
\subsection{Study of stego-security}\label{sec:stego-security-ci-2}


\noindent Let us prove that,

\begin{proposition}
\CID is stego-secure.
\end{proposition}

\begin{proof}
%We suppose that $K,M \sim
%\mathbf{U}\left([0;1]\right)$, and $X \sim
%We consider a strategy in the CIIS scheme, and we
Let us suppose that $x^0 \sim
\mathbf{U}\left(\mathbb{B}^N\right)$ and $m^0 \sim
\mathbf{U}\left(\mathbb{B}^P\right)$ in a \CID setup. 
We will prove by a mathematical induction that $\forall n \in \mathds{N}, x^n
\sim \mathbf{U}\left(\mathbb{B}^N\right)$. The base case is obvious according
to the uniform repartition hypothesis. 

Let us now suppose that the statement $x^n \sim
\mathbf{U}\left(\mathbb{B}^N\right)$ holds for some $n$. 
For a given $k  \in \mathbb{B}^N$,
we denote by $\tilde{k_i} \in \mathbb{B}^N$ the vector defined by: $\forall i
\in  \llbracket 0;\mathsf{N-1}\rrbracket,$
if
$k=\left(k_0,k_1,\hdots,k_i,\hdots,k_{\mathsf{N}-2},k_{\mathsf{N}-1}\right)$,\newline
then $\tilde{k}_i=\left( k_0,k_1,\hdots,\overline{k_i},\hdots,k_{\mathsf{N}-2},k_{\mathsf{N}-1} \right).$


%\begin{equation*}
Let
$E_{i,j}$ be the following events: $$\forall (i,j) \in \llbracket
0;\mathsf{N-1}\rrbracket \times \llbracket 0;\mathsf{P-1}\rrbracket, E_{i,j}=$$
$$S_p^{n+1}=i \wedge S_c^{n+1}=j \wedge  m_j^{n+1}=k_i \wedge \left(x^n=k \vee
x^n=\tilde{k}_i\right),$$ and
$p=P\left(x^{n+1}=k\right).$
So,
\begin{equation*}
p=P\left(\bigvee_{i \in \llbracket
0;\mathsf{N-1}\rrbracket, j \in \llbracket 0;\mathsf{P-1}\rrbracket}
E_{i,j} \right).
\end{equation*}
We now introduce the following notation: $P_1(i)=P\left(S_p^{n+1}=i\right),$
 $P_2(j)=P\left(S_c^{n+1}=j\right),$
 $P_3(i,j)=P\left(m_j^{n+1}=k_i\right),$
and $P_4(i)=P\left(x^n=k \vee x^n=\tilde{k}_i\right).$


These four events are independent in \CID setup, thus:
\begin{equation*}
p=\sum_{i \in \llbracket
0;\mathsf{N-1}\rrbracket, j \in \llbracket 0;\mathsf{P-1}\rrbracket}
P_1(i)P_2(i)P_3(i,j)P_4(i).
\end{equation*}
According to 
Proposition~\ref{prop:CIIS-stego-security}, $P\left(m_j^{n+1}=k_i\right)=\frac{1}{2}$.
As the two events are incompatible:
$$P\left(x^n=k \vee x^n=\tilde{k}_i\right)=P\left(x^n=k\right) + P\left(x^n=\tilde{k}_i\right).$$


Then, by using the inductive hypothesis:
$P\left(x^n=k\right)=\frac{1}{2^N},$
and
$P\left(x^n=\tilde{k}_i\right)=\frac{1}{2^N}.$

Let $S$ be defined by $$S=\sum_{i
\in \llbracket 0;\mathsf{N-1}\rrbracket, j \in \llbracket 0;\mathsf{P-1}\rrbracket}
P_1(i)P_2(j).$$
Then $p =  2 \times \frac{1}{2} \times \frac{1}{2^N} \times S = \frac{1}{2^N} \times S$.

$S$ can now be evaluated:
\begin{equation*}
\begin{array}{lll}
S & = &  \sum_{i
\in \llbracket 0;\mathsf{N-1}\rrbracket, j \in \llbracket 0;\mathsf{P-1}\rrbracket}
P_1(i)P_2(j)\\
  & = &  \sum_{i \in \llbracket 0;\mathsf{N-1}\rrbracket} P_1(i) \times \sum_{j \in \llbracket 0;\mathsf{P-1}\rrbracket}
P_2(j).\\
\end{array}
\end{equation*}

The set of events $\left \{ S_p^{n+1}=i \right \}$ for $i \in \llbracket 0;N-1
\rrbracket$ and the set of events $\left \{ S_c^{n+1}=j \right \}$ for $j \in
\llbracket 0;P-1 \rrbracket$ are both a partition of the universe of possible,
so $S=1$.

%Evaluate now $ P\left(S^n=k\right)$.
%\newline
%We have: $K^0 = M \otimes K \sim
%\mathbf{U}\left([0;1]\right)$, so $K^\prime=F\left( K^0,p \right) \sim
%\mathbf{U}\left([0;1]\right)$, because for PLCMs, ``Uniform input generate
%uniform output''~\cite{Lasota}.
%Well with an immediate proof by recurrence we have:
%$K^n \sim \mathbf{U}\left([0;1]\right)$. As $S^n = \left \lfloor N \times X^n
%\right \rfloor + 1$ we can deduce that: $S^n \sim
%\mathbf{U}\left([1;N]\right)$. Therefore: $ P\left(S^n=k\right)=\frac{1}{N}$.
Finally, 
%\begin{equation*}
%\begin{array}{llr}
%P\left(X^{n+1}=e\right) & =\frac{1}{2^N} \sum_{k=1}^N
% P\left(S^n=k\right) & \\
% & =\frac{1}{2^N} \sum_{k=1}^N \frac{1}{N} & \\
% & =\frac{1}{2^N} & \\
%\end{array}
%\end{equation*}
%\begin{equation*}
%$\begin{array}{llr}
%P\left(X^{n+1}=e\right) & =\frac{1}{2^N} \sum_{k=1}^N
% P\left(S^n=k\right) & \\
%P\left(X^{n+1}=e\right) & =\frac{1}{2^N} \sum_{k=1}^N \frac{1}{N} &
% =\frac{1}{2^N} \\
%\end{array}$.
$P\left(x^{n+1}=k\right)=\frac{1}{2^N}$, which leads to $x^{n+1} \sim
\mathbf{U}\left(\mathbb{B}^N\right)$.
This result is true $\forall n \in \mathds{N}$, we thus have proven that the
stego-content $y$ is uniform in the set of possible stego-content, so
$y \sim \mathbf{U}\left(\mathbb{B}^N\right) \text{
when } x \sim \mathbf{U}\left(\mathbb{B}^N\right).$
\end{proof}


\subsection{Study of topological-security of
$CI_2$}\label{sec:chaos-security-ci2}
\subsubsection{\textbf{Topological model}}
\label{sec:CI2-topological-model}

%\section{THE NEW TOPOLOGICAL SPACE}
\noindent In this section, we prove that \CID can be modeled as a
discret dynamical system in a topological space. 
We will show in the next section that \CID is a case of topological chaos in the sense of Devaney.


