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For a first-order theory $T$ and $A\models T_{\forall }$, an algebraically prime extension is a structure embedding $i:A\to P\models T$ such that every embedding $j:A\to N\models T$ factors as $j=h\circ i$ for some structure embedding $h:P\to N$. The embeddings need not be elementary or unique. A theory has algebraically prime models when every such $A$ has an extension of this kind.

For a first-order theory $T$, suppose every existential quantifier-free formula over a common substructure has a witness in either of two $T$-models exactly when it has one in the other. Then $T$ has quantifier elimination. One may test a single quantified variable, since elimination of these formulas inductively eliminates all quantifiers. A language with a constant avoids empty-generated-substructure and quantifier-free-sentence conventions.

A Henkin construction enlarges a consistent first-order theory by fresh constants, decides sentences while preserving consistency, and supplies constant witnesses for existential sentences. Countably many additional dense requirements can be interleaved, as in the omitting types theorem. A complete consistent extension with witnesses has a term model.

Let $p(\mathbf{x})$ be a nonprincipal partial type over a consistent first-order theory $T$. Suppose $T\cup \{\sigma (\mathbf{c})\}$ is consistent, with $\sigma$ a finite condition in finitely many new constants, and let $\mathbf{t}(\mathbf{c})$ be a tuple of closed terms. Some $\psi \in p$ leaves $T\cup \{\sigma (\mathbf{c}),\neg \psi (\mathbf{t}(\mathbf{c}))\}$ consistent. Otherwise the original-language first-order formulais consistent with $T$ and entails every member of $p$, contradicting nonprincipality. Replace all new constants occurring in the condition or terms by variables. This lemma makes the omission requirements dense in a countable Henkin construction.

A first-order theory locally omits a family $\Gamma (\mathbf{x})$ if every formula $\theta (\mathbf{x})$ consistent with it has a consistent refinement $\theta \wedge \neg \gamma$ for some $\gamma \in \Gamma$. Equivalently no consistent formula implies all of $\Gamma$. In a countable language the omitting types theorem converts this local condition into a countable model with no realization of the family.

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.

The Gödel-Gentzen negative translation transforms $\Gamma {⊢}_{\mathrm{CL}}A$ into ${\Gamma }^N{⊢}_{\mathrm{IL}}{A}^N$. Induct on a natural deduction derivation. The translated logical conjunction, logical implication and universal rules are ordinary intuitionistic rules. Logical disjunction and existential introduction add double logical negation; their elimination rules first derive the double logical negation of the stable translated conclusion, then use its stability. Classical double-negation elimination translates precisely to stability. If logical equality atoms are double-negated, transport a stable translated conclusion through a logical equality by contradiction and then remove the resulting double logical negation. Original nonlogical axioms must also be translated; the result does not say arbitrary untranslated classical first-order theories are intuitionistically valid.

A partial type over a first-order theory $T$ is a set of first-order formulas in a fixed finite tuple of variables which is jointly consistent with $T$. A tuple realizes it when it satisfies every formula in the set. A complete type additionally decides every formula in those variables.

The local omission of a partial type $\Gamma (\mathbf{x})$ by a consistent first-order theory $T$ means that, for every formula $\theta (\mathbf{x})$ consistent with $T$, there is some $\gamma \in \Gamma$ such thatis consistent. Equivalently, no formula consistent with $T$ implies every member of $\Gamma$. Thus every possible local condition has a compatible refinement excluding at least one proposed type formula.

The omitting types theorem states: in a countable first-order language, if $T$ is consistent and locally omits each of countably many finite-arity families ${\Gamma }_j$, then it has a countable model omitting them all. Omission means that no tuple satisfies every formula in the family. In the case of a complete type, local omission is equivalent to the type being nonisolated. Countability of the language and of the list of omission requirements is essential to this formulation.

The Godel completeness theorem is strong completeness: for any first-order theory $T$ and sentence $\sigma$,Equivalently every syntactically consistent theory has a model. Work in the ordinary classical metatheory with choice; the language need not be countable. Soundness is induction on a derivation: logical axioms hold under every assignment, modus ponens preserves truth, and the quantifier rule preserves truth when its variable is not free in the assumptions. We prove the nontrivial converse by a Henkin construction.

Start with a consistent $T$ and add a constant if the language has no closed term. Let $L_0$ be this language. At stage $n$, for every existential sentence $\exists x\phi (x)$ of $L_n$, add a fresh constant $c_{\phi }$ and the witness axiomAn axiom of this form is conservative for consistency: if a contradiction followed after adding it, the old theory would prove its negation, hence both $\exists x\phi (x)$ and $\neg \phi (c_{\phi })$. Since the constant is fresh, replace it by a fresh variable and generalize, obtaining $\forall x\neg \phi (x)$, a contradiction. Every finite set of new witness axioms can be handled one at a time, so each stage remains consistent. Their union $T_H$ is consistent because a finite proof uses only finitely many stages.

Well-order the sentences of $L_H={\bigcup }_n{L}_n$. Successively add a sentence if doing so is consistent, and otherwise add its negation. At least one of those two choices preserves consistency: otherwise the deduction theorem would give both its negation and its double negation in the previous theory. At limits take unions. The deductive closure $\Gamma$ of the resulting union is complete and consistent and retains all witness axioms. Every sentence of $L_H$ uses finitely many symbols and belongs to some $L_n$, so its existential witness was added at the next stage. Thus $\Gamma$ has the Henkin witness property even for formulas using the new constants.

