Show a satisfiable set of quantifier-free sentences containing either $\sigma$ or $\neg \sigma$ for any atomic $\sigma$ has a unique minimal model

Question

A minimal $L$-structure, in the course I'm doing, is defined to be one with no proper substructure, for a language $L$. We also allow structures to be empty.

We have a set of quantifier-free sentences $\Sigma$. Suppose that $\Sigma$ is satisfiable and that for any atomic sentence $\sigma$, either $\sigma \in \Sigma$ or $\neg \sigma \in \Sigma$.

I want to show that there's a unique minimal $L$-structure, up to isomorphism, which is a model of $\Sigma$.

I found a similar looking question here but I'm not sure how to adapt the solution to make mine work. I was able to answer 'exercise 1' the person who responded gave and can see the answer they gave to 'exercise 2', but it's not clear that I can get my conclusion from just what they've given there, as in that question either $\sigma \in \Sigma$ or $\neg \sigma \in \Sigma$ for any quantifier free sentence $\sigma$, whereas the problem I'm trying to solve only requires this for any atomic $\sigma$ and I don't see how the conclusion follows with that weaker condition.

I'd appreciate either any advice on how I might modify that solution or any other approaches anyone can suggest to me - I've just started a model theory course and don't really know what I'm doing yet.

Answer 1

Suppose $\Sigma$ satisfies your criteria. It is indeed not the case that $\Sigma$ need satisfy the stronger criterion that $\sigma\in \Sigma$ or $\neg\sigma\in \Sigma$ for every quantiier-free $\sigma$. However, in a sense $\Sigma$ "morally" satisfies that stronger criterion:

There is a set $\hat{\Sigma}$ which has exactly the same models as $\Sigma$ but satisfies the stronger property of being "complete for quantifier-free sentences."

Don't try too hard to construct this set:

Just look at the quantifier-free deductive closure of $\Sigma$, namely $$\hat{\Sigma}:=\{\varphi\in QuantFree: \Sigma\vdash\varphi\}.$$ You can show that this set has the stronger property by induction on formula commplexity, and by definition its models are exactly the models of $\Sigma$.

The point then is this. By the linked argument, this $\hat{\Sigma}$ has a unique minimal model up to isomorphism; but since $\Sigma$ and $\hat{\Sigma}$ have the same models, this implies the result for $\Sigma$ as well.