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For a finite relational first-order language, with finitely many constants allowed, the Duplicator wins the $n$-round game exactly when the structures are rank-n equivalent. Game-winning implies rank equivalence in arbitrary languages; the converse needs a finiteness condition and can fail even for an infinite unary vocabulary. Over a finite relational vocabulary, winning every announced finite horizon is equivalent to elementary equivalence.

Two first-order structures are elementarily equivalent when they satisfy the same first-order sentences. They need not be isomorphic or embed into one another. Every elementary embedding implies elementary equivalence of its source and target.

An elementary extension is an extension preserving all first-order truth with parameters, hence giving elementary equivalence. An extension $N\supseteq M$ is elementary when every formula with parameters from $M$ has the same truth value in both structures.

Let $T=\mathrm{Th}(\mathbb{C})$ in the finite language of rings: it is the set of all first-order sentences true in the field of complex numbers. It is satisfiable, with first-order model $\mathbb{C}$, so the countable-language Löwenheim-Skolem theorem gives an at most countable first-order model $K$ of $T$.

All field axioms belong to $T$, hence $K$ is a field. For every positive integer $n$, the first-order sentence saying that there are at least $n$ distinct elements is also true in $\mathbb{C}$ and belongs to $T$. Thus $K$ cannot be finite, and is countably infinite.

Finally, for each first-order sentence $\sigma$, either $\sigma \in T$ or $\neg \sigma \in T$. Since $K$ satisfies all of $T$, it satisfies $\sigma$ exactly when $\mathbb{C}$ does. ThereforeElementary equivalence concerns first-order first-order sentences and does not imply an isomorphism; the two fields have different cardinalities. The stronger downward form may even choose $K$ as a countable elementary subfield of $\mathbb{C}$.

Here an ultralimit in the model-theoretic sense is a model-theoretic ultralimit: start with a first-order structure, take successive ultrapowers along their diagonal elementary embeddings, and take the directed limit of elementary embeddings of that chain. The elementary chain theorem, proved by induction on formulas after identifying the embeddings with inclusions, makes every stage elementary in the limit. This definition concerns structures, rather than the topological ultrafilter limit.

We first prove the needed Frayne embedding lemma. If $C\equiv D$, every finite subset of the elementary diagram of a structure $D$ can be realized in $C$: existentially quantify its finitely many named elements, and use elementary equivalence. Index by finite diagram fragments, choose such realizations, and extend the filter of cones requiring each diagram formula to an ultrafilter. For each $d\in D$, let $f_d$ give its chosen coordinate realizations, filling unspecified coordinates arbitrarily. The Łoś theorem shows that $d\mapsto [f_d]$ is an elementary embedding of $D$ into an ultrapower of $C$. The formulas $c_d\ne c_e$ ensure injectivity.

There is a relative version: if $C⪯D$, keep all named elements from $C$ constant in every coordinate. Each finite diagram fragment is realizable in $C$ over those parameters by elementarity, so the resulting embedding extends the diagonal embedding of $C$ into its ultrapower.

Now use alternating ultrapowers give isomorphic ultralimits. First embed $B$ into an ultrapower $A_1$ of $A$. Apply the relative lemma to $B⪯A_1$, embedding $A_1$ into an ultrapower $B_1$ of $B$ over its diagonal copy. Next embed $B_1$ into an ultrapower $A_2$ of $A_1$ over $A_1$, and continue. After coherent identifications this givesEach $A_{n+1}$ is an ultrapower of $A_n$ along its diagonal embedding, and each $B_{n+1}$ is an ultrapower of $B_n$ likewise. The two subsequences are cofinal in the same chain of elementary substructures, so their limits are the same union. Hence $\boxed{A\equiv B⟹A_{\infty }\cong B_{\infty }}$. No isomorphism theorem for single ultrapowers has been assumed.

A model-theoretic ultralimit is the elementary direct limit of a chain of successive ultrapowers along their diagonal embeddings. We will build two such chains with isomorphic limits. This is not a topological ultrafilter limit and does not require the stronger claim that one ultrapower of each structure is already isomorphic.

First prove the Frayne embedding lemma, including the relative form needed to make the diagrams commute. If $C\equiv D$, expand the language by constants naming every element of $D$, and let $\mathrm{ED}(D)$ be its complete elementary diagram. Index by finite subsets $s$ of this diagram. The finitely many constants in $s$ can be interpreted in $C$ so that $s$ holds: existentially quantifying them gives a sentence true in $D$ and hence in $C$. Choose such interpretations, and give unused constants an arbitrary fixed value in the nonempty domain of $C$.

