For an ideal $\mathfrak a \subset A$ we have $R\Gamma_\mathfrak a = R\Gamma_{V(\mathfrak a)}$, see Dualizing Complexes, Lemma \ref{dualizing-lemma-local-cohomology-noetherian}. Next, we observe that $$ R\Gamma_J(M)^\wedge = R\Gamma_I(R\Gamma_J(M))^\wedge = R\Gamma_{I + J}(M)^\wedge = R\Gamma_{I + J'}(M)^\wedge = R\Gamma_I(R\Gamma_{J'}(M))^\wedge = R\Gamma_{J'}(M)^\wedge $$ by Dualizing Complexes, Lemmas \ref{dualizing-lemma-local-cohomology-ss} and \ref{dualizing-lemma-complete-and-local}. This explains how we define the arrow in the statement of the lemma. \medskip\noindent We claim that the hypotheses of Lemma \ref{lemma-kill-colimit-weak-general} are implied by our current hypotheses on $M$. The only thing to verify is hypothesis (3). Thus let $\mathfrak p \not \in V(I)$, $\mathfrak p \in T$. Then $V(\mathfrak p) \cap V(I)$ is nonempty as $I$ is contained in the Jacobson radical of $A$ (Algebra, Lemma \ref{algebra-lemma-radical-completion}). Since $\mathfrak p \in T$ we have $V(\mathfrak p) \cap V(I) = V(\mathfrak p) \cap V(J) \cap V(I)$. Let $\mathfrak q \in V(\mathfrak p) \cap V(I)$ be the generic point of an irreducible component. We have $\text{cd}(A_\mathfrak q, I_\mathfrak q) \leq d$ by Local Cohomology, Lemma \ref{local-cohomology-lemma-cd-local}. We have $V(\mathfrak pA_\mathfrak q) \cap V(I_\mathfrak q) = \{\mathfrak qA_\mathfrak q\}$ by our choice of $\mathfrak q$ and we conclude $\dim((A/\mathfrak p)_\mathfrak q) \leq d$ by Local Cohomology, Lemma \ref{local-cohomology-lemma-cd-bound-dim-local}. \medskip\noindent Observe that the lemma holds for $s < 0$. This is not a trivial case because it is not a priori clear that $H^i(R\Gamma_J(M)^\wedge)$ is zero for $i < 0$. However, this vanishing was established in Dualizing Complexes, Lemma \ref{dualizing-lemma-completion-local}. We will prove the lemma by induction for $s \geq 0$. \medskip\noindent The lemma for $s = 0$ follows immediately from the conclusion of Lemma \ref{lemma-kill-colimit-weak-general} and Dualizing Complexes, Lemma \ref{dualizing-lemma-completion-local-H0}. \medskip\noindent Assume $s > 0$ and the lemma has been shown for smaller values of $s$. Let $M' \subset M$ be the maximal submodule whose support is contained in $V(I) \cup T$. Then $M'$ is a finite $A$-module whose support is contained in $V(J') \cup V(I)$ for some ideal $J' \subset J$ with $V(J') \cap V(I) = V(J) \cap V(I)$. We claim that $$ R\Gamma_{J'}(M') \to R\Gamma_J(M')^\wedge $$ is an isomorphism for any choice of $J'$. Namely, we can choose 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 $I + J'$. Thus it suffices to show that the claim holds for $M_1$, $M_2$, and $N$. 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 $M_1^\wedge = M_1$. This proves the claim for $M_1$ by the initial remarks of the proof. In the case of $M_2$ we see that $H^i_J(M_2) = H^i_{I + J}(M) = H^i_{I + J'}(M) = H^i_{J'}(M_2)$ are annihilated by a power of $I$ and hence derived complete. Thus the claim in this case also. For $N$ we can use either of the arguments just given. 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 $M/M'$. Thus we may assume $\text{Ass}(M) \cap (V(I) \cup T) = \emptyset$. \medskip\noindent Let $\mathfrak p \in \text{Ass}(M)$ be such that $V(\mathfrak p) \cap V(J) \cap V(I) = \emptyset$. Since $I$ is contained in the Jacobson radical of $A$ this implies that $V(\mathfrak p) \cap V(J') = \emptyset$ for any $J' \subset J$ with $V(J') \cap V(I) = V(J) \cap V(I)$. Thus setting $N = H^0_\mathfrak p(M)$ we see that $R\Gamma_J(N) = R\Gamma_{J'}(N) = 0$ for all $J' \subset J$ with $V(J') \cap V(I) = V(J) \cap V(I)$. In particular $R\Gamma_J(N)^\wedge = 0$. Thus we may replace $M$ by $M/N$ as this changes the structure of $M$ only in primes which do not play a role in conditions (4) or (5). Repeating we may assume that $V(\mathfrak p) \cap V(J) \cap V(I) \not = \emptyset$ for all $\mathfrak p \in \text{Ass}(M)$. \medskip\noindent Assume $\text{Ass}(M) \cap (V(I) \cup T) = \emptyset$ and that $V(\mathfrak p) \cap V(J) \cap V(I) \not = \emptyset$ for