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Proof 3.9 If: If [MATH] has a one-in-three model, then using the corresponding region [MATH] of [MATH] defined in Lemma 3.6 and the key region [MATH] of [MATH] introduced in Lemma 3.7 yields a combined region [MATH] of [MATH] that inhibits the key event [MATH] at [MATH] Section presents a series of lemmas that altogeth... |
Only-if: If [MATH] is feasible then the key event is inhibitable at the key state in [MATH] Projecting the respective region to [MATH] , we get an indicator due to Lemmas 3.5 and 3.7 Hence, [MATH] is one-in-three satisfiable. |
Using Lemma 3.2 , Lemma 3.8 , the observation that our construction is in polynomial time, the fact that feasibility is in NP, we have shown Theorem 3.1 |
4. NP-completeness of Feasibility for seven more Petri Net Classes This section presents and proves the following theorem: Theorem 4.1 |
Deciding [MATH] -feasibility as well as [MATH] -language viability is NP-complete for (1) modest TSs and [MATH] or [MATH] (2) modest TSs and [MATH] or [MATH] , and |
(3) general TSs and [MATH] with non-empty [MATH] Notice that Theorem 4.1 does not restrict input to modest TSs. Otherwise, it easy to show that feasibility becomes tractable if [MATH] belongs to these classes: Assume a modest TS [MATH] contains a transition [MATH] If [MATH] , then [MATH] is not feasible because [MATH] ... |
We again present polynomial time reductions of the NP-complete cubic monotone one-in-three [MATH] -SAT problem to the corresponding feasibility problems making sure for every TS [MATH] constructed from a given cubic monotone [MATH] -CNF [MATH] that the ESSP implies the SSP. By Lemma 2.1 the proofs for Theorem 4.1 and 4... |
In every case, we again install a key event [MATH] and a key state [MATH] in [MATH] such that [MATH] is inhibitable at [MATH] by a key region [MATH] if and only if a one-in-three model [MATH] exists. Like before, the variables [MATH] are used as events in [MATH] and their key signature [MATH] tells us how to find [MATH... |
This idea is put into practice by creating six directed labeled paths per clause [MATH] that commonly start at state [MATH] , terminate at [MATH] and consist of three transitions permuting the events [MATH] The TS [MATH] fulfills the following conditions: |
(1) For every [MATH] and every permutation [MATH] of [MATH] there is a path [MATH] in [MATH] (2) If [MATH] is a key [MATH] -region of [MATH] , that is, one that inhibits [MATH] at [MATH] , then [MATH] |
(3) If [MATH] and [MATH] is a key [MATH] -region of [MATH] then exactly one of [MATH] [MATH] [MATH] is different from nop Hence, the signature tells us how to build [MATH] , a one-in-three model of [MATH] |
(4) If [MATH] has a one-in-three model then [MATH] is solvable. (5) If [MATH] is solvable, then [MATH] has the [MATH] -E(SSP). Clearly, having these conditions proves that there is a one-in-three model for [MATH] if and only if [MATH] has the E(SSP). |
Next, we define how [MATH] is constructed from [MATH] See Figure for a visualization of the following concepts. Firstly, we call [MATH] the basic TS with states [MATH] and events [MATH] To omit a lengthy and complex definition of the transitions in [MATH] , we use Figure , which depicts [MATH] with black arcs for all s... |
If [MATH] is one of [MATH] [MATH] , or [MATH] , the construction of [MATH] is more complex. We first require the extended TS [MATH] with extended states [MATH] and extended events [MATH] The transitions of [MATH] are also an extension in the way that [MATH] for basic states [MATH] and basic events [MATH] where [MATH] i... |
While still being depictable, [MATH] is not yet a complete TS [MATH] for extensions of toggle nets. This requires the loop-enhancement |
[MATH] of [MATH] on the same states [MATH] and events [MATH] but with loop-enhanced transitions, that is, for all [MATH] and [MATH] where [MATH] we have [MATH] and [MATH] Now, [MATH] However, for understanding it is mostly better to deal with [MATH] instead of the complicated [MATH] Therefore, Figure desists from showi... |
At this point, we are ready to provide the main piece of our proof. The next lemma shows the equivalence between the one-in-three satisfiability of [MATH] and the inhibitability of [MATH] at [MATH] |
