initial condition $M_{n}(0, x) \equiv 1$ or $$\label{E1.9}
\partial_{t}M_{n}(t, x, y) = (\mathscr{H}M_{n}(t, \cdot, y))(x) + \delta_0(x)g_{n}(M_{1}(t,x, y),\ldots, M_{n-1}(t, x, y))$$ with the initial condition $ M_{n}(0, x, y) = \delta_{x}(y)$, $ y\in \mathbb{Z}^d$, where $$\begin{aligned}
 g_{n}(M_{1}, M_{2},\ldots, M_{n-1}) &=  \sum_{r=1}^{n} \frac{\beta^{(r)}}{r!} \sum_{\stackrel{i_{1},\ldots,i_{r}>0}{i_{1}+\cdots+i_{r}=n}}
\frac{n!}{i_{1}!\cdots i_{r}!} M_{i_{1}}\cdots M_{i_{r}}.\end{aligned}$$

The proof of the lemma is by analogy with [@YarBRW:r]. Again, we confine ourselves to proving only equation .

Let us differentiate $n$ times with respect to $z$ the equation $$\partial_{t}F_{1}(z; t, x, y) = (\mathscr{A}F_{1}(z; t, \cdot, y))(x) + \delta_{0}(x)f(F_{1}(z; t, x, y)),$$ (it is permitted owing to Remark \[Rem1\]). Then we obtain: $$(-1)^n \partial_{t} \partial_{z}^{n}F_{1}(z; t, x, y) = (-1)^{n}(\mathscr{A}\partial_{z}^{n}F_{1}(z; t, \cdot, y))(x) + \delta_{0}(x)\partial_{z}^{n}f(F_{1}(z; t, x, y))$$

Using the Faa di Bruno’s formula [@M:00] we calculate the second summand on the right side of the equation: $$\label{E:df}
\frac{d^{n}f(F(z))}{dz^{n}} = \sum_{r=1}^{n} f^{(r)}(F(z)) \sum_{\stackrel {i_{1}+\cdots+i_{n}=r}{
        i_{1}+2i_{2}+\cdots+ni_{n}=n}}
\frac{n!}{i_{1}!\cdots i_{n}!} \left( \frac{F^{(1)}(z)}{1!} \right)^{i_{1}}\cdots \left( \frac{F^{(n)}(z)}{n!} \right)^{i_{n}}.$$

Let us prove the supportive equation below in order to simplify the inner sum in this formula to the required in the lemma’s statement form: $$\label{E:star} \sum_{\stackrel {i_{1}+\cdots+i_{n}=r}{i_{1}+\cdots+ni_{n}=n}}
\frac{n!}{i_{1}!\cdots i_{n}!}x_{1}^{i_{1}}\cdots x_{n}^{i_{n}} = \sum_{\stackrel {i_{1},\ldots,i_{r}>0}{i_{1}+\cdots+i_{r}=n}} x_{i_{1}}\cdots x_{i_{r}},$$ where $ x_{1},\ldots, x_{n} $ are formal variations. Consider the polynomial $ 