Questions regarding the proof of quantifier elimination of DLO So this is the proof in David Marker's Model theory book, Theorem 3.1.3.
I am a bit confused over the first line of the proof. It reads :
"First suppose $\phi$ is a sentence. If $\mathbb{Q}\models\phi$, then because DLO is complete, DLO $\models \phi$ and 
DLO $\models \phi \leftrightarrow x_1 = x_1$"
My questions is 
1.) Where does completeness come into play here ? $\mathbb{Q}$ by itself is a model of DLO, so $\phi$ is already a logical consequence of DLO, why do I need completeness ?
2.) Why is $\phi$ equivalent to a tautology here ? 
Thanks in advance. Cheers !
 A: Re: your first question, the issue is that different models of a theory might behave differently. Remember that "$\Gamma\models\chi$" means "every model of $\Gamma$ satisfies $\chi$" - this is an easy point to get tripped up on. Basically, the issue is that a specific case doesn't tell you too much about the general case.
Knowing that $\mathbb{Q}\models$ DLO and $\mathbb{Q}\models \phi$ just tells you that $\phi$ is consistent with DLO; other models of DLO might satisfy $\neg\phi$. For example, letting $G$ be your favorite abelian group and Grp be the group axioms, we have that $G\models$ Grp and $G\models x_1*x_2=x_2*x_1$, but we know that $x_1*x_2=x_2*x_1$ isn't a logical consequence of Grp (since there are non-abelian groups).
However, DLO is complete, so any sentence true in some model of DLO is true in every model of DLO. Put another way, anything true in some model of DLO - like $\mathbb{Q}$ - is a logical consequence of DLO; there's no "possible-but-not-necessary" going on here.

Now for your second question, look back at the definition of "equivalent" - two sentences $\psi,\theta$ are equivalent over a theory $T$ if $T$ entails $\psi\leftrightarrow\theta$.
One easy way this could happen is if $T$ entails each sentence separately; that is, if $T$ entails $\psi$ and $T$ entails $\theta$. This is because (think back to truth tables) $\psi\leftrightarrow\theta$ is true in a structure iff either both $\psi$ and $\theta$ are true in that structure or both $\psi$ and $\theta$ are false in that structure; if $T\models\psi$ and $T\models\theta$ then both $\psi$ and $\theta$ are true in every model of $T$, so $\psi\leftrightarrow\theta$ is true in every model of $T$. (And of course we'd get the same behavior of $T$ entails the negation of each sentence separately.) That's what's going on in this case:


*

*$\mathbb{Q}$ is a model of DLO and satisfies $\phi$.

*Since DLO is complete, this means every model of DLO satisfies $\phi$; that is, DLO$\models \phi$.

*And of course DLO $\models x_1=x_1$, since in fact $\emptyset\models x_1=x_1$ (that is, "$x_1=x_1$" is true with respect to every variable assignment in every structure).

*So DLO $\models\phi\leftrightarrow x_1=x_1$.
