[FOM] second-order logic once again

Robert Black mongre at gmx.de
Mon Sep 3 17:27:14 EDT 2012

As everyone on this list knows, the completeness theorem fails for 
second-order logic.

Those of us who, like myself, are card-carrying second-orderists can say 
what this means: the set of second-order validities, *a perfectly 
well-determined set of formulae*, is not r.e. (indeed not even remotely: 
it's not definable in nth-order arithmetic for any n and so on and so on).

But suppose you're not a card-carrying second-orderist, so you don't 
think there's a perfectly determinate 'set of second-order validities'. 
(Perhaps you balk at the consequence that CH must have a truth-value.) 
How do you state the incompleteness theorem?

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.

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).


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