Timeline for Are undecidable consequences of Con recursively enumerable?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 3, 2011 at 0:58 | answer | added | Andreas Blass | timeline score: 10 | |
| May 2, 2011 at 14:21 | vote | accept | Alex Gavrilov | ||
| May 2, 2011 at 12:32 | comment | added | Lucas K. | Do you answer your own question now? | |
| May 2, 2011 at 12:29 | answer | added | Emil Jeřábek | timeline score: 7 | |
| May 2, 2011 at 11:19 | comment | added | Alex Gavrilov | Lucas, If I got you right, you suggest to enumerable the statements which are equivalent to Con. I suspect that there are statements which are strictly weaker then Con but still undecidable. | |
| May 2, 2011 at 11:17 | comment | added | Lucas K. | If you can prove p in PA + Con(PA) and not p in PA + not Con(PA), then p must be independent of PA (or PA is inconsistent). At that moment you can enumerate p in your solution. | |
| May 2, 2011 at 10:45 | comment | added | Alex Gavrilov | What means "opposit results"? | |
| May 2, 2011 at 10:16 | comment | added | Lucas K. | Suggestion. Take also PA + not Con(PA). Then enumerate the theorems of both PA + Con(PA) and PA + non Con(PA). In the final enumeration, take the theorems that give opposite results in the two systems. Could that work? | |
| May 2, 2011 at 10:05 | history | asked | Alex Gavrilov | CC BY-SA 3.0 |