$k^{\rm th}$ equation has order $s+k-1$ in the normal direction, for $k=1,\dots,s$. Since $T_{\alpha,0}^* =\nu^\alpha$ and $-\Delta=-N^2 +T_1 N+T_2$, using formula (\[\*\*\*\]), the operator on the left hand side in the $k^{\rm th}$ equation becomes $$\begin{aligned}
\lefteqn{
\sum_{s-k\le|\alpha|\le s-1} \CP 
T_{\alpha,|\alpha|-s+k}^* ND^\alpha
\biggl( \sum_{j=0}^{s-1-|\alpha|}(-\Delta)^{j}  \biggr)}\\ 
& = N^{s-k+1} \sum_{j=0}^{k-1}(-\Delta)^{j} + \cdots +
\sum_{|\alpha|=s-1} \CP T_{\alpha,k-1}ND^\alpha \\
& = (-1)^{(k-1)}N^{s+k-1} +T_1 N^{s+k-2}+\cdots+ T_{s+k-2}N \end{aligned}$$ as in the statement of the proposition, while the right hand side in the same equation is an operator of order $s+k$ (one order larger than the left hand side), that we denote by $P^{(J)}_{s+k}$. Then we have $$\label{P-s+k}
P^{(J)}_{s+k} (\psi) = 
\sum_{s-k\le|\alpha|\le s} \CP 
T_{\alpha,|\alpha|-s+k}^* (L_\alpha
 \psi)_J .$$ This finishes the proof.

Before proving Corollary \[K-order-1\] we need one more result. Consider the  (\[BV\]) that defines the components of $\K$: $$\label{BVP2}
\begin{cases}
\sum_{j=0}^{s}(-\Delta)^j u =0 & \text{ on }\Omega\\
\sum_{j=0}^{s+\ell}T_j N^{s+\ell-j}u = g_\ell
    & \text{ on }b\Omega ,\, \ell=0,\dots,s-1 \, .
\end{cases}$$ for given $g_\ell \in C^\infty (b\Omega)$, $\ell=0,\dots,s-1$. Notice that the operator $\K$ applied to a form $\psi$ gives rise to the composition of a (non-diagonal) differential operator acting on the components of $\psi$, the restriction to the boundary $b\Omega$, and the solution operator $S$ 