Lemma 4.2 If [MATH] is [MATH] or [MATH] or [MATH] with non-empty [MATH] then the key event [MATH] is [MATH] -inhibitable at the key state [MATH] in [MATH] if and only if [MATH] is one-in-three satisfiable. |
Proof 4.3 Only-if Let [MATH] be a [MATH] -region inhibiting [MATH] at [MATH] in [MATH] We show for every clause [MATH] that there is exactly one variable event [MATH] with [MATH] while the other two have nop -signature. Consequently, the set [MATH] will be a one-in-three model of [MATH] |
For a start, [MATH] or [MATH] As [MATH] is inhibited at [MATH] , assume first that [MATH] and [MATH] But then [MATH] which means that [MATH] Hence, [MATH] and [MATH] This implies [MATH] and [MATH] and, thus, [MATH] By this, we get [MATH] In [MATH] , there is a path [MATH] for every permutation [MATH] of [MATH] As all v... |
Next, let [MATH] with non-empty [MATH] Assume [MATH] and [MATH] , which implies [MATH] and [MATH] and, thus, [MATH] We immediately get [MATH] implying [MATH] and [MATH] and, thus, [MATH] For every [MATH] , we have [MATH] by [MATH] Accordingly, [MATH] leads to [MATH] As [MATH] , there must be at least one event in [MATH... |
The case [MATH] and [MATH] , follows by the same argumentation by inverting the support and interchanging set with res and used with free |
If Let [MATH] be a one-in-three model of [MATH] and, for a start, let [MATH] or [MATH] We define a [MATH] -region [MATH] of [MATH] that inhibts [MATH] at [MATH] by [MATH] and [MATH] [MATH] for all [MATH] and [MATH] for all other events [MATH] in [MATH] |
If [MATH] contains, beside [MATH] , the interaction used then we let [MATH] and create a [MATH] -region [MATH] of [MATH] to inhibit [MATH] at [MATH] by the support [MATH] and [MATH] [MATH] for all [MATH] [MATH] for all [MATH] and [MATH] for all other events [MATH] in [MATH] If used is not in [MATH] then free is and we ... |
Using Lemma 2.1 , we have also covered the types of nets where [MATH] is [MATH] or [MATH] Using the same lemma, we can finish our proof for Theorem 4.1 by the following lemma: |
Lemma 4.4 If [MATH] is [MATH] or [MATH] or [MATH] with non-empty [MATH] and the key event [MATH] is [MATH] -inhibitable at the key state [MATH] in [MATH] then [MATH] has the E(SSP). |
Proof 4.5 We present for each event [MATH] , respectively state [MATH] ), a set of regions which inhibit [MATH] at, respectively separate [MATH] from, all states in [MATH] |
Assume that [MATH] or [MATH] For brevity, we use the following scheme to define [MATH] based on a given support [MATH] If [MATH] and [MATH] for all [MATH] then [MATH] , if [MATH] for all [MATH] then [MATH] and, otherwise, [MATH] Using this, we can define a region simply by defining [MATH] |
The inhibition of [MATH] at [MATH] and [MATH] at [MATH] already follows from Lemma 4.2 Furthermore, if [MATH] inhibits [MATH] at [MATH] then the region [MATH] inhibits [MATH] at [MATH] For [MATH] let [MATH] The region [MATH] inhibits [MATH] at all [MATH] Moreover, [MATH] inhibits [MATH] at the states of [MATH] , where ... |
The inhibition of [MATH] at [MATH] separates [MATH] and [MATH] and the region [MATH] separates [MATH] from all the other states. The region [MATH] completes the separation of [MATH] The regions [MATH] complete the separation of [MATH] |
Now assume [MATH] with non-empty [MATH] For the [MATH] -ESSP of [MATH] , it is sufficient to prove the inhibition of [MATH] at states [MATH] where [MATH] and [MATH] By definition, if [MATH] in [MATH] then [MATH] in [MATH] and [MATH] does not need to be inhibitable at [MATH] We proceed like in the previous case and use ... |
The inhibition of [MATH] at [MATH] is done by the key region of Lemma 4.2 For [MATH] let [MATH] The regions [MATH] complete the inhibition of [MATH] in [MATH] because for every state [MATH] there is an event [MATH] such that [MATH] Moreover, they prove [MATH] inhibitable at all states except for [MATH] |
For every [MATH] [MATH] ) and all [MATH] with [MATH] [MATH] ) we immediately get [MATH] [MATH] ). Hence, for [MATH] the regions [MATH] and [MATH] inhibit [MATH] and its corresponding event [MATH] in [MATH] Moreover, [MATH] inhibits [MATH] at [MATH] proving these events to be inhibitable, too. The regions [MATH] and [MA... |
