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Surely “uncomputable” here means “uncomputable in ZFC”, no? The fact that you’ve found it in a stronger theory doesn’t matter, because the value now depends on
by lilyball 5y ago
Surely “uncomputable” here means “uncomputable in ZFC”, no? The fact that you’ve found it in a stronger theory doesn’t matter, because the value now depends on that stronger theory and may be wrong in ZFC.
- jimhefferon 5y agoWhat's been shown is that, where the number is n, the fact that it equals n cannot be proved in ZFC. Of course it can be proved in a theory with more axioms. In particular, take ZFC and add the statement as an axiom.
- dandanua 5y ago> the value now depends on that stronger theory The value of every BB(n) is well-defined in ZFC. It can't change if we consider a stronger (and consistent) theory than ZFC. Proving that BB(n) equals to a concrete integer is a different matter – it's either possible or not within your theory.
- lilyball 5y agoWhen I say "the value now depends" I mean the value you computed in that stronger theory is not computable in the weaker one, and therefore the computation depends on the stronger theory, and therefore the value is only correct if the stronger theory is consistent.
- SilasX 5y agoNot when you clarify that you mean "not computable at all" by adding the "at all", as the reddit poster did.