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    If PH = PSPACE, PH would have complete problems (since PS... — Carmelics
    Home/Modality & Possibility
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    Supports→It is widely believed that PH lacks complete problems

    If PH = PSPACE, PH would have complete problems (since PSPACE has complete problems)

    Modality & PossibilityTruth & Knowledge
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    Modality & PossibilityTruth & Knowledge

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    Believing PH = PSPACE leads to the counterintuitive conclusion that n-round veri...It is widely believed that PH lacks complete problemsThis collapse runs contrary to expectation, so PH = PSPACE is believed false, an...

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    If PH = PSPACE, then PH would have a complete problem (TWO PLAYER SAT)96%If P ⊊ NP, then NP-complete problems are not in P.88%If NP ≠ coNP, then NP-complete problems cannot be in coNP, because an ...86%If PH = PSPACE, then TWO PLAYER SAT would be complete for PH (since it...86%

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    SEP: computational-complexity
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    \(\sc{TWO}\ \sc{PLAYER}\ \sc{SAT}_n\) may be shown to be complete for the class \(\Sigma^P_n\) in the Polynomial Hierarchy. Note, however, as the value of \(n\) increases, we expect that it should become more difficult to decide membership in \(\sc{TWO}\ \sc{PLAYER}\ \sc{SAT}_n\) in much the same way that it appears to become more difficult to determine whether a given player has a winning strategy for increasingly long games of Go or chess. This observation provides part of the reason why it is

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