Witness and Counterexample Li Tan Oct. 15, 2002

Witness and Counterexample Informal Definition Recall the model checking problem M ² , 1. A witness W should be a subsystem of M s.t. W ² . $W$ should be small. M is a witness of , but it is useless. 2. Witness should show the evidence why M ² . Any system of which $W$ is a subsystem should satisfy . To show $M ²  $, it is enough to show The relation between M and W. 3. Viability. 1. Completeness. 1. Each formula in the targeting logic should have a well-defined witness if it is satisfied by the model. 2. Simple and Efficient. 1. Witness should be verified and analyzed efficiently. 3. Effectiveness. 1. There exists an effective algorithm for generating the witness.

Witness and Counterexample Informal Definition 1. Counterexample is just the dual of witness. 1. C is an counterexample for M ²  iff C is a witness of M ² : . 2. An counterexample always exists if 1. The logic is complete under negation, i.e.,  2 L ) :  2 L 1. The definition of witness is complete in L. 3. The mechanism for generating witness/counterexample are same. 1. For history reason, many model checker like SMV and SPIN only use the notion of counterexample. 2. Our reference to witness/counterexample will be chose depending on the context.

Witness and Counterexample Why we need them? 1. Counterexample can be used for, 1. Debugging the design. 2. Counterexample-based abstract refinement. 1. The abstraction is conservative 1. It may cause the false alarm. 2. A counterexample is generated after checking the abstract system, if it is, 1. A real trace, then model checker terminates with “no”. 2. Not a real trace, then refine the abstract ( add more predicators etc), and re-do the model checking. 2. Witness can be used for, 1. Enable efficiently checking the correctness. 1. Verifying result=Verifying the property on witness+prove that a witness is a subsystem of model. 2. Generating tests. 1. Any system contains the witness should possess the property.

Witness and Counterexample Formal Definition Informal definition is ambiguous, What is a context of definition? => Fix a logic What is a “subsystem”? => Define a preordering Á on Transition system. Definition [Natural Preorder of Logic] Let L be a temporal logic, a preordering Á L on Kripke structures are a nature such that for every f 2 L, if T ² f and T Á L T’, then T’ ² f. Definition [Witness and Counterexample] Let Á be a natural preordering for the logic L ( : L ), C is a witness(or, counterexample) for M ² f s.t. in L if C ²  (C ² :  ) C Á M. Our mission: fix L, then find Á.

Witness and Counterexample Kripke Structure and CTL* Definition [Kripke Structure] Transition System will be modeled as Kripke structure K = where, S is the set of states. s 0 2 S is a starting state. ! µ S £ S is transition relation. A is the set of atomic proposition. V : A ! 2 S is the evulation for atomic proposition.

Witness and Counterexample CTL*: Syntax Formulae in Computational Tree Logic (CTL*) are syntactically constructed by, S ::= a | : a| S Æ S | S Ç S | A P | E P P ::= S | P Æ P | P Ç P | X P | P U P | P R P S is a state formula, and P is a path formula. A (for all the paths), and E (exists a path) are the path quantifier. U and R are the “until” and “release” operators. X is the “next time” operator. CTL* formula is a state formula. Sometimes, we write G P (always hold) for false R P, and F P ( Eventually hold) for true U P

Witness and Counterexample Semantics of path formulae

Witness and Counterexample Semantics of state formulae 1. s ² T a if s 2 V (s) 2. s ² T : a if s  V (s) 3. s ² A P if  ² P for every path  from s. 4. s ² E P if there exists a  from s such that  ² P

Witness and Counterexample LTL, CTL, and ACTL 1. Linear Temporal Logic (LTL) is sublogic of CTL* which contains no path quantifier. 1. Semantically LTL formula f = CTL* formula Af. 2. Intuitively LTL check all the possible paths simultaneously. 2. CTL is sublogic of CTL* such that each temporal operator (X, U, R) must be immediately preceded by a path quantifier. 3. ACTL is a sublogic of CTL in which only path quantifier permitted is A.

