**Proof.**
Consider the obvious morphism $f : (\mathop{\mathit{Sh}}\nolimits (\mathbf{N}), \mathcal{A}) \to (\mathop{\mathit{Sh}}\nolimits (pt), R)$ of ringed topoi and let us consider the adjoint functors $Lf^*$ and $Rf_*$. The first restricts to a functor

\[ F : D_{comp}(R, I) \longrightarrow \mathit{QC}(\mathcal{A}) \]

which sends an object $K$ of $D_{comp}(R, I)$ represented by a K-flat complex $K^\bullet $ to the object $(K^\bullet \otimes _ R R/I^ n)$ of $\mathit{QC}(\mathcal{A})$. The second restricts to a functor

\[ G : \mathit{QC}(\mathcal{A}) \longrightarrow D_{comp}(R, I) \]

which sends an object $(M_ n^\bullet )$ of $\mathit{QC}(\mathcal{A})$ to $R\mathop{\mathrm{lim}}\nolimits M_ n^\bullet $. The output is derived complete for example by More on Algebra, Lemma 15.91.14. Also, it follows from More on Algebra, Proposition 15.94.2 that $G \circ F = \text{id}$. Thus to see that $F$ and $G$ are quasi-inverse equivalences it suffices to see that the kernel of $G$ is zero (see Derived Categories, Lemma 13.7.2). However, it does not appear easy to show this directly!

In this paragraph we will show that $\mathit{QC}(\mathcal{A})$ and $\mathit{QC}(\mathcal{B})$ are equivalent. Write $\mathcal{B} = (B_ n)$ where $B_ n$ is the Koszul complex viewed as a cochain complex in degrees $-r, -r + 1, \ldots , 0$. By Divided Power Algebra, Remark 23.12.2 (but with chain complexes turned into cochain complexes) we can find $1 < n_1 < n_2 < \ldots $ and maps of differential graded $R$-algebras $B_{n_ i} \to E_ i \to R/(f_1^{n_ i}, \ldots , f_ r^{n_ i})$ and $E_ i \to B_{n_{i - 1}}$ such that

\[ \xymatrix{ B_{n_1} \ar[d] & B_{n_2} \ar[d] \ar[l] & B_{n_3} \ar[d] \ar[l] & \ldots \ar[l] \\ E_1 \ar[d] & E_2 \ar[l] \ar[d] & E_3 \ar[l] \ar[d] & \ldots \ar[l] \\ B_1 & B_{n_1} \ar[l] & B_{n_2} \ar[l] & \ldots \ar[l] } \]

is a commutative diagram of differential graded $R$-algebras and such that $E_ i \to R/(f_1^{n_ i}, \ldots , f_ r^{n_ i})$ is a quasi-isomorphism. We conclude

there is an equivalence between $\mathit{QC}(\mathcal{B})$ and $\mathit{QC}((E_ i))$,

there is an equivalence between $\mathit{QC}((E_ i))$ and $\mathit{QC}((R/(f_1^{n_ i}, \ldots , f_ r^{n_ i})))$,

there is an equivalence between $\mathit{QC}((R/(f_1^{n_ i}, \ldots , f_ r^{n_ i})))$ and $\mathit{QC}(\mathcal{A})$.

Namely, for (1) we can apply Lemma 24.35.1 to the diagram above which shows that $(E_ i)$ and $(B_ n)$ are pro-isomorphic. For (2) we can apply Lemma 24.34.2 to the inverse system of quasi-isomorphisms $E_ i \to R/(f_1^{n_ i}, \ldots , f_ r^{n_ i})$. For (3) we can apply Lemma 24.35.1 and the elementary fact that the inverse systems $(R/I^ n)$ and $(R/(f_1^{n_ i}, \ldots , f_ r^{n_ i})$ are pro-isomorphic.

Exactly as in the first paragraph of the proof we can define adjoint functors^{1}

\[ F' : D_{comp}(R, I) \longrightarrow \mathit{QC}(\mathcal{B}) \quad \text{and}\quad G' : \mathit{QC}(\mathcal{B}) \longrightarrow D_{comp}(R, I). \]

The first sends an object $K$ of $D_{comp}(R, I)$ represented by a K-flat complex $K^\bullet $ to the object $(K^\bullet \otimes _ R B_ n)$ of $\mathit{QC}(\mathcal{B})$. The second sends an object $(M_ n)$ of $\mathit{QC}(\mathcal{B})$ to $R\mathop{\mathrm{lim}}\nolimits M_ n$. Arguing as above it suffices to show that the kernel of $G'$ is zero. So let $\mathcal{M} = (M_ n)$ be a good sheaf of differential graded modules over $\mathcal{B}$ which represents an object of $\mathit{QC}(\mathcal{B})$ in the kernel of $G'$. Then

\[ 0 = R\mathop{\mathrm{lim}}\nolimits M_ n \Rightarrow 0 = (R\mathop{\mathrm{lim}}\nolimits M_ n) \otimes _ R^\mathbf {L} B_ t = R\mathop{\mathrm{lim}}\nolimits (M_ n \otimes _ R^\mathbf {L} B_ t) \]

By Lemma 24.35.3 we have $R\mathop{\mathrm{lim}}\nolimits (M_ n \otimes _ R^\mathbf {L} B_ t) = R\mathop{\mathrm{lim}}\nolimits (M_ n \otimes _{B_ n}^\mathbf {L} B_ t)$. Since $(M_ n)$ is an object of $\mathit{QC}(\mathcal{B})$ we see that the inverse system $M_ n \otimes _{B_ n}^\mathbf {L} B_ t$ is eventually constant with value $M_ t$. Hence $M_ t = 0$ as desired.
$\square$

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