Does NP not equal P?
Computational Complexity
2009-11-30 v2
Abstract
Stephen Cook posited SAT is NP-Complete in 1971. If SAT is NP-Complete then, as is generally accepted, any polynomial solution of it must also present a polynomial solution of all NP decision problems. It is here argued, however, that NP is not of necessity equivalent to P where it is shown that SAT is contained in P. This due to a paradox, of nature addressed by both Godel and Russell, in regards to the P-NP system in total.
Cite
@article{arxiv.cs/0209015,
title = {Does NP not equal P?},
author = {C. Sauerbier},
journal= {arXiv preprint arXiv:cs/0209015},
year = {2009}
}
Comments
withdrawn. It was a rediculously absurd notion