**Proof.**
If (2) holds, then $K$ is derived complete with respect to $I$ by Lemma 15.91.16. Conversely, assume that $K$ is derived complete with respect to $I$. Consider the filtrations

\[ K_ n^\bullet \supset \sigma _{\geq -r + 1}K_ n^\bullet \supset \sigma _{\geq -r + 2}K_ n^\bullet \supset \ldots \supset \sigma _{\geq -1}K_ n^\bullet \supset \sigma _{\geq 0}K_ n^\bullet = A \]

by stupid truncations (Homology, Section 12.15). Because the construction $R\mathop{\mathrm{lim}}\nolimits (K \otimes E)$ is exact in the second variable (Lemma 15.87.11) we see that it suffices to show

\[ R\mathop{\mathrm{lim}}\nolimits \left( K \otimes _ A^\mathbf {L} (\sigma _{\geq p}K_ n^\bullet / \sigma _{\geq p + 1}K_ n^\bullet ) \right) = 0 \]

for $p < 0$. The explicit description of the Koszul complexes above shows that

\[ R\mathop{\mathrm{lim}}\nolimits \left( K \otimes _ A^\mathbf {L} (\sigma _{\geq p}K_ n^\bullet / \sigma _{\geq p + 1}K_ n^\bullet ) \right) = \bigoplus \nolimits _{i_1, \ldots , i_{-p}} T(K, f_{i_1}\ldots f_{i_{-p}}) \]

which is zero for $p < 0$ by assumption on $K$.
$\square$

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