If we can prove that  $\mathsf{L}=\mathsf{P}$, does it imply that $\mathsf{NL}=\mathsf{NP}$ ? I thought it is the case, but I cannot prove it (also for the converse).
Implications between $\mathsf{L}=\mathsf{P}$ and $\mathsf{NL}=\mathsf{NP}$?
cc.complexity theory;complexity classes;nondeterminism
No. It is possible that L=P and that P != NP which implies that NL != NP since NL is contained in P.