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Let $T_{\forall }$ denote the universal consequences of a theory: all universal first-order sentences entailed by $T$. A first-order structure satisfies $T_{\forall }$ exactly when it embeds into a model of $T$, by the compactness theorem applied to its diagram of a structure.

The theory has algebraically prime models if, for every $\mathcal{A}\models T_{\forall }$, there are $\mathcal{P}\models T$ and a structure embedding $i:\mathcal{A}\to \mathcal{P}$ such that every structure embedding $j:\mathcal{A}\to \mathcal{N}$, $\mathcal{N}\models T$, factors as $j=h\circ i$ for some structure embedding $h:\mathcal{P}\to \mathcal{N}$. Neither $i$ nor $h$ is required to be elementary.

For $\mathcal{M}\subseteq \mathcal{N}$, simple closure means that every existential quantifier-free formula over $M$ which has a witness in $N$ has one in $M$:Now take two models $\mathcal{M},\mathcal{N}\models T$ with common substructure $\mathcal{A}$. Since $A$ embeds into $M$, it satisfies $T_{\forall }$. Choose its algebraically prime extension $\mathcal{P}$, and embed $P$ into both $M$ and $N$ over $A$.

If $\exists x\theta (x,\bar{a})$ holds in $M$, the image of $P$ in $M$ is a model of $T$. The assumed simple closure of this image transfers a witness from $M$ into $P$. Its embedding into $N$ then transfers the quantifier-free formula and its witness into $N$. Thus the hypothesis of QET1 is satisfied. The second test follows: $T$ has quantifier elimination.

A model companion $T^{\ast }$ of a first-order theory $T$, in the same first-order language, is a model-complete theory with the same universal consequences of a theory as $T$:Equivalently, every model of $T$ embeds into a model of $T^{\ast }$, and every model of $T^{\ast }$ embeds into a model of $T$. The equivalence follows from the diagram embedding criterion for universal theories, which is an application of the compactness theorem. A model companion need not be a syntactic extension of $T$, and uniqueness is understood up to logical equivalence of theories.

Suppose $S$ and $R$ are model companions of $T$. Their universal consequences of a theory agree, so the diagram embedding criterion for universal theories permits embeddings in both directions between their model classes.

Starting from any $M_0\models S$, alternately take such extensions, identifying each model with its image under the embedding:Since $S$ is model-complete, $M_i\preccurlyeq M_{i+1}$; since $R$ is model-complete, $N_i\preccurlyeq N_{i+1}$. The two subsequences have the same union $U$. The elementary chain theorem givesEvery axiom of $R$, as a sentence true in $U$, is therefore true in $M_0$. Thus every model of $S$ is a model of $R$. Reversing their roles proves the converse. The model companion is unique up to logical equivalence. If $T$ is inconsistent, its only possible companion is likewise inconsistent, so the same uniqueness conclusion holds.