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Given a set $\Sigma$ of well-formed formulas(wffs), I want to get a set $\Delta$ with the following conditions: $\Delta$ is a set of wffs such that (i) $\Sigma\subset\Delta$ (ii) it is finitely satisfiable (iii) for every wff $\alpha$, either $\alpha\in\Delta$ or $\neg\alpha\in\Delta$.

To do that one needs to apply Zorn's lemma. But, I don't exactly know how to set up for the Zorn's lemma and to apply the Zorn's lemma. So far I know that we need to deal with a family of sets and the inclusion for the relation to apply the Zorn's lemma. But, how should I set up the sets?; of what form the sets should it be? And how should I apply the Zorn's lemma?

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  • $\begingroup$ Hint: What if you were looking for a set $\Delta$ with properties (i), (ii) and a property (iii') $\Delta$ is $\subseteq$-maximal with properties (i) and (ii)? $\endgroup$ Commented Oct 24, 2020 at 8:22
  • $\begingroup$ This won't work for arbitrary $\Sigma$, though $\endgroup$ Commented Oct 24, 2020 at 8:24
  • $\begingroup$ @JohannesKloos I need the (iii) condition.. $\endgroup$
    – kkkk
    Commented Oct 24, 2020 at 8:28

1 Answer 1

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Let $\mathcal Z$ be the set of candidates for $\Delta$ if we ignore (iii), i.e., the elements of $\mathcal Z$ are all sets $A$ of wffs such that $\Sigma \subseteq A$ and $A$ is finitely satisfiable. Then $\mathcal Z$ is partially ordered by $\subseteq $. Show that it is inductively ordered and conclude.

However, this will work only if $\Sigma$ is nice.

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  • $\begingroup$ I need for arbitray set $\Sigma$ though... $\endgroup$
    – kkkk
    Commented Oct 24, 2020 at 8:29
  • $\begingroup$ And need (iii) condition.. $\endgroup$
    – kkkk
    Commented Oct 24, 2020 at 8:30
  • $\begingroup$ @kkkk Condition (iii) will follow from maximality. Additionally, if $\Sigma$ is inconsistent, you will be in trouble... $\endgroup$ Commented Oct 24, 2020 at 8:50
  • $\begingroup$ I see. By the way what do you mean by inconsistent? $\endgroup$
    – kkkk
    Commented Oct 24, 2020 at 8:57
  • $\begingroup$ To elaborate on the latter point, suppose $\Sigma = \{ p, \neg p \}$ for some proposition $p$. Every set $\Delta \subseteq \Sigma$ will not be finitely satisfiable, since it will contain both $p$ and $\neg p$, so the set you are looking for will not exist. $\endgroup$ Commented Oct 24, 2020 at 8:58

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