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    Quantifying over all subsets naturally licenses the Axiom... — Carmelics
    Home/Philosophy of Language
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    Supports→The Axiom of Choice is innocent in the context of second-order logic and is generally accepted in the second-order logic literature.

    Quantifying over all subsets naturally licenses the Axiom of Choice for families of subsets of any cardinality.

    Modality & PossibilityPhilosophy of Language
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    Related propositions within the same area of thought.
    In set-theoretical terms, all subsets—definable or not—of a set of any cardinali...The Axiom of Choice is innocent in the context of second-order logic and is gene...The basic tenet of second-order logic is that all properties of elements of a fi...

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    In set-theoretical terms, all subsets—definable or not—of a set of any...80%If the power set of the set of all sets were a subset of the set of al...75%A set that contains everything would already include all its own subse...75%Every set has a cardinality strictly less than that of its power set (...74%

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    Let \(\theta_{\le}(P,R)\) be the formula \[ \exists F\left(\forall x\,\forall y\left( (F(x)=F(y)\to x=y) \land(P(x)\to R(F(x)) \right)\right). \] Now \(\mm\models_s\theta_\le(P,R)\) if and only if \(|s(P)|\le |s(R)|\). Let \(\theta_{\textrm{EQ}}(P,R)\) be the formula \(\theta_{{\le}}(P,R)\land \theta_{{\le}}(R,P)\). Now \(\mm\models_s\phi(P,R)\) if and only if \(|s(P)|=|s(R)|\). Let \(\theta'_{\textrm{EC}}(Y)\) be \[ \exists F\left( \forall x\,\forall y((F(x)=F(y)\to x=y)\land R(F(x)))

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