\begin{thrm}\label{prop:balanced}
Let $\mathsf{N}$ in $\Nats$, and $a_{\mathsf{N}}$ be defined by
$a_{\mathsf{N}}= 2 \lfloor \dfrac{2^{\mathsf{N}}}{2\mathsf{N}} \rfloor$.
-Then, there exists a sequence $l$ in
+There exists then a sequence $l$ in
step~(\ref{item:nondet}) of the \emph{Robinson-Cohn extension} algorithm
such that all the transition counts $\textit{TC}_{\mathsf{N}}(i)$
are $a_{\mathsf{N}}$ or $a_{\mathsf{N}}+2$
for any $i$, $1 \le i \le \mathsf{N}$.
+\end{thrm}
-\end{thrm}
-
-
+The proof is done by induction on $\mathsf{N}$. Let us imadialty verify
+that it is established for both odd and even smallest values, \textit{i.e.}
+$3$ and $4$.
+For the initial case where $\mathsf{N}=3$, \textit{i.e.} $\mathsf{N-2}=1$ we successively have: $S_1=1,1$, $l=2$, $u_0 = \emptyset$, and $v=\emptyset$.
+Thus again the algorithm successively produces
+$U= 1,2,1$, $V = 3$, $W= 2, 1, 1,3$, and $W' = 1,2,1,3$.
+Finally, $S_3$ is $1,2,1,3,1,2,1,3$ which obviously verifies the theorem.
+ For the initial case where $\mathsf{N}=4$, \textit{i.e.} $\mathsf{N-2}=2$
+we successively have: $S_1=1,2,1,2$, $l=4$,
+$u_0,u_1,u_2 = \emptyset,\emptyset,\emptyset$, and $v=\emptyset$.
+Thus again the algorithm successively produces
+$U= 1,3,2,3,4,1,4,3,2$, $V = 4$, $W= 3, 1, 2, 1,2, 4$, and $W' = 1, 3, 2, 1,2, 4 $.
+Finally, $S_4$ is
+$
+2,3,4,1,4,3,2,3,1,4,1,3,2,1,2,4
+$
+such that $\textit{TC}_4(i) = 4$ and the theorem is established for
+odd and even initial values.
-\begin{proof}
+For the inductive case, let us first define some variables.
+Let $c_{\mathsf{N}}$ (resp. $d_{\mathsf{N}}$) be the number of elements
+whose transistion count is exactly $a_{\mathsf{N}}$ (resp $a_{\mathsf{N}} +2$).
+These two variables are defined by the system
+
+$$
+\left\{
+\begin{array}{lcl}
+c_{\mathsf{N}} + d_{\mathsf{N}} & = & \mathsf{N} \\
+c_{\mathsf{N}}a_{\mathsf{N}} + d_{\mathsf{N}}(a_{\mathsf{N}}+2) & = & 2^{\mathsf{N}}
+\end{array}
+\right.
+\qquad
+\Leftrightarrow
+\qquad
+\left\{
+\begin{array}{lcl}
+d_{\mathsf{N}} & = & \dfrac{2^{\mathsf{N}} -n.a_{\mathsf{N}}}{2} \\
+c_{\mathsf{N}} &= &\mathsf{N} - d_{\mathsf{N}}
+\end{array}
+\right.
+$$
+Since $a_{\mathsf{N}$ is even, $d_{\mathsf{N}}$ is defined.
+Moreover, both $c_{\mathsf{N}}$ and $d_{\mathsf{N}}$ are obviously positves.
-\end{proof}
\subsection{Toward a local uniform distribution of switches}
