[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).

Robert
```