Here you can ask questions and find or give answers to organizational, academic and other questions about studying computer science.

1.1k questions

1.3k answers

1.7k comments

557 users

Most popular tags

vrs vrs_exam vrs_exercise ltl exam ctl dira omega_automata rosy parallel_computing ltl_model_checking mbes processor_architecture zielonka sequent-calculus local-model-checking se1 lo2 automata bdd mu-calculus simplex temporal rosy_exam greatest-simulation zdd2fdd vrs_lecture kripke studiengang_reform bdd2fdd exists-algorithm tool fourier_motzkin equivalance bdd2zdd studiengangswechsel symbolic_computation ctl_model_checking fdd2bdd floating_point_arithmetic snf induction breakpoint simulation zdd curriculum_change fdd2zdd predicate_logic dynamic_scheduling cmd b_complement_multiplication constrain-algorithm compose-algorithm #parallel_computing minic i2sc format robdd acceptance curriculum_update basic_block_scheduling unit-tests finroc sche ddd2zdd zdd2bdd vrs_cheatsheet hrs infinite_path branch-bound pre-requisite thesis functional state-machine uml-diagram embedded-system ml1 registration pipeline_conflicts vector_computers actiontable modulo_scheduling static_scheduling organisation lo1 conditional #pram restrict ite programming compiler abacus scheduling parallel forall notation shannon important predecessors fdd

Categories

LTL Model Checking

0 votes

Can someone please explain why the answer is it is true for every state, even though our product kripke structure doesn't have any initial state also if we remove deadends then we don't have any transitions in product kripke structure. Since formula asks for all the paths that satisfy the formula why we are showing in the states? why the answer is not for any formula it won't satisfy? 

Exam paper: question 7 (b)

Exam paper: https://es.cs.uni-kl.de/teaching/vrs/exams/2018.02.14.vrs/2018.02.14.vrs.solutions.pdf

in * TF "Vis. and Sci. Comp." by (870 points)

2 Answers

+1 vote
It is because we are asking for a A-quantified formula. Such formulas hold in all states that don't have infinite outgoing paths at all which is the case here.
by (170k points)
Ok, I understand now. Thank you so much.
by (870 points)
+1 vote
As you said. If there are no infinite transitions, then the existential formula cannot be satisfied.

If the formula is not satisfied, then its negation is satisfied. The negation of the existential formula is a universal formula. That isn't surprising. If there are no infinite paths, then all (of the zero) infinite paths satisfy any arbitrary criterion.
by (25.6k points)
Ok, I understand now. Thank you so much.
by (870 points)

Related questions

0 votes
1 answer
LTL Model checking
asked Aug 17, 2023 in # Study-Organisation (Master) by yk (2.7k points)
0 votes
1 answer
LTL Model checking
asked Jan 24, 2023 in # Study-Organisation (Master) by Pavithra rani (850 points)
0 votes
1 answer
LTL model checking doubt
asked Aug 31, 2022 in * TF "Emb. Sys. and Rob." by CS_E (2.9k points)
 | Snow Theme by Q2A Market
Powered by Question2Answer
Imprint | Privacy Policy
...