Let $M$ be a module over Noetherian ring $R$ such that $\operatorname{H}^1_I(M)=0$ for every ideal $I$ of $R$. Show that $\operatorname{H}^1_J(\Gamma_I(M))=0$ for every ideal $J$.
I tried to prove it by Mayer-Vietoris sequence but I can't, unfortunately. Also applied the following sequences $$0\rightarrow \Gamma_I(M)\rightarrow M\rightarrow D_I(M)\rightarrow 0 $$ $$0\rightarrow \Gamma_I(M)\rightarrow M \rightarrow M/\Gamma_I(M)\rightarrow0 .$$ Moreover, since $\operatorname{H}^1_I(M)=0$ for every ideal $I$ of $R$ then Proposition 4.1.3 from Brodmann-Sharp, Local Cohomology, suggests that $\operatorname{H}^i_I(M)=0$ for all $i\geq 0$.
Background: $D_I(M)$ means ideal transform with respect to $I$ and the first exact sequence obtains from Theorem 2.2.4(i)(c) in the same book.
Sketch of proof:
Let $J=(x_1,\dots,x_n)$. If $J\subseteq I$ it's done. Let $J\nsubseteq I$ therefore $\exists x\in J-I $. By 4.1.22 from the same book we have the following exact sequence $$0\rightarrow \Gamma_{I+x}(M)\rightarrow\Gamma_I(M)\rightarrow \Gamma_I(M_x)\rightarrow 0=H^1_{I+x}(M).$$ Applying $\Gamma_J(-)$ we get $$ \cdots\rightarrow \Gamma_J(\Gamma_I(M_x))\rightarrow H^1_J(\Gamma_{I+x}(M))\rightarrow H^1_J(\Gamma_I(M))\rightarrow H^1_J(\Gamma_I(M_x))\rightarrow\cdots $$ Now use the independence Theorem $$\Gamma_J(\Gamma_I(M_x))\cong \Gamma_J(\Gamma_{I_x}(M_x))\cong \Gamma_J(\Gamma_I{(M )})_x\cong \Gamma_{J_x}(\Gamma_I(M))_x=0. ~ (Since~ x\in J)$$ $$H^1_J(\Gamma_I(M_x))\cong H^1_J(\Gamma_{I_x}(M_x))\cong H^1_J(\Gamma_I{(M )})_x\cong H^1_{J_x}(\Gamma_I(M))_x=0.~ (Since~ x\in J)$$ Therefore $H^1_J(\Gamma_{I+x}(M))\cong H^1_J(\Gamma_I(M))$. Continuing as before we have $$H^1_J(\Gamma_I(M))\cong H^1_J(\Gamma_{I+J}(M))=0. ~(Since~ J\subseteq I+J)$$