\paragraph{\textbf{Iteration Function and Phase Space}}\label{Defining}


%\begin{definition}[Function $F$]
Let
\begin{equation*}
\begin{array}{ll}
F: & \llbracket0;\mathsf{N-1}\rrbracket \times
\mathds{B}^{\mathsf{N}}\times \llbracket0;\mathsf{P-1}\rrbracket  \times
\mathds{B}^{\mathsf{P}}
\longrightarrow \mathds{B}^{\mathsf{N}} \\
 & (k,x,\lambda,m)  \longmapsto  \left( \delta
 (k,j).x_j+\overline{\delta (k,j)}.m_{\lambda}\right) _{j\in
 \llbracket0;\mathsf{N-1}\rrbracket}\\
 \end{array}%
\end{equation*}%
%\end{definition}

\noindent where + and . are the boolean addition and product operations.

Consider the phase space $\mathcal{X}_2$ defined as follow:

%\begin{definition}[Phase space $\mathcal{X}_2$]

\begin{equation*}
\mathcal{X}_2 =  \mathbb{S}_N \times \mathds{B}^\mathsf{N} \times \mathbb{S}_P
\times \mathds{B}^\mathsf{P} \times \mathbb{S}_P,
\end{equation*}
where $\mathbb{S}_N$ and $\mathbb{S}_P$ are the sets introduced in Section \ref{section:Algorithm}.
%\end{definition}

We define the map $\mathcal{G}_{f_0}:\mathcal{X}_2  \longrightarrow  \mathcal{X}_2$ by:

\begin{equation*}
\mathcal{G}_{f_0}\left(S_p,x,S_c,m,S_m\right) =   
\end{equation*}
\begin{equation*}
\left(\sigma_N(S_p),
F(i_N(S_p),x,i_P(S_c),m),\sigma_P(S_c),G_{f_0}(m,S_m),\sigma_P(S_m)\right)
\end{equation*}

\noindent Then \CID can be described by the
iterations of the following discret dynamical system:

%\begin{definition}[Discret Dynamical System for \CID]
\begin{equation*}
\left\{
\begin{array}{l}
X^0 \in \mathcal{X}_2 \\
X^{k+1}=\mathcal{G}_{f_0}(X^k).%
\end{array}%
\right.
\end{equation*}%
%\end{definition}%


\paragraph{\textbf{Cardinality of $\mathcal{X}_2$}}

By comparing $\mathcal{X}_2$ and $\mathcal{X}_1$, we have the following result.

\begin{proposition}
The phase space $\mathcal{X}_2$ has, at least, the cardinality of the continuum.
\end{proposition}

\begin{proof}
Let $\varphi$ be the map defined as follow:

\begin{equation*}
\begin{array}{lcll}
\varphi: & \mathcal{X}_1 & \longrightarrow & \mathcal{X}_2 \\
 & (S,x) &  \longmapsto  &(S,x,0,0,0)\\
 \end{array}%
\end{equation*}%

\noindent $\varphi$ is  injective.
So the cardinality of  $\mathcal{X}_2$ is greater than or equal to the
cardinality of $\mathcal{X}_1$. And consequently
$\mathcal{X}_2$ has at least the cardinality of the continuum.
\end{proof}


\begin{remark}
This result is independent on the number of cells of the system.
\end{remark}


\paragraph{\textbf{A New Distance on $\mathcal{X}_2$}}

We define a new distance on $\mathcal{X}_2$ as follow: 
%\begin{definition}[Distance $d_2$ on $\mathcal{X}_2$]
$\forall X,\check{X} \in \mathcal{X}_2,$ if $X =(S_p,x,S_c,m,S_m)$ and
$\check{X} = (\check{S_p},\check{x},\check{S_c},\check{m}, \check{S_m}),$ then:
\begin{equation*}
\begin{array}{lll}
d_2(X,\check{X}) & = &
d_{\mathds{B}^{\mathsf{N}}}(x,\check{x})+d_{\mathds{B}^{\mathsf{P}}}(m,\check{m})\\
& + &
d_{\mathbb{S}_N}(S_p,\check{S_p})+d_{\mathbb{S}_P}(S_c,\check{S_c})+d_{\mathbb{S}_P}(S_m,\check{S_m}),\\
\end{array}
\end{equation*}
where $d_{\mathds{B}^{\mathsf{N}}}$, $d_{\mathds{B}^{\mathsf{P}}}$,
$d_{\mathbb{S}_N}$, and $d_{\mathbb{S}_P}$ are the same distances than in
Definition~\ref{def:distance-on-X1}.
%\end{definition}






\subsubsection{\textbf{Continuity of $CI_2$}}

To prove that \CID is another example of topological chaos in the
sense of Devaney, $\mathcal{G}_{f_0}$ must be continuous on the metric
space $(\mathcal{X}_2,d_2)$.

\begin{proposition}
$\mathcal{G}_{f_0}$ is a continuous function on $(\mathcal{X}_2,d_2)$.
\end{proposition}

\begin{proof}
We use the sequential continuity.

Let $((S_p)^n,x^n,(S_c)^n,m^n,(S_m)^n)_{n\in \mathds{N}}$ be a sequence of the
phase space $\mathcal{X}_2$, which converges to $(S_p,x,S_c,m,S_m)$. We will
prove that $\left( \mathcal{G}_{f_0}((S_p)^n,x^n,(S_c)^n,m^n,(S_m)^n)\right)
_{n\in \mathds{N}}$ converges to $\mathcal{G}_{f_0}(S_p,x,S_c,m,S_m)$. Let us
recall that for all $n$, $(S_p)^n$, $(S_c)^n$ and $(S_m)^n$ are strategies, thus
we consider a sequence of strategies (\emph{i.e.}, a sequence of sequences).


As $d_2(((S_p)^n,x^n,(S_c)^n,m^n,(S_m)^n),(S_p,x,S_c,m,S_m))$ converges to 0,
each distance
$d_{\mathds{B}^{\mathsf{N}}}(x^n,x)$, $d_{\mathds{B}^{\mathsf{P}}}(m^n,m)$,  
$d_{\mathbb{S}_N}((S_p)^n,S_p)$, $d_{\mathbb{S}_P}((S_c)^n,S_c)$, and
$d_{\mathbb{S}_P}((S_m)^n,S_m)$ converges to 0. But
$d_{\mathds{B}^{\mathsf{N}}}(x^n,x)$ and $d_{\mathds{B}^{\mathsf{P}}}(m^n,m)$
are integers, so $\exists n_{0}\in \mathds{N}, \forall
n \geqslant n_0,
d_{\mathds{B}^{\mathsf{N}}}(x^n,x)=0$ and $\exists n_{1}\in \mathds{N},
\forall n \geqslant n_1, d_{\mathds{B}^{\mathsf{P}}}(m^n,m)=0$. 

