Let ${\sf DLO}$ be the theory of dense linear orders without endpoints. Let $c_i$, for $i \in \mathbb N$, be new constant symbols, and let ${\sf DLO}'={\sf DLO}\cup \{ c_i < c_{i+1} \mid i \in \mathbb N\}$. Prove that ${\sf DLO}'$ is a complete theory.
Here, ${\sf DLO}$ has its underlying language as $\mathcal L = \{ < \}$. We know that ${\sf DLO}$ is complete and admits quantifier elimination, and that ${\sf DLO}$ has a model: $\mathbb Q$ with the usual $<$. We also know that any quantifier free formula in ${\sf DLO}$ is precisely a formula that asserts a relative ordering of any variables appearing in it (this includes contradictory formula such as $x_1 < x_2 \land x_2 < x_1$). I am not sure how to show that ${\sf DLO}'$ is complete. I tried proving by contradiction. Let $\mathcal L' := \mathcal L \cup \{ c_i \mid i \in \mathbb N \}$. Suppose there existed an $\mathcal L'$-sentence $\sigma'$ such that ${\sf DLO}' \cup \{ \sigma' \}$ and ${\sf DLO}' \cup \{ \lnot \sigma' \}$ have models, say $\mathcal A'$ and $\mathcal B'$ respectively. Then we want to somehow deduce that there is some $\mathcal L$-sentence $\sigma$ such that ${\sf DLO} \cup \{ \lnot \sigma \}$ and ${\sf DLO} \cup \{ \sigma \}$ both have models. I am not sure if I can prove that the reduct $\mathcal A$ of $\mathcal A'$ to the language $\mathcal L$ satisfies ${\sf DLO}' \cup \{ \sigma \}$ for some $\sigma$ (and similarly for $\mathcal B$ a reduct of $\mathcal B'$ for $\lnot \sigma$). I know that we can use Godel's completeness theorem. How can I proceed?