[FOM] Is PA + ~Con(PA) a complete theory?

Arnon Avron aa at tau.ac.il
Sat Jun 8 11:49:03 EDT 2013

The equivalence of Con_T with Godel's sentence for T is
provable in T if T is an r.e. consistent extension of PA.

This implies that the answer to your question below is
no. Had PA + ~Con(PA) proved its own Goedel sentence
it would has proved its own consistency, and so the 
consistency of PA. This would mean that this theory 
is inconsistent, which is false.


On Tue, Jun 04, 2013 at 06:17:40PM -0400, Richard Heck wrote:
> On 06/04/2013 08:46 AM, Andrew Polonsky wrote:
> >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?
> By Rosser's version of the first incompleteness theorem, the Rosser
> sentence for this theory is not decidable in it, unless the theory
> is inconsistent. Did you have something else in mind?
> Here's an interesting question along these same lines: Does PA +
> ~Con(PA) proves its own Goedel sentence (as opposed to its own
> Rosser sentence)? Goedel's version of the first incompleteness
> theorem does not tell us, since this theory is omega-inconsistent.
> (This is one reason that Rosser's version is a real improvement over
> Goedel's.) And one might think it should since, as you note:
> >(The obvious candidate might be
> >Con(PA + Incon(PA))
> >However, the negation of the above statement can be derived from
> >an axiom.)
> and, typically, the Goedel sentence for a theory is provably
> equivalent, in that theory, to the statement that the theory is
> consistent. But I'm not sure if this holds for omega-inconsistent
> theories. And I'm not near the appropriate references at the moment
> to look this up.
> Richard
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