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[52]1
2We suppose that our concrete model is a composition of several components and each component has been previously verified. Hence, we have a set of verified properties for each component of the concrete model. The main idea of this technique is that we would like to make use of these properties to generate a better abstract model. Properties of the components that appear to be related to the global property to be verified, $\phi$ are selected to generate the abstract model $\widehat{M}_i$. This method is particularly interesting as it gives a possibility to converge quicker to an abstract model that is sufficient to satisfy the global property $\phi$.
3
[56]4\subsection{Properties of good refinement}
5When a counterexample is found to be spurious, it means that the current abstract model $\widehat{M}_i$ is too coarse and has to be refined.
6In this section, we will discuss about the refinement technique based on the integration of more verified properties of the concrete model's components in the abstract model to be generated. Moreover, the refinement step from $\widehat{M}_i$ to $\widehat{M}_{i+1}$ respects the properties below:
[52]7
8%\medskip
9
10\begin{property}
[56]11\begin{enumerate}
12\item The new refinment is an over-approximation of the concrete model: $\widehat{M}_{i+1} \sqsubseteq \widehat{M}$.
13\item The new refinment is more concrete than the previous one:
14$\widehat{M}_{i} \sqsubseteq \widehat{M}_{i+1}$.
15\item The spurious counter-example in $\widehat{M}_i$ is  removed from
16$\widehat{M_{i+1}}$.
17\end{enumerate}
[52]18\end{property}
19
[56]20Moreover, the refinement steps should be easy to compute and ensure a fast
21convergence by the minimizing the number of iteration of the CEGAR loop.
[52]22
23
[56]24A possible refinement : concretization of selected abstract variables. How to choose variables and instants of concretization : introduce new CTL properties. The question is : how to select pertinent CTL properties ???
[52]25
[56]26\TODO{discussion sur comment garantir les points 1/2/3 et le reste du bon
27rafinement}
28\subsection{The Counterexample}
[52]29
[56]30\TODO{Mettre la def avant}
31\TODO{Rafinement par négation du contre-exemple}
32The counterexample at a refinement step $i$, $\sigma$ is a path in the
33abstract model $\widehat{M}_i$ which dissatisfy $\Phi$.  In the counterexample given by the model-checker, the variables' value in each states are boolean.
34The spurious counter-example $\sigma$ is defined such that :
[52]35\begin{definition}
[56]36\textbf{\emph{The counterexample $\sigma$ :}} \\
[52]37\\
[56]38Let $\widehat{M}_i =\langle \widehat{AP}_i, \widehat{S}_i, \widehat{S}_{0i},
39\widehat{L}_i, \widehat{R}_i, \widehat{F}_i \rangle$ and let the length of the
40counterexample, $|\sigma| = n$: $ \sigma = s_{0} \rightarrow s_{1} \ldots
41s_{n}$ with $(s_{k}, s_{k+1}) \in \widehat{R}_i$ $\forall k \in [0..n-1]$.
42\begin{itemize}
43\item All its variables are concrete: $\forall s_i$ and $\forall p\in
44\widehat{AP}_i$, $p$ is either true or false
45(not {\it unknown}).
46\item  $\sigma$ is a counter-example in  $\widehat{M}_i$: $s_0\not\models \Phi$.
47\item  $\sigma$ is not a path of the concrete system $M$: $\exists k$ such
48that $(s_{k}, s_{k+1}) \not\in R$.