Build its term model. The domain consists of closed terms modulo $s\sim t$ iff $\Gamma ⊢s=t$. Equality axioms make this an equivalence relation and a congruence. Interpret functions by forming their corresponding terms, and interpret a relation $R$ by $R([t_1],\dots ,[t_r])$ iff $R(t_1,\dots ,t_r)\in \Gamma$. Congruence makes this independent of representatives.

The truth lemma for a Henkin term model is proved by induction on formulas after substituting closed terms for their free variables. Atomic formulas are the definition. Completeness and consistency of $\Gamma$ give the Boolean steps. For existence, if $\exists x\phi (x,\bar{t})\in \Gamma$, a witness axiom yields a constant $c$ with $\phi (c,\bar{t})\in \Gamma$, and induction supplies a model witness. Conversely every domain element is a term class, so a model witness gives an instance in $\Gamma$, and existential introduction gives the existential sentence. Universality follows by negation and existence. The model therefore satisfies $\Gamma$, hence $T$.

Finally, if $T⊬\sigma$, then $T\cup \{\neg \sigma \}$ is consistent by the deduction theorem and classical logic. Its constructed model is a countermodel to $T\models \sigma$. This proves completeness, with an actual witness model rather than merely a finite-satisfiability assertion.

For each $n$, let ${\sigma }_n$ assert the existence of $n$ distinct elements. If a theory $T$ has arbitrarily large finite models, every finite subset of $T\cup \{{\sigma }_n:n\ge 1\}$ has a model. By the compactness theorem, this extended theory has a model, necessarily infinite. Since fields of size $2^j$ are arbitrarily large, no first-order theory can have exactly the finite fields of characteristic two as its models.

Suppose instead a finite conjunction $\chi$ axiomatized exactly the infinite fields of characteristic two. Let $T_2$ be the field axioms with $1+1=0$. Then $T_2\cup \{\neg \chi \}$ would have exactly the finite characteristic-two fields as models, contradicting the preceding conclusion. Equivalently, compactness of $T_2\cup \{{\sigma }_n\}\models \chi$ would force some finite size threshold, but arbitrarily large finite fields still fail $\chi$. The infinite characteristic-two fields are not finitely axiomatizable.

Here is the existential common-substructure test for quantifier elimination. Use a first-order language with a constant symbol, as in the applications below, or the usual conventions for empty generated substructures and truth constants. Let $T$ be a first-order theory. Suppose that whenever $\mathcal{M},\mathcal{N}\models T$ have a common substructure $\mathcal{A}$, then, for every quantifier-free formula $\theta (x,\bar{y})$ and every finite tuple $\bar{a}\in A$,The condition is symmetric because it applies also with $\mathcal{M}$ and $\mathcal{N}$ exchanged. The conclusion is quantifier elimination for $T$: for every first-order formula $\phi (\bar{y})$ there is a quantifier-free formula $\psi (\bar{y})$ such thatThe common substructure need not itself satisfy $T$, and witnesses are sought in the whole models, not necessarily in $A$. This test is the QET1 version of the common-substructure test for quantifier elimination.

A first-order theory $T$ is model-complete if every structure embedding between models of $T$ is an elementary embedding. Equivalently, whenever $\mathcal{M}\subseteq \mathcal{N}$ and both satisfy $T$, one has $\mathcal{M}\preccurlyeq \mathcal{N}$:for every first-order formula $\phi$ and every finite tuple $\bar{a}\in M$. This is preservation of all formulas with parameters, rather than merely all first-order sentences.

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.

A precise countable-language form of the omitting types theorem is the following. Let $T$ be a consistent first-order theory in a countable first-order language, and let $(p_j({\bar{x}}_j){)}_{j\in \mathbb{N}}$ be a countable family of nonprincipal partial types, each in finitely many variables. Then there is a finite or countably infinite model of $T$ omitting all the $p_j$.

A partial type is a set of formulas jointly consistent with $T$. It is principal if there is a formula $\theta (\bar{x})$ such that $T\cup \{\exists \bar{x}\theta \}$ is consistent andFor a complete type, this is exactly being an isolated type. Nonprincipal means that no such formula exists. A model omits $p$ if no tuple in that model satisfies every formula in $p$; different tuples may fail different members of the type. The model in the theorem need not omit each individual formula everywhere.

For parameter types, use the expanded language naming the parameters. Countability then applies to that expanded language, and hence requires a countable parameter set as well. There is no countability restriction in the finite-satisfiability argument of part (a).

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.

For a first-order theory $T$ in a first-order language $L$, a complete $n$-type is a maximal $T$-consistent set $p(x_1,\dots ,x_n)$ of formulas whose free variables lie among $x_1,\dots ,x_n$. Equivalently, it chooses exactly one of $\phi$ and $\neg \phi$ for every such formula while remaining consistent with $T$.

An isolated type $p$ is isolated by a formula $\theta (\bar{x})$ when $T\cup \{\exists \bar{x}\theta (\bar{x})\}$ is consistent andfor every $\phi \in p$. The type is an omitted type in an $L$-structure $M$ when no tuple $\bar{a}\in M^n$ satisfies every formula in $p$.

The omitting types theorem states that if $T$ is a consistent theory in a countable language and $(p_i{)}_{i\in \mathbb{N}}$ is a countable family of nonisolated finite-arity types, then $T$ has a countable model omitting every $p_i$.