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The countable form of the omitting types theorem is as follows. Let $L$ be a countable first-order language and $T$ a consistent first-order theory in $L$. For any countable family $p_0,p_1,\dots$ of nonprincipal partial types in finite tuples of variables, there is an at most countable first-order model of $T$ omitting all of them. Nonprincipality means that no $T$-consistent first-order formula $\theta (\mathbf{x})$ entails every member of the type modulo $T$. A complete first-order theory with nonisolated complete types is the usual special case; neither an uncountable language nor an arbitrary uncountable family is covered by this statement.

Use the Godel completeness theorem to pass between consistency and the existence of a first-order model. Adjoin a countable stock $C$ of new constants. Construct finite conditions ${\Gamma }_s$ with $T\cup {\Gamma }_s$ consistent, interleaving three countable lists of requirements: decide each $L(C)$-sentence; provide a fresh constant witness for each existential sentence; and, for each $p_i$ and each tuple $\mathbf{t}$ of closed terms of its arity, add $\neg \psi (\mathbf{t})$ for some $\psi \in p_i$. Sentence decisions preserve consistency by choosing a consistent sign. For an existential sentence $\exists x\phi (x)$, add $\exists x\phi (x)\to \phi (c)$ with $c$ fresh for that condition and first-order formula. This is consistent: any first-order model of the old condition can interpret $c$ as a witness if one exists, and otherwise arbitrarily in the nonempty domain.

The omission step is the key Henkin omission extension lemma. Let $\sigma (\mathbf{c})$ be the logical conjunction of the current finite condition, listing every new constant occurring either there or in $\mathbf{t}$. If adding $\neg \psi (\mathbf{t})$ were inconsistent for every $\psi \in p_i$, then $T$ would entail $\sigma (\mathbf{c})\to \psi (\mathbf{t}(\mathbf{c}))$ for each such $\psi$. Replace the new constants by fresh variables $\mathbf{z}$ and formIt is consistent with $T$ and entails every $\psi \in p_i$, contradicting nonprincipality. Hence some omission extension is consistent. The construction need not be computable; countability merely permits all these requirements to be scheduled.

Let $T^{\ast }$ be the deductive closure of $T\cup {\bigcup }_s{\Gamma }_s$. It is consistent by the finite character of formal proofs, complete by sentence decisions, and has the Henkin witness property. Its term model consists of closed terms modulo provable logical equality and has at most countably many elements. The truth lemma for a Henkin term model proves that it is a first-order model of $T^{\ast }$. Every tuple in it is represented by closed terms, whose scheduled omission requirement supplies a negated member of each corresponding type. Thus

A universal sentence has the form $\forall \mathbf{x}\psi (\mathbf{x})$ with quantifier-free $\psi$, including an empty quantifier block. Embeddings preserve and reflect quantifier-free truth, so universal sentences pass to substructures of a first-order structure. For the converse Łoś-Tarski preservation theorem, let $T_{\forall }$ be all universal consequences of an arbitrary first-order theory $T$, and take $A\models T_{\forall }$. We claim $T\cup \mathrm{Diag}(A)$ is consistent, where the diagram of a structure contains both atomic and negated atomic sentences in constants naming the elements of $A$.

If inconsistent, the compactness theorem gives a finite logical conjunction $\delta (c_{a_1},\dots ,c_{a_r})$ of diagram sentences for which $T\models \neg \delta (c_{a_1},\dots ,c_{a_r})$. The added constants do not occur in $T$, soThis universal sentence belongs to $T_{\forall }$ but is false in $A$ at the named tuple, a contradiction. The compactness theorem therefore produces $B\models T$ containing an isomorphic embedded copy of $A$. The signed diagram ensures a genuine substructure of a first-order structure: functions are preserved, constants are included and relations are preserved and reflected. If the first-order model class of $T$ is closed under substructures of a first-order structure, this copy, and hence $A$, is a first-order model of $T$. We have provedNo countability assumption is needed for this second argument. An inconsistent first-order theory is covered as well, with a universally false axiom and an empty first-order model class.

An $L$-substructure of a first-order structure $A\subseteq B$ has a nonempty domain containing all named constants and closed under the functions of $B$, and each relation of $A$ is the restriction of the corresponding relation of $B$. Consequently every quantifier-free formula with parameters in $A$ has the same truth value in $A$ and $B$. This is weaker than being an elementary substructure, which preserves all first-order formulas.

A universal sentence has the form $\forall \mathbf{x}\psi (\mathbf{x})$ with quantifier-free formula $\psi$, including an empty quantifier block. Its truth passes from a first-order structure to every substructure of a first-order structure: tuples from the smaller domain are also tuples of the larger domain, and quantifier-free truth agrees. Therefore a first-order theory of universal sentences has a first-order model class closed under substructures of a first-order structure.