Let $n_3=Max(n_0,n_1)$.
In other words, there exists a threshold $n_{3}\in \mathds{N}$ after which no
cell will change its state:
$\exists n_{3}\in \mathds{N},n\geqslant n_{3}\Longrightarrow (x^n = x) \wedge
(m^n = m).$

In addition, $d_{\mathbb{S}_N}((S_p)^n,S_p)\longrightarrow 0$,
$d_{\mathbb{S}_P}((S_c)^n,S_c)\longrightarrow 0$, and
$d_{\mathbb{S}_P}((S_m)^n,S_m)\longrightarrow 0$, so $\exists n_4,n_5,n_6\in
 \mathds{N},$
 \begin{itemize}
   \item $\forall n \geqslant n_4, d_{\mathbb{S}_N}((S_p)^n,S_p)<10^{-1}$,
   \item $\forall n \geqslant n_5, d_{\mathbb{S}_P}((S_c)^n,S_c)<10^{-1}$,
   \item $\forall n \geqslant n_6, d_{\mathbb{S}_P}((S_m)^n,S_m)<10^{-1}$.
 \end{itemize}
 
 Let $n_7=Max(n_4,n_5,n_6)$. For
 $n\geqslant n_7$, all the strategies $(S_p)^n$, $(S_c)^n$, and $(S_m)^n$ have
 the same first term, which are respectively $(S_p)_0$,$(S_c)_0$ and $(S_m)_0$
 :$\forall n\geqslant n_7,$ $$((S_p)_0^n={(S_p)}_0)\wedge
((S_c)_0^n={(S_c)}_0)\wedge ((S_m)_0^n={(S_m)}_0).
$$

Let $n_8=Max(n_3,n_7)$.
After the $n_8-$th term, states of $x^n$ and $x$ on the one hand, and $m^n$ and $m$ on the other hand, are
identical. Additionally, strategies $(S_p)^n$ and $S_p$, $(S_c)^n$ and $S_c$,
and $(S_m)^n$ and $S_m$  start with the same first term.

Consequently, states of
$\mathcal{G}_{f_0}((S_p)^n,x^n,(S_c)^n,m^n,(S_m)^n)$ and
$\mathcal{G}_{f_0}(S_p,x,S_c,m,S_m)$ are equal, so, after the $(n_8)^{th}$ term,
the distance $d_2$ between these two points is strictly smaller than $3.10^{-1}$, so strictly smaller than 1.

 We now prove that the distance between $\left(
\mathcal{G}_{f_0}((S_p)^n,x^n,(S_c)^n,m^n,(S_m)^n)\right) $ and $\left(
\mathcal{G}_{f_0}(S_p,x,S_c,m,S_m)\right) $ is convergent to 0.
Let $\varepsilon >0$. 

\begin{itemize}
\item If $\varepsilon \geqslant 1$, we have seen that distance
between $\mathcal{G}_{f_0}((S_p)^n,x^n,(S_c)^n,m^n,(S_m)^n)$ and 
$\mathcal{G}_{f_0}(S_p,x,S_c,m,S_m)$ is
strictly less than 1 after the $(n_8)^{th}$ term (same state).
\medskip

\item If $\varepsilon <1$, then $\exists k\in \mathds{N},10^{-k}\geqslant
\frac{\varepsilon}{3} \geqslant 10^{-(k+1)}$. As
$d_{\mathbb{S}_N}((S_p)^n,S_p)$, $d_{\mathbb{S}_P}((S_c)^n,S_c)$ and
$d_{\mathbb{S}_P}((S_m)^n,S_m)$  converges to 0, we have:
\begin{itemize}
 \item $\exists n_9\in \mathds{N},\forall n\geqslant
 n_9,d_{\mathbb{S}_N}((S_p)^n,S_p)<10^{-(k+2)}$,
 \item $\exists n_{10}\in \mathds{N},\forall n\geqslant
 n_{10},d_{\mathbb{S}_P}((S_c)^n,S_c)<10^{-(k+2)}$,
 \item $\exists n_{11}\in \mathds{N},\forall n\geqslant
 n_{11},d_{\mathbb{S}_P}((S_m)^n,S_m)<10^{-(k+2)}$.
\end{itemize}
  
Let $n_{12}=Max(n_9,n_{10},n_{11})$
thus after $n_{12}$, the $k+2$ first terms of $(S_p)^n$ and $S_p$, $(S_c)^n$ and
$S_c$, and $(S_m)^n$ and $S_m$, are equal.
\end{itemize}

\noindent As a consequence, the $k+1$ first entries of the strategies of
$\mathcal{G}_{f_0}((S_p)^n,x^n,(S_c)^n,m^n,(S_m)^n)$ and $
\mathcal{G}_{f_0}(S_p,x,S_c,m,S_m)$ are the same
(due to the shift of strategies) and following the definition of
$d_{\mathbb{S}_N}$ and $d_{\mathbb{S}_P}$:
$$d_2\left(\mathcal{G}_{f_0}((S_p)^n,x^n,(S_c)^n,m^n,(S_m)^n);\mathcal{G}_{f_0}(S_p,x,S_c,m,S_m)\right)$$
is equal to :
$$d_{\mathbb{S}_N}((S_p)^n,S_p) + d_{\mathbb{S}_P}((S_c)^n,S_c) +
d_{\mathbb{S}_P}((S_m)^n,S_m)$$
which is smaller than $3.10^{-(k+1)} \leqslant 3.\frac{\varepsilon}{3} =\varepsilon$.

Let $N_{0}=max(n_{8},n_{12})$. We can claim that
\begin{equation*}
\forall \varepsilon >0,\exists N_{0}\in \mathds{N},\forall n\geqslant N_{0},
\end{equation*}
\begin{equation*}
d_2\left(\mathcal{G}_{f_0}((S_p)^n,x^n,(S_c)^n,m^n,(S_m)^n);\mathcal{G}_{f_0}(S_p,x,S_c,m,S_m)\right)
\leqslant \varepsilon .
\end{equation*}
$\mathcal{G}_{f_0}$ is consequently continuous on $(\mathcal{X}_2,d_2)$.
\end{proof}



% >>>>>>>>>>>>>>>>>>>>>> Discrete chaotic iterations as topological chaos <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
%\subsection{Study of chaos-security}\label{section:chaos-security-ci-2}

\noindent To prove that we are in the framework of Devaney's topological chaos,
we have to check the regularity, transitivity, and sensitivity conditions.
%It will be proven that $\mathcal{G}_{f_0}$ function as introduce in
%definition~\ref{} satisfies these three hypotheses.



