Under the assumption that a PSPACE-hard problem A can be solved in polynomial time, is the following argumentation valid?

Since every problem in PSPACE can be reduced to A in polynomial time, every problem in NP which is a subset of PSPACE is included. That means that each Problem in NP is solvable in polynomial time and therefore P = NP.

Since coNP is a subset of PSPACE as well, we can argue that every problem in coNP can be reduced to A and therefore be solved in polynomial time as well, which means that P = coNP.

Therefore we can say that P = NP = coNP.

I don't see a problem with this argumentation at the moment, but it seems a bit too simplistic to me, so is this a valid way of arguing for the equality above?


Your argument is perfectly fine. However, you actually do not even have to argue with polynomial reduction. Since $\textsf{P} \subseteq \textsf{NP}, \textsf{coNP} \subseteq \textsf{PSPACE}$, under the assummption that there is a $\textsf{PSPACE}$-hard problem solvable in polynomial time (i.e. it is in $\mathsf{P}$), you also have $\textsf{PSPACE} \subseteq \mathsf{P}$. Now $\textsf{NP}, \textsf{coNP}, \textsf{PSPACE}$ are sandwiched in $\mathsf{P}$ and thus, equality holds for the $\subseteq$ relations above.

  • 2
    $\begingroup$ Answer any more complexity theory questions and you'll have to change your name to P/NP Nick. ;-) $\endgroup$ – David Richerby Jun 8 '18 at 13:50

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