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    For all natural numbers n, F proves the negation of Prf_F... — Carmelics
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    For all natural numbers n, F proves the negation of Prf_F(n, ⌜G_F⌝)

    Truth & Knowledge
    ?Rate how convincing each reason is below to see the overall strength.
    1 reason for
    2 reasons against

    Reasons For

    1 perspective
    Reason for
    ?
    • 1.No natural number n is the Gödel number of a proof of G_F in F
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    • 2.The proof relation Prf_F is strongly representable in F
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    • 3.Strong representability entails that unprovability of each instance is provable in F
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    Reasons Against

    2 perspectives
    Reason against 1 of 2
    ?
    • 1.Strong representability of Prf_F presupposes ω-consistency, yet F's ω-consistency cannot be established within F by Gödel's own second theorem.
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    • 2.Without verified ω-consistency, the inference from each numeral-instance unprovability to universal provability of negations conflates object-level and meta-level quantification.
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    • 3.Rosser's 1936 result shows weaker consistency suffices for incompleteness, but the specific claim about provable negations for all n retains sensitivity to the consistency assumption's justificatory source.
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    Reason against 2 of 2
    ?
    • 1.The claim quantifies over all natural numbers using the standard meta-theoretic ℕ, but ultrafinitists like Yessenin-Volpin deny that this totality is well-defined.
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    • 2.If 'all natural numbers' is not a determinate domain, the universal claim lacks a truth-condition independently of a background theory that itself requires justification.
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    Connections

    1 linked claim

    For all n, F proves the negation of Prf_F(n, ⌜G_F⌝)

    Related

    For all n, F proves the negation of Prf_F(n, ⌜G_F⌝)If 'all natural numbers' is not a determinate domain, the universal claim lacks ...No natural number n is the Gödel number of a proof of G_F in FRosser's 1936 result shows weaker consistency suffices for incompleteness, but t...
    +5 moreShow less
    Strong representability entails that unprovability of each instance is provable ...Strong representability of Prf_F presupposes ω-consistency, yet F's ω-consistenc...The claim quantifies over all natural numbers using the standard meta-theoretic ...The proof relation Prf_F is strongly representable in FWithout verified ω-consistency, the inference from each numeral-instance unprova...

    Similar

    For all n, F proves the negation of Prf_F(n, ⌜G_F⌝)96%F cannot prove G_F if F proves the negation of G_F, assuming F is cons...83%F proves the negation of G_F (assumption for conditional proof)81%If F proves both G_F and the negation of G_F, then F is simply inconsi...80%

    Source

    AI-extracted1/3 agreementValid
    SEP: goedel-incompleteness
    View source passageHide passage
    Assume that \(F \vdash \neg G_F\). Then \(F\) cannot prove \(G_F\), for otherwise \(F\) would be simply inconsistent. Hence no natural number \(\boldsymbol{n}\) is the Gödel number of a proof of \(G_F\), and because the proof relation is strongly representable, for all \(\boldsymbol{n}\), \(F \vdash \neg\Prf_F (\underline{n}, \ulcorner G_F\urcorner)\). If also \(F \vdash \exists x\Prf_F (x, \ulcorner G_F\urcorner)\), \(F\) is not 1-consistent, against the assumption. Therefore \(F\) does not pr
    Extraction notes

    Validity: Extracted via Max plan + API grounding/validity checks

    Details

    Type
    claim
    Perspectives
    3 (1 for, 2 against)
    Edits
    1 edit