Timeline for Undecidable easy arithmetical statement
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| Nov 8, 2017 at 0:12 | comment | added | Joel David Hamkins | It seems that one could say, more precisely, that termination of the Ackermann function corresponds to the ordinal $\omega^2$, since the nested recursion has that order type. | |
| Nov 7, 2017 at 22:22 | history | edited | Lucas K. | CC BY-SA 3.0 |
Added note about ordinals.
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| Nov 7, 2017 at 22:12 | history | answered | Lucas K. | CC BY-SA 3.0 |