 letting $d_i'=d_i=0$, we have $$\label{ea16}
        d_i'=\pi(x_i'+ z_i'+w_i')=0.$$ This is an equation in $B\otimes_AR$. Thus there is exactly one independent linear equation among $x_i', z_i', w_i'$.

    5.  The $(2\times 2)$-block is $$1+2 f_i'=1+2 f_i+2\pi(u_i'+\pi x_i').$$ By letting $f_i'=f_i=0$, we have $$\label{ea17}
        f_i'=f_i+\pi(u_i'+\pi x_i')=0. $$ This is an equation in $R$. Thus this equation is trivial.

    6.  We postpone to consider the $(3, 3)$-block later in Step (vi).\

    By combining all five cases (a)-(e), there are exactly $((n_i-2)^2-(n_i-2))/2+2(n_i-2)+1=(n_i^2-n_i)/2$ independent linear equations. Thus $n_i^2-1-(n_i^2-n_i)/2=n_i(n_i+1)/2-1$ entries of $m_{i,i}'$ determine all entries of $m_{i,i}'$ except for $z_i'$ and $(z_i^{\ast})'$.\

6.  Let $i$ be even and $L_i$ be *of type I*. Finally, we consider the $(2, 2)$-block of Equation (\[ea12\]) if $L_i$ is *of type $I^o$* or the $(3, 3)$-block of Equation (\[ea13\]) if $L_i$ is *of type $I^e$* given below: $$\label{ea18}
     2\bar{\gamma}_i+4c_i'=2\bar{\gamma}_i+4c_i+ 2\pi^2z_i'+\pi^4(z_i')^2-\pi^4\delta_{i-2}k_{i-2, i}'-\pi^4\delta_{i+2}k_{i+2, i}'.$$ Here, $k_{i-2, i}'$ and $k_{i+2, i}'$ are as explained in the condition (5) given at the paragraph following (\[ea2\]). Note that the condition (5) yields Equation (\[52\]), $z_i'+\delta_{i-2}k_{i-2, i}'+\delta_{i+2}k_{i+2, i}'=0$, in $R$. Thus, by letting $c_i'=c_i=0$, we have $$\label{ea19}
     0=c_i'=c_i+z_i'=z_i'$$ as an equation in $R$.

    Note that if both $L_i$ and $L_{i+2}$ are *of type I*, then $k_{i+2, i}'=k_{i,i+2}'$ by Equation (\[ea4\]). We now choose an even integer $j$ such that $L_j$ is *of type I* and $L_{j+2}$ is *of type II*. For such $j$, there is a nonnegative integer $m_j$ such that $L_{j-2l}$ is *of type I* for every $l$ with $0\leq l \leq m_j$ and $L_{j-2(m_j+1)}$ is *of type II*. As mentioned in the above paragraph, the condition (5) yields the following equation in $R$: $$\mathcal{Z}_{j-2l}' : z_{j-2l}'+\delta_{j-2l-2}k_{j-2l-2, j-2l}'+\delta_{j-2l+