Hierarchy of persistence
with respect to the length of action’s disability

An algorithm to check if a step $Ma$ is e/l-k-persistent (for some $\mathrm{k}\in\mathbb{N}$) for a given net $\mathrm{S}=(\mathrm{N},M_{0})$:
Let us build the part of the depth of k+1 (we call it the (k+1)-component) of the reachability tree of $(\mathrm{N},M^{\prime})$, where $M^{\prime}$ is a marking obtained from $M$ by execution of $a$. The step $Ma$ is e/l-k-persistent if for every action $b\in T$, such that $a\neq b$ and $b$ is enabled in $M$, there is a path in the (k+1)-component of the reachability tree of $(\mathrm{N},M^{\prime})$ containing an arc labeled by $b$.

For every action $a\in T$ that is enabled in a marking $M$, we check if a step $Ma$ is e/l-k-persistent (for some $\mathrm{k}\in\mathbb{N}$) for a given net $S=(\mathrm{N},M_{0})$, using the algorithm of Theorem 3.1.

Let $M=\mathrm{minE}_{a,b}$. We build a coverability graph for the p/t-net S. We check whether in the graph exists a vertex corresponding to an $\omega$-marking $M^{\prime}$ such that $M^{\prime}$ covers $M$. If so, then actions $a$ and $b$ are simultaneously enabled in some reachable marking of the net S. Otherwise, those transitions are never enabled at the same time.

Let us take the right closure $\mathrm{RE}_{a,b}\uparrow$ of the set $\mathrm{RE}_{a,b}$.
Note that $\mathrm{Min}(\mathrm{RE}_{a,b})=\mathrm{Min}(\mathrm{RE}_{a,b}\uparrow)$. To show that the set of minimal elements of the set $\mathrm{RE}_{a,b}$ is effectively computable, it is enough to demonstrate that the set $\mathrm{RE}_{a,b}\uparrow$ has the property RES (i.e. for any $\omega$-vector $v\in\mathbb{N}_{\omega}^{P}$ the problem "$(\downarrow v\cap RE_{a,b}\uparrow\neq\emptyset)$?" is decidable) and apply Theorem 3.4.
Let $X=\downarrow v\cap E_{a,b}$, where $E_{a,b}=\mathrm{min}E_{a,b}+\mathbb{N}^{P}$. Let us notice, that $\downarrow v$ is a convex set, hence rational (Proposition 1). The set $E_{a,b}$ is also a rational convex set. As an intersection of convex rational sets, the set $X$ is convex and rational (Theorem 2.1) as well.
Hence, putting into work decidability of the Set Reachability Problem for rational convex sets (Theorem 3.5) we decide whether any marking from the set $X$ is reachable in S. Therefore, we can decide whether the set $X^{\prime}=\downarrow v\cap\mathrm{RE}_{a,b}$ is nonempty. (It is the case when at least one marking from the set $X$ is reachable in S.) Let us notice that the set $X^{\prime\prime}=\downarrow v\cap\mathrm{RE}_{a,b}\uparrow$ is nonempty if and only if the set $X^{\prime}$ is nonempty. That is why the set $\mathrm{RE}_{a,b}\uparrow$ has the property RES, and consequently the set $\mathrm{Min}(\mathrm{RE}_{a,b})$ is effectively computable by Theorem 3.4.

Let $M$ be a marking, such that the execution of an action $a$ in $M$ pushes the execution of an action $b$ away for more than k steps (for some $\mathrm{k}\in\mathbb{N}$). Let $M^{\prime}\in\mathrm{Min}(\mathrm{RE}_{a,b})$ such that $M^{\prime}\leq M$. Such a marking has to exist. Suppose that there is a string $w\in T^{*}$, $|w|\leq\mathrm{k}$ such that $M^{\prime}awb$. Then obviously also $Mawb$ (from the monotonicity property - Fact 2). We obtain a contradiction. Hence, the execution of an action $a$ in $M^{\prime}$ postpones the execution of $b$ for more than k steps.

We introduce an algorithm of deciding if an action $a$ pushes the execution of an action $b$ away for more than k steps in some reachable marking $M$.

1. 1.

   We check whether both actions $a$ and $b$ are enabled in some reachable marking (using decidability of Mutual Enabledness Reachability Problem).

   1. (a)

      If not, we answer NO.

   2. (b)

      Otherwise:

      1. i.

         We build the set $\mathrm{Min}(\mathrm{RE}_{a,b})$. This set can be effectively computed by Proposition 2 using Valk/Jantzen algorithm.

      2. ii.

         For all markings $M_{1}\in\mathrm{Min}(\mathrm{RE}_{a,b})$:
         $M_{2}:=M_{1}a$.
         We build an initial part of the depth of k+1 (the (k+1)-component) of the reachability tree of $(\mathrm{N},M_{2})$. If the piece has an edge labeled by $b$, we answer NO. Otherwise we answer YES.

S is e/l-k-persistent iff the algorithm solving EL-k Transition Persistence Problem answers NO for all ordered pairs $(a,b)\in T\times T$, $a\neq b$.

