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The Graph above is the answer from the automata. I have problem to determine some of those transitions. I used a truth table to dertermine which the trasitions. This way works well for other Klausur examples. 

pqϕR
00!(b | a & next(p)) 
01!(b | a & next(p)) & next(q)
10b | a & next(p) & !next(q)
11b | a & next(p) & next(q)

How are we getting the transitions ({p,q}, b , {q}) ; ({p}, b , {}) ; ({}, !b ,{}) ; ({}, !b ,{q}) ?

in * TF "Emb. Sys. and Rob." by (120 points)

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I think that we have the following transition relation:

    (p<->b|a&next(p))&(q<->p&next(q)) 

which reduces to the following sets of transitions per state:

  • p=0;q=0 : !b&!a|!b&!next(p)) 
  • p=0;q=1 : false
  • p=1;q=0 :  (b|a&next(p))&!next(q) 
  • p=1;q=1 :  (b|a&next(p))& next(q) 

So, your truth table seems to be wrong since you ignored some brackets and precedences. With the above formulas, you get the transitions of the given solution.

by (170k points)
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