Timeline for Is strengthening Foundation in NBG sufficient to make it prove Con(ZFC)?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Mar 30 at 18:06 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Mar 29 at 22:35 | history | became hot network question | |||
| S Mar 29 at 19:37 | vote | accept | Zuhair Al-Johar | ||
| Mar 29 at 19:35 | comment | added | Joel David Hamkins | Corrections: (1) I should have been saying "$\in$-induction" in place of "$\in$-recursion" in all my comments above. (2) I should have said "instance of $\Phi(X)$" and not "failing instance of $\Phi(X)$" in my comment. | |
| Mar 29 at 18:27 | answer | added | Julia Williams | timeline score: 8 | |
| Mar 29 at 17:55 | comment | added | Joel David Hamkins | I understood him to be claiming that NBG proves the second-order $\in$-recursion scheme. But perhaps I misunderstood. | |
| Mar 29 at 17:51 | vote | accept | Zuhair Al-Johar | ||
| S Mar 29 at 19:37 | |||||
| Mar 29 at 17:44 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, I thought MonroeEskew was speaking about $\in$-induction in the theory with the Foundation scheme. | |
| Mar 29 at 17:18 | comment | added | Joel David Hamkins | That is, you'd want to take the least-rank instance of the failure of $\Phi(X)$, but you can't form the set of failing instances in which to do this, since you lack second-order separation. | |
| Mar 29 at 16:30 | comment | added | Joel David Hamkins | @MonroeEskew you can't necessarily take the set of least rank, if the formula is second-order, since you don't have second-order separation in NBG. | |
| Mar 29 at 16:27 | answer | added | Joel David Hamkins | timeline score: 8 | |
| Mar 29 at 16:20 | comment | added | Zuhair Al-Johar | @MonroeEskew, although the language of your comment is not clear, but yes I understand that. It is the argument in my mind. That said, then this would prove induction over naturals, which NBG cannot. The point is if that is enough to prove Con(ZFC)? Is it enough to get equi-consistency with MK? | |
| Mar 29 at 15:46 | comment | added | Monroe Eskew | Why not? If there is a set $X$ such that $\Phi(X)$, then take the one of least rank. Otherwise there are only classes $X$ such that $\Phi(X)$, so none of them can possess another. | |
| Mar 29 at 14:59 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, Yes, it is about the second order $\in$ -recursion, since I think it would follow from the foundation scheme. | |
| Mar 29 at 14:47 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Mar 29 at 14:45 | comment | added | Joel David Hamkins | Nice question. For those who use the two-sorted approach to these theories, you are asking whether GBC proves the second-order $\in$-recursion scheme. | |
| Mar 29 at 14:35 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |