[FOM] Is PA + ~Con(PA) a complete theory?
Knight, Joseph W
jknight7 at tulane.edu
Tue Jun 4 17:32:32 EDT 2013
PA is essentially undecidable, T consistently extends PA and is c.e., all complete c.e. theories are decidable, so T is incomplete.
Won't a Rosser sentence for T do the trick? This skirts the omega-inconsistency of T.
________________________________
From: fom-bounces at cs.nyu.edu on behalf of Andrew Polonsky
Sent: Tuesday, June 04, 2013 7:46 AM
To: fom at cs.nyu.edu
Subject: [FOM] Is PA + ~Con(PA) a complete theory?
Let T be the theory obtained by adding to PA the axiom
Incon(PA) = exists n. n is a code of a PA-derivation of Falsum
Since T is a consistent c.e. theory extending PA, one would expect to have undecidable propositions in it. Are there any known examples of such propositions?
(The obvious candidate might be
Con(PA + Incon(PA))
However, the negation of the above statement can be derived from an axiom.)
Cheers,
Andrew
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