[FOM] second-order logic once again

Aatu Koskensilta Aatu.Koskensilta at uta.fi
Tue Sep 4 06:59:31 EDT 2012

Quoting Robert Black <mongre at gmx.de>:

> Of course any r.e. set is determinate, so if the set of second-order  
> validities isn't determinate, it can't be r.e.. But surely that's  
> not what the incompleteness theorem says.

   I'm unsure what you mean by "incompleteness theorem" here.

> So what does it say? (In your answer, don't use expressions like  
> "any model" unless you're sure they aren't presupposing the  
> determinacy of second-order logic).

   Whatever one thinks of the determinacy of second-order logic in  
general, from pretty much any point of view on which it makes sense to  
speak of second-order validity at all, it's an unproblematic theorem  
that the set of second-order validities of the form "the axioms of  
second-order arithmetic imply A" for an arithmetical A is not  
recursively enumerable.

Aatu Koskensilta (aatu.koskensilta at uta.fi)

"Wovon man nicht sprechen kann, darüber muss man schweigen"
  - Ludwig Wittgenstein, Tractatus Logico-Philosophicus

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