Blue Pits Middle Lock No 52 is a minor waterways place minor waterways place place on the waterways major waterways place village town city on the Rochdale Canal between Ducie Street Junction (Junction of Rochdale and Ashton Canals) (9 miles and 6¼ furlongs and 31 locks to the southwest) and Littleborough (5 miles and 3½ furlongs and 3 locks to the northeast).
The nearest place in the direction of Ducie Street Junction is Blue Pits Middle Lock Bridge; a few yards away.
The nearest place in the direction of Littleborough is Manchester Road Bridge No 64; 1 furlong away.
Harp Industrial Estate, Queensway, Rochdale, OL11 2QQ — 715 yards to the northeast.
956 Manchester Road, Castleton, OL112TL — 207 yards to the southeast.
956 Manchester Rd, Rochdale, OL11 2TL — 207 yards to the southeast.
853 Manchester Road, Castleton, OL11 2UY — 196 yards to the northeast.
Rochdale — 1624 yards to the north.
12 Millrace Close, Manchester, OL11 2LY — 269 yards to the south.
2 Kirklee Road, Manchester, OL112PW — 490 yards to the northeast.
1050 Manchester Road Castleton, Rochdale, OL11 2XJ — 589 yards to the southeast.
860 Manchester Road, Rochdale, OL11 2SP — 183 yards to the north.
St. Martins Vicarage Vicarage Rd, Castleton, OL11 2TE — 278 yards to the east.
Rochdale — 1021 yards to the northeast.
Rochelle Rose, Gainsborough, 1A BRITANNIA HOUSE — 1533 yards to the northeast.
Castleton — 805 yards to the southwest.
Northern Revive Magazine, 65-67 Lever Street, Ancoats, Manchester, M1 1FL — 703 yards to the south.
Manchester Road, Rochdale, OL112XZ — 973 yards to the north.
Middleton — 1586 yards to the south.
Cown top business park Rochdale OL11 2 PU, Rochdale, OL11 2PU — 1010 yards to the northeast.
69 Keswick Street, Rochdale, OL11 3AG — 685 yards to the northwest.
23 Cumberland road, Rochdale, 112 — 1460 yards to the east.
castleton, Rochdale — 1021 yards to the northeast.
12 Castleton Hamlett, Manchester, OL112QH — 201 yards to the south.
Castleton — 656 yards to the southeast.
Unit 5 Astra Centre, Royle Barn Road, Rochdale, OL11 3DT — 1295 yards to the north.
86 Carnforth Avenue, Castleton, Rochdale, Manchester, OL11 2YD — 1140 yards to the southeast.
Manchester Road, Rochdale, OL11 3AF — 810 yards to the north.
TeamKarting,Ensor Mill,Queensway, Rochdale, OL11 2NU — 991 yards to the northeast.
Queensway, Rochdale, OL11 2NU — 1080 yards to the northeast.
Unit 10B Harp Business Centre, Rochdale, OL11 2QQ — 712 yards to the northeast.
JD Gym Rochdale and JD Gym Salford, Manchester — 973 yards to the north.
Manchester Road, Castleton, Rochdale, OL11 3HF — 1122 yards to the north.
Rochdale — 1166 yards to the southeast.
Unit D2 to D4 Royle Pennine Trading Estate. Lynroyle Way, Ol11 3EX, Rochdale, OL11 3EX — 1095 yards to the north.
881 Manchester Road, OL11 2ST — 151 yards to the east.
Crown Business Park, Oldham, OL11 2PU — 1010 yards to the northeast.
958 Manchester Rd, Castleton, OL11 2TL — 230 yards to the southeast.
Keswick Street, Rochdale, OL11 3AG — 793 yards to the west.
Unit 10a Ensor Mill Queensway, Rochdale, OL11 2NU — 879 yards to the northeast.
Blue Pit Mill, Rochdale, OL112YW — 478 yards to the northeast.
Manchester Road, Rochdale, OL11 2XZ — 1067 yards to the south.
374 yards to the northeast.


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Debunking 12 Myths About Relationships
John Gottman, Ph.D.	// May 13, 2016
Want to be happy together? Ignore these relationship myths.
Home » The Gottman Relationship Blog » Debunking 12 Myths About Relationships
If you've had or are having trouble in your relationship, you've probably gotten lots of advice. Sometimes it seems like everybody who has ever been married or knows anyone who has ever been married thinks he or she holds the secret to guaranteeing endless love.
As I explain in The Seven Principles for Making Marriage Work, over the years I've found many myths about relationships that are not only false but potentially destructive. They are dangerous because they can lead couples down the wrong path, or worse, convince them that their marriage is a hopeless case. The notion that you can save your relationship just by learning to communicate more sensitively is probably the most widely held misconception about happy marriages, but it's hardly the only one.
1. Marriage is just a piece of paper.
The psychological and physical benefits of actually being married are enormous. After 50 years of social epidemiology, it has been established that in developed countries the greatest source of health, wealth, longevity, and the ultimate welfare of children is a satisfying and healthy marriage.
2. Living alone with occasional relationships is a lifestyle choice that is equivalent in terms of life outcomes to being married.
Again, social epidemiology has shown that – everywhere on the planet – people who live alone die sooner, are less healthy, are less wealthy, and recover from illness slower than people who are married. This is especially true of men, who have much worse social support networks than women. When men are in a committed relationship their social networks increase.
3. Conflict is a sign that you're in a bad relationship.
Conflict is inevitable in all relationships. Furthermore, conflict is there for a reason – to improve our understanding of our partner. Conflict usually arises from missed attempts to communicate, especially in one person attempting to get emotionally closer to the other. Conflict also emerges from discrepancies between partners in expectations. These are worth talking about.
4. Love is enough.
Love is not enough, because in most marriages – especially after a baby arrives – people stop courting one another and they stop making romance, great sex, fun, and adventure a priority. Relationships have a tendency to become endless to-do lists, and conversation becomes limited to errand talk. You need to intentionally make (or keep) these parts of a relationship a priority.
5. Talking about past emotional wounds will only make them worse.
Faulkner once said, "The past is never dead. It's not even past." It is possible to process past emotional injuries. You can't change the past but you can change your recollection and retelling of it.
6. Better relationships are ones in which people are more independent of and less needy of one another.
Interdependence is what relationships are all about. In a great relationships, people try to meet one another's needs. They adopt the motto, "when you're hurting, baby, the world stops and I listen."
7. If you have to work at communication, it's a sure sign that you're not soulmates.
The sure thing is that if you don't work at communication, the relationship will deteriorate over time, just like a car that's not taken care of will fall apart. All relationships require work. The work in relationships is down-regulating your own defensiveness and listening to your partner.
8. If a relationship needs therapy, it's already too late.
There are 900,000 divorces a year in the USA, and fewer than 10% (!) of those couples that divorce ever talk to a professional. Couples therapy is now very effective and many of these couples could have made it work had they sought help.
9. What couples fight about most is sex, money, and in-laws.
The one thing that couples fight about most is nothing. These fights result from failed bids to connect emotionally. In these small moments, it's important to turn towards instead of turning away. I talked about this on Anderson Cooper.
10. All relationship conflicts can be resolved.
Quite the opposite. In fact, 69% of relationship conflicts are perpetual (they keep recurring), so what is required is acceptance of one another's personality differences. Dialogue about these perpetual issues to avoid gridlock and resentment. The goal then is to manage conflict, not resolve it.
11. All relationship conflicts are the same.
Some conflicts are deal breakers and, for those issues, compromise can be very difficult. It's important to understand your non-negotiables when it comes to conflict. What are you willing to give up?
12. It's compatibility that makes relationships work.
It's diversity that makes relationships interesting. We are not looking for our clones. The famous T-shirt study by Claus Wedekind shows that the pheromones we find most sexy are from people most genetically different from us (on the immune system major histocompatibility complex). Agreeability and conscientiousness are the characteristics that people really mean when they talk about "compatibility." These qualities are indexed by a person being able to say things like "Good point," or "That's interesting, tell me more" or, "You may be right, and I may be wrong" during a disagreement.
Have you been told other myths about relationships? Share in the comment section below.
Debunking 12 Myths About Relationships John Gottman, Ph.D. May 13, 2016
John Gottman, Ph.D.
World-renowned for his work on marital stability and divorce prediction, Dr. John Gottman has conducted 40 years of breakthrough research with thousands of couples. He is the author of over 200 published academic articles and author or co-author of more than 40 books, including The New York Times bestseller The Seven Principles for Making Marriage Work.
Summer Romance: Turning Towards
Michael Fulwiler	// July 19, 2013
Today on The Gottman Relationship Blog, we continue our Summer Romancen series by sharing a list of fun, summer-related date ideas for reconnecting with your partner this weekend.
Public Figures Go to Couples Therapy, Too
Michael Fulwiler	// December 20, 2018
As John and Julie Gottman like to say, we're all in the same soup.
3 Key Things That Will Make or Break Your Marriage
Laura Heck, LMFT	// April 5, 2017
Have you ever had a "make-or-break" moment in your marriage?
Timing Is Everything When It Comes To Marriage Counseling
Terry Gaspard, MSW, LICSW	// July 23, 2015
According to relationship and marriage expert Dr. John Gottman, couples wait an average of six years of being unhappy before getting help.

redpajama_set_name: RedPajamaCommonCrawl


text:
ADB powers new Polish service
Another way to view both free-to-air and premium pay-TV is now available in Poland as digital satellite operator ITI Neovision launches its new TNK HD retail offering using the ADB-2850ST hybrid HD platform.
Using the platform, Polish consumers are able to view a range of free-to-air satellite and terrestrial channels, as well as have the option to buy premium content via a smart-card, and access over-the-top services. The new TNK HD receivers are currently being tested in the market since May 21st and will be commercially available throughout the country as of September 1st.
The device comes with ADB's award winning Carbo high definition user interface (pictured), providing an easy and intuitive way for consumers to manage their viewing experience. It also let consumers add a PVR functionality simply by using hard disk drives typically available in Polish retail.
The flexibility of the TNK HD also means that it will be enriched over time with new capabilities such as multi-room functionality, enabling content to be viewed on multiple television sets located in different rooms. A free over-the-air update will enable devices purchased before this date to be automatically upgraded.
www.adbholdings.com
NDSatCom adds comfort to new SNG service
Loft London to provide a new IP feed and capture service
Thomson Video Networks tech powers Italy's first HD DTT service
Discovery, ProSiebenSat.1 launch new streaming service
Sky launches new streaming service in Spain
Amazon launches new livestream service using same tech as Twitch
New HD studio equipped with ADB LED luminaires


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You are at:Home»Features»Jaguar to compete in Formula E
Engine Technology Online gets an exclusive insight into Jaguar's stunning Formula E plans at the team's unveiling of the I-Type 1 racer
Jaguar to compete in Formula E
By Dean Slavnich on	 14th September 2016 Features
Jaguar is to compete in the third season of the Formula E all-electric race series, which begins on October 9 in Hong Kong. Developed in partnership with Panasonic, the I-TYPE 1 will serve as a rolling test bed for developing new advanced powertrain technologies. Drivers are Adam Carroll, Mitch Evans and Ho-Pin Tung.
Gerd Mäuser, chairman of Panasonic Jaguar Racing, said, "Today marks a new chapter in the history of Jaguar Racing. As the first premium car manufacturer in Formula E we are proud to be back in top-level motorsport. The future is changing and we're part of that change. Formula E is the perfect platform to inspire the next generation."
Nick Rogers, executive director for product engineering at Jaguar Land Rover, underlined the important role Formula E will play in developing next-generation electric vehicle technology and the importance of electric vehicles in JLR's future product portfolio.
Rogers commented, "The championship will enable us to engineer and test our advanced technologies under extreme performance conditions. We will apply this vital knowledge as part of our real-world development. At Jaguar Land Rover we employ 9,000 engineers and the team will draw on these engineers to extract data and push the boundaries of electric technology in a race environment."
The Panasonic Jaguar Racing team has successfully completed 21 days of testing, up to and including the last Formula E public test, which recently took place at Donington Park Circuit in the UK.
James Barclay, Panasonic Jaguar Racing team director, said, "Today is a proud moment for myself and the entire Panasonic Jaguar Racing team at our official launch. Everyone has worked very hard over the last nine months to prepare for our debut in the Championship. I am delighted to welcome Adam, Mitch and Ho-Pin, Panasonic, Lear and all of our new fans into the Jaguar family.
"We want to be successful on and off the track. We know that the challenge will be strong – our competitors have a two-year head start. We will be keeping our expectations in check in our first season. Ultimately we're here to win but we're here to innovate too. I would like to thank Jean Todt and his team at the FIA and Alejandro Agag and his team at Formula E for their vision in building this innovative series and for their help and support over the last 12 months with Jaguar's entry."


