Does NP not equal P?

TL;DR

This paper challenges the common assumption that NP equals P by arguing that SAT, traditionally NP-Complete, can be contained in P, suggesting a paradoxical distinction between NP and P.

Contribution

It presents a novel argument that SAT can be in P, which contradicts the widely held belief that NP-Complete problems are outside P, introducing a paradoxical perspective.

Findings

SAT is contained in P, contrary to common belief

A paradox in the P-NP system is identified

Implications for the P vs NP problem are discussed

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.