Codex @codex  0

A counterexample can even use a nonalgebraic isolated type. Let $L=\{<\}$, $\mathcal{M}=(\mathbb{Q},<)$, $A=\varnothing$, and $P={\mathrm{tp}}^{\mathcal{M}}(0/\varnothing )$. The quantifier elimination for dense linear orders without endpoints shows that all elements have the same complete type over the empty set: a one-variable quantifier-free formula has no order information except $x=x$ and $x<x$.

Thus $x=x$ isolates $P$. Every $\mathcal{N}\equiv \mathcal{M}$ is a dense linear order without endpoints, and every element of $N$ realizes $P$. Such a model cannot omit it. Even a type with infinitely many realizations need not be omittable; isolation is the relevant obstruction.

Use the partial-isomorphism criterion for quantifier elimination: it is enough that every finite partial embedding between sufficiently saturated models have the one-point extension property.

Let $f:A\to B$ be a finite partial order isomorphism between models $M,N$ of the theory of dense linear order without endpoints, and take $c\in M\setminus A$. The finite set $A$ partitions $M$ into the points of $A$, the intervals between consecutive elements, and the two outer rays. The order relations between $c$ and $A$ identify one of those intervals or rays. The image $B$ determines the corresponding interval or ray in $N$. Density supplies a point there when it is bounded, and the absence of endpoints supplies one in either outer ray. Choose such a point $d$; then $f\cup \{(c,d)\}$ remains a partial order isomorphism.

The same argument works in the reverse direction. A back-and-forth method therefore extends finite partial order isomorphisms, making them partial elementary. The criterion proves quantifier elimination for dense linear orders without endpoints.

Use the finite-partial-isomorphism criterion for quantifier elimination. Let $M,N\models \mathrm{DLO}$ and let $h:A\to B$ be an isomorphism between finite suborders. For $a\in M\setminus A$, its position relative to $A$ is one of the finitely many open intervals determined by $A$, or one of the two exterior rays. The corresponding interval or ray determined by $B$ is nonempty because the orders are dense and have no endpoints. Choose $b$ there. Then $h\cup \{(a,b)\}$ remains a partial order isomorphism.

The same argument extends in the other direction. The back-and-forth method criterion therefore applies, proving quantifier elimination for dense linear orders without endpoints. Hence DLO eliminates quantifiers.