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There were at least 80 boxes that I needed to move from the kitchen to the bedroom. They belonged to my friend, and she just couldn't bring herself to move them for the past four months. So she paid me to get the job done.
Think back to the last time you had a ton of dirty laundry, a sink full of dirty dishes, projects due for several clients, and a boatload of unread emails. (It was overwhelming for me to just type that.) We've all been there—and then what do we do?
Well, I know what I do: I procrastinate like it's nobody's business. I hop on Instagram, scroll for a while, make a cup of tea, and suddenly experience the need to shop online.
In the face of too many things to do and too many choices, we can experience what research scientist call analysis paralysis. We like options, but research shows the more choice we have, the more we can freeze up.
Research shows the more choice we have, the more we can freeze up.
What I've found that helps: action. As someone who lives with a mental illness, I've found time and time again that action is the antidote to anxiety. Write that down, boo. Pop it on a post-note, and save it for your next overwhelming Monday morning when you're dreading the week ahead.
See, a lot of us go-getters have a billion thoughts swirling through our heads at any moment. If you add being a perfectionist to that, then you're probs like, "Ahhhhhhh, nothing will ever get done a;dlkjfas;dlkjfasl;djfasl;jfd!!!!" But taking action—even a small, small action—can help us start to push past stress.
Taking action—even a small, small action—can help us start to push past stress.
I know it's helped me when I'm stuck. For example: Once I was worrying about a personal budgeting spreadsheet I needed to create. The task felt daunting, but I set my timer for 10 minutes, Googled some templates, and got to work. Once I started chipping away at it, I felt so much better and my Excel doc wasn't too bad either.
Here's what you do the next time you need to move the boxes in your own life.
Diving into a task can be daunting—especially if has lots of steps.
To help clear the fear before you dive in, try and start small. Find the smallest first action you can take, and ask yourself: Can I just get this one little thing done?
Find the smallest first action you can take, and ask yourself: Can I just get this one little thing done?
For example: Instead of writing the entire paper, aim to get the first paragraph written. Instead of cleaning the entire house, take out the trash and pat yourself on the back for doing so. What you'll find: Doing that first small task will create momentum and carry you through more steps in the process.
Fact: Everything might feel important, but everything isn't urgent.
Personally, I use the Eisenhower Box as a guide to help me figure out what exactly needs my attention. It can help you categorize your tasks into the following categories: important and urgent, important not urgent, urgent not important, not important or urgent.
For example, me scrolling on Instagram is not important nor urgent, but grocery shopping is urgent and important when my fridge has nothing but condiments in it. Next time you make a to-do list, take note of how important and urgent tasks are so you everything doesn't feel like it has a due date of ASAP.
These are distractions that feel good at first, but eventually leave you in the land of "OMG, nothing got done"? Those are your quicksand traps.
My quicksand traps in no particular order: texting friends about nothing, Netflix, emailing people when it could have been a phone call, scrolling on social media— you get it. Get clear on those and notice when you're doing them.
I learned this tactic from a counselor years ago. We can prioritize, avoid distractions—and still get caught up in inaction. One of the best things you can do when you're feeling anxious and have a lot do, is to pick something to do, do it, and get curious about how it made you feel later.
For example, I get irrationally anxious when I book travel. Some of the many anxious thoughts that pop in my head: What if there's a better deal? What if my travel plans change?
Thinking about the first step tends to make me more anxious than actually taking the first step.
But, I've learned to load up the reservation page, insert the dates, and go with whatever has the best review or whatever is convenient. Then, I allow myself to feel the feelings later. What I find: Thinking about the first step tends to make me more anxious than actually taking the first step.
So, whatcha waiting for? Don't let this be just another article on productivity. It's the sign you've been waiting for and I'm right with you. My invitation: think about something that needs to get done, go do it, and let your accountability buddy or a friend know that you completed the task.
Download the Shine iOS app and join Sinclair in a 7-day challenge full of tools to curb your anxiety.


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THE VISION ABLAZE
By Anders Ekdahl January 8, 2016 January 8, 2016 Interviews
THE VISION ABLAZE kinda tripped me into thinking that they were anything but Danes but with that corrected I had to interview them. ©2016 Anders Ekdahl
How hard was it to come up with a band name and how does the name fit the music?
-It did take quite some time – a lot of debate. We made these long lists where each of us argued why one particular name would work, and the rest of us would take this into consideration. We were split in two camps – one wanted an obscure and overtly lyrical name, the other part a one word and simple name. In the end we decided on a compromise that made everyone happy. The Vision Ablaze alludes to a dystopic future – a world in flames. A lot of our songs are about the way mankind handle themselves – how we destroy our world, our minds and entrap ourselves in illusions and lies. This will inevitably lead to our self-destruction – and even now, it's ruining our lives. The bandname fits that ill omen perfectly well.
As I am new to your band perhaps a short introduction might be in order?
-We're a melodic metal band from Copenhagen, Denmark. We been playing since 2011, recorded two EP's before releasing our debut "Youtopia". While the genre melodic metal fit us quite well, there's also a certain amount of progressive and death metal in our songs – clean vocals, screams and growls. You'll find clean guitars, chugging riffs and searing solos. We've always wanted to focus on the good melodies – more than trying to fit a certain genre (or subgenre).
As I am no musician I have no idea how it works, but how do you make your own music based on what influences you? What parts do you pick?
-We're all influenced by both metal and other genres. We never actively try to copy a riff or a song. We do, however, try to capture a feeling or the energy of songs we really like. It could be an elated anthem feeling from a great Killswitch Engage song, or a crushing death-metal part from a Decapitated song. It could also be a progressive stint from Between the Buried and Me or a great melodic shift in a Chris Cornell song. We never play cover songs in the band, so similarities to other songs is always incidental. We write songs and piece together the riffs and the composition that brings out the best melody – the song always comes first.
When you are in a band does it feel like you are a part of a worldwide movement?
-Well, being in a metal band will always feel like being part of a big family. Both metal and music, per say, is an experience that defies borders, genders and class. We haven't jumped a bandwagon – we're not trying to play a popular genre because it might boost our own popularity. There's nothing wrong with being part of a wave of something new and exciting – we've just always felt, that the music we write comes quite naturally. And it has just never fitted well with trends.
How important is it that you look the part in promo shots and stuff? How important is the graphic side of the band?
-We actively try to not look super metal. We try not to cross our arms and get tattoos up and down our necks – while it works great for some, we don't feel it defines us. We care a lot about how our visual identity comes across – whenever we can improve on something, we do it. The main challenge has so far been quite a few line-up changes and funds – while we do have a graphic designer in the band, we lack a great photographer. When you send out a promo package, it has to look good. Good photos, great layout and design of the information you want someone to take an interest in. It can change the attitude of a booker, a record company or simply just make the reviewer pick your package before the others. So it's a vital part of being in a band that wants to grow.
When we put out visuals, it's important that it's based on great ideas and that it looks good – our lyric video for 'Subversion' is a good example of this: Pete (our singer) spent countless hours finding the right clips and making the visuals represent the theme of the song. We recorded the video for 'Under the Killing Moon' during the night of a full moon – to get the right night-time feel to it – to us the result is stunning, dark and emotion – and having shot it during the night adds a little extra depth to it.
What would you say influences your lyrics? How important are they?
-The world influences our lyrics. They are a vital part of our music – they convey immense emotions and are the key to unlocking our songs. Vocals are a really important part of our music – and that makes the lyrics important too. Thematically, the lyrics will often criticize situations that create imbalanced situations for man – be it authorities, religion or anything similar that tend to dictate our lives through means of tyranny and lies. Often the songs will cry out – pleading for action or regretting the fact that our world is still so full of hate and abusive authority figures.
Is the album as relevant today as it was in the 70s and 80s? Is digital killing the album?
-It's difficult for us to say – most of us grew up in the 80's and while we all remember our first CD, we can't really say if the album has lost its relevance. One thing that we have come to realize is that as a struggling metal band, no one will take you seriously before you release a full album. It's like the world wants to test your mettle – to see if you have what it takes to captivate an audience for more than 15 minutes. It's about staying power, about craftsmanship and dedication to your music. In that respect, there's no real difference. The real difference seems to be a completely retarded way of distributing the income – the corporations that control the streaming services have immense power. And while older acts that can tour will be able to make a living off it – newer bands will have trouble finding any income from recording an album. So digital is killing the income of the album – though not the importance of it. But it's still relatively new, and artists are still trying to find their footing.
Where will the future of format end – digital verses physical verses whatever?
-We think there'll always be a market for a luxury physical product – consider the revival of the vinyl. But digital is the future – the vast majority of music users use digital platforms to stream and listen to their favorite artists. The digital world is brimming with opportunities for artists, and those who live off artists (promoters, record companies, bookers, etc) – but there's simply so much out there that it becomes increasingly difficult to break through. New tools are being tested and created – some work, others do not. In the end, the format is really about the format of the band – not the platform that the band use to put their music out there.
How much of a touring entity are you guys? What is a live experience with you like?
-We've had plenty of live experience, but no touring experience. We are working to change that this very instance; having the debut album out it's imperative that we play festivals and gain more experience. It's such an important part of playing in a band and for us, it's the reward: being able to share our songs with an audience and connect with them. When we take to the stage, we feel the audience – we move with them and they with us. We love being on stage – we love playing. We're a tight live-band and while we have a very serious take on music, we're also a smiling band – room for jokes and irony. So a live experience with The Vision Ablaze is fun – it's also moving and elating.
What lies in the future?
-We're hard at work booking shows – that's the future for now. We have to play outside Denmark more – we've had many great reviews from Germany, so there's a potential stage for us there. We're also working on new songs already – it takes a long time to write material, and we usually do some new songs along the way to ensure we have enough material once we have to record our follow up. We want to tighten up our show even more – and to meet our fans.
« WRATHRONE
LETHAL STEEL »


Learn how to model and build a five-year financial statement projection and valuation model from scratch. You will be provided with a 10K (Annual Report), a list of advanced Excel shortcuts and customized macros. With a focus on incorporating and fusing your accounting knowledge into the mechanics of Excel, you will input publicly available information and learn various shortcuts to construct a five-year income statement projection model from revenue down to earnings per share. Combined with your knowledge of the basics of corporate finance and valuation, translate the model into a discounted cash flow analysis. Leave the classroom with an easily scalable and customizable model that can be immediately applied to other companies.
Prerequisite: Intermediate proficiency using Excel is required. This Financial Modeling class is a technical workshop. NOTE: Lack of Excel skills will impede your ability to effectively acquire and implement the techniques and shortcuts that are presented in this program.
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Bring your PC laptop with CD-ROM drive with Microsoft Excel installed. (Excel 2007, Vista, and Macs may not be as effective.) Note: Laptops will not be provided.
Cancellations accepted until 5:00 pm, December 6. There is a $5 surcharge for walk-ins.


