This lemma is a special case of Lemma \ref{lemma-algebraize-local-cohomology-general} since condition (5)(c) is implied by condition (4) as $\delta_{max} = \delta_{min} = \delta(\mathfrak m)$. We will give the proof of this important special case as it is somewhat easier (fewer things to check). \medskip\noindent There is no difference between $R\Gamma_\mathfrak a$ and $R\Gamma_{V(\mathfrak a)}$ in our current situation, see Dualizing Complexes, Lemma \ref{dualizing-lemma-local-cohomology-noetherian}. Next, we observe that $$ R\Gamma_\mathfrak m(M)^\wedge = R\Gamma_I(R\Gamma_J(M))^\wedge = R\Gamma_J(M)^\wedge $$ by Dualizing Complexes, Lemmas \ref{dualizing-lemma-local-cohomology-ss} and \ref{dualizing-lemma-complete-and-local} which explains the equality sign in the statement of the lemma. \medskip\noindent Observe that the lemma holds for $s < 0$. This is not a trivial case because it is not a priori clear that $H^s(R\Gamma_\mathfrak m(M)^\wedge)$ is zero for negative $s$. However, this vanishing was established in Lemma \ref{lemma-local-cohomology-derived-completion}. We will prove the lemma by induction for $s \geq 0$. \medskip\noindent The assumptions of Lemma \ref{lemma-kill-colimit-weak} are satisfied by Local Cohomology, Lemma \ref{local-cohomology-lemma-cd-bound-dim-local}. The lemma for $s = 0$ follows from Lemma \ref{lemma-kill-colimit-weak} and Dualizing Complexes, Lemma \ref{dualizing-lemma-completion-local-H0}. \medskip\noindent Assume $s > 0$ and the lemma holds for smaller values of $s$. Let $M' \subset M$ be the submodule of elements whose support is condained in $V(I) \cup V(J)$ for some ideal $J$ with $V(J) \cap V(I) = \{\mathfrak m\}$. Then $M'$ is a finite $A$-module. We claim that $$ R\Gamma_J(M') \to R\Gamma_\mathfrak m(M')^\wedge $$ is an isomorphism for any choice of $J$. Namely, for any such module there is a short exact sequence $0 \to M_1 \oplus M_2 \to M' \to N \to 0$ with $M_1$ annihilated by a power of $J$, with $M_2$ annihilated by a power of $I$ and with $N$ annihilated by a power of $\mathfrak m$. In the case of $M_1$ we see that $R\Gamma_J(M_1) = M_1$ and since $M_1$ is a finite $A$-module and $I$-adically complete we have $M_1^\wedge = M_1$. Thus the claim holds for $M_1$. In the case of $M_2$ we see that $H^i_J(M_2)$ is annihilated by a power of $I$ and hence derived complete. Thus the claim for $M_2$. By the same arguments the claim holds for $N$ and we conclude that the claim holds. Considering the short exact sequence $0 \to M' \to M \to M/M' \to 0$ we see that it suffices to prove the lemma for $M/M'$. This we may assume $\mathfrak p \in \text{Ass}(M)$ implies $V(\mathfrak p) \cap V(I) \not = \{\mathfrak m\}$, i.e., $\mathfrak p$ is a prime as in (4). \medskip\noindent Choose an ideal $J_0$ as in Lemma \ref{lemma-kill-colimit-weak} and an integer $t > 0$ such that $(J_0I)^t$ annihilates $H^s_J(M)$. Here $J$ denotes an arbitrary ideal $J \subset J_0$ with $V(J) \cap V(I) = \{\mathfrak m\}$. The assumptions of Lemma \ref{lemma-kill-completion} are satisfied for every $\mathfrak p \in \text{Ass}(M)$ (see previous paragraph). Thus the annihilator $\mathfrak a \subset A$ of $H^s(R\Gamma_\mathfrak m(M)^\wedge)$ is not contained in $\mathfrak p$ for $\mathfrak p \in \text{Ass}(M)$. Thus we can find an $f \in \mathfrak a(J_0I)^t$ not in any associated prime of $M$ which is an annihilator of both $H^s(R\Gamma_\mathfrak m(M)^\wedge)$ and $H^s_J(M)$. Then $f$ is a nonzerodivisor on $M$ and we can consider the short exact sequence $$ 0 \to M \xrightarrow{f} M \to M/fM \to 0 $$ Our choice of $f$ shows that we obtain $$ \xymatrix{ H^{s - 1}_J(M) \ar[d] \ar[r] & H^{s - 1}_J(M/fM) \ar[d] \ar[r] & H^s_J(M) \ar[d] \ar[r] & 0 \\ H^{s - 1}(R\Gamma_\mathfrak m(M)^\wedge) \ar[r] & H^{s - 1}(R\Gamma_\mathfrak m(M/fM)^\wedge) \ar[r] & H^s(R\Gamma_\mathfrak m(M)^\wedge) \ar[r] & 0 } $$ for any $J \subset J_0$ with $V(J) \cap V(I) = \{\mathfrak m\}$. Thus if we choose $J$ such that it works for $M$ and $M/fM$ and $s - 1$ (possible by induction hypothesis), then we conclude that the lemma is true.