Timeline for For which Millennium Problems does undecidable -> true?
Current License: CC BY-SA 3.0
11 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 19, 2014 at 18:43 | comment | added | Peter LeFanu Lumsdaine | @ChristianRemling: “The truth of a conjecture can’t make a set $\Pi^0_1$ if it wasn't so to start with.” Sure, but the $\Pi^0_1$-ness of a set can be implied by (or even equivalent to) some conjecture. A proof of the conjecture can’t change whether the set is $\Pi^0_1$, but it can tell us whether or not it is. | |
| Apr 18, 2014 at 15:47 | history | bounty ended | John Sidles | ||
| Apr 18, 2014 at 0:44 | history | made wiki | Post Made Community Wiki by Todd Trimble | ||
| Apr 18, 2014 at 0:24 | comment | added | John Sidles | Alex Gavrilov, you have received the bounty. Thank you for taking the time to give such a well-considered answer. | |
| Apr 17, 2014 at 23:07 | vote | accept | John Sidles | ||
| Apr 11, 2014 at 20:48 | comment | added | Bjørn Kjos-Hanssen | Once it's proven, the sentence is proven to be equivalent to a $\Pi^0_1$ statement, namely $0=0$. When we speak of "a sentence" we probably mean "up to provable equivalence" since the Millennium Problems are only defined up to provable equivalence... | |
| Apr 11, 2014 at 19:10 | comment | added | Monroe Eskew | I don't see how assuming a statement to be true makes it $\Pi^0_1$. | |
| Apr 11, 2014 at 17:58 | comment | added | Alex Gavrilov | Yes, but I mean it is proven and assumed to be true. | |
| Apr 11, 2014 at 17:54 | comment | added | Monroe Eskew | What do you mean by, "After all, once a conjecture is proven, it is in $\Pi^0_1$ by definition"? Logically speaking, a sentence $\sigma$ is not in general equivalent to "there is a proof of $\sigma$," as Goedel shows. | |
| Apr 11, 2014 at 17:35 | history | answered | Alex Gavrilov | CC BY-SA 3.0 |