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Question 5(b) on Omega Automata. 

  1. In the first part of the question for the omega languages L1 and L2, the language L1 is expressed by the transition relation (Xq1 ↔ a) ∧ (Xq2 ↔ (q1 → ¬b)) ∧ (Xq3 ↔ (q3 ∧ q2)). I am not quite sure about the Xq3 part. From my understanding, I believe q3 is G(a) which is to be written as q3 ↔ a ∧ Xq3. I am not sure why the expression in the transition relation is that of past(G).
  2. Also in part c) and d) of the same question, why are L1 ∩ L2 = L1 and why does 'G(a → X¬b) ∧ F(a → [a U ¬b]) = G(a → X¬b) // Since G(a → X¬b) implies F(a → [a U ¬b]).' hold?  
in # Study-Organisation (Master) by (300 points)

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The formula G(PWX((PWX a) → ¬b)) is translated first to a safety automaton:

    G(PWX((PWX a) → ¬b))
    = 
    AUTO(exists,{q1,q2},q1∧q2,(Xq1 ↔ a)∧(Xq2 ↔ (q1 → ¬b)), Gq2)

That far that good, I guess. Next, we have to convert this safety automaton to a FG-automaton which is done as explained on slide 60 of the omega-automata chapter. That will introduce a new state variable q3. There is no PG used in that formula.

L1 = G(a → X¬b)  and L2 = F(a → [a WU ¬b]). Note first that we can rewrite L2 as (F ¬a) | F[a WU ¬b].

Assume L1 holds, i.e., whenever a holds, then b is false next time. In a first case, we never have a, but then we certainly have F ¬a, and thus also L2. Second, if at some point of time, we have a, then by L1, we must also have X¬b there, thus, ¬b at the next point of time. This also implies L2 since at the next point of time where ¬b holds, we also have [a WU ¬b], and thus also F[a WU ¬b].

by (91.8k points)
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