# Exercise 4.d problem with reachable states.

+1 vote

Given a FSM: A={2Vin,2Vout,I,2Vstate,R}A={2Vin,2Vout,I,2Vstate,R}
where Vin={a}Vin={a}Vout={o}Vout={o}Vstate={p,q}Vstate={p,q}I=I=q<->(p->q),
R=R=!(a|o)&(next(p)->(p->q))&next(q)

The Kripke structure of the FSM comes,

Isn't the highlighted ones the reachable states? the system is saying that its incorrect. Kindly help me here.

Similar to this one: https://q2a.cs.uni-kl.de/887/exercise-sheet-5-question-1?show=894#a894

In b), we have already updated the initial states (a)) but not yet updated the transition relation regarding deadends (c)) and unreachable states (d)). Hence, deadend state S11 should not be initial, and hence not reachable.

by (25.6k points)
My answer was accepted once I removed S9, S10 and S11.

Can you explain to me that why S8 is considered to be reachable but those three are not? I need to clear my basic in this.
S9, S10, S11 are deadend states. hence, you removed them from the initial states in step a). Since S8 is not a deadend (it has transitions, e.g. to S4), you did not remove this one from the initials. As the said states S8, S9, S10, S11 do not have incoming edges, their only way to be reachable is being initial. Therefore, S8 is the only one among the four that is left reachable.