SolutionsTemporalLogic

SolutionsTemporalLogic - CSE 541 - Logic in Computer...

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CSE 541 - Logic in Computer Science Solutions for Selected Exercises on Temporal Logic Exercise 3.4.9 A CTL formula EFp is true for a state if p is true for that state already, wheras EX EFp need not be true if p is true for the present state. A formula AGp is true for a state s if, and only if, p is true for the present state s and all states reachable from s , wheras AX AGp is true for s and if, and only if, it is true for all states reachable from s . A formula E [ pUq ] is true for a state if q is true for that state already. The formula p EX E [ pUq ], on the other hand, requires that (i) q be true in a future state, not including the present state, and (ii) p be true in all preceding states, including the present state. Exercise 3.4.10 a. EF φ and EG φ are not equivalent. Let φ be p and M be a transition system with States: S = { s 0 , s 1 } Transitions: s 0 s 1 , s 1 s 1 Labels: L ( s 0 ) = { p } , L ( s 1 ) = We have M , s 0 | = EF p but M , s 0 6| = EG p . b. EF φ EF ψ and EF ( φ ψ ) are equivalent. c. AF φ AF ψ and AF ( φ ψ ) are not equivalent. Take φ = p and ψ = q and let M be a transition system with States: S = { s 0 , s 1 , s 2 } Transitions: s 0 s 1 , s 0 s 2 , s 1 s 1 , s 2 s 2 Labels: L ( s 0 ) = , L ( s 1 ) = { p } L ( s 2 ) = { q } Then M , s 0 | = AF ( p q ) but M , s 0 6| = AF p AF q .
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d. AF ¬ φ is equivalent to ¬ EG φ . e. EF ¬ φ and ¬ AF φ are not equivalent. Take the same formula φ = p and transition system M as in 1(a). Then M , s 0 | = EF ¬ p and M , s 0 | = AF p and hence M , s 0 6| = ¬ AF p f. ψ = A [ φ 1 U A [ φ 2 U φ 3 ]] and ψ 0 = A [ A [ φ 1 U φ 2 ] U φ 3 ] are not equivalent. Take φ 1 = p , φ 2 = q , and φ 3 = r ; and let M be a transition system with States: S = { s 0 , s 1 } Transitions: s 0 s 1 , s 1 s 1 Labels: L ( s 0 ) = { p } , L ( s 1 ) = { r } Then M , s 0 | = A [ p U A [ q U r ]] but M , s 0 6| = A [ A [ p U q ] U r ]. Also, let
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SolutionsTemporalLogic - CSE 541 - Logic in Computer...

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