Timeline for In finite model theory, is "invariant FOL with $\varepsilon$-operator" unavoidably second-order?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Aug 29 at 22:03 | history | bounty ended | CommunityBot | ||
| S Aug 29 at 22:03 | history | notice removed | CommunityBot | ||
| S Aug 21 at 20:23 | history | bounty started | Noah Schweber | ||
| S Aug 21 at 20:23 | history | notice added | Noah Schweber | Draw attention | |
| S May 7 at 3:01 | history | bounty ended | CommunityBot | ||
| S May 7 at 3:01 | history | notice removed | CommunityBot | ||
| May 3 at 20:32 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 62 characters in body; edited title
|
| S Apr 29 at 1:41 | history | bounty started | Noah Schweber | ||
| S Apr 29 at 1:41 | history | notice added | Noah Schweber | Draw attention | |
| Apr 24 at 16:36 | comment | added | Noah Schweber | @JoelDavidHamkins Interestingly, no - see the linked paper! Expressions like "$\epsilon\varphi=\epsilon\psi$" add a layer of complexity, and it turns out that (as long as our structure is finite) there are unavoidable "coincidences" that encode meaningful information. This is similar to the fact that (again, over finite structures) there are "FOL-with-any-ordering" sentences which aren't FOL-expressible. | |
| Apr 24 at 12:20 | comment | added | Joel David Hamkins | If the truth of the sentence doesn't depend on the $\varepsilon$ operator that is used, then can't we omit $\varepsilon$ entirely by using universal quantifiers? | |
| Apr 24 at 5:08 | history | asked | Noah Schweber | CC BY-SA 4.0 |