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An atomic first-order formula is a relation applied to terms, or a logical equality of two terms when logical equality is included in the logic. It contains no logical connective or quantifier. Its truth is preserved and reflected by an embedding of first-order structures.

Classical first-order logic has the quantifier and logical equality rules of natural deduction together with unrestricted double-negation elimination, or equivalently the law of excluded middle. It is interpreted in ordinary two-valued first-order structures.

Let $M$ be an infinite first-order structure and $I$ a total order. After choosing a Skolem expansion of $M$, an elementary extension contains distinct elements $(a_i{)}_{i\in I}$ forming an order-indiscernible sequence in that expanded language. Their Skolem hull is an elementary substructure in the expanded language, and its reduct is a first-order model of $\mathrm{Th}(M)$. Every order automorphism of $I$ extends uniquely to a structure automorphism of the hull preserving the chosen Skolem expansion, by transporting terms in the generators.

Uniqueness need not hold among all structure automorphisms of the reduct. For an infinite structure in the pure logical equality language, take two distinct nullary Skolem function values $c_0,c_1$ in the hull. For an infinite order-indiscernible sequence, neither can equal a generator: the expanded formula $x=c_j$ would otherwise hold at every generator, contradicting their distinctness. Swapping $c_0,c_1$ fixes every generator and preserves the pure logical equality reduct, while failing to preserve the Skolem expansion. Thus the identity order automorphism already has two extensions in that reduct.

For an atomic formula $P$ (including logical equality), put $P^N=\neg \neg P$, and put ${\perp }^N=\perp$. Preserve logical conjunction, logical implication and universal quantification recursively, while settingLogical negation is logical implication to logical falsity, so $(\neg A{)}^N=\neg A^N$. The displayed logical disjunction and existence clauses are intuitionistically equivalent to the usual negative forms $\neg (\neg A^N\wedge \neg B^N)$ and $\neg \forall x\neg A^N$. Every translated first-order formula is stable by stability of a formula under double negation. In classical logic the translation is equivalent to the original first-order formula by mathematical induction and double-negation elimination.

Intuitionistic first-order logic has the usual introduction and elimination rules for logical connectives, quantifiers and logical equality, but no unrestricted double-negation elimination or law of excluded middle. Its proofs retain the witness and logical disjunction information that can be lost in classical first-order logic.

In first-order logic with logical equality, $s=t$ asserts that the terms denote the same domain element. Logical equality is interpreted as identity rather than as an arbitrary binary predicate. Its natural deduction rules include reflexivity and substitution: equal terms can replace one another in a first-order formula, subject to capture-avoiding substitution. Double-negating logical equality atoms in a Gödel-Gentzen negative translation preserves these rules when the transported translated first-order formula is stable.

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.

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.

One form of the Ehrenfeucht-Mostowski theorem says that, for every infinite first-order structure $M$ and every total order $I$, an elementary extension contains distinct elements $(a_i{)}_{i\in I}$ forming an order-indiscernible sequence. After a suitable Skolem expansion, their Skolem hull is a first-order model of $\mathrm{Th}(M)$, and every order automorphism of $I$ extends to a structure automorphism of that hull. If $M$ has a definable infinite linearly ordered subset $P$, the generators may lie in $P$ and their order may agree with that definable order. Uniqueness of the extended structure automorphism is asserted in the chosen Skolem expansion, rather than for every structure automorphism of its reduct.

Here is an ultraproduct proof of the Ehrenfeucht-Mostowski theorem. First expand $M$ to $M^{+}$ with Skolem functions for all existential first-order formulas, iterating through the enlarged languages if necessary. This is done in the ordinary classical metatheory; the choice restriction in question 1 is not a restriction on this question. Fix a nonprincipal ultrafilter $\mathcal{U}$ on $\mathbb{N}$.

We can obtain an infinite sequence of distinct elements $b_0,b_1,\dots$ in an elementary extension $M_0$ of $M^{+}$ by one preliminary ultrapower. For each $n\ge 1$ choose a finite list of $n$ distinct elements of $M^{+}$. Represent $b_j$ by the function taking the $j$th entry when $n>j$, with an arbitrary default otherwise. Distinctness holds on a cofinite set, so follows from Łoś theorem. In the definably ordered variant, first name any parameters defining $P$, and choose each finite list increasingly inside $P$ instead; then $M_0\models P(b_j)$ and $b_j<b_k$ for $j<k$. Infinitude of an ordered set supplies every finite increasing chain. This preparatory step uses no Ramsey theorem.

For a finite subset $F=\{i_1<\cdots <i_r\}$ of $I$, define the ultrafilter ${\mathcal{U}}_F$ on ${\mathbb{N}}^F$ by nested membership, in this order:where $(\mathcal{U}n)Q(n)$ means $\{n:Q(n)\}\in \mathcal{U}$. This is the ordered Fubini product of ultrafilters. Closure under intersections and the decision between a set and its complement hold at each nested level, proving it is an ultrafilter. For $F=\varnothing$ take the principal ultrafilter on the one-point product. PutIf $F\subseteq G$, pull a function back along the projection ${\mathbb{N}}^G\to {\mathbb{N}}^F$. A test independent of an omitted coordinate is unchanged by its ultrafilter quantifier, so projection pushes ${\mathcal{U}}_G$ to ${\mathcal{U}}_F$. The induced map $N_F\to N_G$ is therefore an elementary embedding by Łoś theorem. These maps are coherent, giving a directed system indexed by finite subsets of $I$.

