Is it known whether the implication $\mathsf{NEXP} = \Sigma_2 \implies \mathsf{NEXP} = \mathsf{MA}$ holds?

(The question is inspired by well-known $\mathsf{NEXP} \subseteq \mathsf{P/poly} \Leftrightarrow \mathsf{NEXP} = \mathsf{MA}$.)

  • 1
    $\begingroup$ Posted to cstheory $\endgroup$
    – sdcvvc
    Commented May 16, 2013 at 8:01
  • $\begingroup$ Did you get anywhere? If it's an open problem, an answer to that effect may be appropriate. $\endgroup$
    – Raphael
    Commented Apr 7, 2016 at 9:05
  • $\begingroup$ One could get a different (and as far as I know, still open) question by replacing $\Sigma_{\hspace{.03 in}2}$ with ​ SBP $\cap$ S$_{\hspace{.02 in}2}$P . ​ ​ ​ ​ $\endgroup$
    – user12859
    Commented Apr 7, 2016 at 11:52

1 Answer 1


This is not known.

$NEXP = \Sigma_{2}$ means that $NP$ and $NP^{NP}$ are different. But, $MA$ seems too weak to reach that high to catch $NP^{NP}$.


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