Timeline for Set theories that are complete modulo finite-order arithmetic
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 29, 2020 at 5:58 | comment | added | none | @FedorPakhomov what I meant was, unless I misunderstood something, $T^*$ is not effectively axiomitizeable, so I don't see how to build up a consistency formula the way it's done for something like PA. $T^*$ contains many sentences that are not consequenes of ZF if your definition of $T^*$ means what it looks like. I'm still confused but I'm probably missing something silly. | |
| Oct 22, 2020 at 13:11 | comment | added | Fedor Pakhomov | @none Yes, of course. The definition of $T^*$ clearly could be carried out in $\mathsf{ZF}$. Since $T^*\supseteq T\supseteq \mathsf{ZF}$ this definition could be carried out in $T^*$. Now $T^*$ is consistent is simply the sentence "there are no proof of contradiction from the axioms of $T^*$". | |
| Oct 22, 2020 at 11:40 | comment | added | none | @FedorPakhomov Can $T^*$ even state its own consistency? | |
| Oct 22, 2020 at 11:15 | comment | added | Fedor Pakhomov | Do I understand correctly that by theory $T\supseteq \mathsf{ZF}$ "complete modulo finite-order arithmetic" you mean that the theory $T^*=(T+\text{true }\Sigma^{<\omega}_{<\omega}\text{-sentences})$ is complete? Note that Godel's second incompleteness theorem is applicable to the case of $T^*$. If $T^*$ is consistent, it couldn't prove its own consistency. The adaptation of the proof is simply reduced to the verification that provability predicate for $T^*$ satisfies Hilbert-Bernays-Lob conditions. By adapted Rosser's theorem we even get that $T^*$ should be either inconsistent or incomplete. | |
| Oct 22, 2020 at 4:08 | comment | added | BPP | Thanks, I've removed the CH example in that case. | |
| Oct 22, 2020 at 4:07 | history | edited | BPP | CC BY-SA 4.0 |
remove CH example
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| Oct 22, 2020 at 2:42 | comment | added | Andreas Blass | CH is not in the arithmetic hierarchy, which refers to first-order arithmetic. It's a $\Sigma^2_1$ sentence of third-order arithmetic (unless I've overlooked something). | |
| Oct 21, 2020 at 22:49 | comment | added | BPP | I didn't know that CH is a finite-order arithmetic sentence. Where on the arithmetic hierarchy does it lie? | |
| Oct 21, 2020 at 21:10 | comment | added | jeq | Link to OP's previous question: Incompleteness theorems for theories with omega-rule | |
| Oct 21, 2020 at 20:19 | comment | added | Rodrigo Freire | Your example is confusing because $CH$ is a finite-order arithmetic sentence. | |
| Oct 21, 2020 at 19:59 | history | edited | YCor |
edited tags
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| Oct 21, 2020 at 19:33 | history | asked | BPP | CC BY-SA 4.0 |