\subsubsection{\textbf{Regularity}}

\begin{proposition}%[Regularity of $\mathcal{G}_{f_0}$]
\label{prop:regularite}
Periodic points of $\mathcal{G}_{f_0}$ are dense in $\mathcal{X}_2$.
\end{proposition}

\begin{proof}
Let $(\check{S_p},\check{x},\check{S_c},\check{m}, \check{S_m})\in
\mathcal{X}_2$ and $\varepsilon >0$. We are looking for a periodic point
$(\widetilde{S_p},\widetilde{x},\widetilde{S_c},\widetilde{m}, \widetilde{S_m})$
satisfying $d_2((\check{S_p},\check{x},\check{S_c},\check{m},
\check{S_m});(\widetilde{S_p},\widetilde{x},\widetilde{S_c},\widetilde{m},
\widetilde{S_m}))<\varepsilon$.

As $\varepsilon$ can be strictly lesser than 1, we must choose
$\widetilde{x} = \check{x}$ and
$\widetilde{m} = \check{m}$.
Let us define $k_0(\varepsilon) =\lfloor
- log_{10}(\frac{\varepsilon}{3} )\rfloor +1$ and consider the set:
$\mathcal{S}_{\check{S_p},\check{S_c},\check{S_m}, k_0(\varepsilon)} =$
$\left\{ S
\in \mathbb{S}_N \times \mathbb{S}_P \times \mathbb{S}_P/ ((S_p)^k =
\check{S_p}^k) \wedge ((S_c)^k =\check{S_c}^k))\right.$
$\left.\wedge ((S_m)^k=\check{S_m}^k)), \forall k \leqslant k_0(\varepsilon)
\right\}$.


Then, 
$\forall (S_p,S_c,S_m) \in
\mathcal{S}_{\check{S_p},\check{S_c},\check{S_m},k_0(\varepsilon)},$
$d_2((S_p,\check{x},S_c,\check{m},S_m);(\check{S_p},\check{x},\check{S_c},\check{m},
\check{S_m})) < 3.\frac{\varepsilon}{3}=\varepsilon.$
It remains to choose $(\widetilde{S_p},\widetilde{S_c},\widetilde{S_m}) \in
\mathcal{S}_{\check{S_p},\check{S_c},\check{S_m}, k_0(\varepsilon)}$ such that
$(\widetilde{S_p},\widetilde{x},\widetilde{S_c},\widetilde{m}, \widetilde{S_m})
= (\widetilde{S_p},\check{x},\widetilde{S_c},\check{m}, \widetilde{S_m})$ is
a periodic point for $\mathcal{G}_{f_0}$.

Let $\mathcal{J} =$
$\left\{ i \in \llbracket0;\mathsf{N-1}\rrbracket / x_i \neq
\check{x_i}, \text{ where }\right.$
$\left.(S_p,x,S_c,m,S_m) = \mathcal{G}_{f_0}^{k_0}
(\check{S_p},\check{x},\check{S_c},\check{m},\check{S_m}) \right\},$
$\lambda = card(\mathcal{J})$, and
$j_0 <j_1 < ... < j_{\lambda-1}$ the elements of $ \mathcal{J}$. 

\begin{enumerate}
  \item Let us firstly build three strategies: $S_p^\ast$,  $S_c^\ast$, and 
  $S_m^\ast$, as follows.
\begin{enumerate}
  \item 
$(S_p^\ast)^k=\check{S_p}^k$, $(S_c^\ast)^k=\check{S_c}^k$, and
$(S_m^\ast)^k=\check{S_m}^k$, if $k \leqslant k_0(\varepsilon).$ 
\item Let us now explain how to replace $\check{x}_{j_q}$, $\forall q \in
  \llbracket0;\mathsf{\lambda-1}\rrbracket$:\newline
First of all, we must replace $\check{x}_{j_0}$:
\begin{enumerate}
  \item If $\exists \lambda_0 \in \llbracket0;\mathsf{P-1}\rrbracket /
  \check{x}_{j_0}=m_{\lambda_0}$, then we can choose
 $(S_p^\ast)^{k_0+1}=j_0$, $(S_c^\ast)^{k_0+1}=\lambda_0$,
 $(S_m^\ast)^{k_0+1}=\lambda_0$, and so $I_{j_0}$ will be equal to $1$.
 \item If such a $\lambda_0$ does not exist, we choose:\newline
 $(S_p^\ast)^{k_0+1}=j_0$, $(S_c^\ast)^{k_0+1}=0$, $(S_m^\ast)^{k_0+1}=0$,\newline $(S_p^\ast)^{k_0+2}=j_0$,
 $(S_c^\ast)^{k_0+2}=0$, $(S_m^\ast)^{k_0+2}=0$, \newline and 
 $I_{j_0}=2$.\newline

All of the $\check{x}_{j_q}$ are replaced similarly. 
The other terms of $S_p^\ast$,  $S_c^\ast$, and 
  $S_m^\ast$ are constructed identically, and the values of $I_{j_q}$ are defined in
  the same way.\newline
 Let $\gamma = \sum_{q=0}^{\lambda-1}I_{j_q}$.
\end{enumerate}

\item Finally, let $(S_p^\ast)^{k}=(S_p^\ast)^{j}$,
$(S_c^\ast)^{k}=(S_c^\ast)^{j}$, and $(S_m^\ast)^{k}=(S_m^\ast)^{j}$, where
$j\leqslant k_{0}(\varepsilon )+\gamma$ is satisfying $j\equiv k~[\text{mod } (k_{0}(\varepsilon )+\gamma)]$, if
$k>k_{0}(\varepsilon )+\gamma$.
\end{enumerate}
So,
$\mathcal{G}_{f_0}^{k_{0}(\varepsilon
)+\gamma}(S_p^\ast,\check{x},S_c^\ast,\check{m},S_m^\ast)=(S_p^\ast,\check{x},S_c^\ast,m,S_m^\ast).$
Let $\mathcal{K} =\left\{ i \in \llbracket0;\mathsf{P-1}\rrbracket / m_i \neq
\check{m_i}, \text{ where }\right.$\newline
$\left.\mathcal{G}_{f_0}^{k_{0}(\varepsilon
)+\gamma}(S_p^\ast,\check{x},S_c^\ast,\check{m},S_m^\ast)=(S_p^\ast,\check{x},S_c^\ast,m,S_m^\ast)
\right\},$\newline
$\mu = card(\mathcal{K})$, and
$r_0 <r_1 < ... < r_{\mu-1}$ the elements of $ \mathcal{K}$.  
 
