+
+The proof is done by induction on $\mathsf{N}$. Let us immediately 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.
+
+
+For the inductive case, let us first define some variables.
+Let $c_{\mathsf{N}}$ (resp. $d_{\mathsf{N}}$) be the number of elements
+whose transition count is exactly $a_{\mathsf{N}}$ (resp $a_{\mathsf{N}} +2$).
+These two variables are defined by the system
+
+\[
+\left\{
+\begin{array}{lcl}
+c_{\mathsf{N}} + d_{\mathsf{N}} & = & \mathsf{N} \\
+c_{\mathsf{N}}a_{\mathsf{N}} + d_{\mathsf{N}}(a_{\mathsf{N}}+2) & = & 2^{\mathsf{N}}
+\end{array}
+\right.
+\Leftrightarrow
+\left\{
+\begin{array}{lcl}
+d_{\mathsf{N}} & = & \dfrac{2^{\mathsf{N}} -\mathsf{N}.a_{\mathsf{N}}}{2} \\
+c_{\mathsf{N}} &= &\mathsf{N} - d_{\mathsf{N}}
+\end{array}
+\right.
+\]
+
+Since $a_{\mathsf{N}}$ is even, $d_{\mathsf{N}}$ is an integer.
+Let us first proove that both $c_{\mathsf{N}}$ and $d_{\mathsf{N}}$ are positive
+integers.
+Let $q_{\mathsf{N}}$ and $r_{\mathsf{N}}$, respectively, be
+the quotient and the remainder in the Euclidean disvision
+of $2^{\mathsf{N}}$ by $2\mathsf{N}$, \textit{i.e.}
+$2^{\mathsf{N}} = q_{\mathsf{N}}.2\mathsf{N} + r_{\mathsf{N}}$, with $0 \le r_{\mathsf{N}} <2\mathsf{N}$.
+First of all, the integer $r$ is even since $r_{\mathsf{N}} = 2^{\mathsf{N}} - q_{\mathsf{N}}.2\mathsf{N}= 2(2^{\mathsf{N}-1} - q_{\mathsf{N}}.\mathsf{N})$.
+Next, $a_{\mathsf{N}}$ is $\frac{2^{\mathsf{N}}-r_{\mathsf{N}}}{\mathsf{N}}$. Consequently
+$d_{\mathsf{N}}$ is $r_{\mathsf{N}}/2$ and is thus a positive integer s.t.
+$0 \le d_{\mathsf{N}} <\mathsf{N}$.
+The proof for $c_{\mathsf{N}}$ is obvious.
+
+
+For any $i$, $1 \le i \le \mathsf{N}$, let $zi_{\mathsf{N}}$ (resp. $ti_{\mathsf{N}}$ and $bi_{\mathsf{N}}$)
+be the occurence number of element $i$ in the sequence $u_0, \dots, u_{l-2}$
+(resp. in the sequences $s_{i_1}, \dots , s_{i_l}$ and $v$)
+in step (\ref{item:nondet}) of the algorithm.
+
+Due to the definition of $u'$ in step~(\ref{item:u'}),
+$3.zi_{\mathsf{N}} + ti_{\mathsf{N}}$ is the
+ number of element $i$ in the sequence $U$.
+It is clear that the number of element $i$ in the sequence $V$ is
+$2bi_{\mathsf{N}}$ due to step (\ref{item:VW}).
+We thus have the following system:
+$$
+\left\{
+\begin{array}{lcl}
+3.zi_{\mathsf{N}} + ti_{\mathsf{N}} + 2.bi_{\mathsf{N}} + \textit{TC}_{\mathsf{N}-2}(i) &= &\textit{TC}_{\mathsf{N}}(i) \\
+zi_{\mathsf{N}} + ti_{\mathsf{N}} + bi_{\mathsf{N}} & =& \textit{TC}_{\mathsf{N}-2}(i)
+\end{array}
+\right.
+\qquad
+\Leftrightarrow
+$$
+
+\begin{equation}
+\left\{
+\begin{array}{lcl}
+zi_{\mathsf{N}} &= &
+\dfrac{\textit{TC}_{\mathsf{N}}(i) - 2.\textit{TC}_{\mathsf{N}-2}(i) - bi_{\mathsf{N}}}{2}\\
+ti_{\mathsf{N}} &= & \textit{TC}_{\mathsf{N}-2}(i)-zi_{\mathsf{N}}-bi_{\mathsf{N}}
+\end{array}
+\right.