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Q: How should I clean up old mailbox files? I'm using Mail.app to read my gmail via IMAP. I recently realised that, since I have a huge amount of mail in that account, and a fairly small SSD, it would be wise to reconfigure the account the save as much storage space as possible. I've done that, but it seems the old mailbox file hasn't been removed. In particular,
~/Library/Mail/V2/IMAP-{email account}/[Gmail].mbox/All Mail.mbox
is massive: 2.9G. I believe that's a hangover from the old account that I've removed. I'm unsure whether everything else in the [Gmail].mbox is new or old, but it's all much smaller.
There's also a directory named:
~/Library/Mail/V2/IMAP-{email account}/INBOX.mbox which is much smaller (16M) and I believe represents the account I've now configured in Mail.app.
What's the best way of clearing this all up? Should I try to selectively rm individual directories, or should I just remove everything and start over? Is there a nice way to remove everything (or, ideally, purge the stuff not in use) or should I rm the directories in V2 that relate to specific email accounts (I have other accounts in use, but I'm happy to remove them and set them up again if recommended)? I'm wary about ending up with a non-functioning Mail.app - if there's a way to uninstall (including data) and reinstall the app, that might be the best option.

redpajama_set_name: RedPajamaStackExchange


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\section{Introduction}
Cardinal arithmetic has been one of the most important areas in set theory.
Shortly after Cohen devised the method of forcing,
Easton~\cite{easton} proved that the powers of regular cardinals
is subject only to K\"onig's theorem in ZFC.
Easton's theorem left the behavior of the powers of \emph{singular} cardinals
as the Singular Cardinal Problem.
Some time later, Silver~\cite{silver} proved
the first nontrivial result around the problem:
Singular cardinals of uncountable cofinality cannot be the least cardinal
at which the Generalized Continuum Hypothesis (GCH) fails.
Still later,
Shelah~\cite{shelah} developed pcf theory and established a result
that supersedes Silver's theorem:

      \begin{thm}[Shelah]\label{shelah}
    $\aleph_{\omega}^{\aleph_{0}} < (2^{\aleph_{0}})^{+} + \aleph_{\omega_{4}}$.
      \end{thm}

An outline of the proof is as follows.
First we have
$\aleph_{\omega}^{\aleph_{0}} = 2^{\aleph_{0}} + {\rm cf}([\aleph_{\omega}]^{\aleph_0},\subseteq)$.
The crucial claim is that
${\rm cf}([\aleph_{\omega}]^{\aleph_0},\subseteq)
=\max{\pcf{\{\aleph_{n} \mid n < \omega\}}}<\aleph_{\omega_4}$.
Shelah proved it by analyzing the structure of
$\pcf{\{\aleph_{n} \mid n < \omega\}}$.
More specifically, he obtained the latter inequality
by showing the following results
for a progressive interval of regular cardinals $A$:
\begin{itemize}
 \item $\pcf{A}$ is an interval of regular cardinals with a largest element.
 \item $|\pcf{A}| < |A|^{+4}$.
\end{itemize}

Theorem~\ref{shelah} can be generalized for a \emph{non}-fixed point
of the $\aleph$ function.
Let $\kappa$ be a singular cardinal with $\kappa = \aleph_\mu > \mu$.
Shelah proved that
${\rm cf}([\kappa]^{|\mu|},\subseteq)=\max \pcf{A}$ for some progressive interval of regular cardinals $A$ with $\sup A = \kappa$.
As before, this reduces the investigation of the power of $\kappa$
to that of the structure of
${\pcf{A}}$.
Note that we can take $A$ to be progressive because
$\kappa$ is a \emph{non}-fixed point of the $\aleph$ function.
Thus the assumption of $A$ being progressive seems essential in pcf theory.
Now one may ask
     \begin{ques}\label{question}
What if $A$ is a {non}-progressive interval of regular cardinals?
     \end{ques}
Motivated by the question, we prove in this paper
     \begin{thm}\label{maintheorem}
    Suppose $\kappa$ is a measurable cardinal. Then the following hold:
     \begin{enumerate}
      \item[$(1)$] $\pcf{\kappa \cap {\rm Reg}} = (2^{\kappa})^{+}\cap {\rm Reg}$.
      \item[$(2)$] Prikry forcing over $\kappa$ forces that $\pcf{{\kappa} \cap {\rm Reg}} = (2^{{\kappa}})^{+} \cap {\rm Reg}$.
      \end{enumerate}
     \end{thm}

From Theorem~\ref{maintheorem}~(1) we get

     \begin{coro}\label{coro1}
         Suppose $\kappa$ is a measurable cardinal. Then the following hold:
     \begin{enumerate}
      \item[$(1)$]
      $\pcf{\kappa \cap {\rm Reg}}$ has no largest element if $2^{\kappa}$ is singular.
      \item[$(2)$]
      $|\pcf{\kappa \cap {\rm Reg}}|> |\kappa \cap {\rm Reg}|^{+4}$ if $2^{\kappa} > \kappa^{+(\kappa^{+4})}$.
      \end{enumerate}
     \end{coro}

The following corollary of Theorem~\ref{maintheorem}~(2) answers Question~\ref{question}:

     \begin{coro}\label{coro2}
          Suppose there is a supercompact cardinal. Then in some forcing extension
          there is a {non}-progressive interval of regular cardinals $A$
          such that ${\rm sup}(A)$ is singular, $\pcf{A}$ has no largest element
          and $|\pcf{A}|>|A|^{+4}$.
     \end{coro}

The proof of Corollary~\ref{coro2} is as follows.
Let $\kappa$ be a supercompact cardinal.
We may assume that $\kappa$ is indestructibly supercompact
in the sense of Laver~\cite{laver}.
This enables us to get a model in which $\kappa$ is supercompact and
$2^{\kappa}$ is a singular cardinal $> \kappa^{+(\kappa^{+4})}$.
Finally,
Prikry forcing gives a model in which $A=\kappa \cap {\rm Reg}$ is as desired
by Theorem~\ref{maintheorem}~(2).
See Corollary~\ref{coro34} for an additional property of the final model.

Some large cardinal hypothesis is necessary in Theorem~\ref{maintheorem}.
Assume on the contrary that
GCH holds and there is no weakly inaccessible cardinal.
A simple argument shows that if $A \subseteq {\rm Reg}$, then
$\pcf{A} = A \cup \{(\sup B)^{+}\mid B \subseteq A$ has no maximal element$\}$,
so that
$\pcf{A}$ has a largest element and $|\pcf{A}|<|A|^{+4}$.

The structure of this paper is as follows.
In Section~2, we recall basic facts of pcf theory and Prikry forcing.
Theorem~\ref{maintheorem} is proved in Section~3.
We also consider the problem whether $\pcf{\pcf{A}}=\pcf{A}$ holds.
In Section~4, we prove an analogue of Theorem~\ref{maintheorem}~(2)
for Magidor forcing.

    \section{Preliminaries}
    In this section, we recall basic facts of pcf theory and Prikry forcing.
    For more on the topics, we refer the reader to \cite{AM} and \cite{gitik} respectively.
    We also use \cite{kanamori} as a reference for set theory in general.

  Our notation is standard. We let ${\rm Reg}$ denote the class of all regular cardinals.
  Let $A \subseteq {\rm Reg}$.
  Then $\prod A$ is the set
  $\{f:A \to \bigcup A \mid \forall \gamma \in A(f(\gamma)< \gamma )\}$.
 Let $F$ be a filter over $A$. We define a strict order $<_{F}$ on $\prod A$
  by $f <_{F} g$ iff
   $\{\gamma \in A \mid f(\gamma) < g(\gamma)\}\in F$.

 \begin{defi}
 For $A\subseteq {\rm Reg}$,
$$\pcf{A} = \left\{\operatorname{cf}\left(\prod A,<_{D}\right) \mid D\text{ is an ultrafilter over }A
\right\}.$$
 \end{defi}

Note that
$A \subseteq \pcf{A} \subseteq (2^{\sup A})^{+}\cap {\rm Reg}$.
 If there is an increasing and cofinal sequence in
 ${\left(\prod A,<_{F}\right)}$ of length $\theta$ for some filter $F$ over $A$,
 then $\operatorname{cf}(\theta) \in \pcf{A}$.

  A set $A\subseteq {\rm Reg}$ is \emph{progressive} if $\min A > |A|$.
  An interval of regular cardinals is a set of the form
  $[\lambda,\kappa)\cap {\rm Reg}$

  for a pair of cardinals $\lambda<\kappa$.
  Here is the fundamental theorem on progressive intervals of regular cardinals.
 \begin{thm}[Shelah]\label{progressive}
 If $A\subseteq {\rm Reg}$ is a progressive interval, then we have
  \begin{enumerate}
   \item[$(1)$] $\pcf{A}$ has a largest element.
   \item[$(2)$] $|\pcf{A}| < |A|^{+4}$.
   \item[$(3)$] $\pcf{\pcf{A}} = \pcf{A}$.
\end{enumerate}
 \end{thm}

 Theorem~\ref{scaleexists} is known as the scale theorem.
 \begin{thm}[Shelah]\label{scaleexists}
 Suppose $\kappa$ is a singular cardinal.
  Then there is a set
  $A \in [\kappa\cap {\rm Reg}]^{\operatorname{cf}(\kappa)}$ such that $\sup A = \kappa$ and ${\left(\prod A,<_F\right)}$ has an increasing and cofinal sequence of length $\kappa^{+}$. Here, $F$ is the cobounded filter over $A$.
  In particular, $\kappa^{+} \in \pcf{\kappa \cap {\rm Reg}}$.
 \end{thm}

 Next, we recall basic facts of Prikry forcing from \cite{prikry}.
 Let $\kappa$ be a measurable cardinal and $U$ a normal ultrafilter over $\kappa$. Prikry forcing $\mathbb{P}$ is the set $[\kappa]^{<\omega} \times U$ ordered by
 ${\langle b,Y\rangle}\leq {\langle a,X\rangle}$ iff $a\subseteq_{e}b$
 (i.e. $a=b \cap (\max(a)+1)$), $Y \subseteq X$ and $b \setminus a \subseteq X$.

   $\mathbb{P}$ has the $\kappa^{+}$-c.c. and size $2^{\kappa}$.
   Thus, $\mathbb{P}$ does not change the value of $2^{\theta}$
   for any ${\theta}\ge\kappa$.
   $\mathbb{P}$ preserves all cardinals above ${\kappa}$
   but changes the cofinality of ${\kappa}$.
   Let $\dot{g}$ be a $\mathbb{P}$-name such that
   $\mathbb{P}\Vdash \dot{g} = \bigcup \{ a \mid \exists X({\langle a,X\rangle} \in \dot{G})\}$, where $\dot{G}$ is the canonical $\mathbb{P}$-name for a generic filter.
   Then $\dot{g}$ is forced to be a cofinal subset of $\kappa$ of order type $\omega$.
   Moreover, we need
  \begin{itemize}
   \item
   ${\langle a,X\rangle} \Vdash {a} \subseteq_{e}\dot{g} \land \dot{g}\setminus{a} \subseteq {X}$.
   In particular, $\mathbb{P}\Vdash \dot{g} \subseteq^{*} X$ for every $X\in U$.
   \item If $\langle a,X \rangle \Vdash \xi \in \dot{g}$, then $\xi \in a$.
  \end{itemize}
   The latter property follows by $\langle a,X\setminus (\xi + 1) \rangle \leq \langle a,X\rangle$ forces $\dot{g} \setminus a \subseteq X \setminus (\xi + 1)$.

  For subsequent purposes, we present a direct proof of Prikry lemma.
  Suppose $\{X_{b}\mid b \in [\kappa]^{<\omega}\}\subseteq U$.
  The diagonal intersection $\triangle_{b}X_{b}$ is defined to be the set
  $\{\xi < \kappa \mid \forall b \in [\xi]^{<\omega}(\xi \in X_{b})\}$.
  Since $U$ is normal, we have $\triangle_{b}X_{b} \in U$.

  \begin{lem}\label{diagonallemma}
   Suppose $\{X_{b} \mid b \in [\kappa]^{<\omega}\} \subseteq U$
   and $a \in [\kappa]^{<\omega}$.
   Then any extension of ${\langle a,\triangle_{b}X_{b}\rangle}$ is compatible with ${\langle a,X_{a}\rangle}$.
  \end{lem}
  \begin{proof}
   Let
   ${\langle c,Y\rangle}\leq {\langle a,\triangle_{b}X_{b}\rangle}$.
   Then $c\setminus a\subseteq X_{a}$ by $a\subseteq_{e} c$ and
   $c\setminus a\subseteq \triangle_{b}X_{b}$.
   Thus ${\langle c,Y\cap X_{a}\rangle}$ is a common extension of
   ${\langle c,Y\rangle}$ and ${\langle a,X_{a}\rangle}$, as desired.
  \end{proof}

 \begin{lem}[Prikry lemma]\label{prikrycondition}
  Let $a \in [\kappa]^{<\omega}$ and $\sigma$
  be a statement of the forcing language.
  Then there is an $X \in U$ such that ${\langle a,X\rangle}$ decides $\sigma$, i.e. ${\langle a,X\rangle }\Vdash \sigma $ or ${\langle a,X\rangle}\Vdash \lnot \sigma$.
 \end{lem}
 \begin{proof}

  For each $b \in [\kappa]^{<\omega}$ define $X_{b} \in U$ as follows:
  If $a\subseteq_{e} b$, let $X_{b}$ be the unique set from the following
  mutually disjoint sets
\begin{itemize}
 \item $X_{b}^{+} = \{\xi < \kappa \mid {b} \subseteq \xi \land
 \exists Y\in U ({\langle b\cup\{\xi\} ,Y \rangle} \Vdash \sigma) \}$.
 \item $X_{b}^{-} = \{\xi < \kappa \mid {b} \subseteq \xi \land
 \exists Y\in U({\langle b\cup\{\xi\} ,Y \rangle} \Vdash \lnot\sigma)\}$.
 \item $X_{b}^{0} = \kappa \setminus (X_{b}^{+} \cup X_{b}^{-})$.
\end{itemize}
  Otherwise, let $X_{b}= \kappa$.
For each $b \in [\kappa]^{<\omega}$ define $Y_{b} \in U$ as follows:
If there is a $Y \in U$ such that ${\langle b,Y\rangle}$ decides $\sigma$,
let $Y_{b}$ be one such $Y$. Otherwise, let $Y_{b}= \kappa$.
We claim that $X= \triangle_{b}(X_{b}\cap Y_{b})\in U$ is as desired.
Take an arbitrary extension ${\langle c,Y\rangle}\leq {\langle a,X\rangle}$ that decides $\sigma$.
We may assume $c = b\cup\{\xi\}$ with $a \subseteq_{e} b \subseteq \xi$.
Note that
${\langle c,Y_{c}\rangle}$ decides $\sigma$.
We may assume ${\langle c,Y_{c}\rangle } \Vdash \sigma$.
Then ${\langle c,\triangle_{b}Y_{b}\rangle}\Vdash \sigma$
by Lemma~\ref{diagonallemma}.
Thus ${\langle c,X \rangle} \le {\langle c,\triangle_{b}Y_{b}\rangle}$
forces $\sigma$.
We claim that ${\langle b,X\rangle} \Vdash \sigma$,
which completes the proof by repeating the argument.

