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    Carmelics

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

    It is not the case that The theory PA^F, though technically inconsistent, is practically consistent because any proof of a contradiction in PA^F must be infeasibly long.

    ?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.A theory is either consistent or inconsistent; 'practical consistency' conflates epistemic inaccessibility of a proof with the absence of that proof.
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      Think about whether this reason is strong or weak

    • 2.Hilbert's distinction between ideal and real mathematics shows that relaxing consistency standards undermines the foundational role formal systems are meant to play.
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      Think about whether this reason is strong or weak

    • 3.If inconsistency is tolerated whenever its proof is computationally infeasible, then the boundary of 'acceptable' contradiction shifts arbitrarily with technological progress.
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      Think about whether this reason is strong or weak

    Reason for 2 of 2
    ?
    • 1.Feasibility bounds are theory-relative and agent-relative, so 'infeasibly long' lacks the objective standing required for a logical property like consistency.
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      Think about whether this reason is strong or weak

    • 2.Parikh's own work on feasible arithmetic shows that operationally motivated length restrictions require a fully rigorous alternative proof theory, not a reinterpretation of classical inconsistency.
      ?

      Think about whether this reason is strong or weak

    Reasons Against

    1 perspective
    Reason against
    ?
    • 1.PA^F restricts the induction schema to formulas not containing F(x), blocking the inductive form of the sorites argument.
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      Think about whether this reason is strong or weak

    • 2.The conditional sorites argument still applies in PA^F, so PA^F is inconsistent.
      ?

      Think about whether this reason is strong or weak

    • 3.For τ chosen as the super-exponential term 2↑2^k, any proof of a contradiction in PA^F must be on the order of 2^k steps long.
      ?

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