 
 
  \item Let us now build the strategies $\widetilde{S_p}$, 
  $\widetilde{S_c}$, $\widetilde{S_m}$.
\begin{enumerate}
  \item Firstly, let $\widetilde{S_p}^k=(S_p^\ast)^k$,
  $\widetilde{S_c}^k=(S_c^\ast)^k$, and $\widetilde{S_m}^k=(S_m^\ast)^k$, if $k
  \leqslant k_0(\varepsilon) + \gamma.$
\item  How to replace $\check{m}_{r_q}, \forall q \in
  \llbracket0;\mathsf{\mu-1}\rrbracket$:\newline
  
First of all, let us explain how to replace $\check{m}_{r_0}$:
\begin{enumerate}
  \item If $\exists \mu_0 \in \llbracket0;\mathsf{N-1}\rrbracket /
  \check{x}_{\mu_0}=m_{r_0}$, then 
 we can choose $\widetilde{S_p}^{k_0+\gamma +1}=\mu_0$,
 $\widetilde{S_c}^{k_0+\gamma +1}=r_0$, $\widetilde{S_m}^{k_0+\gamma
 +1}=r_0$.\newline In that situation, we define $J_{r_0}=1$.
 \item If such a $\mu_0$ does not exist, then we can choose:\newline
$\widetilde{S_p}^{k_0+\gamma +1}=0$, $\widetilde{S_c}^{k_0+\gamma
 +1}=r_0$, $\widetilde{S_m}^{k_0+\gamma
 +1}=r_0$,\newline 
$\widetilde{S_p}^{k_0+\gamma +2}=0$, $\widetilde{S_c}^{k_0+\gamma
 +2}=r_0$, $\widetilde{S_m}^{k_0+\gamma
 +2}=0$,\newline 
$\widetilde{S_p}^{k_0+\gamma +3}=0$, $\widetilde{S_c}^{k_0+\gamma
 +3}=r_0$, $\widetilde{S_m}^{k_0+\gamma
 +3}=0$.\newline 
 Let $J_{r_0}=3$.

Then the other $\check{m}_{r_q}$ are replaced as previously,
  the other terms of $\widetilde{S_p}$, 
  $\widetilde{S_c}$, and $\widetilde{S_m}$ are constructed in the same way, and
  the values of $J_{r_q}$ are defined similarly.\newline
 Let $\alpha = \sum_{q=0}^{\mu-1}J_{r_q}$.
\end{enumerate}
\item Finally, let $\widetilde{S_p}^{k}=\widetilde{S_p}^{j}$,
$\widetilde{S_c}^{k}=\widetilde{S_c}^{j}$, and
$\widetilde{S_m}^{k}=\widetilde{S_m}^{j}$ where $j\leqslant k_{0}(\varepsilon
)+\gamma +\alpha$ is satisfying $j\equiv k~[\text{mod } (k_{0}(\varepsilon
)+\gamma + \alpha)]$, if $k>k_{0}(\varepsilon )+\gamma + \alpha$.
\end{enumerate}
\end{enumerate}


So, $\mathcal{G}_{f_0}^{k_{0}(\varepsilon
)+\gamma+\alpha}(\widetilde{S_p},\check{x},\widetilde{S_c},\check{m},\widetilde{S_m})=(\widetilde{S_p},\check{x},\widetilde{S_c},\check{m},\widetilde{S_m})$
%\end{enumerate}



Then, $(\widetilde{S_p},\widetilde{S_c},\widetilde{S_m}) \in
\mathcal{S}_{\check{S_p},\check{S_c},\check{S_m},k_0(\varepsilon)}$ defined as
previous is such that $(\widetilde{S_m},\check{x},\widetilde{S_m},\check{m},\widetilde{S_m})$
 is a periodic point, of period $k_{0}(\varepsilon
)+\gamma+\alpha$, which is $\varepsilon -$close to
$(\check{S_p},\check{x},\check{S_c},\check{m}, \check{S_m})$.

As a conclusion, $(\mathcal{X}_2,\mathcal{G}_{f_0})$ is regular.
\end{proof}

\subsubsection{\textbf{Transitivity}}\label{sec:transitivite}

\begin{proposition}%[Transitivity of $ \mathcal{G}_{f_0}$]
\label{prop:transitivite}
$(\mathcal{X}_2,\mathcal{G}_{f_0})$ is topologically transitive.
\end{proposition}

\begin{proof}
Let us define $\mathcal{X}:\mathcal{X}_2\rightarrow \mathbb{B}^{\mathsf{N}},$
such that $\mathcal{X}(S_p,x,S_c,m,S_m)=x$ and
$\mathcal{M}:\mathcal{X}_2\rightarrow \mathbb{B}^{\mathsf{P}},$ such that
$\mathcal{M}(S_p,x,S_c,m,S_m)=m$. Let \linebreak
$\mathcal{B}_{A}=\mathcal{B}(X_{A},r_{A}) $ and
$\mathcal{B}_{B}=\mathcal{B}(X_{B},r_{B})$ be two open balls of $\mathcal{X}_2$,
with\linebreak $X_{A}=((S_p)_{A},x_{A},(S_c)_{A},m_{A},(S_m)_{A})$ and
$X_{B}=((S_p)_{B},x_{B},(S_c)_{B},m_{B},(S_m)_{B})$. We are looking for
$\widetilde{X}=(\widetilde{S_p},\widetilde{x},\widetilde{S_c},\widetilde{m},\widetilde{S_m})$
in $\mathcal{B}_{A} $ such that $\exists n_{0}\in
\mathbb{N},\mathcal{G}_{f_0}^{n_{0}}(\widetilde{X})\in \mathcal{B}_{B}$.\newline
$\widetilde{X}$ must be in $\mathcal{B}_{A}$ and $r_{A}$ can be strictly
lesser than 1, so $\widetilde{x}=x_{A}$ and $\widetilde{m}=m_{A}$. Let
$k_{0}=\lfloor - \log _{10}(\frac{r_{A}}{3})+1\rfloor $.
Let us notice $\mathcal{S}_{X_{A}, k_0} =\left\{ (S_p,S_c,S_m)
\in \mathbb{S}_\mathsf{N}\times (\mathbb{S}_\mathsf{P})^2/ \forall k\leqslant k_{0},\right.$
$\left.(S_p^{k}=(S_p)_{A}^{k})\wedge
(S_c^{k}=(S_c)_{A}^{k})\wedge (S_m^{k}=(S_m)_{A}^{k}))\right\}$.

