Disjunctive Normal Form. X=t1 0 . for a 2 A, Consider Petri Net (p;a;' a ; 0 ), Where ' a Is the Projection of ' on A. Then X=t1 0

[32] H. Yen, A polynomial time algorithm to decide pairwise concurrency of transitions for 1-bounded con BLOCKINict-free Petri nets, Inform. Some complexity bounds for problems concerning nite and 2-dimensional vector addition systems with states, Theoret. Comput. 1) _ (# 1 (t i) 1 ^ # 1 (t j) = 0))). As a result, B-fairness can be detected in P. Acknowledgements: The authors thank the anonymous referees for their comments and suggestions which improved the correctness as well as the presentation of this paper. 27 As a consequence, the nontermination problem for each of the above notions of fairness can be solved in P. 3. Promptness Detection Problem. The concept of promptness was introduced in [30] as a way to deal with systems communicating with the environment. Let T I and T E be two disjoint sets of transitions such that T I [ T E = T. (T I and T E can be viewed as the sets of internal and external transitions, respectively.) A Petri net (P;T;'; 0) is said to be prompt (with respec-w ! =) jwj < a, meaning that for every reachable marking , there is no innite sequence of internal transitions rable in. It is not hard to see that a Petri net is not prompt i 9 1 2 9 0 ; 1 (t 0) > 0))) The polynomial time upper bound for the promptness detection problem follows. 4. Pairwise Synchronization Problem. In [27], the notion of y-distance was introduced. Given a Petri net P and a y 2 Z r , let D(P,y)=sup 0 7 0! (jy # j), where stands for the inner product. P is said to be y-synchronized if D(P,y) is nite. The y-synchronization problem is that of determining, given a Petri net P and a y, whether P is y-synchronized. Given two transitions t i and t j , the pairwise synchronization problem (with respect to t i and t j) is a special case of the y-synchronization problem with y dened as y(i) = 1;y(j) = 01, and y(h) = 0;8h 6 = i; j: P is pairwise synchronized with respect to t i and t j i for every path 0 0 7 0! 1 1 7 0! 2 , if 2 1 , then # 1 (t i) = # 1 (t j). Using our path formulas, P is not pairwise synchronized (t) > 0)) Hence, …