[57]49\end{itemize}
[52]50\end{definition}
51
52%\bigskip
53
[56]54%\begin{definition}
55%\textbf{\emph{Spurious counterexample :}} \\
56%\\
57%Let $\sigma_c = \langle s_{c,0}, s_{c,1}, s_{c,2}, ... , s_{c,k}, s_{c,k+1}, ... , s_{c,n}\rangle$ a path of length $n$ in the concrete model $M$ and in each state of $\sigma_c$ we have $s_{c,k} = \langle v_{c,k}^1, v_{c,k}^2, ... ,  v_{c,k}^{p'}, ... , v_{c,k}^{q'} \rangle$ with $\forall p' \in [1,q'], ~v_{i,k}^{p'} \in V_{c,k}$ and $V_{c,k} \in 2^{q'}$.\\
58%
59%\smallskip
60%
61%If $\forall k$ we have $\widehat{V}_{i,k} \subseteq V_{c,k}$ and $\forall v_{\bar{a}i,k} \in \widehat{V}_{i,k}, ~s_{i,k}|_{v_{\bar{a}i,k}} = s_{c,k}|_{v_{c,k}} $ then $M \nvDash \phi$ else $\sigma_i$ is \emph{spurious}.
62%
63%\end{definition}
[52]64
65
66
[56]67\subsection{Ordering of properties}
[52]68
69Before generating an abstract model to verify a global property $\phi$, the verified properties of all the components in the concrete model are ordered according to their pertinency in comparison to a global property $\phi$. In order to do so, the variable dependency of the variables present in global property $\phi$ has to be analysed. After this point, we refer to the variables present in the global property $\phi$ as \emph{primary variables}.
70
71%\bigskip
72
73The ordering of the properties will be based on the variable dependency graph.
74The variables in the model are weighted according to their dependency level
75\emph{vis-à-vis} primary variables and the properties will be weighted according to the sum of the weights of the variables present in it. We have decided to allocate a supplementary weight for variables which are present at the interface of a component whereas variables which do not interfere in the obtention of a primary variable will be weighted 0. Here is how we proceed:
76
77
78\begin{enumerate}
79
80\item {\emph{Establishment of primary variables' dependency and maximum graph depth}\\
81Each primary variable will be examined and their dependency graph is elobarated. The maximum graph depth among the primary variable dependency graphs will be identified and used to calibrate the weight of all the variables related to the global property.
82Given the primary variables of $\phi$, $V_{\phi} =  \langle v_{\phi_0}, v_{\phi_1}, ... , v_{\phi_k}, ... , v_{\phi_n} \rangle$ and $G{\_v_{\phi_k}}$ the dependency graph of primary variable $v_{\phi_k}$, we have the maximum graph depth $max_{d} = max(depth(Gv_{\phi_0}), depth(Gv_{\phi_1}), ... , depth(Gv_{\phi_k}), ... ,$\\$ depth(Gv_{\phi_n})) $.
83
84}
85
86\item {\emph{Weight allocation for each variables} \\
87Let's suppose $max_d$ is the maximum dependency graph depth calculated and $p$ is the unit weight. We allocate the variable weight as follows:
88\begin{itemize}
89\item{All the variables at degree $max_d$ of every dependency graph will be allocated the weight of $p$.}
90 \\ \hspace*{20mm} $Wv_{max_d} = p$
91\item{All the variables at degree $max_d - 1$ of every dependency graph will be allocated the weight of $2Wv_{max_d}$.}
92\\ \hspace*{20mm} $Wv_{max_d - 1} = 2Wv_{max_d}$
93\item{...}
94\item{All the variables at degree $1$ of every dependency graph will be allocated the weight of $2Wv_{2}$.}
95 \\ \hspace*{20mm} $Wv_{1} = 2Wv_{2}$
96\item{All the variables at degree $0$ (i.e. the primary variables) will be allocated the weight of $10Wv_{1}$.}
97 \\ \hspace*{20mm} $Wv_{0} = 10Wv_{1}$
98\end{itemize}
99
100We can see here that the primary variables are given a considerable
101ponderation due to their pertinency \emph{vis-à-vis} global  property. Furthermore, we will allocate a supplementary weight of $3Wv_{1}$ to variables at the interface of a component as they are the variables which assure the connection between the components if there is at least one variable in the dependency graph established in the previous step in the property. All other non-related variables have a weight equals to $0$.
