# Exam Paper: 2019.02.13 : Question 6b : omega Automata

Can you please explain the below question? How do we get 'true' in the G(!b) formula here? Thank You!

An existential automaton demands that there must be an infinite run that fulfills the acceptance condition. If the acceptance condition is a safety condition, you can always put that into the transition relation which yields then the simple acceptance condition "true". Formally,

Aexists(Q,I,R,Gphi) = Aexists(Q,I,R&phi,true)

In the above case, the transition relation q&!b&q' was used with the initial state q. That means the automaton has only the state {q} with a self-loop that is only enabled if b is false. Hence, all infinite runs are those where b is always false, and therefore nothing else than the existence of such a run is needed.
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Thank you! now I understood the G(!b). There is another doubt- please explain, In FGa, why have we written acceptance condition Fp to FGp directly?
If you consider the transition relation of that automaton, i.e., p -> a & p', you can see with the state transitions that in the initial state {}, you may either stay or leave to state {p} for any inputs. However, once you are in state {p} the only further transitions possible are those with a self-loop on {p} where input a has to be true.  Hence, switching once to {p} with Fp or remaining there forever with FGp is the same due to the special transition relation. Both Fp and FGp demand that we cannot remain in the initial state {} forever, and leaving it has then the same consequence.
yes, now I understood. Thank you very much.