Does NP not equal P?
Charles Sauerbier · arXiv (Cornell University) · 2002
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.