${\rm deg}_{X_n}(r) <s$, such that $f = qg + r$.

Choose $c$ for which $f$ and $g$ are elements of $F[X; Y, \lambda]_c$. Then, choose $c_i' \leq c_i$ such that $g$ is $X_n$-distinguished of degree $s$ in $F[X; Y, \lambda]_{c'}$. Carry out the division in this ring.

To show uniqueness we observe that $|f| = \max\{|q|, |r|\}$. Indeed, without loss of generality we can assume that $\max\{|q|, |r|\} = 1$. Thus, $|f| \leq 1$. Suppose that $|f| < 1$. Then, $0 \equiv qg + r \mod (Y)$. Since $\deg_{X_n} (r) < s = \deg_{X_n} (g \mod (Y))$, we must have that $q = r \equiv 0 \mod (Y)$, contradicting $\max\{|q|, |r|\} = 1$. Thus, if $q'g + r' = qg + r$, then $(q-q')g + (r-r') = 0$, thus, $|q-q'| = |r-r'| = 0$.

Let $f$ be $X_n$-distinguished of degree $s$. Then, there exists a unique monic polynomial $\omega$ in $F[X_1, \cdots, X_{n-1}; Y, \lambda][X_n]$ of degree $s$ in $X_n$ and a unique unit $e$ in $F[X ; Y, \lambda]$ such that $f = e\cdot \omega $. Further, $\omega$ is distinguished of degree $s$.

By the previous theorem there exists an element $q$ in $F[X; Y, \lambda]$ and a polynomial 