ASSIGNMENT 9 - FORMAL MODELS EXERCISES, SUMMER 2011 Selected Solutions to Exercise 35: Formula 4: E [1 U 1] holds in all states 0, 1, 2 and 3. E.g. it holds in state 0 because: we have to find at least one path starting in state 0 (i.e. the "E" quantifier) such that subformula [1 U 1] holds on that path. Let P = 0,3,... be that path. Following the semantics of "U", subformula [1 U 1] requires that we find at least one state s' along P sucht that 1 holds in s'. Once we have such state s', additionally 1 must hold on all states which are predecessors of s' in P. Note that 1 does not have to hold in state s' (although it could). We select s' to be the first state of P, i.e. s' = 0. Then 1 holds in s'. Since s' does not have any predecessors in P, the additional requirement of 1 is vacuously satisfied. Therefore, E [1 U 1] holds in state 0. Checking states 1, 2, 3 is similar, e.g. formula 4 holds in state 3 because: let path P = 3,2,0... Then 1 holds in state s' = 0, and additionally 1 holds in all predecessors of s', i.e. in states 2 and 3. Note that checking 1 is independent of the transitions in path P, i.e. it does not matter that there is a transition from 3 to 2 with symbol 'y' in P. Formula 6: EF (EG 1) holds in all states 0, 1, 2 and 3. E.g. it holds in state 2 because: we have to find a least one path P starting in state 2 (i.e. the first "E" quantifier) such that there is at least one state s' on P (i.e. the "F" operator) where subformula EG 1 holds in state s'. Let P = 2,0,3,3,3,... and s' = 3. We check if subformula EG 1 holds in s': we have to find at least one path P' starting in s' (the "E") such that in all states along P' (the "G" operator) subformula 1 holds. Let P' = 3,3,3,... Then 1 holds in every state of P' because 1 holds in state 3. Therefore, EF (EG 1) holds in state 2. Checking the other states is similiar: we can always choose P' = 3,3,3,... for the second "E" in EF (EG 1), and there is a path P to state 3 from every other state for the first "E" in EF (EG 1).