It suffices to show that any extension of ${\langle b,X\rangle}$
is compatible with a condition forcing $\sigma$.
Let ${\langle d,Z\rangle} \leq {\langle b,X\rangle}$.
We may assume $b \subsetneq_{e} d$.
Note that
$\xi \in X_{b}$ by $\xi \in X$, and hence $X_{b} = X_{b}^{+}$
by ${\langle b\cup\{\xi\},X\rangle}\Vdash \sigma$.
Let $\eta = \min(d\setminus b) \in X$.
Then $\eta \in X_{b} = X_{b}^{+}$, so
${\langle b\cup\{\eta\},Y\rangle}\Vdash\sigma$
for some $Y$, and hence
${\langle b\cup\{\eta\},Y_{b\cup\{\eta\}}\rangle}\Vdash\sigma$.
Note that
$d\setminus (b\cup\{\eta\}) \subseteq Y_{b\cup\{\eta\}}$ by
$b\cup\{\eta\} \subseteq_{e} d$ and
$d\setminus (b\cup\{\eta\}) \subseteq d\setminus b \subseteq X$.
Thus ${\langle d, Z\cap Y_{b\cup\{\eta\}}\rangle}$
is a common extension of
${\langle d,Z\rangle}$ and ${\langle b\cup\{\eta\},Y_{b\cup\{\eta\}}\rangle}$,
as desired.
  \end{proof}

 \begin{coro}
  $\mathbb{P}$ adds no new bounded subsets of $\kappa$.
  In particular, $\mathbb{P}$ preserves all cardinals below $\kappa$.
 \end{coro}

   \section{Prikry Forcing and a Non-progressive Interval}
   The first half of this section is devoted to

   \begin{proof}[Proof of Theorem~\ref{maintheorem}]
  Let $\kappa$ be a measurable cardinal.
  Take a normal ultrafilter $U$ over $\kappa$ and form
  $j:V \to M \simeq \text{Ult}(V,U)$.
  For each $\alpha\le2^{\kappa}$, we can choose
    $f_{\alpha} \in \mbox{}^{\kappa}\kappa$ such that $\alpha =[f_{\alpha}]_{U}$
   by $2^{\kappa}\le(2^{\kappa})^{M}<j(\kappa)$.

    Note that
   ${\kappa \cap {\rm Reg}}\subseteq \pcf{\kappa \cap {\rm Reg}}\subseteq (2^{\kappa})^{+}\cap {\rm Reg}$. To complete the proof, it suffices to show that
   $[\kappa,(2^{\kappa})^{+})\cap{\rm Reg}\subseteq \pcf{\kappa\cap{\rm Reg}}$ in both cases, (1) and (2).

    (1) Let $\theta \in [\kappa,(2^{\kappa})^{+})\cap{\rm Reg}$.
    Then we may assume $f_{\theta} \in \mbox{}^{\kappa}(\kappa\cap{\rm Reg})$.
    Since $\kappa=[{\rm id}]_{U}\le[f_{\theta}]_{U}$, we have
   $$X = \{\xi < \kappa\mid \forall \eta < \xi( f_{\theta}(\eta) < \xi)
   \land \xi \leq f_{\theta}(\xi)\} \in U.$$
   Note that
   $f_{\theta}\upharpoonright X$ is strictly increasing.
   Define an ultrafilter $U_{\theta}$ over $\kappa \cap {\rm Reg}$ by
   $Y \in U_{\theta}$ iff $f_{\theta}^{-1}``Y \in U$.
Then we have
   ${\left(\prod_{\xi\in X}f_{\theta}(\xi),<_{U}\right)}\simeq
   {\left(\prod f_{\theta}``{X},<_{U_{\theta}}\right)}\simeq
   {\left(\prod \kappa\cap {\rm Reg},<_{U_{\theta}}\right)}$.
\medskip

Since $\langle{f_{\alpha}\upharpoonright X \mid \alpha < \theta}\rangle$ is increasing and cofinal in ${\left(\prod_{\xi \in X}f_{\theta}(\xi),<_{U}\right)}$,
we have
   $\theta =\operatorname{cf}\left(\prod_{\xi \in X}f_{\theta}(\xi),<_{U}\right)
   =\operatorname{cf}(\prod \kappa \cap {\rm Reg} ,<_{U_{\theta}})\in \pcf{\kappa\cap {\rm Reg}}$, as desired.
\medskip

  (2)
  Let $\mathbb{P}$ be Prikry forcing defined by $U$.
  Note that the set $(\kappa,(2^{\kappa})^{+})\cap {\rm Reg}$
  remains the same after forcing with $\mathbb{P}$ and $\kappa$ is singular.
   Let $\theta\in(\kappa,(2^{\kappa})^{+})\cap{\rm Reg}$.
  It suffices to prove that
   $\mathbb{P}\Vdash\theta\in \pcf{\kappa\cap{\rm Reg}}$. Again, we may assume $f_\theta \in {^{\kappa}(\kappa \cap {\rm Reg})}$.
First, note that
      $$X = \{\xi < \kappa\mid \forall \eta < \xi( f_{\theta}(\eta) < \xi)
   \land \xi< f_{\theta}(\xi)\} \in U.$$
    Since $\mathbb{P} \Vdash \dot{g}\subseteq^{*} {X}$,
    we have
    \begin{center}
     $\mathbb{P} \Vdash {\left(\prod_{\xi \in \dot{g}}{f}_{\theta}(\xi),<^{*}\right)} \simeq {\left(\prod {f}_{\theta}{``}\dot{g},<_{\dot{F}}\right)}.$    \end{center}
     Here $<^{*}$ and $\dot{F}$ are $\mathbb{P}$-names
     for the order on $\prod_{\xi \in \dot{g}}{f}_{\theta}(\xi)$
 defined by the cobounded filter over $\dot{g}$, and
     the cobounded filter over ${f}_{\theta}{``}\dot{g}$ respectively.
     Thus it suffices to prove

        \begin{itemize}
    \item[(i)]
    $\mathbb{P}\Vdash{\langle{f}_{\alpha}\upharpoonright \dot{g} \mid \alpha < {\theta}\rangle}${ is increasing in }${\left(\prod_{\xi \in \dot{g}} {f}_{\theta}(\xi),<^{*} \right)}$.
    \item[(ii)]
    $\mathbb{P}\Vdash{\langle{f}_{\alpha}\upharpoonright \dot{g} \mid \alpha < {\theta}\rangle}${ is cofinal in }${\left(\prod_{\xi \in \dot{g}} {f}_{\theta}(\xi),<^{*} \right)}$.
   \end{itemize}

(i)
Let $\alpha < \beta$.
Then $Y= \{\xi < \kappa \mid f_{\alpha}(\xi) < f_{\beta}(\xi)\}\in U$.
If ${\langle a,Z\rangle} \in \mathbb{P}$, then
${\langle a,Y \cap Z\rangle} \Vdash\forall \xi\in\dot{g} \setminus{a}
({f}_{\alpha} (\xi)<{f}_{\beta}(\xi))$, as desired.

(ii)
   By the proof of (i), it suffices to show that
   $\left\{h \upharpoonright \dot{g}\mid h\in \prod^{V}_{\xi \in{X}}{f}_{\theta}(\xi) \right\}$
is forced to be cofinal in
${\left(\prod_{\xi \in \dot{g}} {f}_{\theta}(\xi),<^{*} \right)}$.

Assume
$\Vdash \dot{h} \in \prod_{\xi \in \dot{g}}{f}_{\theta}(\xi)$.
   For each $b\in [\kappa]^{<\omega}$
   define $Y_{b}\in U$ and $\eta_b < \kappa$ as follows.
   Note that ${\langle b, X\rangle}$ forces
   $b \subseteq\dot{g}$ and hence $\dot{h}(\max{{b}}) < f_{\theta}(\max{b})$.
     By Prikry lemma, there is a ${\langle b,Y_{b}\rangle}\le{\langle b, X\rangle}$
     that decides
     $\dot{h}(\max{{b}}) ={\eta}$ for every $\eta < f_{\theta}(\max{b})$.
Then we can take an $\eta_b < f_{\theta}(\max{b})$ such that
$${\langle b,Y_{b}\rangle}\Vdash \dot{h}(\max{{b}})=\eta_b.$$
For each $\xi \in{X}$ define
     $$h(\xi) = \sup \{\eta_b+1 \mid b\in [\xi + 1]^{<\omega}\}.$$
Since ${f}_{\theta}(\xi)> \xi$ is regular,
we have $h\in \prod_{\xi \in{X}}{f}_{\theta}(\xi)$ in $V$.
Let $Y = \triangle_{b}Y_{b}\in U$.
We claim that
${\langle a, Y\rangle}\Vdash \forall \xi\in\dot{g}\setminus a(\dot{h}(\xi)<{h}(\xi))$
for every $a\in [\kappa]^{<\omega}$,
which completes the proof.
It suffices to show that any extension of ${\langle a, Y\rangle}$
forcing $\xi\in\dot{g}\setminus a$ is compatible with a condition forcing
$\dot{h}(\xi)<{h}(\xi)$.

Suppose
${\langle b,Z\rangle}\le{\langle a,Y\rangle}$ forces $\xi\in\dot{g}\setminus a$.
By the property we saw in Section 2, we have $\xi\in b\setminus a$. ${\langle b,Z\rangle}$ is compatible with
${\langle b \cap (\xi+1),Y_{b\cap (\xi+1)}\rangle}$ forcing
$\dot{h}(\xi) = \eta_{b\cap(\xi+1)} < h(\xi)$, as in the proof of
Prikry lemma.
  \end{proof}

  Corollary~\ref{coro2} shows that the assumption of $A$ being progressive
  is necessary in Theorem~\ref{progressive}~(1) and (2).
  Corollary~\ref{coro34} does the same for Theorem~\ref{progressive}~(3).
   \begin{coro}\label{coro34}
    One can add ``$A\subsetneq \pcf{A}\subsetneq \pcf{\pcf{A}}$''
    to the list of properties of $A$ in Corollary~\ref{coro2}.
   \end{coro}

   \begin{proof}
   Let $A=\kappa \cap {\rm Reg}$ in the final model for Corollary~\ref{coro2},
   where $2^{\kappa}$ is singular.
   By Theorem~\ref{maintheorem}~(2) we have
   $\pcf{A} = (2^{\kappa})^{+}\cap {\rm Reg} = 2^{\kappa}\cap {\rm Reg} \ne A$,
   which in turn implies that
   $(2^{\kappa})^{+} \in\pcf{\pcf{A}} \setminus\pcf{A}$ by Theorem~\ref{scaleexists}.
   \end{proof}
   The rest of this section is devoted to improving Corollary~\ref{coro34}.
   Define $\operatorname{pcf}^{n}(A)$ for $n<\omega$ by
   $\operatorname{pcf}^{0}(A) = A$
   and
   $\operatorname{pcf}^{n+1}(A) = \operatorname{pcf}(\operatorname{pcf}^{n}(A))$.
   \begin{thm}\label{pcfinc}
   Suppose $\langle\kappa_{i} \mid i < \omega\rangle$ is an increasing sequence of supercompact cardinals. Then the following hold in some forcing extension:
    \begin{enumerate}
     \item[$(1)$] $\kappa_{0}$ is a singular cardinal of cofinality $\omega$.
     \item[$(2)$] $\operatorname{pcf}^{n}(\kappa_{0} \cap {\rm Reg}) \subsetneq \operatorname{pcf}^{n+1}(\kappa_{0} \cap {\rm Reg})$ for every $n < \omega$.
    \end{enumerate}
   \end{thm}
   Lemma~\ref{ccpcf} ensures that sets of the form $\pcf{\theta \cap {\rm Reg}}$
   remain the same throughout forcing extensions for Theorem~\ref{pcfinc}.
   \begin{lem}\label{ccpcf}
Suppose $A\subseteq {\rm Reg}$, and
   $\mathbb{Q}$ has the $\kappa$-c.c. with $\kappa=\min (A)$.
   Then $\mathbb{Q}\Vdash{\rm pcf}^{V}(A)\subseteq\pcf{A}$.
   \end{lem}
   \begin{proof}
    In $V$, let $\theta \in \pcf{A}$ be arbitrary.
    Then there are an ultrafilter $D$ over $A$ and an increasing and cofinal
    sequence
    ${\langle f_{\alpha}\mid \alpha < \theta\rangle}$ in ${\left(\prod A,<_{D}\right)}$.
    Let $\dot{E}$ be a $\mathbb{Q}$-name for the filter generated by ${D}$.
 Since $\theta \geq \kappa$ remains regular after forcing with $\mathbb{Q}$,
    it suffices to prove that
    ${\langle f_{\alpha}\mid \alpha < \theta\rangle}$
    is forced to be increasing and cofinal in ${\left(\prod{A},<_{\dot{E}}\right)}$.

