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From: herbzet on 1 Jul 2010 23:13


Aatu Koskensilta wrote:
> herbzet writes:
>
> > Yeh -- what *exactly* is being assumed has not been grindingly
> > specified at a low level. We're not there yet. Hopefully, we won't
> > have to go there.
>
> We have to! Here goes: a consistent Sigma-1 complete formal theory T
> can't prove for an inconsistent formal theory Q the claim "Q is
> consistent". This is because "Q is inconsistent" is a Sigma-1 statement,
> and so if true provable in T.

Oy ...
From: Aatu Koskensilta on 2 Jul 2010 09:41
herbzet <herbzet(a)gmail.com> writes:

> Aatu says, with certain mild(?) qualifications, no.

If X is Sigma-1 complete -- e.g. if X is an extension of Robinson
arithmetic, either directly or through an interpretation -- X can't
prove an inconsistent Y to be consistent without being inconsistent
itself.

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: herbzet on 3 Jul 2010 11:39


Aatu Koskensilta wrote:
> herbzet writes:
>
> > Aatu says, with certain mild(?) qualifications, no.
>
> If X is Sigma-1 complete -- e.g. if X is an extension of Robinson
> arithmetic, either directly or through an interpretation -- X can't
> prove an inconsistent Y to be consistent without being inconsistent
> itself.

Because "Y is inconsistent" is a Sigma-1 sentence which X, being
Sigma-1 complete, would prove also, were it true -- ok.

And ZFC is an extension of Robinson arithmetic, unless there's
some persnickety reason why it isn't.

--
hz
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