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why does Q2 doesn't have S1 in it's Qf component. If we take intersection of Q component's states with accepting states of the automata we would get S1 but here Qf is empty.
initial condition=q1&!q0|q0&q1, the acceptance condition (!q0&!q1|q0&!q1)

in * TF "Emb. Sys. and Rob." by (250 points)
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Q2 is the successor of Q1 under input {a}, and Q1 = ({s0,s2},{s0}). Since the second component {s0} of Q1 is not empty, we have to compute the first component of Q2 as sucσ∃(Q) and the second component as sucσ∃(Qf)∩F. State s0 has no transition for input {a}, and state s2 has a transition to s1, so that sucσ∃({s0,s2}) = {s1}. For the second component, we already checked that s0 has no transition for input {a}, so sucσ∃({s0}) = {}, and the intersection with the accepting states F does not change this.

So, note that the second component is sucσ∃(Qf)∩F and not sucσ∃(Q)∩F in case that Qf is nonempty.

by (166k points)
Understood. Thank you Professor.

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