Readers may recall that during the 2013 “peak-Elsevier” period, Elsevier made an interesting concession to the mathematical community — they released all their old mathematical content (“old” here means a rolling 4-year embargo) under a fairly permissive licence.

Unfortunately, sometime in the intervening period they have quietly withdrawn some of the rights they gave to that content. In particular, they no longer give the right to redistribute on non-commercial terms. Of course, the 2013 licence is no longer available on their website, but thankfully David Roberts saved a copy at https://plus.google.com/u/0/+DavidRoberts/posts/asYgXTq9Y2r. The critical sentence there is

“Users may access, download, copy, **display, redistribute, adapt**, translate, text mine and data mine the articles provided that: …”

The new licence, at https://www.elsevier.com/about/our-business/policies/open-access-licenses/elsevier-user-license now reads

“Users may access, download, copy, translate, text and data mine (**but may not redistribute, display or adapt**) the articles for non-commercial purposes provided that users: …”

I think this is pretty upsetting. The big publishers hold the copyright on our collective cultural heritage, and they can deny us access to the mathematical literature at a whim. The promise that we could redistribute on a non-commercial basis was a guarantee that we could preserve the literature. If this is to be taken away, I hope that mathematicians will go to war again.

Hopefully Elsevier will soon come out with a “oops, this was a mistake, those lawyers, you know?” but this will only happen if we get on their case.

What to do:

- Elsevier journal editors: please contact your Elsevier representations, and ask that the licence for the open archives be restored to what it was, to assure the mathematical community that we have ongoing access to the old literature.
- Elsevier referees and authors: please contact your journal editors, to ask them to contact Elsevier. If you are currently refereeing or submitting, please bring up this issue directly.
- Everyone: contact Elsevier, either by email or social media (twitter facebook google+).
- Happily, as we have a copy of the 2013 licence, all the Elsevier open mathematics archive up to 2009 is still available for non-commercial redistribution under their terms. You can find these at https://tqft.net/misc/elsevier-oa/.

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Thanks for the mirror! I wasn’t aware that there was such a mirror away from sci-hub.

Hilariously, the [Elsevier user license](https://www.elsevier.com/about/our-business/policies/open-access-licenses/elsevier-user-license) still says:

> Any translations, for which a prior translation agreement with Elsevier has not been established, must prominently display the statement: “This is an unofficial translation of an article that appeared in an Elsevier publication. Elsevier has not endorsed this translation.”

Given how much sense this makes without a right to redistribute, I guess they haven’t been very thorough :)

Note that the license changed to remove explicit permission to redistribute sometime before March 2015, which is when I could get a Wayback machine image of the licence page. We just didn’t notice for some time…

On 22 October 2013 the original license was still in effect:

https://web-beta.archive.org/web/20131022235944/http://www.elsevier.com/about/open-access/oa-and-elsevier/oa-license-policy#open-archive

Why don’t you share this on the Tim Gowers blog at https://gowers.wordpress.com/2016/11/29/time-for-elsexit/

(hi Nicoletta — Tim already knows about this latest development.)

I’ve asked on Twitter about this — see thread and responses from Elsevier

I think you are wrong about materials published before 2009 for the reason that copyright licenses travel with the

copies, not the original works. If you had downloaded a paper before 2013, then your download was subject to the old terms and you may redistribute it according to the license in effect at the time. If I download the same work today, my copy is subject to today’s license terms and I cannot redistribute it, notwithstanding the fact that my copy is word-for-word (or possibly even byte-for-byte) the same as yours.I see — you have a copy of the entire archive — well done! Then yes, your copy of the archive is subject to the license you got it under and copies of your copies can be redistributed.

P versus NP is considered one of the great open problems of science. This consists in knowing the answer of the following question: Is P equal to NP? This incognita was first mentioned in a letter written by John Nash to the National Security Agency in 1955. Since that date, all efforts to find a proof for this huge problem have failed.

I show a solution to that problem as follows:

Given a number x and a set S of n positive integers, MINIMUM is the problem of deciding whether x is the minimum of S. We can easily obtain an upper bound of n comparisons: find the minimum in the set and check whether the result is equal to x. Is this the best we can do? Yes, since we can obtain a lower bound of (n – 1) comparisons for the problem of determining the minimum and another obligatory comparison for checking whether that minimum is equal to x. A representation of a set S with n positive integers is a Boolean circuit C, such that C accepts the binary representation of a bit integer i if and only if i is in S. Given a positive integer x and a Boolean circuit C, we define SUCCINCT-MINIMUM as the problem of deciding whether x is the minimum bit integer which accepts C as input. For certain kind of SUCCINCT-MINIMUM instances, the input (x, C) is exponentially more succinct than the cardinality of the set S that represents C. Since we prove that SUCCINCT-MINIMUM is at least as hard as MINIMUM in order to the cardinality of S, then we could not decide every instance of SUCCINCT-MINIMUM in polynomial time. If some instance (x, C) is not in SUCCINCT-MINIMUM, then it would exist a positive integer y such that y < x and C accepts the bit integer y. Since we can evaluate whether C accepts the bit integer y in polynomial time and we have that y is polynomially bounded by x, then we can confirm SUCCINCT-MINIMUM is in coNP. If any single coNP problem cannot be solved in polynomial time, then P is not equal to coNP. Certainly, P = NP implies P = coNP because P is closed under complement, and therefore, we can conclude P is not equal to NP.

You could read the details in the link below…

https://hal.archives-ouvertes.fr/hal-01509423/document