<p>It would still imply major separations of complexity classes. For example, <span class="math-container">$\mathsf{LOGSPACE \neq NP}$</span> would follow. (If <span class="math-container">$\mathsf{LOGSPACE = NP}$</span> then <span class="math-container">$\mathsf{LOGSPACE = PH}$</span>.)</p>
<p>Also <span class="math-container">$\mathsf{NP \subseteq P/poly}$</span> would imply <span class="math-container">$\mathsf{PSPACE = \Sigma_2 P}$</span> by Karp-Lipton. It follows that <span class="math-container">$\mathsf{NP}$</span> has polysize circuits if and only if <span class="math-container">$\mathsf{PSPACE}$</span> does. And of course, we'd have <span class="math-container">$\mathsf{P = NP}$</span> iff <span class="math-container">$\mathsf{P = PSPACE}$</span>. In any case, the consequences of solving <span class="math-container">$\mathsf{NP}$</span> problems efficiently would be significantly increased.</p>

CC BY-SA 4.0
2018-10-18T20:41:30.793Z
In fact, even NL≠NP follows because $NP^{NL\cap co-NL}=NP$.
domotorp
CC BY-SA 4.0
2021-10-08T00:35:37.793Z
@domotorp NL=coNL.
Turbo
CC BY-SA 4.0
2014-02-22T09:14:46.007Z
2024-07-21T23:53:05.260Z
2024-07-21T23:53:05.260Z
Ryan Williams
<p>$\mathsf{PH}$ 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), $\mathsf{PH} \subseteq \mathsf{\Sigma_k P}$.</p>

CC BY-SA 3.0
2014-02-21T19:51:38.807Z
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 ${\bf NP}  ={\bf PSPACE}$?
Tayfun Pay
CC BY-SA 3.0
2014-02-22T20:55:53.220Z
@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 $\mathsf{PSPACE}$).
Joshua Grochow
CC BY-SA 3.0
2014-02-23T23:13:49.563Z
@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.
Tayfun Pay
CC BY-SA 3.0
2014-02-24T04:48:02.397Z
@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).
Joshua Grochow
CC BY-SA 3.0
2014-02-21T15:19:41.110Z
2014-02-21T15:25:04.467Z
2014-02-21T15:25:04.467Z
Joshua Grochow
<p>As the answers point out,  <span class="math-container">$\mathsf{PH=PSPACE}$</span> would still have significant consequences, even though not as numerous and dramatic ones as <span class="math-container">$\mathsf{NP=PSPACE}$</span>.</p>
<p>Turning the issue  on its head, it could be viewed as &quot;empirical evidence&quot; to support <span class="math-container">$\mathsf{NP\neq PH}$</span>. After all, if <span class="math-container">$\mathsf{NP=PH}$</span>, then the two statements (<span class="math-container">$\mathsf{PH=PSPACE}$</span> and <span class="math-container">$\mathsf{NP=PSPACE}$</span>) 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 <span class="math-container">$\mathsf{NP\neq PH}$</span> (which, in turn, is equivalent to <span class="math-container">$\mathsf{NP\neq coNP}$</span>).</p>

CC BY-SA 4.0
2014-02-23T21:51:15.150Z
2024-07-21T23:52:31.923Z
2024-07-21T23:52:31.923Z
Andras Farago
<p>Using PSPACE-completeness of quantified boolean logic Valerii Sopin claimed to have obtained that PH = PSPACE, see <a href="https://arxiv.org/abs/1411.0628" rel="nofollow noreferrer">https://arxiv.org/abs/1411.0628</a></p>
<p>The idea is the following:
<em>True quantified Boolean formula is indeed a 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 fourth 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 &quot;XOR issue&quot; in the paper because the functional dependence can be expressed by means of an XOR formula. Thus, it is needed to locate these XORs, but there is no need to locate all chains with XORs: any chain includes a XOR of only two variables. The last can be done locally in each iteration (keep in mind the algebraic normal form (ANF)), when all arguments are specified.</em></p>
<p>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 exchange is possible due to finite possibilities for 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.</p>
<p>The consequences can be found in the paper.</p>
<p>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.</p>

