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I see two definitions of Maximally Consistent Set:

  1. $\Phi$ is maximally consistent iff $\Phi$ is consistent and for all formulas $\varphi$, $\Phi\vdash\varphi$ or $\Phi\vdash\neg\varphi$
  2. $\Phi$ is maximally consistent iff $\Phi$ is consistent and for any $\varphi\notin\Phi,\Phi\cup\{\varphi\}$ is not consistent

I think the two definitions are equivalent. It is easy to prove from 2 to 1, but how to prove from 1 to 2?

RobPratt
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William
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    The converse holds if $\Phi$ is deductively closed, but in general it does not (let $\Phi$ be a small set of axioms for any consistent complete theory, eg unbounded dense total orders). Does deductive closure happen to be part of the definition you're using of a theory or whatever you're calling $\Phi$, or was it mentioned somewhere near these definitions? – Izaak van Dongen Nov 11 '22 at 08:24
  • Maybe useful this post. – Mauro ALLEGRANZA Nov 11 '22 at 10:56
  • @IzaakvanDongen It does not mention the deductive closure near the definitions of Maximally Consistent Set, but the deductive closure can be proven according to the second definition. – William Nov 14 '22 at 04:10
  • @IzaakvanDongen Let Φ be a small set of axioms for any consistent complete theory as you mentioned, does there exist a φ, s.t. Φ $\nvdash$ φ and Φ $\nvdash$ ¬φ? – William Nov 14 '22 at 05:51
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    My point was that such a $\Phi$ satisfies (1) but not (2). I expect that the source of these characterisations wanted $\Phi$ to be deductively closed. EG look at https://math.stackexchange.com/q/452396/473276. The correct version of (1) if you're not taking $\Phi$ to be deductively closed is "for all $\phi$, either $\phi \in \Phi$ or $\lnot \phi \in \Phi$". Can you prove that this is equivalent to (2)? And can you prove that under the additional assumption of deductive closure, (1) also implies (2)? – Izaak van Dongen Nov 14 '22 at 14:09
  • I could write an answer, but it would only be a reiteration of what Izaak said in the comments. What is still unclear to motivate a bounty? – spaceisdarkgreen Nov 15 '22 at 00:27
  • @IzaakvanDongen Yes, I can prove from "Φ is consistent, and for all ϕ, either ϕ∈Φ or ¬ϕ∈Φ" to (2). I can also prove from "Φ is consistent, and for all formulas φ, Φ⊢φ or Φ⊢¬φ, and Φ is deductive closure" to (2). I am not familiar with consistent complete theory. Can it prove any formula or its negation? The first definition of Maximally Consistent Set comes from https://www.mcmp.philosophie.uni-muenchen.de/students/math/math_logic_munich.pdf , page 74, Definition 17. – William Nov 15 '22 at 12:53

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