Fix incomplete proof

Thanks to  Elías Guisado
https://stacks.math.columbia.edu/tag/05TC#comment-8399

derived.tex

@@ -5480,14 +5480,18 @@ \section{Higher derived functors}
 
 \begin{proof}
 Assume $K^\bullet$ satisfies the assumptions of (1).
-Let $K^\bullet \to L^\bullet$ be any quasi-isomorphism.
+Let $s : K^\bullet \to L^\bullet$ be any quasi-isomorphism.
 Then it is also true that $K^\bullet \to \tau_{\geq a}L^\bullet$
 is a quasi-isomorphism by our assumption on $K^\bullet$.
 Hence in the category $K^\bullet/\text{Qis}^{+}(\mathcal{A})$ the
-quasi-isomorphisms $s : K^\bullet \to L^\bullet$ with $L^n = 0$ for $n < a$
-are cofinal. Thus $RF$ is the value of the essentially constant
-ind-object $F(L^\bullet)$ for these $s$ it follows that
-$H^i(RF(K^\bullet)) = 0$ for $i < a$.
+quasi-isomorphisms $s : K^\bullet \to L^\bullet$ with $L^n = 0$
+for $n < a$ are cofinal. From Categories, Lemma
+\ref{categories-lemma-cofinal-essentially-constant}
+we deduce that $RF$ is the value of the essentially constant
+ind-object $F(L^\bullet)$ for these $s$. This means that
+$\text{id} : RF(K^\bullet) \to RF(K^\bullet)$ factors through
+$F(L^\bullet)$ for some complex $L^\bullet$ with $L^n = 0$
+for $n < a$. It follows that $H^i(RF(K^\bullet)) = 0$ for $i < a$.
 
 \medskip\noindent
 To prove (2) we use the distinguished triangle