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    Carmelics

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

    It is not the case that The Russell class R cannot be a member of any class (i.e., R must be a proper class).

    ?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.In paraconsistent set theories (e.g., Routley's dialectical sets), R can consistently belong to itself and to a class without deriving explosion.
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    • 2.If contradictions do not entail all propositions, the inference from R∈R ↔ R∉R to R's exclusion from all classes is not logically compelled.
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    • 3.Therefore the proper-class conclusion rests on classical logic's ex contradictione quodlibet, which is a contested assumption, not a logical necessity.
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    Reason for 2 of 2
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    • 1.Quine's NF (New Foundations) permits a universal set V and stratified comprehension, dissolving the distinction between sets and proper classes entirely.
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    • 2.If R can be reconstructed under stratified comprehension as a legitimate set within NF, then labeling R a 'proper class' reflects one axiomatic choice, not an ontological necessity.
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    • 3.The von Neumann axiom invoked in P2 is specific to NBG/MK class theories and cannot ground a theory-neutral claim that R must be a proper class.
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    Reasons Against

    1 perspective
    Reason against
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    • 1.Sets are defined as members, while non-members are labeled 'proper classes'.
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    • 2.If R is assumed to be an element of a class A, then by one of von Neumann's axioms R is not equivalent to V.
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    • 3.R is equivalent to V.
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