[FOM] Arithmetical soundness of ZFC
joeshipman at aol.com
Mon May 25 11:54:59 EDT 2009
Sure, just say "As N --> infinity, the fraction of grammatically
well-formed sentences of length N that are decidable in PA approaches
However, this would probably depend on the precise formulation of the
grammar. Almost all sufficiently large sentences will have Con(PA) as a
conjunct or disjunct, but that is not enough to render the sentence
In fact, for the most obvious and user-friendly formulations of PA,
the above is false because a nonzero fraction of sentences of PA begin
"((0=0) V (" and so are decidably true and a nonzero fraction begin
"((0=S(0)) & (" and so are decidably false. The actual probability that
a well-formed sentence will be decidable (in any reasonable notion of
probability) is likely to be equivalent to Chaitin's number Omega
(which is also coding-dependent) in a strong sense.
As for the arithmetical unsoundness of ZFC: if there is an Inaccessible
Cardinal k, then V(k) models ZFC and Th(V(k)) cannot include any false
arithmetical sentences so ZFC must be arithmetically sound. Therefore
any evidence for AFC's arithmetical unsoundness is also evidence there
are no inaccessibles. In fact the same argument works for any Standard
Model of ZFC because of the absoluteness of arithmetical sentences.
Therefore you won't be able to argue for the arithmetical unsoundness
of ZFC unless you start by assuming that there is no Standard Model.
I think that this is actually a reasonable assumption to make, although
I happen to believe that ZFC *is* arithmetically sound.
From: Timothy Y. Chow <tchow at alum.mit.edu>
Harvey Friedman <friedman at math.ohio-state.edu> wrote:
> WHAT WOULD EVIDENCE OF THE NON ARITHMETICAL SOUNDNESS OF ZFC LOOK
One example would be a proof of "ZFC is inconsistent" in ZFC.
Actually, what I think is more interesting is Nik Weaver's implicit
suggestion that a random sentence in the first-order language of
arithmetic is undecidable in PA. Is there any way to make this precise?
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