    It is easy to see the former.
    For the latter, it suffices to prove that
    $\prod^{V}{A}$ is forced to be cofinal in ${\left(\prod{A},<_{\dot{E}}\right)}$.
Assume $p \Vdash \dot{h} \in \prod{A}$.
For each $\gamma \in A$, define
$$h^*(\gamma) = \sup \{\xi + 1 \mid \exists q \leq p ( q \Vdash \dot{h}({\gamma}) ={\xi})\}.$$
Then
 $p \Vdash \dot{h}({\gamma}) <{h^*}({\gamma})$ for every $\gamma \in A$.
Since $\mathbb{Q}$ has the $\kappa$-c.c. and $\gamma \geq \kappa$ is regular,
we have $h^* \in \prod A$ in $V$, as desired.
   \end{proof}

   \begin{proof}[Proof of Theorem~\ref{pcfinc}]
    We may assume that each $\kappa_i$
    is indestructibly supercompact in the sense of Laver~\cite{laver} and $2^{\kappa_{i}} = \kappa_i^{+}$. We refer the reader to \cite{apter} for more details.

    Let $\mathbb{Q}$ be the full support product
     $\prod_{i < \omega}{\rm Add}(\kappa_i,\kappa_{i + 1})$,
     where ${\rm Add}(\kappa_{i},\kappa_{i + 1})$ is the poset
    adding $\kappa_{i + 1}$ many Cohen subsets of $\kappa_{i}$.
    Standard arguments show that $\mathbb{Q}$ preserves cofinalities and
    forces $2^{\kappa_n} = \kappa_{n + 1}$ for every $n < \omega$.
     We claim that $\mathbb{Q}$ forces
     $\pcf{[\kappa_{n}^{+},\kappa_{n+1}^{+}) \cap {\rm Reg}}\supseteq
     [\kappa_{n}^{+},\kappa_{n + 2}^{+}) \cap {\rm Reg}$
     for every $n < \omega$.

Let $G\subseteq\mathbb{Q}$ be generic.
    Since
    $\mathbb{Q} \simeq \prod_{i > n}\operatorname{Add}(\kappa_{i},\kappa_{i + 1})
  \times\textstyle{\prod_{i\le n}\operatorname{Add}(\kappa_{i},\kappa_{i+1})}$ in $V$,
  we have
     $G\simeq G_n \times H_n$ in $V[G]$.
By Theorem~\ref{maintheorem}~(1),
$\pcf{\kappa_{n}^{+} \cap {\rm Reg}}
=\pcf{\kappa_{n} \cap {\rm Reg}} \cup\{\kappa_{n}\}
=(2^{\kappa_{n}})^{+} \cap {\rm Reg} = \kappa_{n}^{++} \cap {\rm Reg}$
in $V$.
This remains true in $V[G_n]$
by the $\kappa_{n+1}$-closure of the corresponding poset.
Now we work in $V[G_n]$.
Note that $\kappa_{n+1}$ is supercompact and  $2^{\kappa_{n+1}} = \kappa_{n + 2}$.
By Theorem~\ref{maintheorem}~(1) we have
$\pcf{\kappa_{n+1}^{+} \cap {\rm Reg}}
=\pcf{\kappa_{n+1} \cap {\rm Reg}} \cup\{\kappa_{n+1}\}
= (2^{\kappa_{n+1}})^{+} \cap {\rm Reg} = \kappa_{n + 2}^{+} \cap {\rm Reg}$.
Therefore
$\pcf{[\kappa_{n}^{+},\kappa_{n+1}^{+}) \cap {\rm Reg}}
= [\kappa_{n}^{+},\kappa_{n + 2}^{+}) \cap {\rm Reg}$.
Note that
$\left(\prod_{i \le n}\operatorname{Add}(\kappa_{i},\kappa_{i+1})\right)^{V}=
\prod_{i \le n}\operatorname{Add}(\kappa_{i},\kappa_{i+1})$
has the $\kappa_{n}^{+}$-c.c.
By Lemma~\ref{ccpcf} we have
$\pcf{[\kappa_{n}^{+},\kappa_{n+1}^{+}) \cap {\rm Reg}}
\supseteq \pcf{[\kappa_{n}^{+},\kappa_{n+1}^{+}) \cap {\rm Reg}}^{V[G_n]}
= [\kappa_{n}^{+},\kappa_{n + 2}^{+}) \cap {\rm Reg}$
in $V[G]=V[G_n][H_n]$, as desired.

Since $\mathbb{Q}$ is $\kappa_{0}$-directed closed in $V$,
    $\kappa_0$ remains supercompact in $V[G]$.
So we can define Prikry forcing $\mathbb{P}$ over $\kappa_0$.
By Theorem~\ref{maintheorem}~(2), $\mathbb{P}$ forces
$\pcf{\kappa_{0} \cap {\rm Reg}}= (2^{\kappa_{0}} )^{+}\cap {\rm Reg}
    = \kappa_{1}^{+} \cap {\rm Reg}$.
    By Lemma~\ref{ccpcf}, $\mathbb{P}$ forces
$[\kappa_{n}^{+},\kappa_{n + 2}^{+}) \cap {\rm Reg}\subseteq
    \pcf{[\kappa_{n}^{+},\kappa_{n+1}^{+}) \cap {\rm Reg}}\subseteq
    [\kappa_{n}^{+},(2^{\kappa_{n+1}})^{+}) \cap {\rm Reg}
    =[\kappa_{n}^{+},\kappa_{n + 2}^{+}) \cap {\rm Reg}$ for every $n< \omega$.
Let $H\subseteq \mathbb{P}$ be generic. In $V[G][H]$,
we have
   ${\operatorname{pcf}^{n+1}(\kappa_0 \cap {\rm Reg})} ={\kappa_{n+1}^{+} \cap {\rm Reg}}$
   by induction on $n < \omega$.
\end{proof}

   \section{An Analogue for Magidor Forcing}
   Prikry forcing is known for a wealth of variations.
   In this section, we give an analogue of Theorem~\ref{maintheorem}~(2) for one of them.
   Here we take up Magidor forcing from \cite{magidor1978changing}, but the argument works
   equally well for other variations, e.g. the diagonal Prikry forcing as defined in \cite{NU}.

   Magidor forcing uses a sequence of ultrafilters rather than a single ultrafilter, and makes a hypermeasurable cardinal into a singular cardinal of
   uncountable cofinality.
   For normal ultrafilters $U ,U'$ over $\kappa$, $U \vartriangleleft U'$ iff $U\in M \simeq \text{Ult}(V,U')$. Let ${\langle U_{\alpha} \mid \alpha < \lambda\rangle}$ be a $\vartriangleleft$-increasing sequence with $\lambda <\kappa$. Note that there is a such sequence if $\kappa$ is supercompact. For any $\beta< \alpha < \lambda$, we fix a function $F_{\beta}^{\alpha} \in \mbox{}^{\kappa}V$ such that $[F_{\beta}^{\alpha}]_{U_{\alpha}} = U_{\beta}$. For each $\alpha< \lambda$, define
      \begin{align*}
    A_{\alpha} & = \{\delta < \kappa \mid \forall \beta < \alpha \forall \gamma < \beta(F_{\gamma}^{\alpha}(\delta) \vartriangleleft F_{\beta}^{\alpha}(\delta)
    \text{ are normal ultrafilters over }\delta)\}.\\
    B_{\alpha} & = \{\delta \in A_{\alpha}\setminus(\lambda + 1) \mid \forall \beta < \alpha \forall \gamma < \beta([F^{\beta}_{\gamma}\upharpoonright \delta]_{F_{\beta}^{\alpha}(\delta)} = F_{\gamma}^{\alpha}(\delta))\}.
   \end{align*}
   Note that $B_{\alpha} \in U_{\alpha}$.
   Magidor forcing $\mathbb{M}$ is the set of pairs $\langle a,X \rangle$ such that
    \begin{itemize}
     \item $a$ is an increasing function such that
	   \begin{itemize}
	    \item $\operatorname{dom}(a) \in [\lambda]^{<\omega}$ and $\forall \alpha \in \operatorname{dom}(a)(a(\alpha) \in B_{\alpha})$.
	   \end{itemize}
     \item $X$ is a function such that
     \begin{itemize}
      \item $\operatorname{dom}(X) = \lambda \setminus \operatorname{dom}(a)$ and $\forall \alpha \in \operatorname{dom}(X)(X(\alpha) \subseteq B_{\alpha}$),
	    \item For every $\alpha \in \operatorname{dom}(X)$, if $\operatorname{dom}(a) \setminus (\alpha + 1) = \emptyset$, $X(\alpha) \in U_{\alpha}$. Otherwise, $X(\alpha) \in F^{\beta}_{\alpha}(a(\rho))$ where $\beta = \min (\operatorname{dom}(a) \setminus (\alpha + 1))$.
     \end{itemize}
\end{itemize}

    $\mathbb{M}$ is ordered by ${\langle a,X\rangle } \leq {\langle b,Y\rangle }$ iff $b \subseteq a$, $\forall \alpha \in \operatorname{dom}(X)(X(\alpha) \subseteq Y(\alpha))$ and $\forall \alpha \in \operatorname{dom}(a) \setminus \operatorname{dom}(b)(a(\alpha) \in Y(\alpha))$.
    $\mathbb{M}$ has the $\kappa^{+}$-c.c. and size $2^{\kappa}$. Thus, $\mathbb{M}$ does not change the value of $2^{\theta}$ for any $\theta \geq \kappa$. $\mathbb{M}$ preserves all cardinals above $\kappa$ but changes the cofinality of $\kappa$ like Prikry forcing. Let $\dot{g}$ be an $\mathbb{M}$-name such that $\mathbb{M} \Vdash \dot{g} = \bigcup\{a \mid \exists X\langle a,X \rangle \in \dot{G} \}$, where $\dot{G}$ is the canonical $\mathbb{M}$-name for a generic filter. $\dot{g}$ is forced to be an increasing sequence of length $\lambda$ which converges to $\kappa$. As in Prikry forcing, we also have
   \begin{itemize}
    \item $\langle a, X \rangle \Vdash \dot{g}\upharpoonright \operatorname{dom}({a}) = {a} \land \forall \alpha \in \lambda \setminus \operatorname{dom}(a)(\dot{g}(\alpha) \in X(\alpha))$.
   \end{itemize}
    For each $\beta < \lambda$, We let $\mathbb{M}_{\beta}= \{{\langle a,X \rangle}_{\beta}\mid {\langle a,X \rangle} \in \mathbb{M}\}$ and $\mathbb{M}^{\beta} = \{{\langle a,X \rangle}^{\beta} \mid {\langle a,X \rangle} \in \mathbb{M}\}$. Here, ${\langle a,X\rangle}_{\beta}$ and ${\langle a,X\rangle}^{\beta}$ are ${\langle a\upharpoonright (\beta + 1),X\upharpoonright (\beta + 1)\rangle}$ and ${\langle a\upharpoonright (\lambda \setminus (\beta + 1)),X\upharpoonright (\lambda \setminus (\beta + 1))\rangle}$ respectively. The orders on $\mathbb{M}_\beta$ and $\mathbb{M}^{\beta}$ are naturally defined by that on $\mathbb{M}$.
    $\mathbb{M}$ can be factored as follows.
    \begin{lem}
     For every ${\langle a,X\rangle} \in \mathbb{M}$ and $\beta \in \operatorname{dom}(a)$,
     we have
     $$\mathbb{M} / {\langle a,X\rangle} \simeq \mathbb{M}_{\beta}/{\langle a,X\rangle}_{\beta} \times \mathbb{M}^{\beta}/{\langle a,X\rangle}^{\beta}.$$
    \end{lem}

