How Can ZFC/PA do much of Math - it Can't Even Prove PA isConsistent (EASY PROOF)
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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 |