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    Made withinDC&Austin
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    Home/Original/inverse
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    Inverse View

    It is not the case that The Boolean Halting Problem (BHP) is in P only if P equals NP

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

    Reasons For

    2 perspectives
    Reason for 1 of 2
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    • 1.The claim presupposes that NP-completeness is a property stable across all formal systems, but Gödelian incompleteness suggests completeness classifications may be system-relative.
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    • 2.If the NP-completeness of BHP is provable only within systems that already assume P≠NP, the conditional 'BHP ∈ P → P=NP' is trivially vacuous rather than informative.
      ?

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    • 3.A conditional with an unprovably false antecedent (per Scott Aaronson's 'algebrization' barrier results) yields no epistemic traction about the actual structure of P versus NP.
      ?

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    Reason for 2 of 2
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    • 1.Polynomial-time reducibility as a canonical equivalence relation depends on the Church-Turing thesis applied to feasibility, which Wilfried Sieg and others treat as a substantive empirical assumption.
      ?

      Think about whether this reason is strong or weak

    • 2.If 'feasible computation' is not co-extensional with polynomial time across all physically realizable models, the closure property of NP under polynomial reductions loses its unconditional force.
      ?

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    • 3.The logical step from NP-completeness to P=NP therefore inherits the modal fragility of the feasibility thesis rather than standing as a purely formal entailment.
      ?

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    Reasons Against

    1 perspective
    Reason against
    ?
    • 1.BHP is NP-complete
      ?

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    • 2.NP is closed under polynomial-time reductions
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    • 3.If a problem is NP-complete and in P, then P = NP
      ?

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