   Note that $\mathbb{M}_{\beta} / \langle a,X\rangle_{\beta}$ has the $a(\beta)^{+}$-c.c.
    Lemmas~\ref{magidordiagonallemma} and \ref{magidorprikrycondition} are analogues of Lemmas~\ref{diagonallemma} and \ref{prikrycondition} for Magidor forcing respectively.
    See \cite{magidor1978changing} for proofs.
    \begin{lem}\label{magidordiagonallemma}
     Suppose that ${\langle a,X\rangle} \in \mathbb{M}$ and $\{{\langle b,X_{b}\rangle} \mid b \in {\rm LP}\}$ is a set of extensions of $\langle a,X\rangle$ where ${\rm LP} = \{b \mid \exists Y(\langle{b,Y}\rangle \leq \langle{a,X}\rangle )\}$. Then there is a $Z$ such that $\langle a,Z \rangle \in \mathbb{M}$ and every extension of $\langle{b,Y}\rangle$ is compatible with $\langle b,X_{b}\rangle$ if $\langle b,Y \rangle \leq \langle a,Z \rangle$.
    \end{lem}
    \begin{lem}[Prikry lemma]\label{magidorprikrycondition}
     For every $\langle a,X\rangle \in \mathbb{M}$ and statement $\sigma$ of the forcing language, $\beta \in \operatorname{dom}(a)$, there is a $Z$ such that
     \begin{itemize}
      \item $\langle a,Z \rangle \leq \langle a,X \rangle$ and $\langle a,Z \rangle_{\beta} = \langle a,X \rangle_{\beta}$.
      \item If $\langle b,Y \rangle \leq \langle a,Z \rangle$ decides $\sigma$,
      then $\langle b, Y\rangle_{\beta}^\frown \langle a,Z \rangle^{\beta}$ decides $\sigma$.
     \end{itemize}
    \end{lem}
    Here is the fundamental theorem of Magidor forcing:

    \begin{thm}[Magidor]
    The following hold:
     \begin{enumerate}
      \item[$(1)$] $\mathbb{M}$ adds no new subsets of $\lambda$.
      In particular, $\lambda^{+}\cap {\rm Reg}$ remains the same by $\mathbb{M}$.
      \item[$(2)$] $\mathbb{M}$ preserves all cardinals.
      \item[$(3)$] $\mathbb{M}$ forces that ${\kappa}$ is a strong limit singular cardinal of cofinality ${\lambda}$.
     \end{enumerate}
    \end{thm}
  Now we get an analogue of Theorem~\ref{maintheorem}~(2) for Magidor forcing.
  \begin{thm}\label{magidormaintheorem}
  $\mathbb{M}$ forces that $\pcf{{\kappa} \cap {\rm Reg}} = (2^{{\kappa}})^{+} \cap {\rm Reg}$.
  \end{thm}
  \begin{proof}
   By the proof of Theorem~\ref{maintheorem}, it suffices to show that $\mathbb{M} \Vdash (\kappa,{(2^{\kappa})}^{+}) \cap {\rm Reg} \subseteq \pcf{\kappa \cap {\rm Reg}}$. Note that $(\kappa,(2^{\kappa})^{+}) \cap {\rm Reg}$ remains the same after forcing with $\mathbb{M}$. Let $\theta \in (\kappa,(2^{\kappa})^{+})\cap {\rm Reg}$. Let us see that $\mathbb{M} \Vdash \theta \in \pcf{(\kappa,(2^{\kappa})^{+}) \cap {\rm Reg}}$.

 For every $\gamma \leq \theta$ and $\alpha < \lambda$, we fix a function $f^{\alpha}_{\gamma} \in {^{\kappa}}\kappa$ such that $[f^{\alpha}_{\gamma}]_{U_{\alpha}} = \gamma$. We may assume $f_\theta^\alpha \in {^{\kappa}(\kappa \cap {\rm Reg})}$. Let $X' \in \prod_{\alpha < \lambda}U_\alpha$ be a function in $V$ such that $X'({\alpha}) = \{\xi \in B_\alpha \mid \forall \eta < \xi(f^{\alpha}_{\theta}(\eta) < \xi) \land \xi <f_{\theta}^{\alpha}(\xi)\}$ for any $\alpha < \lambda$.

We will show that $\mathbb{M}$ forces ${\left(\prod_{\alpha \in \lambda}{f}^{\alpha}_{\theta}(\dot{g}(\alpha)),<^{*}\right)}$ has an increasing and cofinal sequence of length $\theta$. Here, $<^{*}$ is an $\mathbb{M}$-name for the order on $\prod_{\alpha \in \lambda}{f}^{\alpha}_{\theta}(\dot{g}(\alpha))$ defined by the cobounded filter over $\lambda$. This gives the desired result, as shown by the following argument:

   By a usual density argument, we can find an $\mathbb{M}$-name $\dot{A}$ such that $\mathbb{M}$ forces the following properties:
\begin{itemize}
 \item $\dot{A} \in [\lambda]^{\lambda}$.
 \item $\forall \alpha,\beta \in \dot{A}(\alpha < \beta \to f_\theta^{\alpha}(\dot{g}(\alpha)) < f_\theta^{\beta}(\dot{g}(\beta)))$.
\end{itemize}
   And thus, by the proof of Theorem~\ref{maintheorem}, we have
   \begin{center}
    $\mathbb{M} \Vdash {\left(\prod_{\alpha \in \dot{A}}{f}^{\alpha}_{\theta}(\dot{g}(\alpha)),<^{*} \upharpoonright \dot{A}\right)} \simeq \left(\prod \{{f}_{\theta}^{\alpha}(\dot{g}(\alpha)) \mid \alpha \in \dot{A}\}, <_{\dot{F}}\right)$.
   \end{center}
      Here, $\dot{F}$ is an $\mathbb{M}$-names for the cobounded filter over $\{{f}_{\theta}^{\alpha}(\dot{g}(\alpha)) \mid \alpha \in \dot{A}\}$.  It follows that $\mathbb{M}$ forces $\left(\prod \{{f}_{\theta}^{\alpha}(\dot{g}(\alpha) \mid \alpha \in \dot{A}\}, <_{\dot{F}}\right)$ has an increasing and cofinal sequence of length $\theta$.

   For every $\gamma < \theta$, let $\dot{f}_{\gamma}$ be an $\mathbb{M}$-name for a function $\alpha \mapsto {f}^{\alpha}_{\gamma}(\dot{g}(\alpha))$. It suffices to prove
   \begin{itemize}
    \item[(i)]
	      $\mathbb{M}\Vdash{\langle\dot{f}_{\gamma}\mid \gamma < {\theta}\rangle}${ is increasing in }$\left(\prod_{\alpha < {\lambda}}{f}^{\alpha}_{\theta}(\dot{g}(\alpha)),<^{*}\right)$.
    \item[(ii)]
	       $\mathbb{M} \Vdash \langle\dot{f}_\gamma \mid \gamma < \theta\rangle${ is cofinal in }$\left(\prod_{\alpha < {\lambda}}{f^{\alpha}_{\theta}}(\dot{g}(\alpha)),<^{*}\right)$.
   \end{itemize}

   (i) Let $\gamma < \delta < \theta$. Note that we have $Y(\alpha) = \{\xi < \kappa \mid f_{\gamma}^{\alpha}(\xi) < f_{\delta}^{\alpha}(\xi)\} \in U_\alpha$ for each $\alpha < \lambda$. Let $\langle a,X\rangle \in \mathbb{M}$ be arbitrary. Define $Z = (X \upharpoonright \beta_a)^{\frown}\langle X(\alpha) \cap Y(\alpha) \mid \alpha \geq \beta_a \rangle$. Here, $\beta_a = \max{\operatorname{dom}(a)}$. Then $\langle{a,Z}\rangle \leq \langle a,X\rangle$ forces $f_{\gamma}(\alpha) < f_{\delta}(\alpha)$ for every $\alpha > \beta_a$.

 (ii) Let $\langle a,X \rangle \in \mathbb{M}$ and $\dot{h}$ be arbitrary. Suppose $\langle a,X \rangle \Vdash \dot{h} \in \prod_{\alpha < {\lambda}}{f}^{\alpha}_{\theta}(\dot{g}(\alpha))$. By the proof of (i), we may assume that $X(\alpha) \subseteq X'(\alpha)$ for all $\alpha > \beta_b$. For each $b \in {\rm LP} = \{b \mid \exists Y(\langle b,Y\rangle \leq \langle a,X \rangle)\}$ define $Y_{b}$ and $\eta_b < \kappa$ as follows. If $\beta_b > \beta_a$, by Lemma~\ref{magidorprikrycondition} and $f_{\theta}^{\beta_b}(b(\beta_b)) < \kappa$, there is a $Y_b$ such that
\begin{itemize}
 \item $\langle b,Y_b\rangle \leq \langle a,X\rangle$.
 \item if $\langle c,Z\rangle \leq \langle b,Y_{b}\rangle$ forces $\dot{h}(\beta_b) = {\zeta}$, then $\langle c,Z\rangle_{\beta_b}^{\frown} \langle b,Y_{b}\rangle^{\beta_b}\Vdash \dot{h}({\beta_b}) = {\zeta}$.
\end{itemize}
   Define $\eta_{b}$ by
\begin{center}
 $\eta_{b} = \sup\{\zeta + 1< f^{\beta_b}_{\theta}(b(\beta_b))\mid \exists p \in \mathbb{M}_{\beta_b} / \langle b,Y_{b}\rangle_{\beta_b}(p^{\frown}\langle b,Y_{b}\rangle^{\beta_b} \Vdash \dot{h}({\beta_b}) = {\zeta})\}$.
\end{center}
   Then,
\begin{center}
 $\langle b,Y_{b} \rangle \Vdash \dot{h}({\beta_b}) < \eta_b$.
\end{center}
   For $b \in \mathrm{LP}$ with $\beta_b \leq \beta_a$, $Y_b = X$ and $\eta_b = 0$.

   For each $b \in \mathrm{LP}$, since $\mathbb{M}_{\beta_b} / \langle b,Y_{b}\rangle_{\beta_b}$ has the $b(\beta_b)^{+}$-c.c., $\eta_{b} < f^{\beta_b}_{\theta}(b(\beta_b))$. For every $\alpha > \alpha_b$, define $h^{\alpha}(\xi) = \sup \{\eta_{b}< f^{\alpha}_{\theta}(\xi) \mid b \in {\rm LP} \land b(\beta_b) = \xi \land \beta_b = \alpha\}$. Because of $|\{b \in {\rm LP} \mid b(\beta_b)= \xi \land \beta_b = \alpha\}| = |\alpha| \cdot |\xi|$, we have $h^\alpha(\xi) < f_\theta^\alpha(\xi)$ for every $\xi \in X(\alpha)$. Let $\gamma = \sup_{\alpha > \beta_b}[h^{\alpha}]_{U_{\alpha}} + 1< \theta$. By Lemma~\ref{magidordiagonallemma} and the proof of (i), there is an extension $\langle a,Z \rangle \leq \langle a,X \rangle$ such that
\begin{itemize}
 \item every extension of $\langle b, Y \rangle$ is compatible with $\langle b,Y_{b}\rangle$ if $\langle b,Y \rangle \leq \langle a,Z \rangle$.
 \item $\forall \alpha > \beta_a\forall \xi \in Z(\alpha)(h^{\alpha}(\xi) < f_{\gamma}^{\alpha}(\xi))$.
\end{itemize}

   Lastly, we claim that $\langle a,Z \rangle \Vdash \dot{h}(\alpha) < \dot{f}_{\gamma}(\alpha)$ for all $\alpha > \beta_a$.
   Let $\langle {b,Y}\rangle \leq \langle {a,Z} \rangle$ and $\alpha > \beta_a$ be arbitrary. Extending $\langle b,Y\rangle$ we may assume that $\alpha \in \operatorname{dom}(b) \setminus (\beta_a + 1)$. Now, we can find $\langle c,Y'\rangle$ such that $\langle b,Y\rangle \leq \langle c,Y'\rangle \leq \langle a,Z \rangle$ and $\beta_c = \alpha$. By the certain property of $\langle a,Z \rangle$, $\langle{c,Y_{c}}\rangle$ and $\langle b,Y\rangle$ have a common extension forcing $\dot{h}(\alpha) < \eta_c < h^{\alpha}(c(\alpha)) < f^{\alpha}_{\gamma}(c(\alpha)) = \dot{f}_{\gamma}(\alpha)$, as desired.
  \end{proof}
     Theorem~\ref{magidormaintheorem} enables us to generalize Theorem~\ref{pcfinc}
     as follows, including the case of uncountable cofinality.
  \begin{thm}\label{pcfinc2}
   Suppose $\langle\kappa_{i} \mid i < \omega \rangle$ is an increasing sequence of supercompact cardinals greater than a regular cardinal $\lambda$. Then in some forcing extension the following hold:
   \begin{enumerate}
    \item[$(1)$] $\kappa_{0}$ is a singular cardinal of cofinality $\lambda$.
    \item[$(2)$] $\operatorname{pcf}^{n}(\kappa_{0} \cap {\rm Reg}) \subsetneq \operatorname{pcf}^{n+1}(\kappa_{0} \cap {\rm Reg})$ for all $n< \omega$.
    \item[$(3)$] $\lambda^{+} \cap {\rm Reg} = (\lambda^{+} \cap {\rm Reg})^{V}$.
   \end{enumerate}
  \end{thm}

