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By part (c), $T$ has a first-order model, so it is consistent. By part (d), every first-order sentence is equivalent modulo $T$ to a quantifier-free sentence. In the first-order signature consisting only of $E$ and first-order equality, there are no constant symbols, no ground first-order terms, and no nullary predicates. Thus a closed quantifier-free formula reduces to true or false, using the logical truth constants. All first-order models of $T$ therefore agree on every first-order sentence:

The construction in part (d) is effective. Conversion to disjunctive normal form, enumeration of finite equality partitions and adjacency patterns, and testing the finitely many witness cases are all terminating operations. Apply them recursively to a given first-order sentence and evaluate the final truth constant. This decides whether the first-order sentence belongs to $T$, givingNo useful bound on the running time is needed for decidability; the potentially large enumeration still always terminates.

Both assertions are false. For simple closure, take the inclusionBoth satisfy DAG prime, but the quantifier-free formula $x\ne 0$ has a witness in $N$ and none in $M$. The same example works in the reduced language $\{+,0\}$.

For quantifier elimination, the first-order sentence $\exists x(x\ne 0)$ distinguishes these two models. Every closed group term is zero, so every atomic closed equality is true in both models. Every quantifier-free sentence, being a Boolean combination of such equalities, has the same truth value in both. The distinguishing sentence therefore has no equivalent quantifier-free sentence modulo DAG prime. HenceIncluding the trivial group is exactly what makes the proposed simple closure condition fail.

Use the completeness of the first-order theory of dense linear orders without endpoints and its quantifier elimination. After naming finitely many increasing constants, quantifier elimination still holds in the expanded first-order language: treat these constants as fixed parameters when eliminating quantifiers. Every quantifier-free sentence is a Boolean combination of logical equalities and order comparisons of the named constants, all decided by their stated strict increasing order. Hence ${\mathrm{DLO}}_n$ is complete.

An alternative formal proof of this finite case uses the stated general theorem that a first-order theory in a countable first-order language with only infinite first-order models and a unique countable first-order model is complete: if first-order models disagreed on a first-order sentence, downward Löwenheim-Skolem would give disagreeing countable first-order models. Part (iii) supplies the uniqueness.

Every first-order sentence of $L_{\omega }$ mentions only finitely many constant symbols, hence belongs to some $L_n$. Its truth is decided by ${\mathrm{DLO}}_n$, and every first-order model of ${\mathrm{DLO}}_{\omega }$ has a reduct satisfying that finite first-order theory. Thus the larger first-order theory also decides every first-order sentence. The rational-order examples in (ii) establish first-order consistency. ConsequentlyThere is no contradiction with (ii): completeness is agreement on individual finite first-order sentences, while categoricity requires a global isomorphism respecting all constants simultaneously.