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Apr 21, 2021 at 19:38 comment added Bjørn Kjos-Hanssen OK, sure $\, \, \, \, \, $
Apr 21, 2021 at 19:28 comment added Vladimir Reshetnikov Yes, that version makes sense. But there are subtle issues with those definitions: the outcome depends on which exact representation for each recursive ordinal you choose at each step. See mathoverflow.net/q/67214/9550, mathoverflow.net/a/171723/9550.
Apr 21, 2021 at 19:17 comment added Bjørn Kjos-Hanssen Ah... I guess I meant something like T_1= ZFC+Con ZFC, then T_2 = T_1+Con T_1 ...
Apr 21, 2021 at 18:35 comment added Vladimir Reshetnikov Does ${\sf ZFC}\not\vdash\alpha_n$ represent “${\sf ZFC}$ cannot prove $\alpha_n$”? If so, then $\alpha_1$ represents “${\sf ZFC}$ cannot prove $\operatorname{Con}\left({\sf ZFC}\right)$”, right? Isn’t this statement logically equivalent in ${\sf ZFC}$ (or even ${\sf PA}$) to just $\operatorname{Con}\left({\sf ZFC}\right)$, i.e. $\alpha_0$? And, hence, all $\alpha_\beta$ are equivalent to $\alpha_0$ as well, no?
Mar 8, 2014 at 16:31 history answered Bjørn Kjos-Hanssen CC BY-SA 3.0