Unique minimal model for set of quantifier-free sentences We'll use a language $L$ that has at least one constant symbol. We have a set of quantifier-free sentences ($\Gamma$).
We'll say that an $L$-structure is minimal if it has no proper substructure (there is no strictly smaller domain for the structure, symbols are interpreted in the same way).
We suppose that $\Gamma$ is satisfiable and that for any quantifier-free sentence $\sigma$, either $\sigma \in \Gamma$ or $\neg \sigma \in \Gamma$.
I am trying to prove that there is a unique minimal $L$-structure, up to isomorphism, which is a model of $\Gamma$.
I can't quite get a grasp on this one as I'm discovering Model Theory. I would imagine going for a model with a domain that has just enough elements to give an interpretation to all constants in a way that satisfies the formulas in $\Gamma$. Now with possibly infinitely many formulas, this is hard to do... But for finite subset of $\Gamma$ we can find such a model (can't justify this formally though). Then I'd use Compactness. But I'm stuck for anything that comes after (uniqueness up to isomorphism, minimality). Maybe I should go for an explicit construction?
I'm confused and any help would be very much appreciated
 A: Exercise 1 Show that an $L$-structure is minimal if and only if every element is the interpretation of a term. Moreover, show that, given any $L$-structure, the set of interpretations of terms forms a substructure.
Exercise 2 For any given $L$-structure, define the atomic theory of the structure to be the set of quantifier-free sentences it satisfies. Use the result of Exercise 1 to show that two minimal $L$-structures are isomorphic if and only if they have the same atomic theory.
Exercise 3 From the above two exercises, conclude the desired result.
Extra credit In Exercise 2, show moreover that the isomorphism between the structures is unique. Conclude that, given $\Gamma$ as in the problem statement, $\Gamma$ has a minimal model that is unique up to a unique isomorphism.

Here is a solution to Exercise 2.
Assume we have $L$-structures $M$ and $N$, and an $L$-structure isormophism $f:M \to N$. It is routine to show, by induction on the complexity of the sentences, that $M \models \sigma$ if and only if $N \models \sigma$, for any quantifier free sentence $\sigma$. Therefore $M$ and $N$ have the same atomic theory.
Assume $L$-structures $M$ and $N$ are minimal and have the same atomic theory $\Gamma$. We will build an $L$-structure isomorphism $f:M \to N$. Let $a \in M$. We know that $a = t^M$ for some $L$-term $t$, by Exercise 1. Set $f(a) = t^N$ and note that $t^N$ is the only possible choice of value for $f(a)$ if we want $f$ to be an isomorphism, since isomorphisms preserve the interpretation of terms. Observe that the chosen value of $f(a)$ does not depend on the choice of $t$: if $a = t_1^M$, also, then the sentence $t = t_1$ is in $\Gamma$, so $t^N = t_1^N$ as well. It remains to show that $f$ is injective, surjective, and preserves the interpretation of symbols in $L$.
(Injectivity.) Assume $f(a_1) = f(a_2) = b$. Take terms $t_1$ and $t_2$ such that $a_1 = t_1^M$ and $a_2 = t_2^M$. Then, by the construction of $f$, $t_1^N = t_2^N = b$, so $N \models t_1 = t_2$, so $M \models t_1 = t_2$, so $a_1 = a_2$, as desired.
(Surjectivity.) Let $b \in N$. Since $N$ is minimal, by Exercise 1, we know that $b = t^N$ for some term $t$. Then $f(t^M) = b$, as desired.
(Preservation of symbols.) Let $R(x_1,\ldots,x_n)$ be a relation in $L$, and let $a_1,\ldots,a_n \in M$. Take terms $t_1,\ldots,t_n$ such that $a_i = t_i^M$. Then
\begin{align*}
  M \models R(a_1,\ldots,a_n)
  &\iff M \models R(t_1,\ldots,t_n) \\
  &\iff N \models R(t_1,\ldots,t_n) \\
  &\iff N \models R(f(a_1),\ldots,f(a_n))
  \text{,}
\end{align*}
as desired.
Let $\alpha(x_1,\ldots,x_n)$ be a function in $L$, and let $a_1,\ldots,a_n \in M$. Take terms $t_1,\ldots,t_n$ such that $a_i = t_i^M$. Then
\begin{align*}
  f(\alpha^M(a_1,\ldots,a_n))
  &= f(\alpha(t_1,\ldots,t_n)^M) \\
  &= \alpha(t_1,\ldots,t_n)^N \\
  &= \alpha^N(f(a_1),\ldots,f(a_n))
  \text{,}
\end{align*}
as desired.
We have now shown that $M \cong N$ by the unique isomorphism $f$.
A: Look at not just constants, but all terms (a term is, inductively either a constant symbol or an $n$-ary function symbol applied to $n$ terms).  Define an equivalence relation on terms by saying that $t_1 \sim t_2$ iff the sentence $"t_1=t_2\!"$ is in $\Gamma.$  Now build a structure whose domain is the set of equivalence classes of terms.
