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    Home/Original/inverse
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    Inverse View

    It is not the case that Strong completeness holds for many-sorted logic: if Γ ⊨ φ then Γ ⊢ φ

    ?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.Henkin's completeness proof presupposes a classical, set-theoretically robust metatheory, yet many-sorted logic is often motivated by contexts (e.g., predicative or constructive foundations) where such metatheory is unavailable.
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    • 2.If the metatheory is weakened to predicative or intuitionistic set theory, the Henkin model construction fails because maximal consistent extensions require non-constructive choice principles.
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    • 3.Therefore, strong completeness for many-sorted logic holds only relative to a classical metatheory that many-sorted frameworks were partly designed to avoid presupposing.
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    Reason for 2 of 2
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    • 1.Strong completeness requires that every semantically valid inference over all many-sorted structures is provable, but Lindström's theorem shows any logic stronger than first-order that gains expressive power loses completeness or compactness.
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    • 2.Many-sorted logic, when sorts are allowed to range over proper classes or when sort predicates are defined second-order, exceeds the expressive boundary at which Henkin-style completeness is guaranteed.
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    • 3.The claim therefore conflates the well-behaved finitary fragment of many-sorted logic with richer formulations where incompleteness results analogous to those established by Lindström and later Barwise genuinely apply.
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    Reasons Against

    1 perspective
    Reason against
    ?
    • 1.If Γ ⊨ φ, then Γ ∪ {¬φ} is not satisfiable and has no model
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    • 2.By Henkin's theorem, if Γ ∪ {¬φ} has no model then Γ ∪ {¬φ} is contradictory
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    • 3.The calculus rules allow elimination of ¬φ to infer Γ ⊢ φ
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