  \if0
  We next take up the diagonal Prikry forcing originally introduced by Gitik--Sharon  \cite{GS}. We refer the reader to \cite{NU} for the analogues of Lemmas \ref{diagonallemma} and \ref{prikrycondition}, which imply the following

    \begin{thm}
    The diagonal Prikry forcing over $\kappa$ forces that $\pcf{{\kappa}\cap {\rm Reg}} = (2^{{\kappa}})^{+}\cap {\rm Reg}$.
   \end{thm}
   \fi

   For $A \subseteq {\rm Reg}$, define
   \begin{center}
    $\operatorname{pcf}^{\alpha}(A) = \begin{cases}A & \alpha = 0 \\ \pcf{\operatorname{pcf}^{\beta}(A)} & \alpha = \beta + 1 \\ \bigcup_{\beta < \alpha} \operatorname{pcf}^{\beta}(A) & \alpha \in {\rm Lim}
				      \end{cases}$
  \end{center}
Note that GCH implies $\pcf{\pcf{A}} = \pcf{A}$ for every $A \subseteq {\rm Reg}$.
By Theorem~\ref{pcfinc2}, it is consistent that
${\langle\operatorname{pcf}^{n}(A) \mid n < \omega\rangle}$ is $\subsetneq$-increasing for some $A\subseteq {\rm Reg}$. We conclude this paper with the following
\begin{ques}
   Is it a theorem of {\rm ZFC} that for every $A \subseteq {\rm Reg}$ there is an $\alpha$ such that
   $\operatorname{pcf}^{\alpha + 1}(A) = \operatorname{pcf}^{\alpha}(A)$?
  \end{ques}

   \bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


After discovering your new vegan food groups you may be wondering how to put it all together. This graphic of a vegan food pyramid is a recommendation for the standard vegan diet. It is all about empowering you to create your own nutritious vegan mealplans.


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You are currently searching within the Bipartisan Group of Senators Seek Answers from Health Care Providers and Insurers on Surprise Medical Bills section.
Bill Cassidy, M.D. United States Senator For Louisiana
Bipartisan Group of Senators Seek Answers from Health Care Providers and Insurers on Surprise Medical Bills
WASHINGTON—U.S. Senators Bill Cassidy, M.D. (R-LA), Michael Bennet (D-CO), Todd Young (R-IN), Tom Carper (D-DE), Lisa Murkowski (R-AK), and Maggie Hassan (D-NH) are seeking answers from health care providers and insurers as part of their ongoing effort to develop bipartisan legislation to end surprise medical bills.
The senators' latest questions for the health care industry seek more detailed information to supplement the input they received regarding draft legislation released in September 2018 by Senators Cassidy, Bennet, Young, Carper, Chuck Grassley (R-IA), and former Senator Claire McCaskill (D-MO).
"As we continue our bipartisan effort to lower health care costs and improve price transparency, we seek more detailed information in addition to what we have received thus far. Surprise medical billing is a complex problem, and crafting bipartisan, effective legislation to address it will require greater engagement from the private sector," write the senators. "We want to protect patients from costly surprise bills while preventing undue disruption in the health care system. To meet this goal, it is critical that we receive additional data and more complete feedback in order to refine and inform our legislative proposal."
The full text of the letter is below:
Dear Stakeholders:
As we continue our bipartisan effort to lower health care costs and improve price transparency, we seek more detailed information in addition to what we have received thus far. Surprise medical billing is a complex problem, and crafting bipartisan, effective legislation to address it will require greater engagement from the private sector. We want to protect patients from costly surprise bills while preventing undue disruption in the health care system. To meet this goal, it is critical that we receive additional data and more complete feedback in order to refine and inform our legislative proposal.
With this objective in mind, please send us the following information and data. In order to streamline the compilation of the information we receive, please respond in the following manner, to the extent possible:
With specific state-by-state data, as applicable, for the following states that have a balance billing law in effect: Alaska, California, Colorado, Florida, Maryland, New York, and Texas
An average across remaining states, for those states without any existing balance billing law
For questions pertaining to specific states, please feel free to return data in the manner specified in the questions themselves.
Please supply data and feedback on the following questions:
PLAN QUESTIONS
What do you currently pay for out-of-network care on average, broken down by plan type (e.g., HMO, PPOs, etc.), market type (e.g., individual, small group, large group, ASOs), and provider type, and how do these rates compare to Medicare rates, average in-network rates, and provider charges?
What percentage of your plans' premiums are currently attributable to the following specialty groups: emergency care (ER) physicians, radiologists, anesthesiologists, pathologists, ambulance services, and laboratory services?
For state laws that tie reimbursement to the 80th percentile of provider charges or another percentile of charges, please provide data and analysis on how and by what amount these reimbursement levels have affected premiums in these states. How has this affected contracting rates and charge rates? Can you provide the same information for states that have tied reimbursement to some multiple of Medicare rates and in-network negotiated rates?
Can you provide data and modeling to demonstrate the effect on premiums for the current draft text's proposed changes (e.g., median in-network rate or 125% of allowed in-network amount)? Can you also provide data and modeling to determine the effect on premiums if legislation were to use Medicare rates, as well as if it were to use 80th percentile of charges as payment benchmarks?
For states using independent dispute resolution processes to address balance billing, what has been the effect of the dispute resolution process on premiums and payment rates to providers, both in and out-of-network? How often, and by how much, have the state dispute resolutions processes resulted in payments amounting to greater than or equal to the 80th percentile of charges? In dispute resolution processes, which party has been successful more frequently, and under which scenarios? If federal legislation to address balance billing included a baseball-style dispute resolution process policy, what would be the best entity to handle the process of dispute resolution?
Do you have a process for identifying when providers send balance bills?
? If so, what is it? Do you also track how much is reimbursed to providers by insurers? If so, how do these amounts compare?
? If not, why not? Will you consider a process for identifying when balance bills are sent by providers in the future?
What specific recommendations do you have to facilitate network adequacy and encourage provider participation in health plan networks in the context of federal legislation to address surprise medical billing?
What role do you think hospitals should play in combatting surprise medical billing?
In your view, is there a state model that has worked particularly well at protecting patients from surprise medical billing? If so, why has it worked well? Please provide the details of this model, including its impact on contracting rates and out-of-network payment rates, and describe the data and policy rationale underlying this state legislation.
What share of plans that you administer (either directly or on behalf of another organization) have an out-of-network rate of beneficiary claims of 0-10%, 10-25%, 25-50%, or more than 50% for emergency room care?
What percent of balance bills sent by providers, that you are aware of, are more than $750?
PROVIDER QUESTIONS
What is the average out-of-network payment that your providers receive for emergency services? How does this compare to Medicare and charges, broken down by plan type and market? How does this differ by state?
How does the average out-of-network payment for ancillary providers compare with Medicare reimbursement and physician charges, again broken down by plan type and market?
According to an article published by the Health Care Cost Institute, emergency room spending per person has increased by 98% while overall emergency room utilization remained the same between 2009 and 2016.[1] How do you explain this trend?
What percentage of ER, radiology, anesthesiology, and pathology services are performed by providers that are part of outsourcing firms? For each of these specialties, what are the relative market shares for the large national staffing companies, local or regional physician groups, and hospital-staffed specialists? For providers employed by those firms, what percentage share the network status of the facility where they are practicing?
What percentage of amounts paid for overall emergency care, by both patients and payors, can be attributed to balance billing (dollar amounts and/or percentage amounts)? How about for other specialty departments (e.g., anesthesiology, radiology, pathology, etc.)? If possible, please provide data showing the amounts (or percentages of overall emergency care) paid for services by out-of-network providers at in-network facilities, as well as in-network providers at out-of-network facilities. If possible, please provide data to compare private versus public payments in these scenarios. Please also provide a breakdown of surprise medical bills attributable to each provider specialty.
In situations where the ED or ancillary physician is out-of-network but the facility is in-network, can you provide data to show how often a balance bill is sent to the patient?
What percentage of care provided in the emergency department results in bad debt from patients not paying their part of what is owed from care they received, from missed copayments, denied claims, or other means?
What specific recommendations do you have to facilitate in-network contracting between providers and plans in the context of federal legislation to address surprise medical billing?
What percentage of care provided by providers within each specialty is out-of-network? Broken down by each specialty, what share of providers are out-of-network for 0-10%, 10-25%, 25-50%, or more than 50% of the commercially insured patients (not including Medicaid managed care) they see?
Can you identify specific states where providers have a lower-than-average contracting rate?
What role do you think that hospitals should play in combatting surprise medical billing?
What percentage of balance bills are more than $750?
By providing comprehensive answers to these questions, you will inform our work on this issue and help bring relief from surprise medical bills to Americans in a timely manner. We respectfully request your response to Transparency@cassidy.senate.gov by Feb. 18, 2019. We look forward to reviewing your submissions.
[1] https://www.healthcostinstitute.org/blog/entry/er-spending-among-the-commercially-insured-continued-to-rise-in-2016-driven-by-the-price-and-use-of-high-severity-cases-2009-2016


How many times have you participated in a staff meeting and had a big idea shot down in an instant? Those of us who like to "play devil's advocate" can quickly become the enemy of creativity. While I believe wisdom is vital, a pattern of rushing to identify the holes in the process or the potential for failure can throw cold water on the visionary gifts in other staff members. The long term effect may be that they stop dreaming or sharing their ideas at all.
For this reason, a great church leadership team needs to build a "Why Not" culture.
"Why Not" people allow themselves to embrace new possibilities before countering with detailed questions. They know the importance of sharing in the enthusiasm of an idea before addressing tactical concerns.
"Why" people can drain the team's creativity by focusing on the potential pitfalls, the volume of work or satisfaction with the status quo. Every team needs realists, but focusing more on the "Why" than the "Why Not" can be perceived as negativity that hampers visionary thinking.
Allow yourself to dream before giving your immediate opinion of an idea.
Share someone's zeal and participate in the discussion before focusing on roadblocks.
For major decisions, make sure to give everyone involved time to pray individually after an idea is proposed.
Schedule a follow-up meeting and request a detailed plan. Sometimes an idea will prove its own flaws in the planning stage.
Refuse to let fear of failure win out over faith.


We love how the window graphics at Hot Chocolatier help showcase their brand and tell their story.
Next door to the Stillhouse, the Hot Chocolatier is doing its part to revitalize a Southside block that has long been vacant. Designer and co-founder, Brandon Buckner, asked us to brand their large storefront windows. These graphics are a great example of the big visual impact that can be achieved through simple design and cost effective materials.