CC BY-SA 4.0
2021-10-07T10:40:03.450Z
2022-10-03T15:18:01.843Z
2022-10-03T15:18:01.843Z
user63167
<p>A recent  question (see <a href="https://cstheory.stackexchange.com/questions/21026/consequences-of-np-pspace">Consequences of NP=PSPACE</a>) asked for the &quot;nasty&quot; consequences of <span class="math-container">$\mathsf{NP=PSPACE}$</span>. The answers list quite a few collapse consequences, including <span class="math-container">$\mathsf{NP=coNP}$</span> and others, providing plenty of reasons to believe <span class="math-container">$\mathsf{NP\neq PSPACE}$</span>.</p>
<p>What would be the consequences of the somewhat less dramatic collapse <span class="math-container">$\mathsf{PH=PSPACE}$</span>?</p>

CC BY-SA 3.0
2014-02-21T17:23:14.207Z
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.
Sylvain
CC BY-SA 3.0
2014-02-21T17:54:23.357Z
@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.
András Salamon
<cc.complexity-theory><complexity-classes><conditional-results>
What would be the consequences of $\mathsf{PH=PSPACE}$?
Andras Farago
# What would be the consequences of $\mathsf{PH=PSPACE}$?

(Asked by: username_0 on 2014-02-21)

A recent question (see [Consequences of NP=PSPACE](https://cstheory.stackexchange.com/questions/21026/consequences-of-np-pspace)) asked for the "nasty" consequences of $\mathsf{NP=PSPACE}$. The answers list quite a few collapse consequences, including $\mathsf{NP=coNP}$ and others, providing plenty of reasons to believe $\mathsf{NP\neq PSPACE}$.

What would be the consequences of the somewhat less dramatic collapse $\mathsf{PH=PSPACE}$?

**Tags:** cc.complexity-theory, complexity-classes, conditional-results

> **Comments:**
> **username_1**: 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.  
> **username_2**: @username_1: 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.  

---

## 4 answers

### Answer by username_6 on 2014-02-21. Score: 27

$\mathsf{PH}$ 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), $\mathsf{PH} \subseteq \mathsf{\Sigma\_k P}$.

> **Comments:**
> **username_7**: 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 ${\bf NP}  ={\bf PSPACE}$?  
> **username_6**: @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 $\mathsf{PSPACE}$).  
> **username_7**: @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.  
> **username_6**: @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).  

---

### Answer by username_3 on 2014-02-22. Score: 22

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

Also $\mathsf{NP \subseteq P/poly}$ would imply $\mathsf{PSPACE = \Sigma\_2 P}$ by Karp-Lipton. It follows that $\mathsf{NP}$ has polysize circuits if and only if $\mathsf{PSPACE}$ does. And of course, we'd have $\mathsf{P = NP}$ iff $\mathsf{P = PSPACE}$. In any case, the consequences of solving $\mathsf{NP}$ problems efficiently would be significantly increased.

> **Comments:**
> **username_4**: In fact, even NL≠NP follows because $NP^{NL\cap co-NL}=NP$.  
> **username_5**: @username_4 NL=coNL.  

---

### Answer by username_0 on 2014-02-23. Score: 3

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

Turning the issue on its head, it could be viewed as "empirical evidence" to support $\mathsf{NP\neq PH}$. After all, if $\mathsf{NP=PH}$, then the two statements ($\mathsf{PH=PSPACE}$ and $\mathsf{NP=PSPACE}$) 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 $\mathsf{NP\neq PH}$ (which, in turn, is equivalent to $\mathsf{NP\neq coNP}$).


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### Answer by username_8 on 2021-10-07. Score: -4

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 a 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 fourth 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, but there is no need to locate all chains with XORs: any chain includes a XOR of only two variables. The last can be done locally in each iteration (keep in mind the algebraic normal form (ANF)), when 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 exchange is possible due to finite possibilities for 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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