ is the set of vertices of degree at least $3$ and $V_2$ -- the set of vertices of degree at most $2$ in $H$;
\begin{itemize}
\item every vertex $u$ in $H$ has an initial weight $w(u)=d(u)-\frac{14}{3}$.
\item If we can discharge the weights in $H$ so that:
\begin{enumerate}
\item every vertex in $V_1$ has a non-negative weight;
\item and every vertex $u$ in $V_2$ has a final weight of at least $d(u)-\frac{14}{3}+d_{V_1}(u)$, then\\

for the new weight assignment $\omega'$, we have $\sum_{v \in V_2} (d(v)-\frac{14}{3}+d_{V_1}(v)) \leq \sum_{v \in V_2} \omega'(v)$, as well as\\

$\sum_{v \in V} \omega(v)=\sum_{v \in V} \omega'(v)$ and $\sum_{v \in V_1} \omega'(v) \geq 0$. Therefore,

\begin{eqnarray}
   \sum_{v \in V_1} \left(d_{V_1}(v)-\frac{14}{3}\right) & \geq & \sum_{v \in V_1} \left(d_{V_1}(v)-\frac{14}{3}\right) + \sum_{v \in V_2} (d(v)-\frac{14}{3}+d_{V_1}(v)) - \sum_{v \in V_2} \omega'(v) \nonumber\\
    & \geq & \sum_{v \in V_1} \left(d_{V_1}(v)-\frac{14}{3}\right) + |E(V_1,V_2)|+\sum_{v \in V_2} \left(d(v)-\frac{14}{3}\right) - \sum_{v \in V_2} \omega'(v) \nonumber\\
     & \geq & \sum_{v \in V_1} \left(d(v)-\frac{14}{3}\right) + \sum_{v \in V_2} \left(d(v)-\frac{14}{3}\right) - \sum_{v \in V_2} \omega'(v) \nonumber\\
     & \geq & \sum_{v \in V} \omega(v) - \sum_{v \in V_2} \omega'(v) \nonumber\\
   & \geq & \sum_{v \in V_1} \omega'(v) \nonumber\\
    & \geq & 0. \nonumber
\end{eqnarray}
\end{enumerate}

Thus we can conclude that ${\rm mad}(H) \geq {\rm mad}(H[V_1]) \geq \frac{14}{3}$.

\end{itemize}

\end{observation}

In other words, the vertices in $V_2$ can be seen but, in a way, do not contribute to the sum analysis.
\bigskip

In order to finish the proof of Theorem~\ref{th:summad}, it suffices to obtain a contradiction, \emph{e.g.} with the fact that ${\rm mad}(H)<\frac{14}{3}$,
implying that in fact no counterexample to its thesis may exist.
By Observation~\ref{obs1v}, it is thus enough to
redistribute the weight (defined by $\omega$ above) in $H$
so that every vertex of degree at least $3$ has a non-negative resulting weight, every vertex of degree $2$ has weight at least $2-\frac{14}{3}+2$ and every vertex of degree $1$ has weight at least $1-\frac{14}{3}+1$.

\medskip

The discharging rules we shall use for this aim are defined as follows:

\begin{enumerate}
\item[(R1)] A vertex of degree $d\geq 6$ gives $1$ to every adjacent $1$-vertex and to every adjacent $2$-vertex.
\item[(R2)] A vertex of degree $d\geq 6$ gives $\frac59$ to every adjacent $3$-vertex.
\item[(R3)] A vertex of degree $d\geq 5$
gives $\frac{1}{6}$ to every adjacent $4$-vertex.
\end{enumerate}

Let $v$ be a vertex in $H$. We consider different cases depending on the degree of $v$.

\begin{itemize}
\item Assume $d(v)=1$. By $(C1)$, $v$ is adjacent to a vertex of degree at least $6$. Thus, by $(R1)$, $v$ receives $1$. So every vertex of degree $1$ in $H$ has an initial weight of $-\frac{11}{3}$, gives nothing according to our rules and receives $1$, hence has the final weight of $-\frac{8}{3}$.

\item Assume $d(v)=2$. By $(C1)$, $v$ is adjacent to two vertices of degree at least $6$, and thus receives $2 \times 1$ by $(R1)$ and gives away nothing according to the rules above. So every vertex of degree $2$ in $H$ has an initial weight of $-\frac{8}{3}$ and has the final weight of $-\frac{2}{3}$.

\item Assume $d(v)=3$. By $(C3)$, $v$ is adjacent to three vertices of degree at least $6$, and thus receives $3 \times \frac59$ by $(R2)$ and gives away nothing according to the rules above. So every vertex of degree $3$ in $H$ has
    the final weight $0$.

\item Assume $d(v)=4$. By $(C2)$, $v$ is adjacent to four vertices of degree at least $5$, and thus receives $4 \times \frac16$ by $(R3)$ and gives away nothing according to the rules above. So every vertex of degree $4$ in $H$ has
    the final weight $0$.

\item Assume $d(v)=5$. By $(C3)$, $v$ is not adjacent to a vertex with degree at most $3$, and by $(C4)$, $v$ is adjacent to at most two vertices of degree $4$, and thus gives at most $2 \times \frac16$ by $(R3)$. Hence, $\omega'(v) \ge 5-\frac{14}{3}-2 \times \frac16=0$.

