Timeline for Can second-order logic identify "amorphous satisfiability"?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jul 9, 2021 at 17:42 | vote | accept | Noah Schweber | ||
| Jul 9, 2021 at 16:31 | comment | added | Noah Schweber | @HarryWest I've addressed the completeness issue above - you're right, I should have been clearer. | |
| Jul 9, 2021 at 16:30 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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| Jul 9, 2021 at 13:41 | history | became hot network question | |||
| Jul 9, 2021 at 9:55 | comment | added | Harry West | @EmilJeřábek: the question doesn't restrict to consistent theories though - that's all I wanted to point out. (I thought we could just complete a theory using Henkin semantics, though I haven't thought this through.) | |
| Jul 9, 2021 at 9:32 | comment | added | Emil Jeřábek | @HarryWest That does not make much sense to me. Second-order logic has no sensible proof system, and it is not compact, thus consistency in second-order logic is defined by the existence of a model. Also, due to lack of compactness, second-order theories cannot be completed by the usual inductive procedure; the way to complete a consistent second-order theory is to pick its model, and take the complete theory of that model. | |
| Jul 9, 2021 at 8:50 | comment | added | Harry West | There are complete second-order theories with no models at all in ZF. To be concrete, take Asaf's example here and complete it (which doesn't require choice): math.stackexchange.com/questions/1005283/…). | |
| Jul 9, 2021 at 8:25 | answer | added | Harry West | timeline score: 6 | |
| Jul 9, 2021 at 8:18 | comment | added | Asaf Karagila♦ | The second question implies the first, by the way. (And not only in the sense that both have a negative answer...) | |
| Jul 9, 2021 at 7:53 | answer | added | Joel David Hamkins | timeline score: 5 | |
| Jul 9, 2021 at 5:39 | history | asked | Noah Schweber | CC BY-SA 4.0 |