FG automaton

Here, we have to create a FG automaton,

[a SU b] = Aexists({q},1, q<->b|a&next(q), q & FG!q) , can we use this formula to solve it? I am not sure how the !q came as the initial state?

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The equivalence [a SU b] = Aexists({q},1, q<->b|a&next(q), q & FG!q) is not valid: Look, for all points of time, you may have a=0, b=1 and q=1. Then,  [a SU b] is always true, but this is not an accepting run of the automaton on the right hand side. Why do you think that the two formulas are equivalent?
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We need to convert [a SU b] to FG automaton, but in that Temporal logic chapter slide 115 has the formula for SU to a FG automaton. Then how can i convert [a SU b] to a FG automaton?
Check slide 115: [a SU b] = Aexists({q},q, q->b|a&next(q), FG!q). Note that this is not valid for <-> in the transition relation but it does hold for -> as explained in the lecture notes. The remaining formula, in the above case q, is then the set of initial states.
[a SU b] = Aexists({q},q, q->b|a&next(q), FG!q), but in the solution the it's given as initial state as !q and acceptance state as FGq . How !q came to initial state and what about FG!q?
The solution is a different automaton, yes. It has other transitions and initial states, but that may still be equivalent to [a SU b]. It is not the case that there is a unique automaton.
Thanks for the clarification, i understood it.