Timeline for Is there a consistent, unsound, $\omega$-inconsistent, effective theory that doesn't prove its own inconsistency?
Current License: CC BY-SA 4.0
14 events
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| Sep 10, 2023 at 16:27 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, True! But, unlike what was presented in the other question, here we can apply that to theories that are not necessarily true nor are necessarily written in the language of arithmetic! So, $T$ whether true or not, whether written in the language of arithmetic or not, once it's arithmetically sound then it is arithmetically $\omega$-consistent. This is a broader result. | |
| Sep 10, 2023 at 16:20 | comment | added | Joel David Hamkins | Yes, because the truth is $\omega$-consistent. This is the same issue we discussed recently on your other question. | |
| Sep 10, 2023 at 16:15 | comment | added | Zuhair Al-Johar | @Joel David Hamkins, so, is every arithmetically sound theory an arithmetically $\omega$-consistent theory? Where the latter is $\omega$-consistency for arithmetic properties. | |
| Sep 10, 2023 at 15:09 | answer | added | Zuhair Al-Johar | timeline score: 1 | |
| Sep 10, 2023 at 13:55 | comment | added | Joel David Hamkins | No, because the theory $\text{PA}+\neg\omega\text{-Con}(\text{PA})$ in your link is unsound but $\omega$-consistent. | |
| Sep 10, 2023 at 13:47 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Sep 10, 2023 at 13:43 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, so is arithmetically sound and being $\omega$-consistent equivalent if the theory is expressed in the language of arithmetic? | |
| Sep 10, 2023 at 12:53 | comment | added | Joel David Hamkins | If the theory is expressed in the language of arithmetic, then it is redundant. An $\omega$-inconsistent theory proves every instance $\varphi(n)$ of some formula, but also $\neg\forall x\ \varphi(x)$. This violates soundness, since either some $\varphi(n)$ is not actually true or the universal is true. | |
| Sep 10, 2023 at 12:15 | vote | accept | Zuhair Al-Johar | ||
| Sep 10, 2023 at 11:58 | comment | added | Zuhair Al-Johar | @bof, Not it is not redundant. See: $\omega$-consistent theory | |
| Sep 10, 2023 at 3:58 | comment | added | bof | "both unsound and $\omega$-inconsistent"? Isn't that redundant? What would be an example of a theory that's sound and $\omega$-inconsistent? Maybe I don't understand what "unsound" means (or what $\omega$-inconsistent means), I'm not a logician. | |
| Sep 10, 2023 at 1:43 | answer | added | Noah Schweber | timeline score: 10 | |
| Sep 10, 2023 at 0:36 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Sep 10, 2023 at 0:24 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |