Check my proof of P=NP for errors: Let X be NP. Then by https://t.co/Eea8PD8M1Q exists X’ in P such that X’(Z)=Y. But X’ in NP. Therefore applying it again, exists X’’ in P such that X’’(Y)=Z. So X’’ produces the same result as X in polynomial time. So, P=NP. #complexity

— Victor Porton (@victorporton) April 12, 2021