In the course of the proof of Lemma 6.5 in Yuri Manin's "A Course in Mathematical Logic", the author states that a certain set of first-order logic formulas $\mathcal{E}$ is consistent, complete and contains $\text{Ax }L$ (by assumption). $\text{Ax }L$ is the set of logical axioms of $L$, which consists of the tautologies and quantifier axioms. Moreover, he shows that certain formulas $Q_1$ and $\neg Q_2$ are deducible from $\mathcal{E}$ (using the modus ponens rule). Both $Q_1$ and $Q_2$ are assumed to be closed, i.e. without any free variables.
He then concludes that in fact $Q_1$ and $\neg Q_2$ must be contained in $\mathcal{E}$ because it is consistent and complete.
I can see that this claim sounds reasonable, but for some reason I am unable to justify it. In particular, I am unable to reach a contradiction from the assumption that $\mathcal{E}$ does not contain $Q_1$.
Let's assume that $\mathcal{E}$ does not contain $Q_1$. Since $\mathcal{E}$ is complete and does not contain $Q_1$, it must contain its negation $\neg Q_1$. On the other hand, $Q_1$ is deducible from $\mathcal{E}$ by an application of modus ponens (it was shown earlier that $\mathcal{E}$ contained formulas $A$ and $A \implies Q_1$).
But this does not immediately prove that $Q_1$ is in $\mathcal{E}$, since we do not a priori assume that $\mathcal{E}$ is closed under applications of the deduction rules. Hence we do not reach a contradiction with the assumed consistency of $\mathcal{E}$.
Could someone clarify this point of the proof?
If $\mathcal E$ contains $\lnot Q_1,$ then you can derive $\lnot Q_1$ from it. If in addition you can derive $Q_1$ from it, then you can derive $Q_1\land\lnot Q_1$ from it, so it is inconsistent.