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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 extension $N\supseteq M$ is elementary when every formula with parameters from $M$ has the same truth value in both structures.

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.