[FOM] Strong Hypotheses and the Theory of N

Robert Solovay solovay at gmail.com
Sat Mar 27 23:51:09 EDT 2010

I do not accept Richard Heck's amendment.

ZFC can not prove that "ZFC is $n$-consistent".

 Indeed, if ZFC is $n$-consistent, it is consistent. If ZFC could
prove its own consistency, it would be inconsistent by Godel's second
incompleteness theorem.

The proposed answer to Shipman's first question was intended to be a
simple system of true arithmetic statements that included all the
consequences of ZFC in arithmetic. It was not required for this first
question that they are precisely the theorems of arithmetic provable
in ZFC. My second answer was designed to achieve this latter goal.

--Bob Solovay
On Sat, Mar 27, 2010 at
6:53 PM, Richard Heck <rgheck at bobjweil.com> wrote:
> On 03/25/2010 12:36 AM, Robert Solovay wrote:
>> In a recent posting (March 20th) Joe Shipman asks the following question:
>> Would the problem be easier if I asked for a (presumably consistent)
>> “natural” set of arithmetical axioms which implied all the arithmetical
>> consequences of ZF but was not necessarily limited to them?
>> This relates to his prior question (in a posting on March 15th):
>> A related question: is there a natural way to represent the
>> "arithmetical content" of ZF by arithmetical axioms; in other words, a
>> natural decidable set of arithmetical statements which have the same
>> arithmetical consequences as ZF?
>> [snip]
>> Then the answer to the first of Shipman's questions that I propose is:
>> 1) The axioms of Peano Arithmetic;
>> 2) For each positive integer $n$, the arithmetical formulation of "ZFC
>> is $n$-consistent".
>> That all these arithmetical formulas are true can be proved, e. g., in
>> ZFC + "There is an inaccessible cardinal".
> And that EACH of these is true can be proved in ZFC itself, of course, since
> ZFC is reflexive.
> rh

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