Then $\forall (S_p,S_c,S_m) \in \mathcal{S}_{X_{A}, k_0},
(S_p,\widetilde{x},S_c,\widetilde{m},S_m)\in \mathcal{B}_{A}.$


Let $\mathcal{J} =\left\{i\in
\llbracket0,\mathsf{N-1}\rrbracket/\check{x}_{i}\neq
\mathcal{X}(X_{B})_{i}, \text{ where }\right.$\newline
$\left.(\check{S_p},\check{x},\check{S_c},\check{m},\check{S_m})=\mathcal{G}_{f_0}^{k_{0}}(X_{A})\right\},$
$\lambda = card(\mathcal{J})$,\newline and
$j_0 <j_1 < ... < j_{\lambda-1}$ the elements of $ \mathcal{J}$.

\begin{enumerate}
  \item Let us firstly build three strategies: $S_p^\ast$,  $S_c^\ast$, and 
  $S_m^\ast$ as follows.
\begin{enumerate}
  \item 
$(S_p^\ast)^k=(S_p)_A^k$, $(S_c^\ast)^k=(S_c)_A^k$, and
$(S_m^\ast)^k=(S_m)_A^k$, if $k \leqslant k_0.$
\item Let us now explain how to replace $\mathcal{X}(X_{B})_{j_q}$, $\forall q \in
  \llbracket0;\mathsf{\lambda-1}\rrbracket$:\newline
First of all, we must replace $\mathcal{X}(X_{B})_{j_0}$:
\begin{enumerate}
  \item If $\exists \lambda_0 \in \llbracket0;\mathsf{P-1}\rrbracket /
  \mathcal{X}(X_{B})_{j_0}=\check{m}_{\lambda_0}$, then we can choose
 $(S_p^\ast)^{k_0+1}=j_0$, $(S_c^\ast)^{k_0+1}=\lambda_0$,
 $(S_m^\ast)^{k_0+1}=\lambda_0$, and so $I_{j_0}$ will be equal to $1$.
 \item If such a $\lambda_0$ does not exist, we choose:\newline
 $(S_p^\ast)^{k_0+1}=j_0$, $(S_c^\ast)^{k_0+1}=0$,
 $(S_m^\ast)^{k_0+1}=0$,\newline $(S_p^\ast)^{k_0+2}=j_0$,
 $(S_c^\ast)^{k_0+2}=0$, $(S_m^\ast)^{k_0+2}=0$ \newline and so let us notice
 $I_{j_0}=2$.\newline

All of the $\mathcal{X}(X_{B})_{j_q}$ are replaced similarly.
The other terms of $S_p^\ast$,  $S_c^\ast$, and $S_m^\ast$ are constructed
identically, and the values of $I_{j_q}$ are defined on the same way.\newline Let $\gamma = \sum_{q=0}^{\lambda-1}I_{j_q}$.
\end{enumerate}

\item $(S_p^\ast)^{k}=(S_p^\ast)^{j}$, $(S_c^\ast)^{k}=(S_c^\ast)^{j}$
and $(S_m^\ast)^{k}=(S_m^\ast)^{j}$ where $j\leqslant k_{0}+\gamma$
is satisfying $j\equiv k~[\text{mod } (k_{0}+\gamma)]$, if
$k>k_{0}+\gamma$.
\end{enumerate}
So,$\mathcal{G}_{f_0}^{k_{0}+\gamma}((S_p^\ast,x_A,S_c^\ast,m_A,S_m^\ast))=(S_p^\ast,x_B,S_c^\ast,m,S_m^\ast)$

Let $\mathcal{K} =\left\{ i \in \llbracket0;\mathsf{P-1}\rrbracket / m_i \neq
\mathcal{M}(X_{B})_{i}, \text{ where }\right.$\newline
$\left.(S_p^\ast,x_B,S_c^\ast,m,S_m^\ast)=\mathcal{G}_{f_0}^{k_{0}+\gamma}((S_p^\ast,x_A,S_c^\ast,m_A,S_m^\ast))\right\},$\newline
$\mu = card(\mathcal{K})$ and $r_0 <r_1 < ... < r_{\mu-1}$ the elements of $
\mathcal{K}$.
 
 
  \item Let us secondly build three other strategies:  $\widetilde{S_p}$, 
  $\widetilde{S_c}$, $\widetilde{S_m}$ as follows.
\begin{enumerate}
  \item $\widetilde{S_p}^k=(S_p^\ast)^k$, $\widetilde{S_c}^k=(S_c^\ast)^k$, and
$\widetilde{S_m}^k=(S_m^\ast)^k$, if $k \leqslant k_0 + \gamma.$
\item  Let us now explain how to replace $\mathcal{M}(X_{B})_{r_q}, \forall q
\in \llbracket0;\mathsf{\mu-1}\rrbracket$:
  
First of all, we must replace $\mathcal{M}(X_{B})_{r_0}$:
\begin{enumerate}
  \item If $\exists \mu_0 \in \llbracket0;\mathsf{N-1}\rrbracket /
  \mathcal{M}(X_{B})_{r_0}=(x_B)_{\mu_0}$, then we can choose 
 $\widetilde{S_p}^{k_0+\gamma +1}=\mu_0$, $\widetilde{S_c}^{k_0+\gamma
 +1}=r_0$, $\widetilde{S_m}^{k_0+\gamma
 +1}=r_0$, and $J_{r_0}$ will be equal to $1$.
 \item If such a $\mu_0$ does not exist, we choose:
$\widetilde{S_p}^{k_0+\gamma +1}=0$, $\widetilde{S_c}^{k_0+\gamma
 +1}=r_0$, $\widetilde{S_m}^{k_0+\gamma
 +1}=r_0$,\newline 
$\widetilde{S_p}^{k_0+\gamma +2}=0$, $\widetilde{S_c}^{k_0+\gamma
 +2}=r_0$, $\widetilde{S_m}^{k_0+\gamma
 +2}=0$,\newline 
$\widetilde{S_p}^{k_0+\gamma +3}=0$, $\widetilde{S_c}^{k_0+\gamma
 +3}=r_0$, $\widetilde{S_m}^{k_0+\gamma
 +3}=0$,\newline 
 and so let us notice $J_{r_0}=3$.\newline

All the $\mathcal{M}(X_{B})_{r_q}$ are replaced similarly.
The other terms of $\widetilde{S_p}$, 
$\widetilde{S_c}$, and $\widetilde{S_m}$ are constructed identically, and the
values of $J_{r_q}$ are defined on the same way.\newline
Let $\alpha = \sum_{q=0}^{\mu-1}J_{r_q}$.
\end{enumerate}
\item $\forall k \in \mathbb{N}^\ast, \widetilde{S_p}^{k_0+ \gamma + \alpha
+ k}=(S_p)_B^{k}$, $\widetilde{S_c}^{k_0+ \gamma + \alpha + k}=(S_c)_B^{k}$, and
$\widetilde{S_m}^{k_0+ \gamma + \alpha + k}=(S_m)_B^{k}$.
\end{enumerate}
\end{enumerate}

