[FOM] Con(ZFC) is trivial

Harvey Friedman hmflogic at gmail.com
Tue Jan 22 21:37:38 EST 2013

I never dismissed Con(ZFC) as not interesting, or trivial. On the
contrary, it is of great importance to have varied reasons for
believing it or its negation. The same assertions hold for Con(T),
where T is various well known strong extensions of ZFC.

As Goedel pointed out, it cannot be proved in ZFC, but might be proved
using extramathematical notions.

These Eight Supernatural Consistency Proofs are new ways of
accomplishing just this.

As you can see, they overshoot Con(ZFC). That's just the way it came
out. If I had an equally natural way of accomplishing this along these
lines, I would have included it, and the number would be greater than

Thanks for the typos.

Harvey Friedman

On Tue, Jan 22, 2013 at 12:09 PM,  <MartDowd at aol.com> wrote:
> In respnse to a recent posting of Vaugh Pratt, I have been reading over
> Harvey's manuscript "A Divine Consistency Proof for Mathematics", to
> see why he might have dismissed Con(ZFC) as not of much interest.
> ZFC+"there exists a Ramsey cardinal" is interpretable in T_5.  Ramsey
> cardinals are strongly inaccessible.  Thus, Con(ZFC) is a triviality.
> Ramsey cardinals have not been strigently justified, and mathematicians
> such as myself suspect that they never will be.  Axioms which have been
> stringently justified imply that the class of strongly inaccessible
> cardinals is stationary.  Thus, Con(ZFC) is a triviality even in this
> much more limited setting.
> I noted the following typos in Harvey's manuscript:
> p.10, proof of lemma 1.1.  Shouldn't the second sentence be
>  We want $\psi\leftrightarros\pi_1(\pi_2(\psi))E provable in S_2.
> p.17, definition 3.1.ii, shouldn't this be
>  $X\in K$ iff $X\A\notin K$
> p.17, theorem 3.1, second sentence; shouldn't this be
>  If $A\subseteq B$ and $A\in K$ then $B\in K$,
> Si99 seens to be omitted from the references.
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