$0$ as $|\mu|$ approaches $\infty$. Let $c_n' = \frac{c_n}{2(d+1)}$. Then, $c_1\mu_1 + \cdots + c_n\mu_n > c_1\mu_1 + \cdots + c_n'\mu_{n}(d+1)$, so $|f_\mu(Y)|p^{c_1\mu_1 + \cdots + c_n'\mu_{n}(d+1)}$ converges to $0$ as $|\mu|$ approaches $\infty$. Therefore, this map is well defined with inverse, $\sigma^{-1}(X_n) = X_n - X_1^d$ and $\sigma^{-1}(X_j) = X_j$, if $j \neq n$.

Suppose $f(X,Y) = \sum_{\mu}f_\mu(Y) X^\mu$ is an element of $F[X; Y,\lambda]$. If $|f_\mu(Y)| = 1$, for some $\mu$, where $\mu_n > 0$, then there exists an automorphism $\sigma$ of $F[X; Y, \lambda]$ such that $\sigma(f)$ is $X_n$-distinguished.

Let $f(X,Y) = \sum_{\mu} f_\mu(Y) X^\mu = \sum_{\mu} f_\mu(Y)X_1^{\mu_1}\cdots X_n^{\mu_n}$. Let $\nu = (\nu_1, \cdots, \nu_n)$ be the maximal $n$-tuple, with respect to lexicographical ordering, such that $f_\nu(Y)$ is not an element of $(Y)$. Let $t \geq \max_{1\leq i\leq n} \mu_i$ for all indices $\mu$ such that $f_\mu(Y)$ is not an element of $(Y)$, e.g., let $t$ be the total $X$-degree of $f(X,Y) \mod (Y)$. Now, define an automorphism $\sigma(X_i) = X_i + X_n^{d_i}$ for $i = 1, \cdots, n-1$, and $\sigma(X_n) = X_n$, where $d_n = 1$, and $d_{n-j} = 1 + t\sum_{k=0}^{j-1} d_{n-k}$, for $j = 1, \cdots, n-1$. We see that this map is just 