3 and 4, without loss of generality, we may assume that either $a_{k,r_{k}}\in I$ or $a_{k,j}+a_{k,r_{k}}\in I$ for some $1\leq j < r_{k}$.

Suppose first that $a_{k,r_{k}} \in I$. Then by Lemma \[ideal\], for all $t'_{k+1} =(v_1, \dots , v_{m_{k+1}})\in T_{k+1}$, we get that $$\begin{aligned}
\lefteqn{(\lambda_{a_{k,r_{k}}}-{\operatorname{id}}) ((0,\dots , 0, t'_{k+1}, 0,\dots , 0),(0,\dots , 0))}\\
 &=&
 ((0,\dots, 0, f_{e_{r_{k}}^{(k)}}^{(k,k+1)}(t'_{k+1})-t'_{k+1}, 0, \dots , 0),(0,\dots,s'_{k+1},0,\dots, 0))\in
 I\cap A_{k+1}\end{aligned}$$ for some $s'_{k+1}\in S_{k+1}$ and $$f_{e_{r_{k}}^{(k)}}^{(k,k+1)}(t'_{k+1}) -t'_{k+1} =
((f_{k+1}-{\operatorname{id}})(v_{1}), \dots , (f_{k+1}-{\operatorname{id}})(v_{m_{k+1}})).$$ Since, by assumption, $f_{k+1} -{\operatorname{id}}$ is bijective, by Claim 1 and Claim 2, we get that $$a_{k+1,r_{k+1}},\;  a_{k+1,j}+ a_{k+1,r_{k+1}} \in I$$ for all $1\leq j < r_{k+1}$. Thus, all $a_{k+1,j}\in I\cap
A_{k+1}$ (for $1\leq j \leq r_{k+1}$). Consequently, $$((0,\dots, 0),(0,\dots,0,s'_{k+1},0,\dots ,0))\in I\cap
A_{k+1}.$$ Therefore $$\begin{aligned}
 \lefteqn{((0,\dots, 0,
f_{e_{r_{k}}^{(k)}}^{(k,k+1)}(t'_{k+1})-t'_{k+1}, 0, \dots ,
0),(0,\dots, s'_{k+1},0,\dots, 0))}\\
&&-((0,\dots, 0),(0,\dots,0,s'_{k+1},0,\dots ,0))\\
&=& ((0,\dots, 0, f_{e_{r_{k}}^{(k)}}^{(k,k+1)}(t'_{k+1})-t'_{k+1},
0, \dots , 0),(0,\dots,0))\in I\cap A_{k+1}.\end{aligned}$$ Since $$f_{e_{r_{k}}^{(k)}}^{(k,k+1)}(t'_{k+1}) -t'_{k+1} =
((f_{k+1}-{\operatorname{id}})(v_{1}), \dots , (f_{k+1}-{\operatorname{id}})(v_{m_{k+1}}))$$ and $f_{k+1} -{\operatorname{id}}$ is bijective, in this case, $$A_{k+1}\subseteq I.$$

Now, consider the second case, namely suppose that $a_{k,j}+a_{k,r_{k}}\in I$ for some $1\leq j < r_{k}$. Then by Lemma \[ideal\], for all $t'_{k+1} =(v_1, \dots , v_{m_{k+1}})\in T_{k+1}$, $$\begin{aligned}
 \lefteqn{
 (\lambda_{a_{k,j}+a_{k,r_{k}}}-{\operatorname{id}}) ((0,\dots , 0,t'_{k+1}, 0,\dots , 0),(0,\dots, 