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A finite forbidden relational pattern can be excluded by a universal sentence. If a finite structure $F$ has distinct vertices $1,\dots ,m$, let ${\delta }_F(x_1,\dots ,x_m)$ say that the variables are distinct and have the required relations. Thenforbids that pattern. Taking one sentence for each forbidden finite pattern gives a universal first-order theory.

For a triangle-free graph, the forbidden diagram asserts that three distinct vertices are pairwise adjacent. To forbid an induced pattern, the diagram must specify absent relations as well as present ones. Forbidding an ordinary path on three vertices allows extra edges between its vertices; forbidding an induced path on three vertices does not. Thus the meaning of pattern determines the exact diagram. The construction explains why many local graph constraints are preserved under taking substructures.

The values of all closed terms in a first-order structure form a substructure when the first-order language has at least one constant. The constant ensures nonemptiness, and applying a function symbol to closed-term values again gives a closed-term value. Relations are restricted from the ambient first-order structure. Every universal sentence true in the ambient first-order structure remains true in this substructure.

A first-order theory has its class of first-order models closed under substructures of a first-order structure exactly when it is equivalent to its universal consequences $T_{\forall }$. If $A\models T_{\forall }$, then $T\cup \mathrm{Diag}(A)$ is consistent. Otherwise a finite logical conjunction $\delta$ of signed diagram of a structure sentences yields a universal consequence $\forall \mathbf{x}\neg \delta (\mathbf{x})$ false in $A$. The compactness theorem gives a first-order model of $T$ containing an embedded copy of $A$, and closure under substructures of a first-order structure implies $A\models T$. The converse is preservation of universal sentences. The diagram of a structure must include negated as well as positive atomic sentences so that the resulting map is an embedding, not just a homomorphism.

Name every element $a\in A$ by a new constant $c_a$, distinct from the symbols already in $L^{\prime }$. Let $\mathrm{Diag}(A)$ be its full atomic and negated atomic diagram of a structure. A model of this diagram contains an isomorphic copy of $A$ as an $L$-substructure. We claimis satisfiable.

Otherwise compactness theorem gives a finite diagram conjunction $\delta (c_{a_1},\dots ,c_{a_m})$ inconsistent with $T^{\prime }$. Replace these new constants by variables. Since $T^{\prime }$ says nothing about the new names, it then entails the universal sentenceThe conjunction $\delta$ uses only symbols of $L$, so this is one of the universal $L$-consequences assumed true in $A$. But $A$ satisfies $\delta (a_1,\dots ,a_m)$, a contradiction. A model of the consistent enlarged theory gives an embedding of $A$ into the $L$-reduct of a model of $T^{\prime }$. Negated atomic facts and disequalities in the diagram ensure that this is a structure embedding, not merely a homomorphism.

Let $U$ be all universal sentences true in $C$. The theory $T\cup U$ is consistent. Indeed, if a finite conjunction $u$ of sentences from $U$ were inconsistent with $T$, then $T\models \neg u$. A finite conjunction of universal sentences is universal, so its negation is equivalent to an existential sentence. By the hypothesis $C$ would satisfy $\neg u$, whereas it satisfies $u$.

Choose $A\models T\cup U$. Now expand the language by names for every element of $C$ and consider its elementary diagram $\mathrm{ElDiag}(C)$. Its universal consequences in the original language are exactly $U$: any such consequence is true in the named copy of $C$, while any original-language sentence true in $C$ already belongs to its elementary diagram. Part (b), applied to $A$ and this expanded theory, gives a structure embedding of $A$ into a model $D$ of $\mathrm{ElDiag}(C)$. The named copy of $C$ is an elementary substructure of $D$. Identifying that copy with $C$, we obtainThus the required target is an elementary extension of $C$, although $C$ itself need not be a model of $T$.

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.

Here $\mathcal{B}\subseteq \mathcal{A}$ means a substructure, so the smaller carrier contains all named constants, is closed under every function symbol, and has each relation restricted from the larger first-order structure. First-order term evaluation on tuples from $B$ agrees in the two first-order structures. Consequently atomic formulas agree, and induction on Boolean connectives makes every quantifier-free formula agree as well.

Write an arbitrary universal sentence of $T$ as $\forall \vec{x}\theta (\vec{x})$ with quantifier-free matrix. Every tuple from $B$ is also a tuple from $A$, so $\theta$ holds at that tuple in $\mathcal{A}$. Quantifier-free agreement transfers it to $\mathcal{B}$. Since the tuple was arbitrary, $\mathcal{B}$ satisfies the universal sentence. ThusAn arbitrary subset without the substructure operations and relation restrictions would not be the hypothesis used by this formal proof.

If $B\models T$ and $B\subseteq A^{\ast }\equiv A$, every universal sentence of $A$ holds in $A^{\ast }$ and is preserved by passage to the substructure $B$. Thus $B$ witnesses satisfiability of the joint theory.

For universal theory and substructures of elementarily equivalent models, conversely let $B$ satisfy $T$ together with the universal theory of $A$. Every existential sentence true in $B$ must be true in $A$: if an existential $\sigma$ were false in $A$, its universal negation would belong to ${\mathrm{Th}}_{\forall }(A)$ and hold in $B$, a contradiction. Hence ${\mathrm{Th}}_{\exists }(B)\subseteq \mathrm{Th}(A)$. In particular $\mathrm{Th}(A)\cup {\mathrm{Th}}_{\exists }(B)$ is satisfiable, with model $A$. Apply the preceding embedding result to the theory $\mathrm{Th}(A)$ and structure $B$. It supplies a structure $A^{\ast }\models \mathrm{Th}(A)$ containing $B$. Therefore $A^{\ast }\equiv A$ and $B\subseteq A^{\ast }$, as required. Elementary equivalence here does not assert that $A$ itself contains a copy of $B$.

Existential sentences are preserved under extensions: witnesses remain present and quantifier-free truth is unchanged by an embedding. Therefore any existential axiomatization implies the specified extension-closure property.

For existential axiomatization from extension closure, let $E$ be all existential consequences of $T$, and suppose $A\models E$. We claim that $T\cup {\mathrm{Th}}_{\forall }(A)$ is satisfiable. Otherwise compactness supplies finitely many universal sentences ${\theta }_1,\dots ,{\theta }_m$ true in $A$ such that $T\models \neg ({\theta }_1\wedge \cdots \wedge {\theta }_m)$. The finite conjunction is universal after renaming bound variables, so its negation is existential. It belongs to $E$ but is false in $A$, a contradiction.

The preceding universal-theory embedding result now yields $B\models T$ and $A^{\ast }\equiv A$ with $B\subseteq A^{\ast }$. By the assumed extension closure, $A^{\ast }\models T$; elementary equivalence implies $A\models T$. Thus $E$ axiomatizes exactly the models of $T$, proving existential axiomatizability is equivalent to closure under extensions. This includes the inconsistent case, whose empty model class is axiomatized by an existential contradiction.

No. Universal sentences are preserved under substructures. Every infinite $K$-vector space has its zero subspace as an $L$-substructure, since zero is named and every scalar multiplication preserves it. This one-element structure is not infinite and is not a model of $T$. Thus the model class is not closed under substructures and has no universal axiomatization, even relative to the vector-space axioms.

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.