What is wrong with this conditional proof of P=NP?
Suppose L=P. Let A be a problem in NP. By the verifier definition of NP, each positive solution to A has a witness that can be verified in polynomial time. Since P=L, the same solution can be verified in logarithmic space. Thus NP=NL.


You haven't shown that NP$\,=\,$NL. The only method you've derived to show that $x\in A$, is to nondeterministically "guess" a witness&nbsp;$y$ and use the deterministic logspace algorithm to verify it. However, $y$ itself may be polynomially long, whereas your NL machine can only guess logarithmically many bits: your NL machine doesn't have enough space to store the witness, so you don't have an NL algorithm.

Also, note that we already know that witnesses to NP-complete problems can be verified in a complexity class smaller than&nbsp;L. Fagin's theorem says that NP is exactly the set of problems that can be defined in the existential fragment of second-order logic. This is equivalent to saying that you can verify a polynomial-length witness using first-order logic. FO is strictly weaker than L, since first-order logic can't solve simple logspace problems like telling you if your witness string contains an even number of&nbsp;$1$s.