\item Assume $d(v)=6$.
Then consider the following subcases:
\begin{itemize}
\item if $v$ is adjacent to a $3^-$-vertex, then by $(C5)$, $v$ is adjacent to five vertices of degree at least $5$. Hence, by $(R1)$ and $(R2)$, $\omega'(v) \geq 6-\frac{14}{3}-1 \times \max \{1 ; \frac59\}=\frac13 \ge 0$.
\item if $v$ is not adjacent to a $3^-$-vertex, then $v$ is adjacent to at most six $4$-vertices. Hence, by $(R3)$, $\omega'(v) \ge 6-\frac{14}{3}-6 \times \frac16=\frac13 \ge 0$.
In both cases $v$ has a non-negative final weight.

\end{itemize}

\item Assume $d(v) \ge 7$. Recall that by Corollary~\ref{obs2v}, $n_{2^-}(v) \le d(v)-5$. Then consider the following subcases:
\begin{itemize}
\item If $n_{2^-}(v) = d(v)-5$ then by $(C6)$ and $(C7)$, $v$ is not adjacent to another $4^-$-vertex. Hence, by $(R1)$, $\omega'(v) \ge d(v)-\frac{14}{3}-(d(v)-5)\times 1 \ge 0$.
\item If $n_{2^-}(v) = d(v)-6$ then either $v$ is adjacent to no $3$-vertex or, by $(C6)$ and $(C7)$,
to exactly
one $3$-vertex and to no $4$-vertices. Hence, by $(R1)$, $(R2)$ and $(R3)$, $\omega'(v) \ge d(v)-\frac{14}{3}-(d(v)-6)\times 1 - \max\{6\times \frac16;1\times \frac59\} \ge 0$.

\item If $n_{2^-}(v) = d(v)-7$ then $v$ is adjacent to at most three $3$-vertices;
for $d(v)\geq 8$ it follows by $(C7)$, while for $d(v)=7$ by Lemma~\ref{lemma4v}, which implies then that
$n_{4^+}(v) \ge 1+ n_{3^-}(v)(k - d(v))\geq 1+n_{3^-}(v)$.
Hence,  by $(R1)$, $(R2)$ and $(R3)$, $\omega'(v) \ge d(v)-\frac{14}{3}-(d(v)-7)\times 1-3\times \frac59 - 4 \times \frac16 \ge 0$.

\item If $n_{2^-}(v) \le d(v)-8$ then by $(C8)$:

\begin{itemize}
\item if $v$ is not adjacent to any $4$-vertex then $v$ is adjacent to
$(d(v)-8-\alpha)$ $2^-$-vertices and to at most $(\alpha+6)$ $3$-vertices for some $\alpha \ge 0$.  Hence, by $(R1)$, $(R2)$ and $(R3)$, $\omega'(v) \ge d(v)-\frac{14}{3}-(d(v)-8-\alpha)\times 1-(\alpha+6)\times \frac59=\frac49 \alpha \ge 0$.

\item if $v$ is adjacent to at least one $4$-vertex then $v$ is adjacent to
$(d(v)-8-\alpha)$ $2^-$-vertices,
to
$(\alpha+6-\beta)$ $3$-vertices and to at most $(\beta+2)$ $4$-vertices for some $\alpha \ge 0$ and some $\beta \ge 1$. Hence, by $(R1)$, $(R2)$ and $(R3)$, $\omega'(v) \ge d(v)-\frac{14}{3}-(d(v)-8-\alpha)\times 1-(\alpha+6-\beta)\times \frac59 - (\beta+2)\times \frac16=\frac49 \alpha + \frac{7}{18}\beta - \frac13 \ge 0$.
\end{itemize}

\end{itemize}

\end{itemize}

In all cases $v$ has a non-negative final weight.

This, by Observation~\ref{obs1v}, completes the proof of Theorem~\ref{th:summad}. $\blacksquare$

redpajama_set_name: RedPajamaArXiv


text:
Q: How can I find a covariant vector in different coordinates Suppose $B$ is a covariant vector in $\mathbb{R}^{2}$ and that $B = <x^{2},x^{1}+3x^{2}>$ in the $x^{1},x^{2}$ coordinate system. Find $B$ in the new coordinate system $\bar{x}^{1},\bar{x}^{2}$ if
$$ \bar{x}^{1} = (x^{2})^{3}$$
$$ \bar{x}^{2} = x^{1} + 2x^{2}$$
How can I do this in terms of $\bar{x}^{1},\bar{x}^{2}$?

A: When you say that $B$ is a covariant tensor field, you are considering the $1$-form $\omega = x^2\,dx^1 + (x^1+3x^2)dx^2$. Note that $x^2 = (\bar x^1)^{1/3}$ and $x^1=\bar x^2 - 2(\bar x^1)^{1/3}$, so
$$dx^1 = d\bar x^2 - \tfrac23(\bar x^1)^{-2/3}d\bar x^1 \quad\text{and}\quad dx^2 = \tfrac13(\bar x^1)^{-2/3}d\bar x^1.$$
Therefore
$$\omega = (\bar x^1)^{1/3}\left(d\bar x^2 - \tfrac23(\bar x^1)^{-2/3}d\bar x^1\right) + \left(\bar x^2 - 2(\bar x^1)^{1/3}+3(\bar x^1)^{1/3}\right)\left(\tfrac13(\bar x^1)^{-2/3}d\bar x^1\right).$$
Now collect terms and find the coefficients of $d\bar x^1$ and $d\bar x^2$. These will be your new coordinates $\bar B$ of your covariant tensor field. (If I may be permitted to say so, yuck.)
If I haven't messed up, the answer is $\bar B = \langle \frac13(\bar x^1)^{-2/3}\big(\bar x^2 - (\bar x^1)^{1/3}\big), (\bar x^1)^{1/3}\rangle$.