Note that the first assumption implies that $RF : D^+(\mathcal{A}) \to D^+(\mathcal{B})$ exists, see Proposition \ref{proposition-enough-acyclics}. Let $A$ be an object of $\mathcal{A}$. Choose an injection $A \to A'$ with $A'$ acyclic. Then we see that $R^{n + 1}F(A) = R^nF(A'/A) = 0$ by the long exact cohomology sequence. Hence we conclude that $R^{n + 1}F = 0$. Continuing like this using induction we find that $R^mF = 0$ for all $m \geq n$. \medskip\noindent We are going to use Lemma \ref{lemma-replace-resolution} with the function $d : \Ob(\mathcal{A}) \to \{0, 1, 2, \ldots \}$ given by $d(A) = \max \{0\} \cup \{i \mid R^iF(A) \not = 0\}$. The first assumption of Lemma \ref{lemma-replace-resolution} is our assumption (1). The second assumption of Lemma \ref{lemma-replace-resolution} follows from the fact that $RF(A \oplus B) = RF(A) \oplus RF(B)$. The third assumption of Lemma \ref{lemma-replace-resolution} follows from the long exact cohomology sequence. Hence for every complex $K^\bullet$ there exists a quasi-isomorphism $K^\bullet \to L^\bullet$ into a complex of objects right acyclic for $F$. This proves statement (3). \medskip\noindent We claim that if $L^\bullet \to M^\bullet$ is a quasi-isomorphism of complexes of right acyclic objects for $F$, then $F(L^\bullet) \to F(M^\bullet)$ is a quasi-isomorphism. If we prove this claim then we get statements (1) and (2) of the lemma by Lemma \ref{lemma-find-existence-computes}. To prove the claim pick an integer $i \in \mathbf{Z}$. Consider the distinguished triangle $$ \sigma_{\geq i - n - 1}L^\bullet \to \sigma_{\geq i - n - 1}M^\bullet \to Q^\bullet, $$ i.e., let $Q^\bullet$ be the cone of the first map. Note that $Q^\bullet$ is bounded below and that $H^j(Q^\bullet)$ is zero except possibly for $j = i - n - 1$ or $j = i - n - 2$. We may apply $RF$ to $Q^\bullet$. Using the second spectral sequence of Lemma \ref{lemma-two-ss-complex-functor} and the assumed vanishing of cohomology (2) we conclude that $H^j(RF(Q^\bullet))$ is zero except possibly for $j \in \{i - n - 2, \ldots, i - 1\}$. Hence we see that $RF(\sigma_{\geq i - n - 1}L^\bullet) \to RF(\sigma_{\geq i - n - 1}M^\bullet)$ induces an isomorphism of cohomology objects in degrees $\geq i$. By Proposition \ref{proposition-enough-acyclics} we know that $RF(\sigma_{\geq i - n - 1}L^\bullet) = \sigma_{\geq i - n - 1}F(L^\bullet)$ and $RF(\sigma_{\geq i - n - 1}M^\bullet) = \sigma_{\geq i - n - 1}F(M^\bullet)$. We conclude that $F(L^\bullet) \to F(M^\bullet)$ is an isomorphism in degree $i$ as desired. \medskip\noindent Part (4)(a) follows from Lemma \ref{lemma-negative-vanishing}. \medskip\noindent For part (4)(b) let $E$ be represented by the complex $L^\bullet$ of objects right acyclic for $F$. By part (2) $RF(E)$ is represented by the complex $F(L^\bullet)$ and $RF(\sigma_{\geq c}L^\bullet)$ is represented by $\sigma_{\geq c}F(L^\bullet)$. Consider the distinguished triangle $$ H^{b - n}(L^\bullet)[n - b] \to \tau_{\geq b - n}L^\bullet \to \tau_{\geq b - n + 1}L^\bullet $$ of Remark \ref{remark-truncation-distinguished-triangle}. The vanishing established above gives that $H^i(RF(\tau_{\geq b - n}L^\bullet))$ agrees with $H^i(RF(\tau_{\geq b - n + 1}L^\bullet))$ for $i \geq b$. Consider the short exact sequence of complexes $$ 0 \to \Im(L^{b - n - 1} \to L^{b - n})[n - b] \to \sigma_{\geq b - n}L^\bullet \to \tau_{\geq b - n}L^\bullet \to 0 $$ Using the distinguished triangle associated to this (see Section \ref{section-canonical-delta-functor}) and the vanishing as before we conclude that $H^i(RF(\tau_{\geq b - n}L^\bullet))$ agrees with $H^i(RF(\sigma_{\geq b - n}L^\bullet))$ for $i \geq b$. Since the map $RF(\sigma_{\geq b - n}L^\bullet) \to RF(L^\bullet)$ is represented by $\sigma_{\geq b - n}F(L^\bullet) \to F(L^\bullet)$ we conclude that this in turn agrees with $H^i(RF(L^\bullet))$ for $i \geq b$ as desired. \medskip\noindent Proof of (4)(c). Under the assumption on $E$ we have $\tau_{\leq a - 1}E = 0$ and we get the vanishing of $H^i(RF(E))$ for $i \leq a - 1$ from part (4)(a). Similarly, we have $\tau_{\geq b + 1}E = 0$ and hence we get the vanishing of $H^i(RF(E))$ for $i \geq b + n$ from part (4)(b).