102}
103
104
105\item {\emph{Ordering of the properties} \\
106Properties will be ordered according to the sum of the weight of the variables in it. Therefore, given a property $\varphi_i$ which contains $n+1$ variables, $V_{\varphi_i} =  \langle v_{\varphi_{i0}}, v_{\varphi_{i1}}, ... , v_{\varphi_{ik}}, ... , v_{\varphi_{in}} \rangle$, the weight  of $\varphi_i$ , $W_{\varphi_i} = \sum_{k=0}^{n} Wv_{\varphi_{ik}}$ .
107After this stage, we will check all the properties with weight $>0$ and allocate a supplementary weight of $3Wv_{1}$ for every variable at the interface present in the propery. After this process, the final weight of a property is obtained and the properties will be ordered in a list with the weight  decreasing (the heaviest first). We will refer to the ordered list of properties related to the global property $\phi$ as $L_\phi$.
108
109
110}
111
112\end{enumerate}
113
114%\bigskip
115
116\emph{\underline{Example:}}  \\
117
118For example, if a global property $\phi$ consists of 3 variables: $ p, q, r $ where:
119\begin{itemize}
120\item{$p$ is dependent of $a$ and $b$}
121\item{$b$ is dependent of $c$}
122\item{$q$ is dependent of $x$}
123\item{$r$ is independent}
124\end{itemize}
125
126Example with unit weight= 50.
127The primary variables: $p$, $q$ and $r$ are weighted $100x10=1000$ each. \\
128The secondary level variables : $a$, $b$ and $x$ are weighted $50x2=100$ each. \\
129The tertiary level variable $c$ is weighted $50$. \\
130The weight of a non-related variable is $0$.
131
132So each verified properties available pertinency will be evaluated by adding the weights of all the variables in it. It is definitely not an exact pertinency calculation of properties but provides a good indicator of their possible impact on the global property.
133
134\bigskip
135\begin{figure}[h!]
136   \centering
137%   \includegraphics[width=1.2\textwidth]{Dependency_graph_weight_PNG}
138%     \hspace*{-15mm}
139     \includegraphics{Dependency_graph_weight_PNG}
140   \caption{\label{DepGraphWeight} Example of weighting}
141\end{figure}
142
143%Dans la figure~\ref{étiquette} page~\pageref{étiquette},  
144
145
146
147After this pre-processing phase, we will have a list of properties $L_\phi  $ ordered according to their pertinency in comparison to the global property.
148
149
150
151
152\subsection{Initial abstraction generation}
153
154In the initial abstraction generation, all primary variables have to be represented. Therefore the first element(s) in the list where the primary variables are present will be used to generate the initial abstraction, $\widehat{M}_0$ and we will verify the satisfiability of the global property $\phi$ on this abstract model. If the model-checking failed and the counterexample given is found to be spurious, we will then proceed with the refinement process.
155
156
157
158\subsection{Abstraction refinement}
159
160The refinement process from $\widehat{M}_i$ to $\widehat{M}_{i+1}$ can be seperated into 2 steps:
161
162\begin{enumerate}
163
164\item {\emph{\underline{Step 1:}} \\
165
166As we would like to ensure the elimination of the counterexample previously found, we filter out properties that don't have an impact on the counterexample $\sigma_i$ thus won't eliminate it. In order to reach this obective, a Kripke Structure of the counterexample $\sigma_i$, $K(\sigma_i)$ is generated. $K(\sigma_i)$ is a succession of states corresponding to the counterexample path which dissatisfies the global property $\phi$.