+\label{eq:sys:zt1}
+\end{equation}
+
+In this set of 2 equations with 3 unknown variables, let $b_i$ be set with 0.
+In this case, since $\textit{TC}_{\mathsf{N}}$ is even (equal to $a_{\mathsf{N}}$
+or to $a_{\mathsf{N}}+2$), the variable $zi_{\mathsf{N}}$ is thus an integer.
+Let us now prove that the resulting system has always positive integer
+solutions $z_i$, $t_i$, $0 \le z_i, t_i \le \textit{TC}_{\mathsf{N}-2}(i)$
+and s.t. their sum is equal to $\textit{TC}_{\mathsf{N}-2}(i)$.
+This latter consraint is obviously established if the system has a solution.
+We thus have the following system.
+
+
+
+\begin{equation}
+\left\{
+\begin{array}{lcl}
+zi_{\mathsf{N}} &= &
+\dfrac{\textit{TC}_{\mathsf{N}}(i) - 2.\textit{TC}_{\mathsf{N}-2}(i) }{2}\\
+ti_{\mathsf{N}} &= & \textit{TC}_{\mathsf{N}-2}(i)-zi_{\mathsf{N}}
+\end{array}
+\right.
+\label{eq:sys:zt2}
+\end{equation}
+
+The definition of $\textit{TC}_{\mathsf{N}}(i)$ depends on the value of $\mathsf{N}$.
+When $3 \le N \le 7$, values are defined as follows:
+\begin{eqnarray*}
+\textit{TC}_{3} & = & [2,2,4] \\
+\textit{TC}_{5} & = & [6,6,8,6,6] \\
+\textit{TC}_{7} & = & [18,18,20,18,18,18,18] \\
+\\
+\textit{TC}_{4} & = & [4,4,4,4] \\
+\textit{TC}_{6} & = & [10,10,10,10,12,12] \\
+\end{eqnarray*}
+It is not hard to verify that all these instanciations verify the aformentioned contraints.
+
+When $N \ge 8$, $\textit{TC}_{\mathsf{N}}(i)$ is defined as follows:
+\begin{equation}
+\textit{TC}_{\mathsf{N}}(i) = \left\{
+\begin{array}{l}
+a_{\mathsf{N}} \textrm{ if } 1 \le i \le c_{\mathsf{N}} \\
+a_{\mathsf{N}}+2 \textrm{ if } c_{\mathsf{N}} +1 \le i \le c_{\mathsf{N}} + d_{\mathsf{N}}
+\end{array}
+\right.
+\label{eq:TCN:def}
+\end{equation}
+
+
+We thus have
+\[
+\begin{array}{rcl}
+\textit{TC}_{\mathsf{N}}(i) - 2.\textit{TC}_{\mathsf{N}-2}(i)
+&\ge&
+a_{\mathsf{N}} - 2(a_{\mathsf{N}-2}+2) \\
+&\ge&
+\frac{2^{\mathsf{N}}-r_{\mathsf{N}}}{\mathsf{N}}
+-2 \left( \frac{2^{\mathsf{N-2}}-r_{\mathsf{N-2}}}{\mathsf{N-2}}+2\right)\\
+&\ge&
+\frac{2^{\mathsf{N}}-2N}{\mathsf{N}}
+-2 \left( \frac{2^{\mathsf{N-2}}}{\mathsf{N-2}}+2\right)\\
+&\ge&
+\frac{(\mathsf{N} -2).2^{\mathsf{N}}-2N.2^{\mathsf{N-2}}-6N(N-2)}{\mathsf{N.(N-2)}}\\
+\end{array}
+\]
+
+A simple variation study of the function $t:\R \rightarrow \R$ such that
+$x \mapsto t(x) = (x -2).2^{x}-2x.2^{x-2}-6x(x-2)$ shows that
+its derivative is strictly postive if $x \ge 6$ and $t(8)=224$.
+The integer $\textit{TC}_{\mathsf{N}}(i) - 2.\textit{TC}_{\mathsf{N}-2}(i)$ is thus positive
+for any $\mathsf{N} \ge 8$ and the proof is established.
