[FOM] Is PA + ~Con(PA) a complete theory?
Andrew Polonsky
andrew.polonsky at gmail.com
Wed Jun 5 13:53:36 EDT 2013
Dear Richard,
Thanks for the clear and concise answer.
Andrew
On Wed, Jun 5, 2013 at 12:17 AM, Richard Heck <richard_heck at brown.edu>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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