# A theory $T$ is model-complete if the union of $T$ with an atomic diagram is complete

Let $$T$$ be a theory in first order logic over some language $$L$$. Let $$\mathfrak A$$ be some structure over $$L$$ with $$\mathfrak A \models T$$ and with $$A$$ be its universe. Then consider every $$a \in A$$ as a constant and look at the enriched language $$L(A) = L \cup A$$ with the $$L(A)$$-structure $$\mathfrak A_A = (\mathfrak A, a)_{a\in A}$$. A formula over $$L(A)$$ is called basic if it is an atomic formula. The set $$\operatorname{Diag}(\mathfrak A) = \{ \varphi \mbox{ is a basic L(A)-sentence } \mid \mathfrak A_A \models \varphi \}$$ is called the atomic diagram of $$\mathfrak A$$.

A theory $$T$$ is called model-complete if every substructure relation between two models is actually an elementary embedding.

Then $$T$$ is model-complete if and only if for any $$\mathfrak A \models T$$ the theory $$T \cup \operatorname{Diag}(\mathfrak A)$$ is complete.

These definitions are from A course in model theory by K.Tent/M.Zeigler.

I do not understand the quoted statement. For let $$t_1 = t_2$$ and $$t_3 \ne t_4$$ be two atomic sentences for terms $$t_1, t_2,t_3,t_4$$ in $$L(A)$$. Then Set $$\varphi = (t_1 = t_2) \land (t_3 \ne t_4)$$. Now suppose in the terms we have some constants from $$A$$. Then neither $$\varphi$$ nor $$\neg \varphi$$ is in $$T \cup \operatorname{Diag}(\mathfrak A)$$ as it is not in $$\operatorname{Diag}(\mathfrak A)$$ for it is not atomic, nor is it in $$T$$ as it is a statement over the enrichted language $$L(A)$$, but not over $$L$$. Could someone please explain the above statement (and what I oversee here...)?

Complete means for any sentence $$\varphi,$$ either $$T\vdash \varphi$$ or $$T\vdash \lnot \varphi,$$ not $$\varphi\in T$$ or $$\lnot\varphi\in T.$$

If $$\mathfrak A,\mathfrak B\models T$$ and $$\mathfrak A\subseteq \mathfrak B,$$ then both are models of $$T\cup Diag(\mathfrak A).$$ If $$T\cup Diag(\mathfrak A)$$ is complete, then they must agree on all $$L(A)$$-sentences, and hence the embedding is elementary. On the other hand if $$T\cup Diag(\mathfrak A)$$ is not complete for some $$\mathfrak A\models T,$$ then we can find a $$\mathfrak B\models T$$ with $$\mathfrak A\subseteq \mathfrak B$$ that differs from $$\mathfrak A$$ on some $$L(A)$$-sentence (just let it be a model of the negation of some $$\varphi$$ that is true in $$\mathfrak A$$ but that $$T\cup Diag(\mathfrak A)$$ does not decide). Hence this embedding is not elementary.

• What makes you sure that such a model $\mathfrak B$ of $T \cup Diag(\mathfrak A) \cup \{\neg \varphi\}$ exists that contains $\mathfrak A$ as an $L$-structure? – StefanH Dec 15 '18 at 17:58
• @StefanH The whole point of the diagram of $A$ is that any model of it contains an isomorphic copy of $A$ as a substructure (via the interpretation of the constant symbols). – Alex Kruckman Dec 15 '18 at 18:12
• Got it, just needs additionally that completeness of a theory is equivalent with the property that all models are elementary equivalent, then surely non-completeness lets us choose such a model. – StefanH Dec 16 '18 at 21:09

You have two errors here:

1. "$$T$$ is complete" means that for every sentence $$\varphi$$, $$T\models \varphi$$ or $$T\models \lnot\varphi$$. It does not mean that $$\varphi\in T$$ or $$\lnot \varphi\in T$$.

2. A basic formula is not just an atomic formula. It is an atomic formula or a negation of an atomic formula.

In your example, if $$\mathfrak{A}\models \varphi$$, then $$t_1 = t_2\in \text{Diag}(\mathfrak{A})$$ and $$t_3\neq t_4\in \text{Diag}(\mathfrak{A})$$, so $$T\cup \text{Diag}(\mathfrak{A})\models \varphi$$. Otherwise, if $$\mathfrak{A}\not\models \varphi$$, then either $$t_1\neq t_2\in \text{Diag}(\mathfrak{A})$$ or $$t_3 = t_4\in \text{Diag}(\mathfrak{A})$$, and in either case $$T\cup \text{Diag}(\mathfrak{A})\models \lnot \varphi$$.

It's an easy exercise to show that for any structure $$\mathfrak{A}$$ and any quantifier-free $$L(A)$$-sentence $$\varphi$$, either $$\text{Diag}(\mathfrak{A})\models \varphi$$ or $$\text{Diag}(\mathfrak{A})\models \lnot \varphi$$. The interesting fact is that $$T$$ is model complete if and only if $$T\cup\text{Diag}(\mathfrak{A})$$ decides the truth of all $$L(A)$$-formulas, even those with quantifiers. The answer by spaceisdarkgreen explains the easy proof of this fact.

• Guess you mean $T \cup \operatorname{Diag}(\mathfrak A)$ decides the truth of all $L(A)$-formulas, not just $\operatorname{Diag}(\mathfrak A)$? – StefanH Dec 15 '18 at 17:52
• @StefanH "All extensions of $\mathfrak A$ are elementary" means exactly the same thing as "$Diag(\mathfrak A)$ is complete." If this holds and $\mathfrak A\models T,$ then all the extensions are models of $T$ as well. "$T$ is model complete" means all models of $T$ have this property. – spaceisdarkgreen Dec 15 '18 at 23:54
• @StefanH Yes, that was a typo! I misunderstood what you were referring to at first. Fixed. – Alex Kruckman Dec 16 '18 at 1:52