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Consider $\Sigma \subseteq \Sigma'$ and $L \subseteq L'$. Then the theorem is:

If every model of $\Sigma$ has an L'-expansion, then $\Sigma'$ is conservative over $\Sigma$.

Recall conservative means: $\forall \sigma$ that are L-sentences,

$$ \Sigma ' \vdash_{L'} \sigma \iff \Sigma \vdash_L \sigma $$

Recall that an L'-expansion means we extend the language only and not the set of the structures so: $\mathcal A = (A,L)$ and $ \mathcal A' = (A,L')$ is true and $L \subseteq L'$.

It's clear that $\Leftarrow$ is easy, since to form a proof we just copt the same proof from $\Sigma$ and ignore any additional assumptions in our axioms $\Sigma'$. Also, $L'$ is irrelevant since $\sigma $ is an L-sentence.

I have a hint that we should do this proof by completeness. So this is what I tried:

WTS, $\forall \sigma$ L-sentences:

$$ \Sigma ' \vdash_{L'} \sigma \Rightarrow \Sigma \vdash_L \sigma $$

So assume

1) $\Sigma ' \vdash_{L'} \sigma$ is true.

By completeness we have every model $\mathcal A'$ of $\Sigma'$ models $\sigma$ i.e. $\mathcal A' \models \sigma$.

Ok but consider some model $\mathcal A \models \Sigma$. Then by assumption there is an $L'$-extension $\mathcal A' \models \Sigma' \implies \mathcal A' \models \sigma$.

Here is where I get stuck and can't connect to back to $\mathcal A$. Why would $\mathcal A \models \sigma$?

To me it seems I've reduced the problem to showing something like:

$$ \mathcal A' \models \sigma \implies \mathcal A \models \sigma $$

which I would assume is always trivially true if $\mathcal A'$ only contains additional symbols and the sets didn't change and if $\sigma$ is an L-sentence. Since, $\sigma$ only contain original symbols and the interpretations of those symbols didn't change so $ \mathcal A' \models \sigma \implies \mathcal A \models \sigma $ must hold.


context:

https://faculty.math.illinois.edu/~vddries/main.pdf

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    $\begingroup$ You mean if every model of $\Sigma$ has an $L'$-expansion to a model of $\Sigma'$, then $\Sigma'$ is conservative over $\Sigma.$ $\endgroup$ – spaceisdarkgreen Dec 18 '18 at 5:09
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    $\begingroup$ And yes, your reasoning in the last paragraph is correct: Since the domains are the same, the reduct of $\mathcal A'$ to $L$ is $\mathcal A,$ so they agree on $L$-sentences. I can't tell if you have a question beyond that. $\endgroup$ – spaceisdarkgreen Dec 18 '18 at 5:26

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