# How powerful is PA+Con(PA)+Con(PA+Con(PA))… etc?

From what i remember from Godel encoding there was alot of freedom in how one chooses to expresses the statement Con(PA), my question is if one can classify all statements, or some subclass of all statements equivalent to Con(PA)?

And if we add Con(PA+Con(PA)) and Con(PA+Con(PA)+Con(PA+Con(PA))) etc we get alot of statements about polyonomials, which are quite central in mathematics, my question is if this new powerful theory with all possible formulations of Con(PA) etc have any uses in pure number theory or other mainstream mathematics, is there a connection here? Can any "interesting" mathematics be encoded as Con(PA) ?

Also, does the sequence of polynomials Con(PA), Con(PA+Con(PA)) etc, converge in any sense? Is there a limiting statement which is approached as its iterated towards infinity?

• I am not an expert in this subject, but I think that Con(PA) is simply a construction to prove that PA cannot prove its own consistency. So it seems to me that the answer to your question is no. – Peter Dec 30 '13 at 14:45
• Con(PA) says that PA is consistent, which is a reasonable thing to assume and thus add as axiom, and repeat. – user117658 Dec 30 '13 at 15:02
• To work with PA, it is anyway necessary to assume, PA is consistent. So I do not see what changes if Con(PA) is added to PA. If PA is consistent, then Con(PA) and PA is equivalent to PA. Or do I oversee something ? – Peter Dec 30 '13 at 15:08
• @tomasz They are not equiconsistent. – Andrés E. Caicedo Dec 30 '13 at 15:54
• $\mathsf{PA}+\mathrm{Con}(\mathsf{PA})$ is stronger: It proves the consistency of $\mathsf{PA}$, while $\mathsf{PA}$ cannot prove the consistency of $\mathsf{PA}+\mathrm{Con}(\mathsf{PA})$: There are nonstandard models of $\mathsf{PA}$ where "$\mathsf{PA}$ is inconsistent" holds. The theorem you recall is true. – Andrés E. Caicedo Dec 30 '13 at 16:20