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    LoyalLoyalJusticeJustice
    Made withinDC&Austin
    Home/Original/inverse
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    Inverse View

    It is not the case that At least one of the inclusions L ⊆ NL, NL ⊆ P, P ⊆ NP, NP ⊆ PSPACE must be proper

    ?Set your confidence on the premises below to see your aggregate.

    Reasons For

    2 perspectives
    Reason for 1 of 2
    ?
    • 1.The proof that L ⊊ PSPACE relies on the space hierarchy theorem, which itself presupposes the robustness of the Turing machine model as the canonical computational substrate.
      ?

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    • 2.If computational complexity classes are model-relative rather than model-invariant, the 'proper containment' of L in PSPACE may reflect artifact of the Turing formalism rather than an absolute mathematical fact.
      ?

      Think about whether this reason is strong or weak

    • 3.Cobham-Edmonds thesis disputes that Turing machine resource bounds track an objective notion of feasibility, undermining the claim that hierarchy theorems reveal mind-independent structural truths.
      ?

      Think about whether this reason is strong or weak

    Reason for 2 of 2
    ?
    • 1.The disjunctive conclusion 'at least one inclusion is proper' is a theorem of classical logic applied to set-theoretic containment, but its epistemic status depends on whether mathematical existence is constructively or platonistically interpreted.
      ?

      Think about whether this reason is strong or weak

    • 2.Under strict constructivism in the tradition of Brouwer and Bishop, a proof that not-all-are-equalities does not constructively exhibit which specific inclusion is proper, leaving the disjunction epistemically inert.
      ?

      Think about whether this reason is strong or weak

    • 3.A disjunction lacking a constructive witness—where we cannot currently prove any single disjunct—fails to constitute genuine mathematical knowledge by intuitionist standards, even if classically valid.
      ?

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    Reasons Against

    1 perspective
    Reason against
    ?
    • 1.L is properly contained in PSPACE
      ?

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    • 2.If all four inclusions were equalities, L would equal PSPACE, contradicting the proper containment
      ?

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