Timeline for The lattice of analogues of Robinson's $Q$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| May 3, 2020 at 19:52 | comment | added | Fedor Pakhomov | I think that my argument about non-existence of the greatest element could be adopted to this case with almost no changes. We just need to note that the interpretability relation $\psi^{\star}\triangleright \varphi_0$ is $\Sigma_1$. Although I don't know enough about the lattice of interpretability degrees of arbitrary sentences to say whether it would be enough to figure out the answer to your question. | |
| May 3, 2020 at 19:29 | comment | added | Noah Schweber | (I can get $\Pi^0_2$-hardness of essential incompleteness assuming this question has an affirmative answer - basically attach finite pieces of $\mathbb{N}$ to "stages," where the pieces grow each time we see another piece of evidence for the $\Pi^0_2$ property - but I haven't convinced myself that the answer given there is correct yet.) | |
| May 3, 2020 at 19:26 | comment | added | Noah Schweber | A potentially-interesting followup question: what if we look at the ordering of essentially incomplete sentences (possibly not true of $\mathbb{N}$) ordered by interpretability? Given any $\varphi$ the set of e.i.-sentences interpreting $\varphi$ is d.c.e. (interpreting $\varphi$ is c.e. and consistency is co-c.e.) while a priori essential incompleteness looks $\Pi^0_3$, so there should be no minimal element in this order either. But now d.c.e. seems a bit too big to get this kind of argument to work. | |
| May 3, 2020 at 19:23 | comment | added | Noah Schweber | Yes, I think that's right. (Certainly we can't go much higher and get a naive index set argument: $Th_\Gamma(\mathbb{N})$ quickly becomes too complicated as $\Gamma$ gets bigger.) | |
| May 3, 2020 at 19:21 | comment | added | Fedor Pakhomov | I think that something like $\Pi_1^{-}$ is essential for this kind of argument. | |
| May 3, 2020 at 19:10 | history | bounty ended | Noah Schweber | ||
| May 3, 2020 at 19:10 | comment | added | Noah Schweber | Lovely! My original attempt to show that $\mathfrak{Q}$ had no greatest element was along these lines but not focusing on as small a syntactic class as $\Pi^-_1$, which made coding the necessary complexity into the sentence no longer easy (or doable for me). | |
| May 3, 2020 at 19:07 | vote | accept | Noah Schweber | ||
| May 3, 2020 at 18:58 | history | answered | Fedor Pakhomov | CC BY-SA 4.0 |