First note that, by the monotonicity property (Fact 2), the set
$\mathrm{E}_{a}\cap\mathrm{E}_{b}\cap(\mathbb{N}^{|\mathrm{P}|}-\mathrm{E}_{a(k)b})$ is convex, thus rational (by Proposition 1). The rational expressions for $E_{a}$ and $E_{b}$ are $E_{a}=^{\bullet}a+\mathbb{N}^{\mathrm{k}}$ and $E_{b}=^{\bullet}b+\mathbb{N}^{\mathrm{k}}$. Clearly, the set $\mathrm{E}_{a(k)b}$ is right-closed, by the monotonicity property. We shall prove that it has the property RES. Namely, $\downarrow\!v$ ($v\in\mathbb{N}_{\omega}^{P}$)intersects $\mathrm{E}_{a(k)b}$ if and only if ${}^{\bullet}a\leq v$ (i.e. $a$ is enabled in $v$) and there is a path in the reachability tree, limited to (k+1) first levels, of the net $(\mathrm{N},v^{\prime})$, where $v^{\prime}$ is an $\omega$ -marking obtained from $v$ by the execution of $a$, containing an arc labelled by $b$. It is obviously decidable. Hence, the set $\mathrm{E}_{a(k)b}$ has the property RES, thus (by Theorem 3.4) the set $\mathrm{Min}(\mathrm{E}_{a(k)b})$ is effectively computable. As $\mathrm{E}_{a(k)b}$ is right-closed, we get the rational expression for it: $\mathrm{E}_{a(k)b}=\mathrm{Min}(\mathrm{E}_{a(k)b})+\mathbb{N}^{P}$. Finally, using Theorem 2.1 of Ginsburg/Spanier [8], we compute rational expression for $\mathrm{E}_{a}\cap\mathrm{E}_{b}\cap(\mathbb{N}^{P}-\mathrm{E}_{a(k)b})$ and Theorem 3.4 yields decidability of the problem.

Suppose that the lemma does not hold for some action $a\in T$. It means that for each $\mathrm{k}\in\mathbb{N}$ there is a marking $M$ such that $a$ is not k-enabled but not dead. This means that $a$ is $\mathrm{k}^{\prime}$-enabled for some $\mathrm{k}^{\prime}>\mathrm{k}$. Thus, there exists an infinite sets of markings $M_{1},M_{2},\ldots$ and integers $\mathrm{k}_{1}<\mathrm{k}_{2}<\ldots$, such that the action $a$ is live in each marking $M_{\mathrm{i}}$ and it is not $\mathrm{k}_{\mathrm{i}}$-enabled in $M_{\mathrm{i}}$ for all $\mathrm{i}=1,2,\ldots$. Let us choose (by Fact 6) an infinite increasing sequence of markings $M_{\mathrm{i}1}\leq M_{\mathrm{i}2}\leq\ldots$. Since the action $a$ is live in $M_{\mathrm{i}1}$, it is k-enabled in $M_{\mathrm{i}1}$, for some $\mathrm{k}\in\mathbb{N}$. As the strictly increasing sequence $\mathrm{k}_{1}<\mathrm{k}_{2}<\ldots$ is infinite, $\mathrm{k}<\mathrm{k}_{\mathrm{ij}}$ for some j. By the monotonicity property (Fact 2), the action $a$ is k-enabled, hence $\mathrm{k}_{\mathrm{ij}}$-enabled in the marking $M_{\mathrm{ij}}$. Contradiction.

If the net is e/l-persistent, then no action kills another enabled one. From the Lemma 2 we know, that if an action $a\in T$ is not dead then it is $\mathrm{k}_{a}$-enabled. Let us take $\mathrm{K}=\mathrm{max}\{\mathrm{k}_{a}|a\in T\}$, for the numbers $\mathrm{k}_{a}$ from the Lemma 2. One can see that every action in the net that is not dead, is K-enabled. Thus, the execution of any action may postpone the execution of an action $a$ for at most K steps. So we have the implication: if a p/t-net is e/l-persistent, then it is e/l-K-persistent, for K defined above.

• •

  We check whether both actions $a$ and $b$ are enabled in some reachable marking (using decidability of Mutual Enabledness Reachability Problem). If not, $\mathrm{k}_{a,b}$ does not exist (actions $a$ and $b$ are never enabled at the same time).  Otherwise:

  • –

    We ask whether $a$ kills an enabled $b$ (EL-Persistence Problem).
    If YES then $\mathrm{k}_{a,b}$ does not exist ($a$ kills $b$)
    else:

    • *

      We compute the set $\mathrm{Min}(\mathrm{RE}_{a,b})$.This set can be effectively computed by Proposition 2 using Valk/Jantzen algorithm.

    • *

      We build the initial part of reachability tree of the net $S$ as long as from every $M\in\mathrm{Min}(\mathrm{RE}_{a,b})$ we get a marking $M^{\prime}$ with the property that a path leads to a vertex $M^{\prime}$ (it can be an empty path) such that $M^{\prime}b$. Clearly, such part of the tree is finite, as we get the whole $\mathrm{Min}(\mathrm{RE}_{a,b})$ and for any $M\in\mathrm{Min}(\mathrm{RE}_{a,b})$ a finite path leading from $M$ to a vertex $M^{\prime}$ such that $M^{\prime}b$. The maximum length of such paths is the desired number $\mathrm{k}_{a,b}$.

For every pair $(a,b)$ of transitions we find $\mathrm{k}_{a,b}$ defined above. The number we are looking for is $\mathrm{k}=\mathrm{max}(\mathrm{k}_{a,b}:a,b\in T)$.