all $\mathfrak p \in \text{Ass}(M)$. Let $\mathfrak p \in \text{Ass}(M)$. We want to show that we may apply Lemma \ref{lemma-kill-completion-general}. It is in the verification of this that we will use the supplemental condition (5). Choose $\mathfrak p' \subset \mathfrak p$ and $\mathfrak q' \subset V(\mathfrak p) \cap V(J) \cap V(I)$. \begin{enumerate} \item If $M_{\mathfrak p'} = 0$, then $\text{depth}(M_{\mathfrak p'}) = \infty$ and $\text{depth}(M_{\mathfrak p'}) + \dim((A/\mathfrak p')_{\mathfrak q'}) > d + s$. \item If $\text{depth}(M_{\mathfrak p'}) < s$, then $\text{depth}(M_{\mathfrak p'}) + \dim((A/\mathfrak p')_{\mathfrak q'}) > d + s$ by (4). \end{enumerate} In the remaining cases we have $M_{\mathfrak p'} \not = 0$ and $\text{depth}(M_{\mathfrak p'}) \geq s$. In particular, we see that $\mathfrak p'$ is in the support of $M$ and we can choose $\mathfrak p'' \subset \mathfrak p'$ with $\mathfrak p'' \in \text{Ass}(M)$. \begin{enumerate} \item[(a)] Observe that $\dim((A/\mathfrak p'')_{\mathfrak p'}) \geq \text{depth}(M_{\mathfrak p'})$ by Algebra, Lemma \ref{algebra-lemma-depth-dim-associated-primes}. If equality holds, then we have $$ \text{depth}(M_{\mathfrak p'}) + \dim((A/\mathfrak p')_{\mathfrak q'}) = \text{depth}(M_{\mathfrak p''}) + \dim((A/\mathfrak p'')_{\mathfrak q'}) > s + d $$ by (4) applied to $\mathfrak p''$ and we are done. This means we are only in trouble if $\dim((A/\mathfrak p'')_{\mathfrak p'}) > \text{depth}(M_{\mathfrak p'})$. This implies that $\dim(M_\mathfrak p) \geq s + 2$. Thus if (5)(a) holds, then this does not occur. \item[(b)] If (5)(b) holds, then we get $$ \text{depth}(M_{\mathfrak p'}) + \dim((A/\mathfrak p')_{\mathfrak q'}) \geq s + \delta(\mathfrak p') - \delta(\mathfrak q') \geq s + 1 + \delta(\mathfrak p) - \delta_{max} > s + d $$ as desired. \item[(c)] If (5)(c) holds, then we get \begin{align*} \text{depth}(M_{\mathfrak p'}) + \dim((A/\mathfrak p')_{\mathfrak q'}) & \geq s + \delta(\mathfrak p') - \delta(\mathfrak q') \\ & \geq s + 1 + \delta(\mathfrak p) - \delta(\mathfrak q') \\ & = s + 1 + \delta(\mathfrak p) - \delta(\mathfrak q) + \delta(\mathfrak q) - \delta(\mathfrak q') \\ & > s + 1 + (s + d + \delta_{max} - \delta_{min} - 2) + \delta(\mathfrak q) - \delta(\mathfrak q') \\ & \geq  2s + d - 1 \geq s + d \end{align*} as desired. Observe that this argument works because we know that a prime $\mathfrak q \in V(\mathfrak p) \cap V(J) \cap V(I)$ exists. \end{enumerate} Now we are ready to do the induction step. \medskip\noindent Choose an ideal $J_0$ as in Lemma \ref{lemma-kill-colimit-weak-general} and an integer $t > 0$ such that $(J_0I)^t$ annihilates $H^s_J(M)$. The assumptions of Lemma \ref{lemma-kill-completion-general} are satisfied for every $\mathfrak p \in \text{Ass}(M)$ (see previous paragraph). Thus the annihilator $\mathfrak a \subset A$ of $H^s(R\Gamma_J(M)^\wedge)$ 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 $H^s(R\Gamma_J(M)^\wedge)$ 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 $$ \xymatrix{ H^{s - 1}_{J'}(M) \ar[d] \ar[r] & H^{s - 1}_{J'}(M/fM) \ar[d] \ar[r] & H^s_{J'}(M) \ar[d] \ar[r] & 0 \\ H^{s - 1}(R\Gamma_J(M)^\wedge) \ar[r] & H^{s - 1}(R\Gamma_J(M/fM)^\wedge) \ar[r] & H^s(R\Gamma_J(M)^\wedge) \ar[r] & 0 } $$ for any $J' \subset J_0$ with $V(J') \cap V(I) = V(J) \cap V(I)$. Thus if we choose $J'$ such that it works for $M$ and $M/fM$ and $s - 1$ (possible by induction hypothesis -- see next paragraph), then we conclude that the lemma is true. \medskip\noindent To finish the proof we have to show that the module $M/fM$ satisfies the hypotheses (4) and (5) for $s - 1$. Thus we let $\mathfrak p$ be a prime in the support of $M/fM$ with $\text{depth}((M/fM)_\mathfrak p) < s - 1$ and with $V(\mathfrak p) \cap V(J) \cap V(I)$ nonempty. Then $\dim(M_\mathfrak p) = \dim((M/fM)_\mathfrak p) + 1$ and $\text{depth}(M_\mathfrak p) = \text{depth}((M/fM)_\mathfrak p) + 1$. In particular, we know (4) and (5) hold for $\mathfrak p$ and $M$ with the original value $s$. The desired inequalities then follow by inspection.