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During the proof of the compactness theorem, from an arbitrary finitely satisfiable set $\Sigma$ of WFFs, we construct a finitely satisfiable set $\Delta\supseteq \sigma$ such that for every WFF $\alpha$, either $\alpha\in\Delta$ or $\lnot\alpha \in\Delta$. Show that $\Delta$ need not be unique by describing an infinite, finitely satisable set $\Sigma$ of WFFs such that there is more than one possible extension $\Delta$.

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  • $\begingroup$ If $\beta\notin \Sigma$ then $\Sigma\cup\{\beta\}$ and $\Sigma\cup\{\lnot\beta\}$ can be extended maximal finitely satisfiable set and they are not same. $\endgroup$
    – Hanul Jeon
    Commented Oct 6, 2013 at 1:08
  • $\begingroup$ @tetori: Not so fast. How about $\Sigma=\{\alpha, \alpha\to\beta\}$? Then $\Sigma\cup\{\neg\beta\}$ will not be satisfiable. $\endgroup$
    – hmakholm left over Monica
    Commented Oct 6, 2013 at 2:12
  • $\begingroup$ @HenningMakholm Oh, it is my mistake :( $\endgroup$
    – Hanul Jeon
    Commented Oct 6, 2013 at 2:40

1 Answer 1

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Hint. Suppose the language contains some unary predicate that is not mentioned in $\Sigma$ at all ...

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