Timeline for Why do all non-empty ω-regular languages have periodic members?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 13, 2015 at 9:38 | comment | added | Louis | Does this answer really offer anything to the OP? | |
| Mar 13, 2015 at 9:34 | history | edited | J.-E. Pin | CC BY-SA 3.0 |
added 194 characters in body
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| Mar 13, 2015 at 9:20 | comment | added | David Richerby | Please don't patronize me. I know what "by definition" means. | |
| Mar 13, 2015 at 9:15 | comment | added | J.-E. Pin | I just say "by definition", a rather precise mathematical term. | |
| Mar 13, 2015 at 8:12 | comment | added | David Richerby | It doesn't seem to be what you're saying at all. You say nothing about the state $q$ in the question and you don't say that the proof in the question is incorrect. So, in fact, you say neither of the things that I said in my comment. | |
| Mar 13, 2015 at 8:03 | comment | added | J.-E. Pin | This is exactly what I am saying. | |
| Mar 13, 2015 at 7:55 | comment | added | David Richerby | But that doesn't mean that $q$ is accepting. Actually, the proof in the question is incorrect, precisely because it doesn't say that $q$ is accepting. | |
| Mar 13, 2015 at 7:30 | history | answered | J.-E. Pin | CC BY-SA 3.0 |