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    LoyalLoyalJusticeJustice
    Made withinDC&Austin
    Statements
    321,452
    Perspectives
    108,905
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    42
    Home/Original/inverse
    See Original
    Inverse View

    It is not the case that A function f(x) is in FP if and only if it is definable by a Σ^B_1-formula relative to which it is provably total in V^1

    ?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.Provability in V^1 is a syntactic property, while membership in FP is a semantic/extensional property about computational complexity.
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    • 2.The identification of syntactic derivability with extensional complexity classes conflates proof-theoretic and model-theoretic notions, a distinction Kreisel's 'unwinding' program treats as non-trivial.
      ?

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    • 3.A function could be provably total in V^1 under one formalization yet fail FP membership under a different but equivalent axiomatization, undermining biconditional stability.
      ?

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    Reason for 2 of 2
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    • 1.Zambella's characterization presupposes a fixed encoding of finite objects into binary strings, but Cook and Nguyen's 'Logical Foundations of Proof Complexity' (2010) acknowledges that Σ^B_1-definability is encoding-sensitive.
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    • 2.If the biconditional holds only relative to a privileged encoding scheme, it expresses a representation artifact rather than an intrinsic mathematical equivalence, undermining its status as a foundational characterization.
      ?

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    Reasons Against

    1 perspective
    Reason against
    ?
    • 1.The second-order theories V^i introduced by Zambella (1996) characterize levels of the polynomial hierarchy
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    • 2.At level i=1, V^1 characterizes FP via Σ^B_1-definability
      ?

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