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I rewrote the sentence to read as above. Formerly it said "any finite von Neumann algebra". In this context, "any" is ambiguous. Is it the case that "any finite von Neumann algebra" can be so embedded? If so, then "every" is true. But "asks whether any finite von Neumann algebra" can be so embedded could be construed as: "asks whether there is any finite von Neumann algebra" that can be so embedded. If there is one, then the answer is "yes", even if "every" is not true. [[User:Michael Hardy|Michael Hardy]] ([[User talk:Michael Hardy|talk]]) 01:15, 30 November 2009 (UTC)
== This is proven to be false as MIP*=RE ==
I have no idea about this topic, but as stated on https://www.scottaaronson.com/blog/?p=4512:
(5) The Connes embedding conjecture, a central conjecture from the theory of operator algebras dating back to the 1970s, is false.
The paper proving this is https://arxiv.org/abs/2001.04383.
I guess you need to adjust this article.
Coming form this article BTW (German): https://www.spektrum.de/news/sind-quantencomputer-wirklich-besser-als-klassische-computer/1572490
--[[User:Rugk|rugk]] ([[User talk:Rugk|talk]]) 09:25, 27 January 2020 (UTC)
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