167
168\bigskip
169
170\begin{definition}
171\textbf{\emph{The counterexample $\sigma_i$ Kripke Structure $K(\sigma_i)$ :}} \\
172Let a counterexample of length $n$, $ \sigma_i = \langle s_{\bar{a}i,0}, s_{\bar{a}i,1},\\ s_{\bar{a}i,2}, ... , s_{\bar{a}i,k}, s_{\bar{a}i,k+1}, ... , s_{\bar{a}i,n}\rangle $ with $ \forall k \in [0,n-1]$, we have \\
173$K(\sigma_i) = (AP_{\sigma_i}, S_{\sigma_i}, S_{0\sigma_i}, L_{\sigma_i}, R_{\sigma_i})$ a 5-tuple consisting of :
174
175\begin{itemize}
176\item { $AP_{\sigma_i}$ : a finite set of atomic propositions which corresponds to the variables in the abstract model $\widehat{V}_{i}$ }     
177\item { $S_{\sigma_i} = \{s_{\bar{a}i,0}, s_{\bar{a}i,1}, s_{\bar{a}i,2}, ... , s_{\bar{a}i,k}, s_{\bar{a}i,k+1}, ... , s_{\bar{a}i,n}\}$}
178\item { $S_{0\sigma_i} = \{s_{\bar{a}i,0}\}$}
179\item { $L_{\sigma_i}$ : $S_{\sigma_i} \rightarrow 2^{AP_{\sigma_i}}$ : a labeling function which labels each state with the set of atomic propositions true in that state. }
180\item { $R_{\sigma_i}$ = $ (s_{\bar{a}i,k}, s_{\bar{a}i,k+1})$ }
181\end{itemize}
182\end{definition}
183
184%\bigskip
185All the properties available are then model-checked on $K(\sigma_i)$.
186
187If:
188\begin{itemize}
189\item {\textbf{$K(\sigma_i) \vDash \varphi  \Rightarrow \varphi $ will not eliminate $\sigma_i$}}
190\item {\textbf{$K(\sigma_i) \nvDash \varphi  \Rightarrow \varphi $ will eliminate $\sigma_i$}}
191\end{itemize}
192
193%\bigskip
194
195
196\begin{figure}[h!]
197   \centering
198%   \includegraphics[width=1.2\textwidth]{K_sigma_i_S_PNG}
199%     \hspace*{-15mm}
200     \includegraphics{K_sigma_i_S_PNG}
201   \caption{\label{AKSNegCex} Kripke Structure of counterexample $\sigma_i$, $K(\sigma_i)$}
202\end{figure}
203
204%Dans la figure~\ref{étiquette} page~\pageref{étiquette},  
205
206%\bigskip
207
208
209\begin{figure}[h!]
210   \centering
211
212\begin{tikzpicture}[->,>=stealth',shorten >=1.5pt,auto,node distance=1.8cm,
213                    thick]
214  \tikzstyle{every state}=[fill=none,draw=blue,text=black]
215
216  \node[initial,state] (A)                            {$s_{\bar{a}i,0}$};
217  \node[state]           (B) [below of=A]     {$s_{\bar{a}i,1}$};
218
219  \node[state]           (C) [below of=B]        {$s_{\bar{a}i,k}$};
220
221  \node[state]           (D) [below of=C]       {$s_{\bar{a}i,n-1}$};
222  \node[state]           (E) [below of=D]       {$s_{\bar{a}i,n}$};
223
224  \path (A) edge              node {} (B)
225            (B) edge       node {} (C)
226            (C) edge             node {} (D)
227            (D) edge             node {} (E);
228
229\end{tikzpicture}
230
231   \caption{\label{AKSNegCex} Kripke Structure of counterexample $\sigma_i$, $K(\sigma_i)$}
232\end{figure}
233
234
235Therefore all properties that are satisfied won't be chosen to be integrated in the next step of refinement. At this stage, we already have a list of potential properties that will definitely eliminate the current counterexample $\sigma_i$ and might converge the abstract model towards a model sufficient to verify the global property $\phi$.
236
237}
238%\bigskip
239
240\item {\emph{\underline{Step 2:}} \\
241
242The property at the top of the list (not yet selected and excluding the properties which are satisfied by $K(\sigma_i)$) is selected to be integrated in the generation of $\widehat{M}_{i+1}$.
243%\bigskip
244
245}
246\end{enumerate}
247
248$\widehat{M}_{i+1}$ is model-checked and the refinement process is repeated until the model satisfies the global property or there is no property left to be integrated in next abstraction.
249
250
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