LTL model checking - state transition

Question

August 2021 Problem 6a

Can you please show me how q2 = 0 is ¬q1 & ¬q0  & ¬a & ¬q' from transition relation.

I m arriving at ¬a & ¬q0' & q1'

Thank you in advance for answering our questions patiently. 

Answer 1

I get the following:

      (q0 <-> a | a & Xq0) & (q1 <-> a | ¬a & Xq1) & (q2 <-> q1 | q0 & Xq2) & !q2
    = (q0 <-> a | a & Xq0) & (q1 <-> a | ¬a & Xq1) & !(q1 | q0 & Xq2) & !q2
    = (q0 <-> a | a & Xq0) & (q1 <-> a | ¬a & Xq1) & !q1 & !(q0 & Xq2) & !q2
    = (q0 <-> a | a & Xq0) & !(a | ¬a & Xq1) & !q1 & !(q0 & Xq2) & !q2
    = (q0 <-> a | a & Xq0) & !a & !(¬a & Xq1) & !q1 & !(q0 & Xq2) & !q2
    = (q0 <-> 0) & !a & !Xq1 & !q1 & !(q0 & Xq2) & !q2
    = !q0 & !a & !Xq1 & !q1 & !(0 & Xq2) & !q2
    = !q0 & !a & !Xq1 & !q1 & !q2

Comments

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