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On iPlayer: King Kong (1933)
Taking a giant monkey to New York is up there with showering at a remote motel and visiting a Transylvanian castle on the list of things which established movie wisdom tells us not to do. It's strange to think of a time, then, in cinema's infancy, when there seemed nothing risky about taking a giant monkey to New York at all.
This is the classic tale of man vs nature as a film crew, led by the ambitious film director Carl Denham (Robert Armstrong) to a mysterious island in search of footage.
"These tickets cost me 20 bucks!"
In a sense this is a film about film. Denham's desire for a spectacle is what sends him to such a dangerous location, a pursuit that ultimately ends in destruction. Here is the contradiction at the heart of Kong: it criticises the unending pursuit of spectacle while creating a film based almost completely on the desire to see spectacle.
Similarly Denham, during his search for a woman to be in his film, says that "the public must have a pretty face to look at" before hiring/abducting the inexperienced Ann Darrow (Fay Wray). Wray is a terrible actor, who seems to be there primarily so the audience has a pretty face to look at as she lurches haplessly from disaster to disaster. For most of the film all she's required to do is scream, and even this sounds like a bad Olive Oil impression. Yet when she talks it's even worse. As Denham points out, the public just needs a pretty face.
It's not clear whether these self referential aspects are intended, but it plays very much like an early version of Wes Craven's Scream. Denham's belief that audiences require outlandish spectacle and attractive women seems reflective of his low opinion of the cinema-going public, even though this is exactly what Kong provides. Are the film's writers subtly berating their audience? Perhaps.
In terms of spectacle, however, it's very hard to argue with, as from about half an hour in it's essentially non stop action which still looks impressive now. It's easy to see how it would have been terrifying on its original release.
It's dated in its sexual politics and the presentation of the natives, although not compared to Peter Jackson's 2005 remake, where the natives were made into weird zombie people in a way that was actually more offensive. Admittedly the sexism is more overt here, as the ship's crew openly bemoan the presence of a woman on board, but the notion of Kong, who has been immune to the charms of island's natives, taking a liking to a pretty blond American girl is equally offensive in both incarnations.
In many ways this film feels hokey and old, but as the classic monster attack film it really can't be beaten.
King Kong is on iPlayer until 22 September.
Tags: 1930s, 1933, Alex Chafey, Ann Darrow, cinema, disaster, Fay Wray, film, films, iPlayer, King Kong, man vs nature, movie, movies, New York, Peter Jackson, Retro review, review
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2 responses to "On iPlayer: King Kong (1933)"
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I am trying to download richfaces-extension for Portet Bridge 3.1.1.Final. I am having this problem that looks like it's fixed in 3.1.1.Final.
I see 3.1.1.Final at the Downloads page, posted 11/27/12. But when I download the richfaces extension, I keep getting 3.1.0.Final (not 3.1.1.Final).
I was able to download 3.1.1.Final of the impl and api. Is there a 3.1.1 of the richfaces extension?
1. Re: Where to download Portlet Bridge 3.1.1.Final - Richfaces-Extension?
If you go to the downloads page again it should now be fixed.
Also, in the next day or so I'll be releasing 3.1.2.Final, which includes a few more fixes.
2. Re: Where to download Portlet Bridge 3.1.1.Final - Richfaces-Extension?
OK, thanks. I see 3.1.2.Final just came out. I'll download that. Are there release notes for 3.1.2?
3. Re: Where to download Portlet Bridge 3.1.1.Final - Richfaces-Extension?


text:
Okrzeszyn is een dorp in de Poolse woiwodschap Neder-Silezië. De plaats maakt deel uit van de gemeente Lubawka en telt 278 inwoners.

Verkeer en vervoer
 Station Okrzeszyn

Galerij

Plaats in Neder-Silezië

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text:
La Gaeltacht Corca Dhuibhne es troba a l'oest del final de la península de Dingle (Corca Dhuibhne) al Comtat de Kerry. Es tracta d'una àrea on predomina el gaèlic irlandès. S'estén des d'Abhainn an Scail a Dún Chaoin i An Clochán a An Daingean. Les principals viles de la zona són Abhainn an Scáil, Lios Póil, An Daingean, Ceann Trá, Dún Chaoin, Baile an Fheirtéaraigh, Baile na nGall i An Clochán. Hi ha 6.000-7.000 persones vivint a la zona dels quals 3.000 parlen irlandès.

L'estació de ràdio en irlandès RTE Raidió na Gaeltachta té uns estudis a Baile na nGall.

Aquesta és una llista de la població i el percentatge de parlants d'irlandès als districtes electorals:

 Cill Chuain (438) (79%)
 Cill Maoilcheadair (536) (77%)
 Dun Urlann (407) (77%)
 Marthain (238) (67%)
 Dun Chaoin (207) (63%)
 Ceann Tra (448) (55%)
 Na Beathacha (15) (46%)
 An Mhin Aird (373) (43%)
 Na Gleannta (1,419) (41%)
 Cinn Aird (357) (39%)
 Ce Bhreanainn (127) (36%)
 An Clochan (258) (32%)
 An Daingean (1,593) (24%)
 An Baile Dubh (131) (32%)

Gaeltacht Ui Rathach
És una àrea Gaeltacht poc poblada situada en la península d'Iveragh al sud de la de Dingle.

Llista d'EDs:
 Cathair Donall (97) (10%)
 Doire Fhionain (151) (13%)
 Trian Iarthach (126) (15%)
 An tImleach (319) (14%)
 An Baile Breac (64) (18%)
 Maistir Gaoithe (83) (20%)
 Doire Ianna (206) (21%)
 Ceannuigh (143) (26%)
 Loch Luioch (16) (25%)
 Baile an Sceilg (346) (30%)
 Toghroinn Fhionain (134) (31%)
 Na Beathacha (15) (46%)

Vegeu també
 Gaeltacht ciutat de Galway
 Gaeltacht Cois Fharraige
 Conamara Theas
 Illes Aran
 Joyce Country
 Gaoth Dobhair
 Na Rosa
 Cloch Cheann Fhaola
 Gaeltacht an Láir
 Gaeltacht Iorrais agus Acaill

Enllaços externs
 Gaeltacht Corca Dhuibhne definition

Gaeltacht


text:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">

<title>RegExValidator (esoco-business 1.3.0 API)</title>
<meta name="date" content="2018-10-29">
<link rel="stylesheet" type="text/css" href="../../../../stylesheet.css" title="Style">

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de.esoco.data.validate

Class RegExValidator

java.lang.Object

de.esoco.data.validate.RegExValidator

All Implemented Interfaces:

Validator<java.lang.String>, java.io.Serializable

public class RegExValidator
extends java.lang.Object
implements Validator<java.lang.String>

A validator that validates string values against a regular expression
 pattern.

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RegExValidator(java.lang.String sPattern)

Creates a new instance.

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boolean
equals(java.lang.Object rObj)

protected java.lang.String
getPattern()

Returns the regular expression pattern of this validator.

int
hashCode()

boolean
isValid(java.lang.String sValue)

Validates the string value by matching it against the regular expression
 pattern of this instance.

Methods inherited from class java.lang.Object

clone, finalize, getClass, notify, notifyAll, toString, wait, wait, wait

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RegExValidator

public RegExValidator(java.lang.String sPattern)

Creates a new instance.

Parameters:

sPattern - The regular expression pattern to validate against

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equals

public boolean equals(java.lang.Object rObj)

Overrides:

equals in class java.lang.Object

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Object.equals(Object)

hashCode

public int hashCode()

Overrides:

hashCode in class java.lang.Object

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Object.hashCode()

isValid

public boolean isValid(java.lang.String sValue)

Validates the string value by matching it against the regular expression
 pattern of this instance.

Specified by:

isValid in interface Validator<java.lang.String>

Parameters:

sValue - The value to validate

Returns:

TRUE if the value is valid according to this validator's rules

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protected final java.lang.String getPattern()

Returns the regular expression pattern of this validator.

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The regular expression pattern

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Prof. Cai Wei Chairman, School Administration Committee of NCIST and Director Professor Mr. Liu Zhongjian at North China Institute of Science & Technology and a team of esteemed scientist members at the University hosted a meeting with Chief Executive Officer and Global President of Gexcon Mr. Sturle H Pedersen and GM in Gexcon China Mr. Bin Xie in Beijing last week.
The meeting was focusing upon strengthen the working relationship between the two organization both in the R&D area, but also to establish a new exchange program for Master students and new PhD program. NCIST and Gexcon has already developed a close relationship with exchange and transfer of technology, expertise and knowhow in safety and risk management and both parties see a great opportunity to develop the relationship further.


text:
You are at:Home»older»KRCU Holding Fund Raising 'Power Morning'
KRCU Holding Fund Raising 'Power Morning'
By southeast on	 January 1, 1990 older
CAPE GIRARDEAU, Mo., Nov. 15, 2005 — KRCU 90.9FM at Southeast Missouri State University will hold a "Power Morning" of fund raising from 6 to 9 a.m. on Friday, Nov. 18.
The event's goal is to raise $10,000 in three hours.
All listeners who call in with a pledge, will have it matched dollar for dollar thanks to a group of concerned KRCU listeners.
"New or renewing members will have their pledge matched," said KRCU General Manager Dan Woods.
The morning will also include giveaways — three complete turkey dinnersfrom Schnucks and KRCU sweatshirts.
"It will be a fun, fund-raising event," Woods said.
For more information, call (573) 651-5070 or visit krcu.org.


The USB Payment Card Reader is a kind of transfer payment tool used for bank equipment or personal transfer payments. It includes a high performance microcontroller (MCU), LCD display, Keypad, payment card slot, Secure Access Module (SAM), Tamper detection and USB2.0 interface. The USB Payment Card Reader needs to pass both Europay/Master Card/ Visa (EMV) International Organizations standard EMV 2000 and China PBOC3.0 of Level-1 contact approval certification. Since the Nano120 series MCU has passed EMV 2000 and China PBOC3.0 of Level-1 certification, it is very suitable for USB Payment Card Reader applications.
The Nano120 series provides high performance and low cost ultra-low power 32-bit ARM® Cortex™-M0 core, runs up to 42 MHz, and supports 32-bit hardware multiplier, Low operating voltage 1.8V to 3.6V, Ultra-low power operating mode (< 200uA/MHz), Power-down mode (RTC, RAM retention< 2.5uA), Deep Power-down mode, RAM retention (< 1uA), Flash ROM up to 128 KB and RAM up to 16 KB, high integrated RTC, 12-bit ADC, 2 MSPS conversion rate, internal RC 12 MHz OSC, and USB2.0 to reduce system BOM cost, rich peripheral I²Cx2 , SPIx3, UARTx5,ISO7816-3x3, Tamper detection ,Anti-theft lock code bit, 96-bit Unique ID and 128-bit Unique customer ID, and strong anti-jamming ability (ESD 8kV, EFT 4kV).


text:
WINES TASTING
ROSÈ WINES
Torre Quarto winery was built by the French Dukes de la Rochefoucauld, in 1847. They had large estates in southern Italy, and the one in Cerignola was their greatest one. It covered an area of about 5000 hectares of which 3000 were vineyards.
That's why they built their winery in the area surrounding Cerignola. It was called "Quarto Ducale" (that means "duke quarter"), because the estates was a quarter of the Cerignola district. Here the Dukes de la Rochefoucauld produced red wine that they sent to France, where it was used for blending and bottled to be sold as French wine. They had a private railway line to link their winery to the nearest railway station, to send their wine to France regularty.
The wine was aged in French oak casks, over 150 of these huge casks (containing each about 290 hl) are still kept the cellars of Torre Quarto. In the 1930s the Belgian family Solvay Bautier became owner of this winery and renamed it "Torre Quarto", after the tower which part of the building, which was one, a schnapps distillery. Torre Quarto began marketing and selling the wine in Italy and all over the world and doing so, were among the first wine producers to dissolve the belief that wines from Puglia were used for blending wine only. Passing the time the name "Torre Quarto" became synonym of quality; in fact it was receveid a lot of rewards, medals and certificates for its excellence and it is also know as one of the
most famous wines from Puglia. Nowadays the still owner of the winery, attorney Angela Tarantino, thanks to family tradition producing excellent quality wines.
© 2013 Torrequarto Cantine tutti i diritti riservati / credits


text:
Events in Mallorca on 31 May 2021
Discover & book the top Mallorca events
There are loads of events in Mallorca organised year round. Find loads of top tips and local insights in our Mallorca Main Events.
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Deia International Music Festival 2021 3 June 2021, All day
TotalTri Mallorca International Triathlon 2021, Cala Millor 6 June 2021, All day
Origen Fest 2021, Palma de Mallorca 6 June 2021, 15:00 - 00:00
ATP Mallorca Tennis Championships 2021, Santa Ponsa 19 June 2021, All day
International Yoga Day at Nikki Beach, Calvia 21 June 2021, 11:00 - 20:00
Sant Pere Festival 2021, Port d'Alcudia 22 June 2021, All day
Fiesta de Sant Joan, Mallorca 23 June 2021, All day
San Juan at Nikki Beach, Calvia 23 June 2021, 20:00 - 00:00
Rockin Matxin Festival in Porto Colom 26 June 2021, All day
Friday Like You Mean It!, Magaluf 2 July 2021, 11:00 - 20:00
4th of July Party at Nikki Beach, Calvia 4 July 2021, 11:00 - 20:00
Make It Happy, Magaluf 10 July 2021, 11:00 - 20:00
Amazing Sunday's Brunch, Magaluf 11 July 2021, 12:00 - 16:00
Friday Like You Mean It!, Magaluf 16 July 2021, 11:00 - 20:00
Reggaeton Beach Festival 2021, Inca 17 July 2021, All day
It's The Weekend, Magaluf 17 July 2021, 11:00 - 20:00
International Tequila Day Celebration, Magaluf 24 July 2021, 11:00 - 20:00