Witness and Counterexample An example 1. Does T satisfy LTL F(G : y)? 2. Does T satisfy ACTL AF( : y Æ AX : X)? x, y y s0s0 s2s2 s1s1

Witness and Counterexample Counterexample in LTL Customizing the definition of counterexample, 1. Fix the logic: LTL 2. Fix the preordering relation Á : language inclusion 2 Definition Let f be a LTL formula, a path  is a linear counterexample for model-checking problem M ² f if  ² : f and  2 L(M). Is the definition viable?  2 L(T) => T ² : f, why? Is the definition complete? If M ² f, then we can always find a  2 L(M) as a counterexample, why?

Witness and Counterexample LTL model checking via Büchi automaton A generalized Büchi automaton is a tuple where, 1. Q is the set of states with q 0 as the starting state. 2. ! µ Q £ Q is the transition relation. 3. l: Q ! {a, : a, Æ, Ç, <>, []} 4. F µ 2 Q is Büchi acceptance condition.

Witness and Counterexample Büchi automaton as the temporal specification 1. G T, B ={ S, !, L} is the product graph for B and T if, 1. S = S £ Q U {true, false} 2. If q ! q’ and 1. l(q) 2 { Ç, Æ }, !. 2. l(q) = <> ([]), ! for some (all) s ! s’. 3. l(q) = a ( : a), ! (s 2 a ?)( : (s 2 a ?)) 3. L( )= Ç for l(q) 2 { Ç, <>} and L( )= Æ otherwise.

Witness and Counterexample Büchi automaton-based model checking 1. B accepts T if there is a successful run in G T, B, 1. A run of G T, B is a subgraph such that keeps one (all) child(ren) if L(q)= Ç (L(q)= Æ ). 2. A run is successful if, 1. All the leaf are true, and 2. Any infinite path in a run visits some nodes in F infinitely often, for every F 2 F. 2. Searching for a (un)successful run, 1. Mark all the leaves with true/false depending on the labeling, then propagate the values. 2. For the nodes in a strong connected component, marking the nodes as true/false depending on the coverage of SCC on F, then propagate the values. 3. Eventually, all the nodes will be marked as true/false.

Witness and Counterexample An example <> Æ Ç y q0q0 q 1 q2q2 q3q3 q4q4 q5q5 Æ F ={{q 1, q 2, q 4 }} x, y y s0s0 s2s2 s1s1

Witness and Counterexample s 0, q 2 \or true s 1, q 0 Æ s 0,q 0 Æ s 0, q 1 Ç s 0, q 3 Ç s 0, q 4 Æ s 1, q 1 Ç s 1, q 3 Æ s 1, q 4 Ç s 1, q 2 Ç s 2,q 0 Æ s 2, q 1 Ç s 2, q 3 Æ s 2, q 4 Ç false s 2, q 2 Ç

Witness and Counterexample s 0, q 2 \or true s 1, q 0 Æ s 0,q 0 Æ s 0, q 1 Ç s 0, q 3 Ç s 0, q 4 Æ s 1, q 1 Ç s 1, q 3 Æ s 1, q 4 Ç s 1, q 2 Ç s 2,q 0 Æ s 2, q 1 Ç s 2, q 3 Æ s 2, q 4 Ç false s 2, q 2 Ç

Witness and Counterexample Translate a LTL to a “never claim” BTA Construct BTA for : f by,

Witness and Counterexample Translate a LTL (cont.) 2. For each  = P 1 U P 2, there is a F 2 F such that F={q| (   q Æ X   q) or P 2 2 q} Basically F won’t contain any nodes of a loop on which P 1 U P 2 will produce itself. ) \cP_2 will eventually be satisfied.

Witness and Counterexample Translate F( G : y) : A (F (G : y))=E(G(F y)) <> Æ Ç y q0q0 q 1 q2q2 q3q3 q4q4 q5q5 Æ F ={{q 1, q 2, q 4 }} E(G(F y)) = E(G(F y)) E(XG(F y), F y ) E(XG(F y), y) E(XG(F y), XF y ) E(XG(F y))

Witness and Counterexample Step 1: find a successful run true s 0,q 0 Æ s 0, q 1 Ç s 0, q 4 Æ s 1, q 1 Ç s 1, q 3 Æ s 1, q 2 Ç s 2,q 0 Æ s 2, q 1 Ç s 2, q 4 Ç

