Timeline for On the optimal strength of Goodstein's theorem
Current License: CC BY-SA 4.0
3 events
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| Dec 12 at 17:51 | comment | added | Hanul Jeon | @none I don't think your guess is true. If my question holds, then Goodstein's theorem implies the consistency of $\mathsf{PA+Con(PA)}$. I think there is a more direct proof, but I have no idea how to prove it. | |
| Dec 12 at 10:02 | comment | added | none | I thought PA+CON(PA) proves Goodstein's theorem. So PA is just barely not strong enough. | |
| Dec 10 at 18:20 | history | asked | Hanul Jeon | CC BY-SA 4.0 |