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Take $I=\{0,1\}$ and the proper filter on a set $F=\{I\}$. In the first-order language with unary predicates $P,Q$, let both factors be one-element first-order structures. Set $P$ true and $Q$ false in the first factor, and reverse these truth values in the second.

The reduced product also has one element $a$. Neither $P(a)$ nor $Q(a)$ holds there: their truth sets are $\{0\}$ and $\{1\}$, neither belonging to $F$. ThusLogical disjunction can therefore break the equivalence. A union can belong to a filter on a set without either summand belonging to it; the corresponding union property does hold for an ultrafilter.

Let $T_{\forall }$ denote the universal consequences of a theory: all universal first-order sentences entailed by $T$. A first-order structure satisfies $T_{\forall }$ exactly when it embeds into a model of $T$, by the compactness theorem applied to its diagram of a structure.

The theory has algebraically prime models if, for every $\mathcal{A}\models T_{\forall }$, there are $\mathcal{P}\models T$ and a structure embedding $i:\mathcal{A}\to \mathcal{P}$ such that every structure embedding $j:\mathcal{A}\to \mathcal{N}$, $\mathcal{N}\models T$, factors as $j=h\circ i$ for some structure embedding $h:\mathcal{P}\to \mathcal{N}$. Neither $i$ nor $h$ is required to be elementary.

For $\mathcal{M}\subseteq \mathcal{N}$, simple closure means that every existential quantifier-free formula over $M$ which has a witness in $N$ has one in $M$:Now take two models $\mathcal{M},\mathcal{N}\models T$ with common substructure $\mathcal{A}$. Since $A$ embeds into $M$, it satisfies $T_{\forall }$. Choose its algebraically prime extension $\mathcal{P}$, and embed $P$ into both $M$ and $N$ over $A$.

If $\exists x\theta (x,\bar{a})$ holds in $M$, the image of $P$ in $M$ is a model of $T$. The assumed simple closure of this image transfers a witness from $M$ into $P$. Its embedding into $N$ then transfers the quantifier-free formula and its witness into $N$. Thus the hypothesis of QET1 is satisfied. The second test follows: $T$ has quantifier elimination.

The quantifier rank measures maximum nesting depth:Two first-order structures are rank-n equivalent if they agree on all first-order sentences of quantifier rank at most $n$. For tuples one similarly compares all such formulas evaluated at the tuples.

In the $n$-round Ehrenfeucht-Fraïssé game, player I, the Spoiler, chooses an element from either structure each round; player II, the Duplicator, chooses an element in the other. After $n$ rounds the corresponding chosen tuples must define a partial isomorphism of structures: all atomic formulas, including equality and the interpretations of constants, have matching truth values. Repetitions must be respected. Failure of this condition is a loss for II.

The precise finite-vocabulary Ehrenfeucht-Fraïssé theorem is: for a finite relational language, allowing finitely many constants,The tuple version has the same conclusion when the corresponding tuples are fixed before play. The implication from a winning strategy to rank equivalence holds for arbitrary languages, by induction on formulas. For the converse, finite relational vocabulary ensures only finitely many rank-bounded types in a fixed tuple of variables; their defining formulas let one transfer the type of the next move. Without a finiteness hypothesis, the converse for the full atomic-diagram game need not hold. This qualification matters in the next part.

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.

For parameters $a_1,\dots ,a_n\in M$, set-theoretic absoluteness means agreement of the two membership first-order structures:This must hold for every such parameter tuple. Equivalently, formula relativization to a class produces two formulas ${\phi }^M$ and ${\phi }^N$ whose truth values agree on tuples from $M$. Agreement for parameters only in $N\setminus M$ is not part of the definition. For a proper class in set theory, this notation is understood separately for each fixed first-order formula, rather than as an unrestricted truth predicate for the entire universe.

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.

A quantifier-free first-order formula is built from atomic formulas using logical connectives, with no quantifiers. An embedding of first-order structures preserves and reflects its truth, by mathematical induction on the first-order formula. Both positive and negated atomic information are needed for this conclusion in a relational language.

Two first-order structures are rank-n equivalent if they agree on all sentences of quantifier rank at most $n$. The tuple version compares formulas at fixed tuples. For finite relational vocabulary this is equivalent to a Duplicator winning strategy in the $n$-round Ehrenfeucht-Fraïssé game.

Given nonempty first-order structures ${\mathcal{M}}_i$ and a proper filter on a set $F$, a reduced product identifies product functions $f,g$ when $\{i:f(i)=g(i)\}\in F$. Operations are interpreted coordinatewise, and relations hold when their coordinate truth sets belong to $F$. For an ultrafilter this is an ultraproduct. Primitive positive formulas transfer through an arbitrary proper filter, whereas logical disjunction and logical negation can fail to transfer.

The satisfaction relation says that the finite code $e$ of a first-order formula is true in a set-sized first-order structure $A$ at assignment $\vec{a}$. It is defined uniformly in set theory by recursion on the finite syntax tree: evaluate atomic relations, Boolean operations and quantifiers over the given domain. A code includes only finitely many subformulas, and their truth relations are sets. This does not define a truth predicate for the ambient universe, whose domain is not a set. Finite relation closure for set-theoretic coding gives an explicit closure criterion ensuring that these truth relations are computed correctly in a transitive set.

Expand an infinite first-order structure by Skolem functions, choose distinct elements $b_n$, and fix a nonprincipal ultrafilter $\mathcal{U}$ on $\mathbb{N}$. For each finite subset $F$ of the required index order, form the ultrapower over ${\mathbb{N}}^F$ using the ordered Fubini product of ultrafilters. Pullback along coordinate projections gives coherent elementary maps by the Łoś theorem. In their directed limit of elementary embeddings, the classes of the coordinate functions $\mathbf{n}\mapsto b_{n_i}$ are distinct and form an order-indiscernible sequence; every first-order formula on increasing coordinates has the same nested ultrafilter test. Their Skolem hull gives an Ehrenfeucht-Mostowski model. An index-order automorphism sends a term in the generators to the same term in the transported generators; indiscernibility makes this well defined. Uniqueness is in the specified Skolem expansion, not necessarily in its reduct. If generators must respect a definable infinite order, first produce an increasing sequence by an ultrapower of finite increasing chains.

Universal quantification forms $\forall x\phi (x,\mathbf{y})$. In a first-order structure, it is true at an assignment to $\mathbf{y}$ when every domain element assigned to $x$ makes the instance true. Its introduction rule requires that the generalized variable not occur freely in undischarged assumptions.

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.