Assume that $\mathrm{SAT} \in \mathrm{PSIZE}$, does it imply that $\mathrm{NP} = \mathrm{coNP}$ ?

I think that I've managed to show that if $\mathrm{SAT} \in \mathrm{PSIZE}$, then both $\mathrm{NP}$ and $\mathrm{coNP}$ are contained in $\mathrm{PSIZE}$, but I can't see how does help me. Any ideas ?

  • $\begingroup$ I wonder whether space-complexity would be more appropriate here. $\endgroup$
    – Raphael
    Commented Mar 24, 2014 at 22:44
  • $\begingroup$ Space complexity is something different. This is circuit complexity or non-uniform complexity. $\endgroup$ Commented Mar 25, 2014 at 1:22
  • $\begingroup$ could someone define or ref PSIZE? afaik this is not a real complexity class. $\endgroup$
    – vzn
    Commented Mar 25, 2014 at 20:24
  • $\begingroup$ @vzn: $L\in P/poly$ if and only if $L$ has polynomial circuit complexity ($L \in PSIZE$) $\endgroup$
    – Vor
    Commented Mar 25, 2014 at 20:47
  • $\begingroup$ @Vor ok was guessing that. do you know of a std ref pref book (paper ok if not) that uses it? $\endgroup$
    – vzn
    Commented Mar 25, 2014 at 20:51

1 Answer 1


If $SAT \in PSIZE$ then the polynomial hierarchy collapses to the second level: $\Sigma_2 = \Pi_2$ (see Karp-Lipton theorem ); but $NP=coNP$ (i.e. $\Sigma_1 = \Pi_1$) is stronger (the PH collapses to the first level).


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