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I have one problem and I m sure that can be solved by using compactness theorem but I cant solve it.

Let $T$ be an $L$-theory and $\{F_i (x) \mid i\in I\}$ family of $L$-formulas. Suppose further that every element of every model of theory $T$ satisfies at least one of $F_i (x)$. Prove that exist finite $J \subseteq I$ and $T \vDash \forall x (\lor F_i(x))$. Disjunction is on finite $J$.

I try to suppose that for every finite $J \subseteq I$, $T \vDash \forall x (\lor F_i(x))$ doesn't hold. So, exists some model $m \vDash T$ and $m \vDash \lnot \forall x (\lor_J F_i(x))$. So, $m \vDash \exists x\lnot (\lor_J F_i(x))$.

So, for every finite $J \subseteq I$, exists $m \vDash T$ and exists some $b\in M$ and $m \vDash \lnot (\lor_J F_i[b])$.......$m \vDash (\land_J \lnot F_i[b])$........$(\forall i\in J)$ $m \vDash \lnot F_i[b]$ .

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  • $\begingroup$ It is very hard to understand what you're asking. Maybe if you try to use some LaTeX, it will be easier. Here a tutorial for basic MathJax. $\endgroup$ Commented Sep 2, 2015 at 14:07
  • $\begingroup$ Sorry,I ll try to change immediatly $\endgroup$
    – Jelena
    Commented Sep 2, 2015 at 14:12
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    $\begingroup$ I hope that it's clearer now. $\endgroup$
    – Jelena
    Commented Sep 2, 2015 at 14:54

1 Answer 1

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1) We have that every model $M$ of $T$ (i.e. $M \vDash T$) satisfy at least one $F_i$.

2) Consider the set of formulae $\{ \lnot F_i \}_{i∈I}$; no model of $T$ can satisfy it, becuase for each model $M$ of $T$ there is at least one $\lnot F_i$ that is not satisfied by it.

Thus, applying Compactness theorem to the set of models of $T$, we have that exists finite $J \subseteq I$ such that $\{ \lnot F_j \}_{j∈J}$ is not satisfiable by any model of $T$.

3) Let $N$ the largest $j \in J$ and consider $\{ \lnot F_1, \lnot F_2, \ldots, \lnot F_N \}$ that is again unsatisfiable by any model of $T$. But this means that every $M$ satisfy at least one of $F_1, F_2, \ldots, F_N$ i.e. satisfy $\lor_{j∈J} F_j$ (it is a finite disjunction, and thus a well-formed formula).

4) In conclusion, we have that every model $M$ of $T$ is a model of $\lor_{j∈J} F_j$, i.e.

if $M \vdash T$, then $M \vDash \lor_{j∈J} F_j$.

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  • $\begingroup$ Thank you very much, this helps me. $\endgroup$
    – Jelena
    Commented Sep 2, 2015 at 15:14
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    $\begingroup$ I hove some questions: what's about quantifier and how we can be sure in 2) that {¬Fi}i∈I doesn't have any model to apply compactness? $\endgroup$
    – Jelena
    Commented Sep 2, 2015 at 15:31
  • $\begingroup$ @Jelena -{¬Fi}i∈I has no model of $T$ ... $\endgroup$ Commented Sep 2, 2015 at 16:02
  • $\begingroup$ Well, yes I understand. I didn't know that then I apply compactness in terms just of model of T. This is then enough for me for contradiction because I already proved that {¬Fj}j∈J is satisfiable for every J by some model of T upper in question. $\endgroup$
    – Jelena
    Commented Sep 2, 2015 at 16:11

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