LTL - SNF -State transition

Question

Thank you so much for answering my question. Its intereseting to see that

August 31,2021 Problem 6a
1) (q0 <-> a | a & Xq0) & (q1 <-> a | ¬a & Xq1) & (q2 <-> q1 | q0 & Xq2) & !q2 - why is !q2 added in the end?
2) in equivalence 4 (q0 <-> a | a & Xq0) is written as in next line (q0 <-> 0). I m considering this state is a self loop and cannot understand why it is negated to 0.

Answer 1

I have added a conjunction with !q2 since we want to derive the transition relation for the states where !q2 holds. Note that phi&!q2=phi[q2<-0]&!q2 holds, where phi[q2<-0] denotes the formula phi where all occurrences of q2 are replaced with 0.

For the same reason, (q0 <-> a | a & Xq0) & !a is equivalent to (q0 <-> 0 | 0 & Xq0) & !a which is again equivalent to (q0 <-> 0) & !a.