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Given the LTL formula: r  pU(qOr), Write down the closure of the formula

Assignment #3 - Solution. Given the LTL formula: r  pU(qOr), Write down the closure of the formula Write down the corresponding Atoms

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Given the LTL formula: r  pU(qOr), Write down the closure of the formula

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  1. Assignment #3 - Solution Given the LTL formula: r  pU(qOr), • Write down the closure of the formula • Write down the corresponding Atoms • Develop the corresponding graph (it is recommended to employ the Tableau method), indicate the partition into acyclic s.c. sub-graphs, and a specific sub-graph that is self-fulfilling (namely makes the formula satisfiable). Closure: Cl{r  pU(qOr)} = {rpU(qOr), r(pU(qOr)), r, r, pU(qOr), (pU(qOr)), p, p, qOr, qOr, q, q, Or, Or}

  2. Extended , , and ‘next’ tables

  3. The partition into Atoms and Next relation A:{rpU(qOr), r} B:{rpU(qOr), pU(qOr), p, qOr, q} C:{rpU(qOr), pU(qOr), p, qOr, Or} D:{rpU(qOr), pU(qOr), qOr, q, Or} Next(A)=A1:{True} Next(B)=B1:{pU(qOr), p, qOr, q} B2:{pU(qOr), p, qOr, Or} B3:{pU(qOr), qOr, q, Or} Next(B1)= {B1,B2,B3} Next(B2)=B21:{pU(qOr), r, p, qOr, q} B22:{pU(qOr), r, p, qOr, Or} B23:{pU(qOr), r, qOr, q, Or} Next(B21)= {B1,B2,B3} Next(B22)= {B21,B22,B23} Next(B23)={D1} Next(B3)={D1} Next(C)= {B21,B22,B23} Next(D)=D1:{r}, Next(D1)={A1}

  4. The Tableau graph of rpU(qOr)

  5. The partition into acyclic s.c. sub-graphs, and self-fulfilling of rpU(qOr) A, A1, D1 are self-fulfilling, {B1,B2,B21,B22} is not self-fulfilling

  6. Note: to save space the events START_N_MSG, END_N_MSG, START_N_MSG, END_N_MSG were replaced by @Ntx, Ntx!, @Utx, Utx!, respectively Communication Controller XCTL Specification Assertions: Message transmission, if not interrupted, takes 3 seconds: (@Ntxx=T)(¬@Utx(O¬(@Ntx@Utx)U(T=x+3)) (x=T)(Ntx! T=x+3) This formula does not eliminate scenario where Ntx! occurs indistinct stateswith same time tag. So based on the physics of the systemwe may want to add anexplicit assumptions: (x=T¬Ntx!)  ¬◊(Ntx!  x=T) –if occurs, Ntx! must occur at the first state where x=T (Ntx!  x=T) => O(¬◊(Ntx!  x=T)) -- Ntx! may occur at most once with x=T Compare with the LTL formulae @Ntx(O0,2 @Utx O1,2@Ntx)  O3Ntx! @Utx(O0,2 @Ntx O1,2 @Utx )  O3 Utx! Subsequent N, U messages arrive at CC with a minimal delay of 15, 10 sec. (resp.) of each other. Nin  (x=T)  O(¬◊(Nin(T< x+15))), Uin  (x=T)  O(¬UinU(T= x+10)) Compare with the LTL formulae: Nin O1,14Nin, Uin O1,9Uin

  7. Requirements Transmit every N message before the next N message arrives. Nin  O(NinUNtx!) Every U message is transmitted (finish) within 4 seconds after its arrival (Uin  x=T)  ◊(Utx!  Tx+4) Compare with the LTL formula: Uin  O[1,4]Ntx! At most one message may be started at each time instant: (@Utx  (x=T)) ◊(@Ntx  (x=T)), (@Ntx  (x=T)) ◊(@Utx  (x=T)) Compare with the LTL formula: (@Utx  @Ntx)

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