[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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