Let $\varphi(x_1, \cdots, x_n)$ be a formula of $n$ variables that is consistent with a complete theory $T$. In several proofs I read, it follows that
$$
T\models \exists x_1\cdots\exists x_n\:\varphi(x_1, \cdots, x_n)\tag1
$$
This seems incomplete with a complete theory because for a complete theory $T$, either $T\models\varphi$ or $T\models\neg\varphi$, but not both.
Considering the (formal) definition of consistency, (1) is still incomplete. Formally, $T$ is inconsistent iff for any $\varphi$, $T\vdash \varphi\land\neg\varphi$ or $T\vdash \varphi\land T\vdash\neg\varphi$. So $T$ is consistent (with $\varphi$) iff $\neg (T\vdash \varphi)\lor \neg (T\vdash\neg\varphi)$ or $T\nVdash\varphi\lor T\nVdash\neg\varphi$.
Best Answer
That there is no sentence that $T\vdash \varphi\land\lnot\varphi$ is the definition of a theory being consistent. But if a theory is inconsistent, then $T\vdash \varphi\land\lnot \varphi$ for every sentence $\varphi.$ To isolate individual instances is useless since they either all hold or none of them do.
Instead, when we say some sentence $\varphi$ is consistent with $T$, we mean that $T\nvdash\lnot \varphi.$ So in other words, the sentence is not incompatible with $T$... it will hold in some models.
When we go from sentences to formulas, the natural thing to do is to say a formula $\varphi(\vec x)$ is consistent with $T$ if it not impossible that there be a tuple of elements satisfying it. In other words, if $T\nvdash \lnot\exists \vec x \varphi(\vec x).$
So if $T$ is complete and $\varphi(\vec x)$ is consistent with it, we must have $T\vdash \exists \vec x \varphi(\vec x),$ since $T\nvdash \lnot\exists \vec x \varphi(\vec x)$ and one of the two must hold.