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NETW 703. Network Protocols. Symbolic Model Checking Revision Slides. Dr. Eng. Amr T. Abdel-Hamid. Slides based on slides of: Jim Kurose, Keith Ross, “Computer Networking: A Top Down Approach Featuring the Internet”, 2nd edition, Addison-Wesley, July 2002.
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NETW 703 Network Protocols Symbolic Model Checking Revision Slides Dr. Eng. Amr T. Abdel-Hamid • Slides based on slides of: • Jim Kurose, Keith Ross, “Computer Networking: A Top Down Approach Featuring the Internet”, 2nd edition, Addison-Wesley, July 2002. • Jiangchuan (JC) Liu , Assistant Professor, SFU & others Winter 2012
Functionals • Now, we can think of all temporal operators also as functions from sets of states to sets of states • For example: or if we use the set notation AX p = (S - EX(S - p)) • Logic Set • p q p q • p q p q • p S – p False True S
Fixpoint Characterization Equivalences AG p = y . p AX y AG p = p AX AG p EG p = y . p EX y EG p = p EX EG p AF p = y . p AX y AF p = p AX AF p EF p = y . p EX y EF p = p EX EF p A(pUq) = y . q A (p X (y)) A(pUq)=q (p AX (p AU q)) E(pUq) = y . q E (p X (y)) E(pUq) = q (p EX (p EU q)) Fixpoint Characterizations
EF Fixpoint Computation EF p = y . p EX y is the limit of the sequence: , pEX , pEX(pEX ) , pEX(pEX(p EX )) , ... which is equivalent to , p, p EX p , p EX (p EX (p) ) , ...
EF Fixpoint Computation p s1 s2 s3 s4 p Start 1st iteration pEX = {s1,s4} EX()= {s1,s4} ={s1,s4} 2nd iteration pEX(pEX ) = {s1,s4} EX({s1,s4})= {s1,s4} {s3}={s1,s3,s4} 3rd iteration pEX(pEX(p EX )) = {s1,s4} EX({s1,s3,s4})= {s1,s4} {s2,s3,s4}={s1,s2,s3,s4} 4th iteration pEX(pEX(pEX(p EX ))) = {s1,s4} EX({s1,s2,s3,s4})= {s1,s4} {s1,s2,s3,s4} = {s1,s2,s3,s4}
EF Fixpoint Computation EF(p)states that can reach p p EX(p) EX(EX(p)) ... p • • • EF(p)
Greatest Fixpoint Given a monotonic function F, its greatest fixpoint is the least upper bound (lub) of all the extensive elements: y. F y = { y | F y y } The greatest fixpoint y . F y is the limit of the following sequence (assuming F is -continuous): S, F S, F2 S, F3 S, ... If S is finite, then we can compute the greatest fixpoint using the above sequence
EG Fixpoint Computation Similarly, EG p = y . p EX y is the limit of the sequence: S, pEX S, pEX(p EX S) , pEX(p EX (p EX S)) , ... which is equivalent to S, p, p EX p , p EX (p EX (p) ) , ...
EG Fixpoint Computation p p s1 s2 s3 s4 p Start S = {s1,s2,s3,s4} 1st iteration pEX S = {s1,s3,s4}EX({s1,s2,s3,s4})= {s1,s3,s4}{s1,s2,s3,s4}={s1,s3,s4} 2nd iteration pEX(pEX S) = {s1,s3,s4}EX({s1,s3,s4})= {s1,s3,s4}{s2,s3,s4}={s3,s4} 3rd iteration pEX(pEX(pEX S)) = {s1,s3,s4}EX({s3,s4})= {s1,s3,s4}{s2,s3,s4}={s3,s4}
EG Fixpoint Computation EG(p) states that can avoid reaching pp EX(p) EX(EX(p)) ... • • • EG(p)
Example • For the FSM below, formally check the following properties, using Fixpoint Theorm: • AG(a ∨c ∨b) • AF(ab) • If failed show the subset of the design the property holds for as well as the counter example 1 2 4 3 c a a,b b,c d c 6 S = {1,2,3,4,5,6}, AP = {a,b,c,d}, R = {(1,2), (1,3),(2,3), (3,4), (4,4), (4,5), (5,2), (2,6), (6,1)} L(1) = {a,b}, L(2) = {c}, L(3) = {b,c}, L(4) = {a}, L(5) = {c}, L(6) = {d} 5
Example (cont.) • Remember that: • H(a∪b) = H(a) ∪ H(b) ∪ H(c) ={1,4} ∪ {2,3,5} ∪ {1,3} = {1,2,3,4,5} • AG(a ∨c ∨b) = AG p = y . p AX y = y . p AX y • AX p = EX(p) • I0 S = {1,2,3,4,5,6} • I1 {1,2,3,4,5} ∩ S = {1,2,3,4,5} ∩ {1,3,4,5,6} = {1,2,3,4,5} • I2 {1,2,3,4,5} ∩ AX(1,2,3,4,5) = {1,2,3,4,5} ∩ {1,3,4,5,6} = {1,3,4,5} • This is because that : AX(1,2,3,4,5) = EX((1,2,3,4,5)) = EX(6) = (2) = S- {2 } = {1,3,4,5,6} • I3 {1,2,3,4,5} ∩ AX(1,3,4,5) = {1,3,4,5} • This is because that : AX(1,3,4,5) = EX((1,3,4,5)) = **** • I3 = I2 H(AG(a∨b ∨ c)) = {1,3,4,5} • The property does not hold, except for the above states, and it is clear that states {2,6} can be considered as counter examples. • state 6 does not contain neither a,c,b and state 2 does not have a proceeding one on one of its pathes path (2,6)