# What would be the consequences of PH=PSPACE?

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 $PH=PSPACE$?

• 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. Feb 21, 2014 at 17:23
• @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. Feb 21, 2014 at 17:54
• coNP=NP=PSPACE=#P=#Q=QSPACE regardless of how small the formalism is. +daniel GREs2380 Nov 9, 2021 at 14:45

$\mathsf{PH}$ 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), $\mathsf{PH} \subseteq \mathsf{\Sigma_k P}$.

• 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 ${\bf NP} ={\bf PSPACE}$? Feb 21, 2014 at 19:51
• @TayfunPay : $\:$ I don't see how such an implication could be shown. $\;\;\;\;$
– user6973
Feb 22, 2014 at 0:52
• @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 $\mathsf{PSPACE}$). Feb 22, 2014 at 20:55
• @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. Feb 23, 2014 at 23:13
• @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). 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.

• In fact, even NL≠NP follows because $NP^{NL\cap co-NL}=NP$. Oct 18, 2018 at 20:41
• @domotorp NL=coNL. Oct 8, 2021 at 0:35
• @ Mr. I know that. Oct 8, 2021 at 5:04

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:

1. the quantified Boolean formula problem is equivalent to existing of boolean functions for each variable with the quantifier ∃, which make the formula to be Tautology (Skolem functions).
2. size of boolean functions can be exponential, but it is considered in frames of full DNF. It allows to not construct precisely the boolean functions. Namely, iterations of ∀x∃y reduce to “long” conjunctions of separated ∀x∃y (XOR is the only issue here and it can be handled).

This way PSPASE-complete problem is solved using alternating Turing machine with only 4 alternations.

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.