+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.