Skip to content
Carmelics
TopicsThinkersChangesContributorsLoading account…

    Carmelics

    A reasoning platform. Break down any belief into clear reasons, explore both sides, and weigh the evidence honestly.

    Navigate

    • Topics
    • Search
    • Recent Changes
    • Contribute
    • How It Works
    • Glossary
    • Thinkers
    • Contributors
    • About
    • Statistics
    • Terms
    • Privacy

    Database

    Statements
    —
    Perspectives
    —
    Topics
    —

    Press ? for keyboard shortcuts

    LoyalLoyalJusticeJustice
    Made withinDC&Austin
    L is a proper subset of PSPACE and P is a proper subset o... — Carmelics
    Statements
    321,452
    Perspectives
    108,905
    Topics
    42
    Home/Modality & Possibility
    HistoryEditSee Inverse

    Part of a larger discussion

    Supports→At least one of the inclusions among L, P, NP, PSPACE, and EXP must be proper

    L is a proper subset of PSPACE and P is a proper subset of EXP

    Modality & PossibilityTruth & Knowledge
    ?Rate how convincing each reason is below to see the overall strength.

    No one has weighed in yet. Be the first to share reasons for or against this statement.

    Sign in or register to share your perspective on this statement.

    Topics

    Modality & PossibilityTruth & Knowledge

    Connections

    1 linked claim

    P is a proper subset of EXP

    Related

    At least one of the inclusions among L, P, NP, PSPACE, and EXP must be proper

    Next step

    Based on where you are in your exploration

    Browse more in Modality & Possibility
    Related propositions within the same area of thought.
    P is a proper subset of EXP
    The chain L ⊆ P ⊆ NP ⊆ PSPACE ⊆ EXP must include at least one proper inclusion t...

    Similar

    NP is a proper subset of NEXP96%P is a proper subset of EXP96%L is a proper subset of PSPACE96%It is widely believed that PH is a proper subset of PSPACE91%

    Source

    AI-extracted
    SEP: computational-complexity
    View source passageHide passage
    As \(\phi \in \sc{SAT}\) just in case a satisfying valuation exists, this is a correct method for deciding \(\sc{SAT}\) relative to conventions (i)–(iii) from above. This means that \(\sc{SAT}\) can be solved in polynomial time relative to \(\mathfrak{N}\). This example also illustrates why adding non-determinism to the original deterministic model \(\mathfrak{T}\) does not enlarge the class of decidable problems. [12] It is evident that if \(N\) has time complexity \(f(n)\), then \(T_N\) must

    Details

    Type
    premise
    Perspectives
    0 (0 for, 0 against)
    Edits
    1 edit

    Open for perspectives

    This idea is waiting for its first supporting or challenging perspective.

    Share the first perspective