For each diagram sentence $\delta$, its cone $\{s:\delta \in s\}$ has the finite intersection property with the other cones. Extend the generated filter to an ultrafilter $U$. For $d\in D$, let $f_d(s)$ be its chosen interpretation at coordinate $s$. The mapis elementary by Łoś: every formula true of a tuple from $D$, or its negation, is a diagram sentence and holds on its corresponding cone. In particular distinct named elements stay distinct.

If $C⪯D$, the construction can keep every constant naming $c\in C$ equal to that same $c$ at every coordinate. For a finite diagram fragment, existentially quantify just the names outside $C$; the resulting formula with parameters in $C$ is true in $D$, hence true in $C$. Thus the finite realizations still exist with all those old names fixed. ConsequentlyThis relative extension of the diagonal map is the essential strengthening.

Now let $A_0\equiv B_0$. The nonrelative lemma embeds $B_0$ into an ultrapower $A_1$ of $A_0$. Identify the image with a copy of $B_0$ inside $A_1$. Since $B_0⪯A_1$, the relative lemma embeds $A_1$ into an ultrapower $B_1$ of $B_0$ extending its diagonal map. Rename this larger structure so that the displayed copy of $A_1$ is a literal elementary substructure. Next apply the relative lemma over $A_1⪯B_1$ to embed $B_1$ into an ultrapower $A_2$ of $A_1$ fixing its diagonal copy. Continue alternately. We obtain a coherent elementary chainwith $A_0⪯A_1$ as well. Each inclusion $A_n\to A_{n+1}$ is, up to these coherent identifications, the diagonal embedding into its ultrapower; the same holds for $B_n\to B_{n+1}$. It is the relative clause, not merely the existence of arbitrary elementary embeddings, which guarantees this.

By the elementary chain theorem, the unions of the two subsequences are elementary limits. They are the same structure: each $A_n$ lies in a subsequent $B_n$, and each $B_n$ lies in $A_{n+1}$. Hence the alternating ultrapowers give isomorphic ultralimits construction provesConversely each ultrapower and each elementary direct limit has the same first-order theory as its starting structure. Isomorphic ultralimits therefore imply $A_0\equiv B_0$ as well. Thus the conclusion exactly characterizes elementary equivalence at the level of ultralimits.

Under the finite relational vocabulary hypothesis of the Ehrenfeucht-Fraïssé theorem, II wins the timed game exactly when the structures are elementarily equivalent. If II wins, then for every choice of $n$ by I, its strategy restricts to a winning strategy in $G_n$. Hence the models agree on every sentence of quantifier rank at most $n$. Every first-order sentence has finite quantifier rank, so they agree on all sentences.

Conversely, elementary equivalence gives $\mathcal{M}{\equiv }_n\mathcal{N}$ for every $n$. Choose a winning strategy for each $G_n$, and use that strategy once I announces $n$. Unlike the infinite game, the timed game permits the strategy to depend on the announced horizon.

The PDF does not explicitly impose finite vocabulary. If an arbitrary language and the full partial isomorphism of structures winning condition are intended, the reverse assertion is false. Here is a failure of finite-round game equivalence in infinite languages. Let $L=\{P_i:i\in \mathbb{N}\}$ be unary relational. Letand obtain $N$ by adjoining one point $b$ satisfying every $P_i$.

For any finite sublanguage, every Boolean cell of predicates has infinitely many elements in both structures. Every finite partial isomorphism of structures of these reducts can therefore be extended in either direction. Induction on formulas, or a back-and-forth method, shows their reducts are elementarily equivalent. Since every sentence uses finitely many symbols, $M\equiv N$ in the whole language.

Nevertheless I wins $G_1(M,N)$ by choosing $b$: every point of $M$ fails some $P_i$, so no response preserves all atomic formulas. Thus the printed unrestricted equivalence requires the finite-vocabulary convention, or a game which additionally restricts each play to a finite vocabulary fragment.

Use the first-order language with one constant $c$, take $\mathcal{M}=(\mathbb{N},c=0)$ with equality only, and let $A=\varnothing$. SetThis complete type contains $x=c$ and is isolated by that formula. In any first-order structure, at most one element can satisfy $x=c$. In every $\mathcal{N}\equiv \mathcal{M}$, its interpretation of $c$ realizes $P$, by elementary equivalence. Hence no elementarily equivalent model has infinitely many realizations of $P$. An algebraic type can have a fixed finite number of realizations in every model.

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.