As explained in the solution of the exam problem, the formula [a SU [b WU a]] is equivalent to [a WU [b WU a]] so that we can replace the SU operator in the formula by a WU operator to obtain the second equivalent formulation. If that one is translated, no fairness constraints are generated and we get the following automaton:
A({p,q},q,(p <-> a ⋁ b ⋀ Xp) & (q <-> p ⋁ a ⋀ Xq), true)
Why is that so? Note that [phi SU psi] and [phi WU psi] only differ in the case where G(phi&!psi) holds. In that case, [phi SU psi] is false and [phi WU psi] is true. Hence, the formulas [a SU [b WU a]] and [a WU [b WU a]] may only differ if G(a&![b WU a]) holds. The latter is however a contradiction: G(a&![b WU a]) is equivalent to (Ga)&(G![b WU a]), so we must have Ga to satisfy this formula. But then also [b WU a] always holds, and therefore ![b WU a] is always false. Thus, we cannot have (Ga)&(G![b WU a]), and therefore, we cannot have G(a&![b WU a]), and thus, [a SU [b WU a]] and [a WU [b WU a]] never differ!
Therefore, we translate the equivalent formula [a WU [b WU a]] to a safety automaton:
A({p,q},q,(p <-> a ⋁ b ⋀ Xp) & (q <-> p ⋁ a ⋀ Xq), true)
This automaton formula just requires that a word has an accepting run of infinite length.
If we translate the original formula [a SU [b WU a]], we would get
A({p,q},q,(p <-> a ⋁ b ⋀ Xp) & (q <-> p ⋁ a ⋀ Xq), GF(q->p))
In that case, a run is accepted, if it infinitely often satisfies q->p which is equivalent to p&q|p&!q|!p&!q. Thus, the set of the three states p&q, p&!q, !p&!q must be visited infinitely often, which means that at least one of these states must be visited infinitely often.
