[FOM] Arithmetical soundness of ZFC

[Timothy Y. Chow]

Joe Shipman wrote:
> I cannot think of any other possible way in which we could come to 
> believe in the non-arithmetical soundness of ZFC than a proof that ZFC 
> is omega-inconsistent.

Well, it's conceivable that we could come to believe in the truth of some 
arithmetic statement, either by heuristic arguments or (more exotically) 
by some mathematical reasoning that is not formalizable in ZFC, and 
continue to hold that belief firmly even in the face of a ZFC-disproof.
Not likely, I grant you, but conceivable.

Harvey Friedman wrote:
> But in order to prove in ZFC that "ZFC is inconsistent", we are going  
> to have to find an inconsistency in ZFC + "an inaccessible cardinal".  
> This is just a variant of the program to find inconsistencies.
> 
> So it remains unclear to me how "the finding arithmetical unsoundness  
> adventure" differs from "the finding inconsistencies adventure".

Suppose we discount my exotic suggestion above, so that indeed there is no 
essential difference between the two adventures.  So what?  Nik Weaver's 
main claim isn't that there's some essential difference between the two 
adventures.  It's simply that ZFC is probably consistent but probably not 
arithmetically sound.  If it turns out that he's essentially just saying 
that ZFC is probably consistent but ZFC + "an inaccessible cardinal" 
isn't, so be it.

Nik Weaver wrote:
> Tim Chow wrote:
> 
> > Apparently you believe that "experience" with ZFC can
> > legitimately give one confidence that ZFC is consistent
> > ... I would argue that we have analogous grounds for
> > believing that ZFC is arithmetically sound.  We've looked
> > hard for false theorems of ZFC and haven't found any.
> 
> How would we know?

So for instance we've looked for inconsistencies in ZFC + "inaccessible 
cardinal" and haven't found any, so the candidate "ZFC is inconsistent" 
(for a false theorem of ZFC) seems to be ruled out.

Besides, as Friedman has pointed out, the question "how would we know?" 
cuts both ways.  It's still totally unclear what grounds you have for 
claiming to know that ZFC is probably arithmetically unsound.

Tim