# Why is it if every model of $\Sigma$ has an L'-expansion then $\Sigma'$ is conservative over $\Sigma$?

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

• You mean if every model of $\Sigma$ has an $L'$-expansion to a model of $\Sigma'$, then $\Sigma'$ is conservative over $\Sigma.$ – spaceisdarkgreen Dec 18 '18 at 5:09
• 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. – spaceisdarkgreen Dec 18 '18 at 5:26