Take its directed limit of elementary embeddings $N$. One can verify elementarity directly: representatives of a finite tuple occur at a common stage; functions and atomic relations are interpreted there. In the existential step, a witness in the limit occurs with the parameters at some later common stage, and elementarity pulls the existence statement back. Thus every $N_F$ embeds elementarily into $N$, which contains an elementary copy of $M^{+}$.

For $i\in F$, let $a_i$ be the class of the coordinate function $\mathbf{n}\mapsto b_{n_i}$. Projection coherence makes this independent of $F$. If $i<j$, then for each fixed $n_i$ the set $\{n_j:n_j\ne n_i\}$ is cofinite. The nested test therefore gives $a_i\ne a_j$. In the ordered variant, the same argument with $n_j>n_i$ gives $P(a_i)$ and $a_i<a_j$.

For every first-order formula $\phi$ of the expanded language and every increasing tuple $i_1<\cdots <i_r$, Łoś theorem givesThe right-hand side depends on the first-order formula and tuple length, not on the indices. This proves order indiscernibility in the expanded language. Let $H$ be the Skolem hull of the generators in $N$. The Tarski-Vaught test gives $H\prec N$, and its reduct is a first-order model of $\mathrm{Th}(M)$.

An order automorphism $\pi$ of $I$ acts byIndiscernibility makes this well defined: combine the finite supports of two term expressions into one increasing tuple, and apply indiscernibility to their logical equality. Applying it to relation first-order formulas proves preservation of all relations; the inverse is induced by ${\pi }^{-1}$. Every element of the hull is such a term, so the extension is unique among structure automorphisms preserving the Skolem expansion. MoreoverThis proves the theorem, including its ordered version and structure automorphism conclusion, using ultrapowers and a directed limit throughout.

Use the Gödel-Gentzen negative translation, writing $A^N$ for the translated first-order formula. On atomic formulas $P$, including logical equality, put $P^N=\neg \neg P$, and put ${\perp }^N=\perp$. Extend recursively byLogical negation abbreviates logical implication to logical falsity, so $(\neg A{)}^N=\neg A^N$. The displayed logical disjunction and existential clauses are intuitionistically equivalent to the usual negative clauses $\neg (\neg A^N\wedge \neg B^N)$ and $\neg \forall x\neg A^N$, respectively. Thus these clauses specify the same negative interpretation. Translation commutes with capture-avoiding substitution of terms for free variables.

First prove by structural induction the stability of a formula under double negation:Every double logical negation is stable, since intuitionistic first-order logic proves $\neg \neg \neg \neg C\to \neg \neg C$; logical falsity is stable as well. Stability passes to logical conjunction by obtaining double logical negation of each component. For a logical implication $C\to D$ with stable $D$, assume $\neg \neg (C\to D)$ and $C$. An assumption $\neg D$ would give $\neg (C\to D)$, a contradiction, so $\neg \neg D$ and then $D$. For a universal first-order formula, $\neg \neg \forall xC(x)$ implies $\neg \neg C(y)$ for arbitrary fresh $y$; use stability pointwise and generalize. Logical disjunction and existential translations are already double negations. This proves all cases.

Now regard classical first-order natural deduction as intuitionistic natural deduction plus unrestricted double-negation elimination, and induct on a classical derivation. The rules for logical conjunction, logical implication, universal quantification and logical falsity translate directly. Logical disjunction introduction gives $A^N\vee B^N$ and then its double logical negation. For logical disjunction elimination, the translated premise is $\neg \neg (A^N\vee B^N)$ and the two translated branches yield the stable conclusion $C^N$. Assuming $\neg C^N$ makes each branch contradictory, giving $\neg A^N$ and $\neg B^N$, hence $\neg (A^N\vee B^N)$, which contradicts the premise. We have $\neg \neg C^N$ and remove it by stability.

Existential introduction likewise adds a double logical negation to the ordinary witness introduction. In existential elimination, the witness branch $A^N(y)⊢C^N$ has the original fresh-variable condition. Assuming $\neg C^N$ makes that branch give $\neg A^N(y)$; generalization gives $\forall y\neg A^N(y)$ and therefore $\neg \exists y{A}^N(y)$, contradicting the translated premise. Again stability supplies $C^N$. The universal-rule side conditions are preserved because the translation introduces no new free variables.

For logical equality, reflexivity gives $t=t$ and then its double logical negation. For substitution, from $\neg \neg (s=t)$ and $A^N(s)$, temporarily assume $\neg A^N(t)$. An assumption $s=t$ would transport $A^N(s)$ to $A^N(t)$ by intuitionistic logical equality substitution, so it gives a contradiction and hence $\neg (s=t)$. The first premise contradicts this. Thus $\neg \neg A^N(t)$ holds and stability gives $A^N(t)$. Finally a classical double-negation elimination step translates precisely to $\neg \neg A^N\to A^N$, already proved. Every rule is therefore covered, giving negative translation of a classical proof:In particular a classical thesis has an intuitionistic, constructive translated proof. If nonlogical assumptions are present, they must be translated too.

In classical logic, double-negation elimination gives $P^N\leftrightarrow P$ on atoms. Structural induction propagates equivalence through logical conjunction, logical implication and both quantifiers, and removes the added double negations on logical disjunction and existence. Therefore