Let $T$ and $\mathcal{M}_t \to \mathcal{M}'_t$ be as in Lemma \ref{lemma-characterize-graded-injectives-in-dg}. Let $S$ and $\mathcal{M}_s$ be as in Lemma \ref{lemma-small-acyclics}. Choose an injective map $\mathcal{M}_s \to \mathcal{M}'_s$ of acyclic differential graded $\mathcal{A}$-modules which is homotopic to zero. This is possible because we may take $\mathcal{M}'_s$ to be the cone on the identity; in that case it is even true that the identity on $\mathcal{M}'_s$ is homotopic to zero, see Differential Graded Algebra, Lemma \ref{dga-lemma-id-cone-null} which applies by the discussion in Section \ref{section-conclude-triangulated}. We claim that $R = T \coprod S$ with the given maps works. \medskip\noindent The implication (1) $\Rightarrow$ (2) holds by Lemma \ref{lemma-first-property-dg-injective}. \medskip\noindent Assume (2). First, by Lemma \ref{lemma-characterize-graded-injectives-in-dg} we see that $\mathcal{I}$ is graded injective. Next, let $\mathcal{M}$ be an acyclic differential graded $\mathcal{A}$-module. We have to show that $$ \Hom_{K(\text{Mod}_{(\mathcal{A}, \text{d})})}(\mathcal{M}, \mathcal{I}) = 0 $$ The proof will be exactly the same as the proof of Injectives, Lemma \ref{injectives-lemma-characterize-K-injective}. \medskip\noindent We are going to construct by induction on the ordinal $\alpha$ an acyclic differential graded submodule $\mathcal{K}_\alpha \subset \mathcal{M}$ as follows. For $\alpha = 0$ we set $\mathcal{K}_0 = 0$. For $\alpha > 0$ we proceed as follows: \begin{enumerate} \item If $\alpha = \beta + 1$ and $\mathcal{K}_\beta = \mathcal{M}$ then we choose $\mathcal{K}_\alpha = \mathcal{K}_\beta$. \item If $\alpha = \beta + 1$ and $\mathcal{K}_\beta \not = \mathcal{M}$ then $\mathcal{M}/\mathcal{K}_\beta$ is a nonzero acyclic differential graded $\mathcal{A}$-module. We choose a differential graded $\mathcal{A}$ submodule $\mathcal{N}_\alpha \subset \mathcal{M}/\mathcal{K}_\beta$ isomorphic to $\mathcal{M}_s$ for some $s \in S$, see Lemma \ref{lemma-small-acyclics}. Finally, we let $\mathcal{K}_\alpha \subset \mathcal{M}$ be the inverse image of $\mathcal{N}_\alpha$. \item If $\alpha$ is a limit ordinal we set $\mathcal{K}_\beta = \colim \mathcal{K}_\alpha$. \end{enumerate} It is clear that $\mathcal{M} = \mathcal{K}_\alpha$ for a suitably large ordinal $\alpha$. We will prove that $$ \Hom_{K(\text{Mod}_{(\mathcal{A}, \text{d})})}(\mathcal{K}_\alpha, \mathcal{I}) $$ is zero by transfinite induction on $\alpha$. It holds for $\alpha = 0$ since $\mathcal{K}_0$ is zero. Suppose it holds for $\beta$ and $\alpha = \beta + 1$. In case (1) of the list above the result is clear. In case (2) there is a short exact sequence $$ 0 \to \mathcal{K}_\beta \to \mathcal{K}_\alpha \to \mathcal{N}_\alpha \to 0 $$ By Remark \ref{remark-why-graded-injective} and since we've seen that $\mathcal{I}$ is graded injective, we obtain an exact sequence $$ \Hom_{K(\text{Mod}_{(\mathcal{A}, \text{d})})}(\mathcal{K}_\beta, \mathcal{I}) \to \Hom_{K(\text{Mod}_{(\mathcal{A}, \text{d})})}(\mathcal{K}_\alpha, \mathcal{I}) \to \Hom_{K(\text{Mod}_{(\mathcal{A}, \text{d})})}(\mathcal{N}_\alpha, \mathcal{I}) $$ By induction the term on the left is zero. By assumption (2) the term on the right is zero: any map $\mathcal{M}_s \to \mathcal{I}$ factors through $\mathcal{M}'_s$ and hence is homotopic to zero. Thus the middle group is zero too. Finally, suppose that $\alpha$ is a limit ordinal. Because we also have $\mathcal{K}_\alpha = \colim \mathcal{K}_\alpha$ as graded $\mathcal{A}$-modules we see that $$ \Hom_{\text{Mod}^{dg}_{(\mathcal{A}, \text{d})}} (\mathcal{K}_\alpha, \mathcal{I}) = \lim_{\beta < \alpha} \Hom_{\text{Mod}^{dg}_{(\mathcal{A}, \text{d})}} (\mathcal{K}_\beta, \mathcal{I}) $$ as complexes of abelian groups. The cohomology groups of these complexes compute morphisms in $K(\text{Mod}_{(\mathcal{A}, \text{d})})$ between shifts. The transition maps in the system of complexes are surjective by Remark \ref{remark-why-graded-injective} because $\mathcal{I}$ is graded injective. Moreover, for a limit ordinal $\beta \leq \alpha$ we have equality of limit and value. Thus we may apply Homology, Lemma \ref{homology-lemma-ML-over-ordinals} to conclude.