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We know there are statements that are undecidable/independent of ZFC. Can there be a statement S, such that (ZFC $\not\vdash$ S and ZFC $\not\vdash$ ~S) is undecidable?

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    Can you define exactly what you mean by "$\mathrm{ZFC} \vdash S$ is undecidable"? It's meaningless to say that ask whether "$\mathrm{ZFC} \vdash S$" is independent, because you haven't asked what it's independent of.2011-01-12
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    Con(T) => (T $\not\vdash$ **(Con(T) => T $\not\vdash$ S $\land$ T $\not\vdash$ ~S)** $\land$ T $\not\vdash$ **(Con(T) $\land$ (T $\vdash$ S $\lor$ T $\vdash$ ~S))**)2011-01-14
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    T $\not\vdash$ (Con(T) $\land$ (T $\vdash$ S $\lor$ T $\vdash$ ~S)) is always true because T $\not\vdash$ Con(T).2011-01-14
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    Version 2: Con(T) => (Con(T + **(Con(T) => (Con(T + S) $\land$ Con(T + ~S)))**) $\land$ Con(T + **(Con(T) $\land$ (~Con(T + S) $\lor$ ~Con(T + ~S)))**))2011-01-14

4 Answers 4