So,
$\mathcal{G}_{f_0}^{k_{0}+\gamma+\alpha}(\widetilde{S_p},x_A,\widetilde{S_c},m_A,\widetilde{S_m})=X_B$,
with $(\widetilde{S_p},\widetilde{S_c},\widetilde{S_m}) \in \mathcal{S}_{X_{A},
k_0}$.
Then $\widetilde{X} = (\widetilde{S_p},x_A,\widetilde{S_c},m_A,\widetilde{S_m})
\in \mathcal{X}_2$ is such that $\widetilde{X} \in \mathcal{B}_A$ and
$\mathcal{G}_{f_0}^{k_0+\gamma+\alpha}( \widetilde{X}) \in \mathcal{B}_B$.
Finally we have proven the result.
\end{proof}

\subsubsection{\textbf{Sensibility to the Initial
Condition}}\label{sec:sensibilite}

\begin{proposition}%[Sensitivity of $,\mathcal{G}_{f_0}$]
\label{prop:sensibilite}
$(\mathcal{X}_2,\mathcal{G}_{f_0})$ has sensitive dependence on initial conditions.
\end{proposition}

\begin{proof}
$\mathcal{G}_{f_0}$ is regular and transitive. Due to Theorem~\ref{theo:banks}, $\mathcal{G}_{f_0}$ is sensitive.
\end{proof}

%In a following  section, we will show that this result can be stated independently of regularity,
%which must be redefined in the context of the finite set of
%machine numbers (see section \ref{Concerning}).

%In addition, the constant of sensitivity will be computed.


%This sensitive dependence could be stated as a consequence of regularity and
%transitivity (by using the theorem of Banks~\cite{Banks92}). However, we
%prefer to prove this result independently of regularity, because the
%notion of regularity must be redefined in the context of the finite set of
%machine numbers (see section \ref{Concerning}).

%In addition, the constant of sensitivity has been computed.

\subsubsection{\textbf{Devaney's topological chaos}}



In conclusion, $(\mathcal{X}_2,\mathcal{G}_{f_0})$ is topologically transitive, regular,
and has sensitive dependence on initial conditions. Then according to
Devaney's topological chaos
defintition~\ref{def:chaos-devaney}~\vpageref{def:chaos-devaney}, we have the result.

\begin{theorem}%[$\mathcal{G}_{f_0}$ is a chaotic map]
\label{theo:chaotic}
$\mathcal{G}_{f_0}$ is a chaotic map on $(\mathcal{X}_2,d_2)$ in the sense of Devaney.
\end{theorem}

So we can claim that:

\begin{theorem}%[topological-security of \CID]
\label{theo:CID-chaosecurity}
\CID is chaos-secure.
\end{theorem}


\subsection{Quantitative property of \CID}\label{sec:quantitative-properties-ci2}

In this section, it is studied one of the quantitative properties of the
topological-security of the steganographic process \CID: the constant of
sensitivity evaluation.

\begin{proposition}%[Constant of sensitivity of $ \mathcal{G}_{f_0}$]
\label{prop:sensitivity-constant-evaluation}

$(\mathcal{X}_2,\mathcal{G}_{f_0})$ has sensitive dependence on initial
conditions, and it constant of sensitivity is equal to $N-1$.
\end{proposition}

\begin{proof}
Let us define $\mathcal{X}:\mathcal{X}_2\rightarrow \mathbb{B}^{\mathsf{N}},$
such that $\mathcal{X}(S_p,x,S_c,m,S_m)=x$ and
$\mathcal{M}:\mathcal{X}_2\rightarrow \mathbb{B}^{\mathsf{P}},$ such that
$\mathcal{M}(S_p,x,S_c,m,S_m)=m$.

Let $\check{X}=(\check{S_p},\check{x},\check{S_c},\check{m}, \check{S_m})\in
\mathcal{X}_2$. We are looking for
$\widetilde{X}=(\widetilde{S_p},\widetilde{x},\widetilde{S_c},\widetilde{m},\widetilde{S_m})$
in $\mathcal{X}_2$ such that $d_2\left(\check{X},\widetilde{X} \right) \leq
\delta$ and $\exists n_0 \in \mathds{N},
d_2\left(\mathcal{G}_{f_0}^{n_0}(\check{X}),\mathcal{G}_{f_0}^{n_0}(\widetilde{X})
\right) \geq N-1$.
Let $k_{0}=\lfloor - \log _{10}(\frac{\delta}{3}\rfloor +1$,
and consider the set:
$\mathcal{S}_{\check{S_p},\check{S_c},\check{S_m}, k_0} =$
$\left\{ S
\in \mathbb{S}_N \times \mathbb{S}_P \times \mathbb{S}_P/ ((S_p)^k =
\check{S_p}^k) \wedge ((S_c)^k =\check{S_c}^k))\right.$
$\left.\wedge ((S_m)^k=\check{S_m}^k)), \forall k \leqslant k_0(\varepsilon)
\right\}$.
Then, 
$\forall (S_p,S_c,S_m) \in
\mathcal{S}_{\check{S_p},\check{S_c},\check{S_m},k_0},$
$d_2((S_p,\check{x},S_c,\check{m},S_m);(\check{S_p},\check{x},\check{S_c},\check{m},
\check{S_m})) < 3.\frac{\delta}{3}=\delta.$

Let $\mathcal{J} =\left\{i\in
\llbracket0,\mathsf{N-1}\rrbracket/
\mathcal{X}\left(\mathcal{G}_{f_0}^{k_{0}}(\check{S})\right)_{i} \neq
\mathcal{X}\left(\mathcal{G}_{f_0}^{k_{0}+N}(\check{S})\right)_{i}\right\}$ the
set of indexes of cells of the second component different between the two states
$\mathcal{G}_{f_0}^{k_{0}}(\check{X})$ and
$\mathcal{G}_{f_0}^{k_{0}+N}(\check{X})$.

Let $\lambda = card(\mathcal{J})$ the number of differences.

\begin{enumerate}
  \item If $\lambda = N$, then the point
  $(\widetilde{S_p},\widetilde{x},\widetilde{S_c},\widetilde{m},\widetilde{S_m})$
  defined by: \begin{enumerate}
    \item $\widetilde{x}$ = $\check{x}$ and $\widetilde{m}$ = $\check{m}$ in
    order to have the same cells that $\check{X}$, and thus be at a distance less than 1 of $\check{X}$. 
  \item $\forall k \leqslant k_0, \widetilde{S_p}^k= \check{S_p}^k$,
  $\widetilde{S_c}^k= \check{S_c}^k$, and $\widetilde{S_m}^k= \check{S_m}^k$.
\item Let us now explain how to replace $\mathcal{X}(X_{B})_{j_q}$, $\forall q \in
  \llbracket0;\mathsf{\lambda-1}\rrbracket$:\newline
First of all, we must replace $\mathcal{X}(X_{B})_{j_0}$:
\begin{enumerate}
  \item If $\exists \lambda_0 \in \llbracket0;\mathsf{P-1}\rrbracket /
  \mathcal{X}(X_{B})_{j_0}=\check{m}_{\lambda_0}$, then we can choose
 $(S_p^\ast)^{k_0+1}=j_0$, $(S_c^\ast)^{k_0+1}=\lambda_0$,
 $(S_m^\ast)^{k_0+1}=\lambda_0$, and so $I_{j_0}$ will be equal to $1$.
 \item If such a $\lambda_0$ does not exist, we choose:\newline
 $(S_p^\ast)^{k_0+1}=j_0$, $(S_c^\ast)^{k_0+1}=0$,
 $(S_m^\ast)^{k_0+1}=0$,\newline $(S_p^\ast)^{k_0+2}=j_0$,
 $(S_c^\ast)^{k_0+2}=0$, $(S_m^\ast)^{k_0+2}=0$ \newline and so let us notice
 $I_{j_0}=2$.\newline

