most recent 30 from http://mathoverflow.net 2013-05-25T07:10:04Z http://mathoverflow.net/feeds/user/11307 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48248/what-do-named-tricks-share/48263#48263 none 2010-12-04T07:42:46Z 2010-12-04T07:42:46Z

<p><a href="http://en.wikipedia.org/wiki/Rosser%27s_trick" rel="nofollow">http://en.wikipedia.org/wiki/Rosser%27s_trick</a></p> <p>"A technique is a trick that works twice"</p> <p>Note that Grothendieck never published his proof of the Grothendieck-Riemann-Roch theorem because he felt that the proof depended on an "astuce" (trick) rather than flowing naturally.</p>

http://mathoverflow.net/questions/3776/when-are-two-proofs-of-the-same-theorem-really-different-proofs/48258#48258 none 2010-12-04T06:53:12Z 2010-12-04T06:53:12Z

<p>This might be of interest: <a href="http://arxiv.org/pdf/cs.LO/0610123" rel="nofollow">http://arxiv.org/pdf/cs.LO/0610123</a></p> <ul> <li>Straßburger, Lutz (20 October 2006), "Proof Nets and the Identity of Proofs", Technical Report 6013, INRIA</li> </ul>

http://mathoverflow.net/questions/45844/hahn-banach-without-choice/48257#48257 none 2010-12-04T06:48:14Z 2010-12-04T06:48:14Z

<p><a href="http://en.wikipedia.org/wiki/Hahn-Banach_theorem#Relation_to_the_axiom_of_choice" rel="nofollow">http://en.wikipedia.org/wiki/Hahn-Banach_theorem#Relation_to_the_axiom_of_choice</a></p> <blockquote> <p>As mentioned earlier, the axiom of choice implies the Hahn–Banach theorem. The converse is not true. One way to see that is by noting that the ultrafilter lemma, which is strictly weaker than the axiom of choice, can be used to show the Hahn–Banach theorem, although the converse is not the case. The Hahn–Banach theorem can in fact be proved using even weaker hypotheses than the ultrafilter lemma.[4] For separable Banach spaces, Brown and Simpson proved that the Hahn–Banach theorem follows from WKL0, a weak subsystem of second-order arithmetic.[5]</p> </blockquote>