[FOM] I\Sigma_1 + Con(I\Sigma_1)
V.Yu. Shavrukov
v.yu.shavrukov at gmail.com
Sun Aug 25 20:51:50 EDT 2013
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
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