Witness and Counterexample Step 2: Get the skeleton Remove branches like (s, q) ! (s’, q’) such that (s, q) is a leaf and l(q’)  {<>, []} s 0,q 0 Æ s 0, q 1 Ç s 0, q 4 Æ s 1, q 1 Ç s 1, q 3 Æ s 1, q 2 Ç s 2,q 0 Æ s 2, q 1 Ç s 2, q 4 Ç

Witness and Counterexample Step 3: Unroll the skeleton and get a path. (s 0, q 0 ) (s 0, q 1 ) (s 0, q 4 ) {(s 1,q 1 )(s 1, q 3 )(s 1, q 2 )(s 2, q 0 )(s 2, q 1 )(s 2, q 4 )}  S 0 S 0 S 0 S 1 S 1 S 1 S 2 S 2 S 2

Witness and Counterexample Step 4: Eliminate the redundancy Remove (s’, q’) from ……(s, q)(s’, q’) …… if l(q)  {<>, []} or (s, q) is the head of the path. (s 0, q 0 ) (s 0, q 1 ) (s 0, q 4 ) {(s 1,q 1 )(s 1, q 3 )(s 1, q 2 )(s 2, q 0 )(s 2, q 1 )(s 2, q 4 )}  (s 0, q 0 ) {(s 1,q 1 ) (s 2, q 0 ) }   S 0 { S 1 S 2 } 

Witness and Counterexample Is there a witness /counterexamples for CTL* Any Kripke structure which are not bi- similar are distinguishable by a CTL* formula [Mil71]. => Let Á be a nature order for CTL, then T 1 Á T 2 iff T 1 bisimulates T 2. => The witness(counterexample) for any $T ² f$ must bisimulate T. Since bisimularity on KS is similar to is isomorphism in graphics. => The witness and counterexample is trivial in CTL*

Witness and Counterexample Is there a witness /counterexamples for CTL*? Any Kripke structure which are not bi- similar are distinguishable by a CTL* formula [Mil71]. => Let Á be a nature order for CTL, then T 1 Á T 2 iff T 1 bisimulates T 2. => The witness(counterexample) for any T ² f must also bisimulate T. Since bisimularity on KS is basically the isomorphism in graphics. => The witness and counterexample is trivial in CTL*

Witness and Counterexample Counterexample for ACTL* Simulation preordering is a nature preordering for ECTL* => Counterexample for ACTL* is a well-defined. Most of desirable safeness and fairness properties can be expressed in ACTL* Definition [Simulation Preordering Á sim ] Á sim is a simulation preordering on T iff for every s Á sim s 1 and s ! s’, there exists a s 1 ! s’ 1 such that s’ Á sim s’ 1 T 1 Á sim T 2 if s 1 0 Á sim s 2 0, where s 1 0 and s 2 0 are the starting states of T 1 and T 2

Witness and Counterexample Tree-like Counterexample for ACTL* Let f be a ACTL* formula and M ² : f, then there exists a tree-like Kripke structure C Á sim M such that C ² : f [CJLV02]. C is a tree-like Kripke structure if, Its (strongly-connected) component graph is a tree Only strongly-connected components permitted in C are circles.

Witness and Counterexample An example ² : AF( : y Æ AX : X)? x, y y s0s0 s2s2 s1s1 y y s’ 0 s’ 2 s’ 1 x s’ 3 ² : AF( : y Æ AX : X)? Á sim

Witness and Counterexample Generating Tree-like Counterxample for M ² f …Similar to generate a linear counterexample for LTL! 1. Constructing a BTA B for : f. 2. Find a successful run in G M, B. 3. Get the skeleton of the run by cutting out any node (s’, q’) s.t. (s’, q’) is the only child of its father (s, q) and l(q)  { Æ, Ç }. 4. Eliminate the redundancy and project the skeleton to states.

Witness and Counterexample If we’ve know, ² AF( : y Æ AX : X)? x, y y s0s0 s2s2 s1s1 y y s’ 0 s’ 2 s’ 1 x s’ 3 is a tree-like counterexample for

Witness and Counterexample … and want to prove, ² : AF( : y Æ AX : X) x, y y s0s0 s2s2 s1s1 y

Witness and Counterexample We only need to show, x, y y s0s0 s2s2 s1s1 y y y s’ 0 s’ 2 s’ 1 x s’ 3 Á sim