Timeline for Is the union of two conservative extensions of a theory conservative?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 15, 2023 at 15:47 | vote | accept | Giacomo Cozzi | ||
| Apr 14, 2023 at 22:19 | history | became hot network question | |||
| Apr 14, 2023 at 15:41 | answer | added | Holo | timeline score: 7 | |
| Apr 14, 2023 at 15:21 | comment | added | godelian | @SimonHenry You might be interested in the categorical version of Robinson's theorem treated in Makkai's "On Gabbay's proof of the Craig interpolation theorem for intuitionistic predicate logic". There he proves that if in a pushout of Boolean categories one leg is conservative, so is the parallel leg. Interestingly, this fails for coherent categories, unless the legs are Heyting functors. | |
| Apr 14, 2023 at 15:14 | history | edited | Giacomo Cozzi | CC BY-SA 4.0 |
Added the requirement $Σ_1 \cap Σ_2 = Σ$ I previously forgot
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| Apr 14, 2023 at 15:12 | comment | added | Giacomo Cozzi | @EmilJeřábek thank you, yes, I did mean that Σ_1∩Σ_2=Σ and forgot to explicitly mention it. I will edit the question to fix this, and I apologize if editing such an important detail in hindsight is considered bad form | |
| Apr 14, 2023 at 15:09 | comment | added | Emil Jeřábek | @SimonHenry All right, done. | |
| Apr 14, 2023 at 15:08 | answer | added | Emil Jeřábek | timeline score: 11 | |
| Apr 14, 2023 at 14:56 | comment | added | Simon Henry | @EmilJeřábek I agree it is not clearly said, but the paragraph explaining what $T'$ gave me the impression that it was the intent (at least that's how I read the question before you commented - but that might be my category theory oriented mind that always assume all sets that are not clearly introduced as subset of something are disjoints) - in any case your second comment is a much more interesting answer to the question which I think you should post as an answer. | |
| Apr 14, 2023 at 14:50 | review | Close votes | |||
| Apr 14, 2023 at 15:27 | |||||
| Apr 14, 2023 at 14:47 | comment | added | Emil Jeřábek | The question does not state anything like that. But if $T_1$ and $T_2$ are conservative extensions of $T$ in languages $\Sigma_1$ and $\Sigma_2$ (resp.) such that $\Sigma_1\cap\Sigma_2=\Sigma$, then $T_1\cup T_2$ is indeed a conservative extension of $T$. This is a variant of Robinson’s joint consistency theorem, as I mentioned. | |
| Apr 14, 2023 at 14:43 | comment | added | Simon Henry | I was implicitely reading the question with $\Sigma_1$ and $\Sigma_2$ disjoint extention of $\Sigma$ ( as in $\Sigma = \Sigma_1 \cap \Sigma_2$ ), which rules out the most obvious counter-example, is it what was intended ? | |
| Apr 14, 2023 at 14:32 | answer | added | Arno | timeline score: 0 | |
| Apr 14, 2023 at 14:31 | comment | added | Emil Jeřábek | Certainly not. There is any number of trivial counterexamples; the simplest I can think of is let $T$ be the empty theory in the empty language, $T_1$ the theory $\exists x\,R(x)$ in language $\{R(x)\}$, and $T_2$ the theory $\neg\exists x\,R(x)$ in the same language. If you want some positive results, you should look at variants of Robinson’s joint consistency theorem. | |
| S Apr 14, 2023 at 14:15 | review | First questions | |||
| Apr 14, 2023 at 14:48 | |||||
| S Apr 14, 2023 at 14:15 | history | asked | Giacomo Cozzi | CC BY-SA 4.0 |