Timeline for Is DNC/DNR stronger than "prompt" non-computability?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 24, 2015 at 6:27 | vote | accept | Eric Astor | ||
| Sep 24, 2015 at 6:13 | vote | accept | Eric Astor | ||
| Sep 24, 2015 at 6:17 | |||||
| Sep 23, 2015 at 21:19 | answer | added | Bjørn Kjos-Hanssen | timeline score: 7 | |
| Sep 23, 2015 at 17:57 | history | edited | Eric Astor | CC BY-SA 3.0 |
Clarified totality, suggested by Joel's comment
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| Sep 23, 2015 at 17:57 | comment | added | Eric Astor | Ah. Good point, and that's part of why we restrict to total functions... to avoid the debate of whether if $U(e)\uparrow$ and $f(e)\uparrow$, we can say that $U(e)=f(e)$. Thanks for making me clarify. | |
| Sep 23, 2015 at 17:54 | comment | added | Joel David Hamkins | Well, that function wouldn't have $U(e)\neq f(e)$ in the case $U(e)\uparrow$, since both sides would diverge equally, so I don't take it as a "counterexample". I think the concept makes fine sense when $f$ is partial, provided that you really have $\varphi_e(m)\neq f(m)$, either because one side diverges and the other doesn't, or both converge, but to different values. But it is also fine to require that $f$ is total. | |
| Sep 23, 2015 at 17:51 | comment | added | Eric Astor | @JoelDavidHamkins Yes, for the same reason we do so for standard DNC functions; if we allow partial functions to be DNC, there's a trivial computable example. (Specifically, U(e) + 1). | |
| Sep 23, 2015 at 17:40 | comment | added | Joel David Hamkins | Could you clarify: you insist that $f$ is a total function? | |
| Sep 23, 2015 at 17:31 | history | asked | Eric Astor | CC BY-SA 3.0 |