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.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 144 / 3 / a / Solution

The theory of dense linear orders without endpoints eliminates quantifiers. A finite partial order isomorphism extends by placing each new point in the corresponding interval or ray.


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