All of the $\mathcal{X}(X_{B})_{j_q}$ are replaced similarly.
The other terms of $S_p^\ast$,  $S_c^\ast$, and $S_m^\ast$ are constructed
identically, and the values of $I_{j_q}$ are defined on the same way.\newline Let $\gamma = \sum_{q=0}^{\lambda-1}I_{j_q}$.
\end{enumerate}

\item $(S_p^\ast)^{k}=(S_p^\ast)^{j}$, $(S_c^\ast)^{k}=(S_c^\ast)^{j}$
and $(S_m^\ast)^{k}=(S_m^\ast)^{j}$ where $j\leqslant k_{0}+\gamma$
is satisfying $j\equiv k~[\text{mod } (k_{0}+\gamma)]$, if
$k>k_{0}+\gamma$.
\end{enumerate}
So,$\mathcal{G}_{f_0}^{k_{0}+\gamma}((S_p^\ast,x_A,S_c^\ast,m_A,S_m^\ast))=(S_p^\ast,x_B,S_c^\ast,m,S_m^\ast)$

Let $\mathcal{K} =\left\{ i \in \llbracket0;\mathsf{P-1}\rrbracket / m_i \neq
\mathcal{M}(X_{B})_{i}, \text{ where }\right.$\newline
$\left.(S_p^\ast,x_B,S_c^\ast,m,S_m^\ast)=\mathcal{G}_{f_0}^{k_{0}+\gamma}((S_p^\ast,x_A,S_c^\ast,m_A,S_m^\ast))\right\},$\newline
$\mu = card(\mathcal{K})$ and $r_0 <r_1 < ... < r_{\mu-1}$ the elements of $
\mathcal{K}$.
 
 
  \item Let us secondly build three other strategies:  $\widetilde{S_p}$, 
  $\widetilde{S_c}$, $\widetilde{S_m}$ as follows.
\begin{enumerate}
  \item $\widetilde{S_p}^k=(S_p^\ast)^k$, $\widetilde{S_c}^k=(S_c^\ast)^k$, and
$\widetilde{S_m}^k=(S_m^\ast)^k$, if $k \leqslant k_0 + \gamma.$
\item  Let us now explain how to replace $\mathcal{M}(X_{B})_{r_q}, \forall q
\in \llbracket0;\mathsf{\mu-1}\rrbracket$:
  
First of all, we must replace $\mathcal{M}(X_{B})_{r_0}$:
\begin{enumerate}
  \item If $\exists \mu_0 \in \llbracket0;\mathsf{N-1}\rrbracket /
  \mathcal{M}(X_{B})_{r_0}=(x_B)_{\mu_0}$, then we can choose 
 $\widetilde{S_p}^{k_0+\gamma +1}=\mu_0$, $\widetilde{S_c}^{k_0+\gamma
 +1}=r_0$, $\widetilde{S_m}^{k_0+\gamma
 +1}=r_0$, and $J_{r_0}$ will be equal to $1$.
 \item If such a $\mu_0$ does not exist, we choose:
$\widetilde{S_p}^{k_0+\gamma +1}=0$, $\widetilde{S_c}^{k_0+\gamma
 +1}=r_0$, $\widetilde{S_m}^{k_0+\gamma
 +1}=r_0$,\newline 
$\widetilde{S_p}^{k_0+\gamma +2}=0$, $\widetilde{S_c}^{k_0+\gamma
 +2}=r_0$, $\widetilde{S_m}^{k_0+\gamma
 +2}=0$,\newline 
$\widetilde{S_p}^{k_0+\gamma +3}=0$, $\widetilde{S_c}^{k_0+\gamma
 +3}=r_0$, $\widetilde{S_m}^{k_0+\gamma
 +3}=0$,\newline 
 and so let us notice $J_{r_0}=3$.\newline

All the $\mathcal{M}(X_{B})_{r_q}$ are replaced similarly.
The other terms of $\widetilde{S_p}$, 
$\widetilde{S_c}$, and $\widetilde{S_m}$ are constructed identically, and the
values of $J_{r_q}$ are defined on the same way.\newline
Let $\alpha = \sum_{q=0}^{\mu-1}J_{r_q}$.
\end{enumerate}
\item $\forall k \in \mathbb{N}^\ast, \widetilde{S_p}^{k_0+ \gamma + \alpha
+ k}=(S_p)_B^{k}$, $\widetilde{S_c}^{k_0+ \gamma + \alpha + k}=(S_c)_B^{k}$, and
$\widetilde{S_m}^{k_0+ \gamma + \alpha + k}=(S_m)_B^{k}$.
\end{enumerate}
\end{enumerate}

So,
$\mathcal{G}_{f_0}^{k_{0}+\gamma+\alpha}(\widetilde{S_p},x_A,\widetilde{S_c},m_A,\widetilde{S_m})=X_B$,
with $(\widetilde{S_p},\widetilde{S_c},\widetilde{S_m}) \in \mathcal{S}_{X_{A},
k_0}$.
Then $\widetilde{X} = (\widetilde{S_p},x_A,\widetilde{S_c},m_A,\widetilde{S_m})
\in \mathcal{X}_2$ is such that $\widetilde{X} \in \mathcal{B}_A$ and
$\mathcal{G}_{f_0}^{k_0+\gamma+\alpha}( \widetilde{X}) \in \mathcal{B}_B$.
Finally we have proven the result.
\end{proof}


\subsection{Qualitative property of
\CID}\label{sec:qualitative-properties-ci2}

In this section, it is studied one of the qualitative properties of the
topological-security of the steganographic process \CID: the strong
transitivity.



\textbf{TODO}

\subsection{Consequences in attacks
frameworks}\label{sec:qualitative-properties-ci2}

\textbf{TODO} In this section, it is given the implication of the qualitative
and quantitative properties of $CI_2$ in KOA, KMA,CMA and WOA setups.


