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I'm taking a beginner course in mathematical logic, and I was struggling to understand some applications of the downward Löwenheim–Skolem theorem. The version of Löwenheim–Skolem theorem that was presented in the lectures is:

Suppose $L$ is countable first order language, and $\mathcal{A}$ is an $L$-structure. Let $T$ be the theory of $\mathcal{A}$, i.e. $T = Th(\mathcal{A}) =\{ \theta: \mathcal{A} \models \theta \text{ where } \theta \text{ is an L-sentence}\}$. Then, there is a countable model $\mathcal{B}$ such that $\mathcal{B}\models T$.

$\Delta$ was defined to be the sentences which describe a dense linear order with no end points, i.e. $\Delta = \{ \varphi_1, \dots , \varphi_6\}$ where for example $\varphi_1 = \forall x_1, \exists x_2, \ x_2 < x_1$. The proof of the "Los-Vaught test" was given.

(Los-Vaught Test) For every $L$-sentence $\theta$, $\Delta \vdash \theta$ or $\Delta \vdash \lnot \theta$.

The proof starts: Suppose not. Then, $\Delta \cup \{\theta\}$ and $\Delta \cup \{ \lnot\theta\}$ are both consistent. Godel's completeness theorem tells us (the model existence lemma tells us?) that if it is consistent, then it should have a model. Then Löwenheim–Skolem theorem tells us, that we should have a countable model.

I don't understand the last sentence, namely, why the Löwenheim–Skolem theorem tells us that. How can I know that $\Delta \cup \{\theta\}$ and $\Delta \cup \{ \lnot\theta\}$ are theories of some $L$-structure?

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If $M$ is a model of $\Delta\cup\{\theta\}$, then $\Delta\cup\{\theta\}\subseteq\mathrm{Th}(M)$. So any model of $\mathrm{Th}(M)$ is a model of $\Delta\cup\{\theta\}$.

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  • $\begingroup$ Ah, so am I correct to understand that $\Delta \cup \{\theta\}$ has a model $M$ by Model existence lemma, and $Th(M)$ has a countable model $\mathcal{B}$ due to Löwenheim–Skolem. As $\Delta \cup \{\theta\} \subset Th(M)$, we conclude $\mathcal{B} \models \Delta \cup \{\theta\}$? $\endgroup$
    – Albert
    Commented Mar 9, 2021 at 3:50
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    $\begingroup$ Yep, that's correct! $\endgroup$ Commented Mar 9, 2021 at 3:52
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    $\begingroup$ (Assuming that by "model existence lemma", you mean the special case of Gödel's Completeness Theorem that says "every consistent theory has a model".) $\endgroup$ Commented Mar 9, 2021 at 3:53

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