Is the theory of linear dense orders with distinct endpoints complete? I'm trying to solve some problems about elementarily equivalent structures. For example, I know that the structures $\langle\mathbb{Q},<\rangle$ and $\langle\mathbb{R},<\rangle$ are elementarily equivalent because they are both models of the same complete theory: the theory of linear dense orders with no endpoints.
So in order to show that structures of the form $\langle[a,b],<\rangle$ are elementarily equivalent to structures of the form $\langle[a,b]\cap\mathbb{Q},<\rangle$, I want to know if the theory of linear dense orders with distinct endpoints is complete. If it's the case, I would highly appreciate it if someone can give me a hint on how to show that this theory is in fact complete.
Update: I'm trying to show that the theory $T$ of linear dense orders with distinct endpoints has only one countably infinite  model up to isomorphism. $\langle[0,1]\cap\mathbb{Q},<\rangle$ is one of such models. Let $\langle A,<\rangle$ be a countably infinite model of $T$. Let $m=\min A$ and $M=\max A$. I know that there exists an isomorphism $\varphi$ from $\langle(m,M),<\rangle$ to $\langle (0,1)\cap\mathbb{Q},<\rangle$. So let $\psi:A\to[0,1]\cap\mathbb{Q}$ be defined by $\psi(m)=0$, $\psi(M)=1$ and $\psi(a)=\varphi(a)$ for all $a\in(m,M)$. Is $\psi$ an isomorphism between $\langle[0,1]\cap\mathbb{Q},<\rangle$ and $\langle A,<\rangle$?
 A: Yes, this theory is complete, and your argument works. A bit more abstractly: if $A,B$ are linear orders with distinct endpoints and $A',B'$ are the suborders of $A,B$ consisting of everything except the endpoints, then any isomorphism $i$ between $A'$ and $B'$ can be extended to an isomorphism $i'$ between $A$ and $B$. In fact, this is "information-preserving" in a precise sense: $i$ determines $i'$ completely, any isomorphism $j$ between $A$ and $B$ restricts to an isomorphism $\hat{j}$ between  $A'$ and $B'$, and these processes invert each other in the sense that $$i=\hat{i'}\quad\mbox{and}\quad j=\hat{j}'$$ for all $i:A'\cong B'$ and all $j:A\cong B$.

Incidentallly, the notion of interpretations (in the model-theoretic sense) gives a far-reaching version of the argument above. Roughly speaking, if $X$ is a linear order with endpoints and $X'$ is the suborder without endpoints, we can "unqiuely construct" $X$ from $X'$ and vice-versa. This lets us transfer model-theoretic properties about $X$ to $X'$ and vice-versa, and similarly for their theories. I've said a bit about this in some earlier answers; see e.g. here.
