[FOM] I\Sigma_1 + Con(I\Sigma_1)

[V.Yu. Shavrukov]

On Aug 23, 2013, at 20:52, Richard Heck <richard_heck at brown.edu> wrote:

> Obviously, I\Sigma_1 + Con(I\Sigma_1) is a sub-theory of I\Sigma_2, since I\Sigma_2 proves Con(I\Sigma_1). I am guessing (a) that it is a proper sub-theory and (b) that it does not even interpret I\Sigma_2. Yes? References?

Yes.

When n > 0,  ISigma{n+1}  is equivalent to  ISigma{n} + RFN{Sigma_{n+2}}(ISigma{n})  [see Theorem 7 in

L.Beklemishev. Russ.Math.Surveys 60 (2005) 197–268  (not the earliest source, but a good survey/compendium),

and note that  RFN{Sigma_{n+2}}(EA)  is equivalent to  RFN{Sigma_{n+2}}(ISigma{n})  modulo  ISigma{n} ],

so  ISigma2 proves Con(ISigma1 + Con(ISigma1))  which  ISigma1 + Con(ISigma1)  doesn't (by 2G).  Neither can the latter interpret  ISigma1 + Con(ISigma1 + Con(ISigma1))  because no finitely axiomatized sequential theory  T  interprets  T + Con T — see Corollary 3.5 in

P.Pudlák. JSL 50 (1985) 423–441.


> Does the same hold for I\Sigma_n + Con(I\Sigma_{n+k})?


ISigma_n + Con(ISigma_{n+k})  interprets  ISigma_{n+k}  (Pudlák, Theorem 3.1) and fails to interpret  ISigma_{n+k+1}  by the same argument as before.

If k > 0, then  ISigma_n + Con(ISigma_{n+k})  is incomparable with  ISigma{n+1} — see Corollary 4.4(ii) in Beklemishev.


regards,
V.Shavrukov