Showing a Theory $T$ is Substructure Complete

Question

Let $T$ be a (complete and consistent) theory. Suppose $T$ exhibits the following two properties:

(1) model-completeness: if $\mathcal{M} \models T$ and $\mathcal{A} \subseteq \mathcal{M}$ s.t. $\mathcal{A} \models T$, then $\mathcal{A} \cup D(\mathcal{A})$ is complete.

(2) amalgmation property: If $\mathcal{A}$ is a joint substructure of the models $\mathcal{M}, \mathcal{N} \models T$, then there is a model $\mathcal{M}' \supseteq \mathcal{M}$ of $T$ and an embedding $g : \mathcal{N} \hookrightarrow \mathcal{M}'$ such that $(id_M)(id_A) = (g)(id_A)$.

Now I am trying to show that $T$ must -- under these two conditions -- also admit substructure completness, which is just model-completeness minus the condition that $\mathcal{A} \models T$ (that is, all substructures $\mathcal{A}$ of $\mathcal{M}$ -- and not just submodels -- exhibit that $T \cup D(\mathcal{A})$ is complete).

My attempt so far has been to let $\mathcal{M} \models T$ and suppose that $\mathcal{N} \subseteq \mathcal{M}$. Then let $\mathcal{A}$ from (2) just be $\mathcal{N}$. However, it is still not obvious to me why $T \cup D(\mathcal{N})$ must be complete (i.e., it's still not obvious to me why $T$ must be substructure complete).

Answer 1

Let $\cal M$ be a model of $T$, $\cal A \subseteq \cal M$ and assume that $T \cup D(A)$ is not complete. Let $\phi$ be an $L(A)$-sentence undecided by $T \cup D(A)$. Wlog assume that $\cal M \models \phi$. Now pick a model $\cal N$ satisfying $T \cup D(A) \cup \{\lnot \phi\}$. Observe that since $T \cup D(M)$ and $T \cup D(N)$ are complete, we have $T \cup D(M) \models \phi$ and $T \cup D(N) \models \lnot \phi$. Since ${\cal N} \models D(A)$ we may identify $\cal A$ with a substructure of $\cal N$. Now we are in a position of amalgamating $\cal M$ and $\cal N$ over $\cal A$. Let $\cal K$ be model of $T$ such that there are embeddings of $\cal M$ and $\cal N$ to $\cal K$ that agree on $A$. Then ${\cal K} \models T \cup D(M)$ and ${\cal K} \models T \cup D(N)$. Hence ${\cal K} \models \phi$ and ${\cal K} \models \lnot \phi$.

As a side note, there are alternative (more commonly used) characterisations of model completeness and substructure completeness. The following are equivalent.

• $T$ is model complete.
• For every models $\cal M$ and $\cal N$ of $T$ if $\cal M \subseteq N$ then $\cal M \preceq N$.
• For every formula $\phi(\bar x)$ there is a quantifier-free formula $\psi(\bar x, \bar y)$ such that $T \models \forall \bar x (\phi(\bar x) \leftrightarrow \exists \bar y \psi (\bar x, \bar y))$.
• For every formula $\phi(\bar x)$ there is a quantifier-free formula $\psi(\bar x, \bar y)$ such that $T \models \forall \bar x (\phi(\bar x) \leftrightarrow \forall \bar y \psi (\bar x, \bar y))$.

Similarly the following are equivalent.

• $T$ is substructure complete
• For every formula $\phi(\bar x)$ there is a quantifier-free formula $\psi(\bar x)$ such that $T \models \forall \bar x (\phi(\bar x) \leftrightarrow \psi (\bar x))$.