**Proof.**
If (2) holds, then we may compute $K^\bullet \otimes _ R^\mathbf {L} M = E^\bullet \otimes _ R M$ and it is clear that (1) holds. Assume that (1) holds. We may replace $K^\bullet $ by a projective resolution. Let $n$ be the largest integer such that $K^ n \not= 0$. If $n > b$, then $K^{n - 1} \to K^ n$ is surjective as $H^ n(K^\bullet ) = 0$. As $K^ n$ is projective we see that $K^{n - 1} = K' \oplus K^ n$. Hence it suffices to prove the result for the complex $(K')^\bullet $ which is the same as $K^\bullet $ except has $K'$ in degree $n - 1$ and $0$ in degree $n$. Thus, by induction on $n$, we reduce to the case that $K^\bullet $ is a complex of projective $R$-modules with $K^ i = 0$ for $i > b$.

Set $E^\bullet = \tau _{\geq a}K^\bullet $. Everything is clear except that $E^ a$ is flat which follows immediately from Lemma 15.63.2 and the definitions.
$\square$

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