Derived Hom and tensor of chain complexes homologically concentrated in degree zero Let $R$ be a commutative ring, and let $X,Y\in \mathcal D_0(R)$ (i.e. $X,Y$ are represented by chain complexes with only non-zero homology at the $0$-th spot).
Then, is it true that
$$\text{H}_n \mathbf R\text{Hom}_R(X,Y) =\text{H}_{-n} \left(X \otimes_R^{\mathbf L}Y\right)=0,\forall n>0\ ?$$
I think this is true because in $\mathcal D(R)$ we have the isomorphisms $X\cong \text H_0(X)$ and $Y\cong \text H_0(Y)$, hence essentially $X,Y\in \text{Mod} (R)$.
Am I correct?
 A: First off, as you've notice in $\mathcal D(R)$ you have $X\simeq H_0(X)$ and $Y\simeq H_0(Y)$, so you can assume that $X$ and $Y$ are just $R$-modules considered as complexes concentrated in degree $0$.
Now, according to the definition of $X\otimes_R^\mathbf{L}Y$ I am going off of, i.e. here, you can calculate $X\otimes_R^\mathbf{L}Y$ by taking a "K-flat resolution" $ K_\bullet\to Y$ and then $X\otimes_R^\mathbf{L}Y$ is represented by the complex $\operatorname{Tot}(X[0]\otimes_R K_\bullet)$, which you can see from the definition is just $X\otimes_R K_n$ in the $n$-th position. In particular we can take $P_\bullet\to Y$ to be a projective resolution of $Y$ (remember $Y$ is just an $R$-module), for which we know classically we can arrange for $P_n=0$ for $n<0$, then by Lemma $15.58.9$ on the link I've provided (replacing "bounded above" by "bounded below”, etc.) this is a K-flat resolution of $Y$, and then you can see our complex representing $X\otimes_R^\mathbf{L}Y$ is concentrated in degree $\ge0$, resulting in the claim $H_{-n}(X\otimes_R^\mathbf{L}Y)=0$ for $n>0$.
Similarly one can compute RHom by taking a K-injective resolution $Y\to I_\bullet$ and then $\mathbf{R}\mathrm{Hom}_R(X,Y)$ is represented by the complex $\operatorname{Hom}_\bullet(X[0],I_\bullet)$. Lemma $13.31.4$ in the second link says one can take $Y\to I_\bullet$ to be an injective resolution of the $R$-module $Y$, for which one knows we can always arrange for $I_n=0$ for $n>0$, and then you can see again that our complex representing $\mathbf{R}\mathrm{Hom}_R(X,Y)$ is concentrated in degree $\le0$.
