= \|f_s(X,Y)X_n^{s}\|_{\lambda, c} =
    p^{c_ns} > \|f_k(X,Y)X_n^k\|_{\lambda, c}$ for all $k \neq s$.

If an element $f$ in $F[X; Y, \lambda]$ is $X_n$-distinguished of degree $s$ in $F[X; Y, \lambda]$, then it is $X_n$-distinguished in $F[X; Y, \lambda]_c$ for some $c$ in ${\mathbf{R}}^n.$ Indeed, suppose that $f$ is an element of $F[X; Y,\lambda]_c$. Since $f_s(X,Y)$ is a unit, we can write $f_s(X,Y) = u + h$, where $u$ is a unit in $F[[Y]]$, and $h$ is an element of $(Y)$. By choosing $c_1, \cdots, c_{n-1}$ small enough, we can make $\|h\|_{\lambda, c} < 1$, and so $\|f_s(X,Y)\|_{\lambda,  (c_1, \cdots, c_{n-1})} = 1$. We can reduce $c_1, \cdots, c_{n-1}$ even further to ensure that $\|f_k(X,Y)X_n^{k}\|_{\lambda, c} < \|f_s(X,Y)X_n^s\|_{\lambda, c} = p^{c_ns}$, because $f_k$ is an element of $(Y)$ for all $k > s$. Now, to ensure that $\|f_k(X,Y)X_n^{k}\|_{\lambda, c} < \|f_s(X,Y)X_n^s\|_{\lambda, c} = p^{c_ns}$, for $k < s$, we can shrink $c_1, \cdots, c_{n-1}$ once again so that $\|f_k(X,Y)\|_{\lambda, c} < p^{c_n}$. In this way we find that for all $k < s$, $\|f_k(X,Y)X_n^k\|_{\lambda, c} < p^{c_n} p^{c_n(s-1)} =  p^{c_ns} = \|f_s(X,Y)X_n^{s}\|_{\lambda, c}$ as desired.

We can use the notion of $X_n$-distinguished elements to derive a Euclidean algorithm for $F[X; Y, 