Timeline for Transfinitely extending $\sf PA$ — can we get stronger than $\sf ZFC$?
Current License: CC BY-SA 3.0
8 events
when toggle format | what | by | license | comment | |
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Jun 16, 2014 at 15:36 | vote | accept | Vladimir Reshetnikov | ||
Jun 13, 2014 at 15:01 | comment | added | Noah Schweber | Emil, that's nice - I knew $\Sigma^0_1$ soundness was necessary, but I didn't know it was sufficient. | |
Jun 13, 2014 at 9:48 | comment | added | Emil Jeřábek | ... the uniform reflection principle along an ordinal is, wrt $\Pi^0_1$ consequences, equivalent to iterating consistency along a suitably longer ordinal. Beklemishev’s papers such as sciencedirect.com/science/article/pii/0168007295000074 are highly relevant on these matters. | |
Jun 13, 2014 at 9:43 | comment | added | Emil Jeřábek | @VladimirReshetnikov: Yes. @ NoahS: (Assuming something like PA as a metatheory.) As every true $\Pi^0_1$ sentence is provable along some path, this obviously implies that PA is $\Sigma^0_1$-sound. Less obviously, the converse also holds. For iterated local reflection principle, this follows easily enough from an exercise in provability logic showing that if $T+\mathrm{Rfn}_T$ proves a $\Sigma^0_1$-sentence $\phi$, then $T$ proves $\Box^n_T\phi$ for some $n$. For iterated uniform reflection principles, one needs the “fine-structure theorem” by Schmerl, which basically says that iterating ... | |
Jun 13, 2014 at 1:57 | comment | added | Vladimir Reshetnikov | I meant, if we formally assume axioms of $\sf ZFC$, can we prove that all paths yield only consistent theories? | |
Jun 13, 2014 at 1:36 | comment | added | Noah Schweber | Certainly not if $PA$ is true; I'm sure there's a weaker and less-Platonist criterion guaranteeing that all paths yield consistent theories, but I don't know one off the top of my head. | |
Jun 13, 2014 at 0:33 | comment | added | Vladimir Reshetnikov | Can we get an inconsistent theory on at least one path? | |
Jun 13, 2014 at 0:20 | history | answered | Noah Schweber | CC BY-SA 3.0 |