Let $T=\mbox{Th}(\mathbb{Q};<)$ . According to Theorem 31F, we only need to show that for a formula $\phi$ of the form $\exists x(\alpha_{0}\wedge\ldots\wedge\alpha_{n})$ , where each $\alpha_{i}$ is atomic or the negation of an atomic formula, there is a quantifier-free formula $\psi$ such that $T\vDash\phi\leftrightarrow\psi$ .

I am not sure whether the equality symbol is assumed to be a part of the language, but in any case, we can assume that the language does not have the equality symbol ($T\vDash t_{1}=t_{2}\leftrightarrow\neg t_{1}<t_{2}\wedge\neg t_{2}<t_{1}$ and $T\vDash t_{1}\neq t_{2}\leftrightarrow t_{1}<t_{2}\vee t_{2}<t_{1}$ where we can regroup the terms and use the fact that $\exists x(\alpha\vee\beta)$ iff $\exists x\alpha\vee\exists x\beta$ ). In fact, instead of removing the equality symbol, it would be more convenient to get rid of the negations of atomic formulas instead (see Exercise 4 of Section 3.2), but, as I said, we are not explicitly given that the language has the equality symbol. There are also no function or constant symbols in the language. Therefore, each atomic formula is of the form $v_{i}<v_{j}$ where each variable can be $x$ .

1. If there is an atomic formula that contains $x$ on both sides, then, if the formula itself is used in the conjunction then $\vDash\phi\leftrightarrow x<x$ , and if its negation is used then we can safely omit this negation.
2. If there is an atomic formula that does not contain $x$ , then we can safely move it out of the scope of the quantifier symbol, as $\vDash\exists x(v_{i}<v_{j}\wedge\phi')\leftrightarrow v_{i}<v_{j}\wedge\exists x\phi'$ and $\vDash\exists x(\neg v_{i}<v_{j}\wedge\phi')\leftrightarrow\neg v_{i}<v_{j}\wedge\exists x\phi'$ . Indeed, $\exists x(\alpha\wedge\phi')$ , where $\alpha$ does not contain $x$ , is logically equivalent to $\neg\forall x(\alpha\rightarrow\neg\phi')$ , and we can use Example Q2a, page 121.
3. If all the remaining atomic formulas have $x$ on the same side and the negations of atomic formulas have $x$ on the other side, i.e. $v_{i}<x$ and $\neg x<v_{j}$ , or $x<v_{i}$ and $\neg v_{j}<x$ , then $T\vDash\phi'\leftrightarrow\neg x<x$ , as $\mathbb{Q}$ is unbounded.
4. If there are two atomic formulas containing $x$ on the right side, i.e. $v_{i}<x$ and $v_{j}<x$ , then $T\vDash\exists x(v_{i}<x\wedge v_{j}<x\wedge\phi')$ $\leftrightarrow(v_{i}<v_{j}\wedge\exists x(v_{j}<x\wedge\phi'))\vee(\neg v_{i}<v_{j}\wedge\exists x(v_{i}<x\wedge\phi'))$ . Indeed, if $\mathfrak{A}$ is a model of $T$ and $s:V\rightarrow|\mathfrak{A}|$ , then $\vDash_{\mathfrak{A}}\exists x(v_{i}<x\wedge v_{j}<x\wedge\phi')[s]$ iff for some $d\in|\mathfrak{A}|$ , $\vDash_{\mathfrak{A}}v_{i}<x\wedge v_{j}<x\wedge\phi'[s(x|d)]$ iff for some $d\in|\mathfrak{A}|$ , either $<s(v_{i}),s(v_{j})>\in<^{\mathfrak{A}}$ and $\vDash_{\mathfrak{A}}v_{j}<x\wedge\phi'[s(x|d)]$ , or $<s(v_{i}),s(v_{j})>\notin<^{\mathfrak{A}}$ and $\vDash_{\mathfrak{A}}v_{i}<x\wedge\phi'[s(x|d)]$ iff $\vDash_{\mathfrak{A}}(v_{i}<v_{j}\wedge\exists x(v_{j}<x\wedge\phi'))\vee(\neg v_{i}<v_{j}\wedge\exists x(v_{i}<x\wedge\phi'))[s]$ .
5. Similarly, if there are two atomic formulas containing $x$ on the left side, i.e. $x<v_{i}$ and $x<v_{j}$ , then $T\vDash\exists x(x<v_{i}\wedge x<v_{j}\wedge\phi')$ $\leftrightarrow(v_{i}<v_{j}\wedge\exists x(x<v_{i}\wedge\phi'))\vee(\neg v_{i}<v_{j}\wedge\exists x(x<v_{j}\wedge\phi'))$ .
6. Similar to cases 4 and 5, we consider cases of two formulas of the form $\neg x<v_{i}$ and $v_{j}<x$ ; $\neg x<v_{i}$ and $\neg x<v_{j}$ (4); $\neg v_{i}<x$ and $x<v_{j}$ ; and $\neg v_{i}<x$ and $\neg v_{j}<x$ (5).
7. The last case is when there are only two formulas left in the expression $\phi'$ , which is now $\exists x(v_{i}<x\wedge x<v_{j})$ or $\exists x(\neg x<v_{i}\wedge x<v_{j})$ or $\exists x(\neg x<v_{i}\wedge\neg v_{j}<x)$ . In the first two cases, $T\vDash\phi'\leftrightarrow v_{i}<v_{j}$ , and in the third case $T\vDash\phi'\leftrightarrow v_{i}<v_{j}\vee v_{i}=v_{j}$ . Indeed, if $\mathfrak{A}$ is a model of $T$ and $s:V\rightarrow|\mathfrak{A}|$ , then $\vDash_{\mathfrak{A}}\exists x(v_{i}<x\wedge x<v_{j})[s]$ iff for some $d\in|\mathfrak{A}|$ , $\vDash_{\mathfrak{A}}v_{i}<x\wedge x<v_{j}[s(x|d)]$ iff for some $d\in|\mathfrak{A}|$ , $s(v_{i})<d$ and $d<s(v_{j})$ iff (density, see page 159) $s(v_{i})<s(v_{j})$ iff $\vDash_{\mathfrak{A}}v_{i}<v_{j}[s]$ . The other two cases are similar.

Example. $\exists x(x=v_{1}\wedge v_{2}<v_{1}\wedge\neg v_{3}<x)$ (intended to say $v_{2}<v_{1}\le v_{3}$ ) becomes $\exists x(\neg x<v_{1}\wedge\neg v_{1}<x\wedge v_{2}<v_{1}\wedge\neg v_{3}<x)$ becomes $v_{2}<v_{1}\wedge\exists x(\neg x<v_{1}\wedge\neg v_{1}<x\wedge\neg v_{3}<x)$ becomes $v_{2}<v_{1}\wedge((v_{1}<v_{3}\wedge\exists x(\neg x<v_{1}\wedge\neg v_{1}<x))$ $\vee(\neg v_{1}<v_{3}\wedge\exists x(\neg x<v_{1}\wedge\neg v_{3}<x)))$ becomes $v_{2}<v_{1}\wedge((v_{1}<v_{3}\wedge(v_{1}<v_{1}\vee v_{1}=v_{1}))$ $\vee(\neg v_{1}<v_{3}\wedge(v_{1}<v_{3}\vee v_{1}=v_{3})))$ . This last formula can, of course, be simplified to $v_{2}<v_{1}\wedge(v_{1}<v_{3}\vee v_{1}=v_{3})$ corresponding to its initial meaning.