A recent question (see Consequences of NP=PSPACE) asked for the "nasty" consequences of $NP=PSPACE$. The answers list quite a few collapse consequences, including $NP=coNP$ and others, providing plenty of reasons to believe $NP\neq PSPACE$.

What would be the consequences of the somewhat less dramatic collapse $PH=PSPACE$?

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    $\begingroup$ Am I the only person bored with the surge of "Consequences of $A=B$" questions these days? Granted, they can lead to interesting answers, but the question should at least ask for unexpected, surprising, etc. consequences. $\endgroup$
    – Sylvain
    Feb 21, 2014 at 17:23
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    $\begingroup$ @Sylvain: some of those are actually old questions that have risen from the dead because I added the "conditional-results" tag to them. You can then choose to ignore that tag to make such questions less visible to you. $\endgroup$ Feb 21, 2014 at 17:54
  • $\begingroup$ coNP=NP=PSPACE=#P=#Q=QSPACE regardless of how small the formalism is. +daniel GREs2380 $\endgroup$ Nov 9, 2021 at 14:45

4 Answers 4


$\mathsf{PH}$ collapses. A $\mathsf{PSPACE}$-complete problem must be in some level of $\mathsf{PH}$, say it's in $\mathsf{\Sigma_k P}$. Since it's $\mathsf{PSPACE}$-complete$=\mathsf{PH}$-complete (by assumption), $\mathsf{PH} \subseteq \mathsf{\Sigma_k P}$.

  • $\begingroup$ Isn't ${\bf PSPACE}$ closed under complement and low for itself? That is ${\bf PSPACE}$ = ${\bf PSPACE}^{\bf PSPACE}$ So wouldn't that imply ${\bf NP} = {\bf CoNP}$ and ${\bf NP} ={\bf PSPACE}$? $\endgroup$
    – Tayfun Pay
    Feb 21, 2014 at 19:51
  • $\begingroup$ @TayfunPay : $\:$ I don't see how such an implication could be shown. $\;\;\;\;$ $\endgroup$
    – user6973
    Feb 22, 2014 at 0:52
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    $\begingroup$ @TayfunPay: Note that $\mathsf{PH}$ - when considered as the single class defined by alternating poly-time TM's with $O(1)$ alternations - is also closed under complement and self-low (even without assuming it's equal to $\mathsf{PSPACE}$). $\endgroup$ Feb 22, 2014 at 20:55
  • $\begingroup$ @JoshuaGrochow Doesn't the existence of a PH-Complete imply that ${\bf PH}$ collapses? I remember something like this being in the old Papadimitriou book. I will check it out tonight. $\endgroup$
    – Tayfun Pay
    Feb 23, 2014 at 23:13
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    $\begingroup$ @TayfunPay: Yes, using the same proof as in my answer (but that doesn't, and seemingly can't, say what level it collapses to under that assumption). $\endgroup$ Feb 24, 2014 at 4:48

It would still imply major separations of complexity classes. For example, $\mathrm{LOGSPACE \neq NP}$ would follow. (If $\mathrm{LOGSPACE = NP}$ then $\mathrm{LOGSPACE = PH}$.)

Also $\mathrm{NP \subseteq P/poly}$ would imply $\mathrm{PSPACE = \Sigma_2 P}$ by Karp-Lipton. It follows that $\mathrm{NP}$ has polysize circuits if and only if $\mathrm{PSPACE}$ does. And of course, we'd have $\mathrm{P = NP}$ iff $\mathrm{P = PSPACE}$. In any case, the consequences of solving $\mathrm{NP}$ problems efficiently would be significantly increased.

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    $\begingroup$ In fact, even NL≠NP follows because $NP^{NL\cap co-NL}=NP$. $\endgroup$
    – domotorp
    Oct 18, 2018 at 20:41
  • $\begingroup$ @domotorp NL=coNL. $\endgroup$
    – Turbo
    Oct 8, 2021 at 0:35

As the answers point out, $PH=PSPACE$ would still have significant consequences, even though not as numerous and dramatic ones as $NP=PSPACE$.

Turning the issue on its head, it could be viewed as "empirical evidence" to support $NP\neq PH$. After all, if $NP=PH$, then the two statements ($PH=PSPACE$ and $NP=PSPACE$) must have the same consequences. As the second hypothesis has noticeably more and stronger known consequences, that can be viewed as empirical evidence to support that the left-hand sides in the equations must be different, that is $NP\neq PH$ (which, in turn, is equivalent to $NP\neq coNP$).


Using PSPACE-completeness of quantified boolean logic Valerii Sopin claimed to have obtained that PH = PSPACE, see https://arxiv.org/abs/1411.0628

The idea is the following: True quantified Boolean formula is indeed generalisation of the Boolean Satisfiability Problem, where determining of interpretation that satisfies a given Boolean formula is replaced by existence of Boolean functions that makes a given QBF to be tautology. Such functions are called the Skolem functions. The essential idea of the proof is to show that for any (fully) quantified Boolean formula ϕ we can obtain a formula ϕ′ which is in the forth level of the polynomial hierarchy, no more than polynomial in the size of a given ϕ, such that the truth of ϕ can be determined from the truth of ϕ′. The idea is to skolemize, and then use additional formulas from the second level of the polynomial hierarchy inside the skolemized prefix to enforce that the skolem variables indeed depend only on the universally quantified variables they are supposed to. However, some dependence is lost when the quantification is reversed. It is called "XOR issue" in the paper because the functional dependence can be expressed by means of an XOR formula. Thus, it is needed to locate these XORs. The last can be done locally for each leaf/ branch/ iteration (keep in mind the algebraic normal form (ANF)), i.e. in polynomial time, since all arguments are specified.

Relativization is defeated due to the well-known fact: PH = PSPACE iff second-order logic over finite structures gains no additional power from the addition of a transitive closure operator. Boolean algebra is finite. The exchanging quantifiers is possible only for finite possibilities of arguments. So, the theorems with oracles are not applicable since a random oracle is an arbitrary set. And that’s why Polynomial Hierarchy is infinite relative to a random oracle with probability 1.

The consequences can be found in the paper.

A related point to consider is L. Gordeew, E. H. Haeusler, Proof Compression and NP Versus PSPACE, Studia Logica, 107:1, 2019, 55–83; L. Gordeew, E. H. Haeusler, Proof Compression and NP Versus PSPACE II, Bulletin of the Section of Logic , 49:3, 2020, 213–230.


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