It is easy to see that the state separating regions defined above for the consumer nets can be used here to separate the same states simply by replacing inp by res Moreover, the states [MATH] are clearly separable from the states [MATH] as [MATH] is itself a [MATH] -support. The remaining SSP-atoms are solved by [MATH]... |
5. Polynomial Time Net Synthesis for 36 Types of Nets This section proves the tractability of synthesis for the 36 types of nets given in the following theorem: |
Theorem 5.1 There is a polynomial time algorithm, that, on input TS [MATH] , synthesizes a [MATH] -net [MATH] with state graph isomorphic to [MATH] or rejects [MATH] if [MATH] does not exist, for every |
(1) [MATH] with [MATH] (2) [MATH] with [MATH] (3) [MATH] with [MATH] (4) [MATH] with [MATH] Notice that input is not limited to modest TSs for any case of Theorem 5.1 Hence, our efficient methods are robust with respect to general input and do not depend on any restrictions. |
The following subsection introduces a new polynomial time algorithm for Theorem 5.1 and 5.1 The basic idea is to compute a region set [MATH] solving all (E)SSP atoms of a given TS [MATH] , if such a set exists. Using [MATH] , the sought net [MATH] can easily be computed. |
For Theorem 5.1 , Section 5.2 shows how to extend the algorithm of Schmitt in order to efficiently synthesize nets. Before going into the two subsections, we turn towards Theorem 5.1 which covers types of nets with a rather trivial synthesis problem. A related polynomial time algorithm is established in the proof of th... |
Lemma 5.2 If [MATH] is a type of nets with [MATH] then [MATH] -feasibility of a given TS [MATH] can be decided in [MATH] time. [MATH] -feasible TS [MATH] can be synthesized into a [MATH] -net [MATH] with [MATH] isomorphic to [MATH] in [MATH] time. |
Proof 5.3 Let [MATH] be any transition of [MATH] with [MATH] To separate these states, we require a [MATH] -region [MATH] with [MATH] However, [MATH] implies that [MATH] Hence, [MATH] and [MATH] are not [MATH] -separable. As all states have to be reachable, this implies that input TSs with more than one state cannot be... |
Being reduced, [MATH] is the only single-state TS. For this input, [MATH] has a state graph isomorphic to [MATH] and is, thus, written to the output in [MATH] time. |
5.1. Net Synthesis by Incremental Region Growing The result of this section is the following contribution to Theorem 5.1 Lemma 5.4 |
If [MATH] is a type of nets with [MATH] or [MATH] with [MATH] then any given TS [MATH] can be synthesized into a [MATH] -net [MATH] with [MATH] isomorphic to [MATH] , respectively rejected if the net [MATH] does not exist, in [MATH] time. |
Before we can go into the proof of this lemma, we introduce the respective algorithm. The core subroutine of this method is Algorithm According to Lemma 5.5 , this method accepts a given set of states [MATH] and returns a minimal superset [MATH] that, together with a matching signature, forms a region of [MATH] Later, ... |
Data: TS [MATH] and set of states [MATH] Result: A support [MATH] for a region of [MATH] while [MATH] do [MATH] end while return |
[MATH] Algorithm 1 Given [MATH] , the algorithm minimally extends [MATH] to the support of a [MATH] -region [MATH] of [MATH] for all reset types of nets [MATH] extended with [MATH] |
Lemma 5.5 If [MATH] is a type of nets with [MATH] and [MATH] is a TS and [MATH] then the result [MATH] of Algorithm started on [MATH] forms a [MATH] -region [MATH] of [MATH] with |
[EQUATION] for all [MATH] Moreover, for all [MATH] -regions [MATH] of [MATH] with [MATH] it is true that even [MATH] Algorithm terminates after [MATH] time. |
Proof 5.6 That the algorithm terminates is trivial as every iteration extends [MATH] , which is possible for at most [MATH] times. After termination, [MATH] obviously contains input [MATH] Moreover, there are no events [MATH] participating in a transition [MATH] with [MATH] On the other hand, if there is transition [MA... |
Now let [MATH] be any [MATH] -region of [MATH] with [MATH] We show by induction that the set [MATH] resulting from [MATH] while-iterations of Algorithm fulfills [MATH] For a start, [MATH] Assume that [MATH] and [MATH] and let [MATH] which, thus, fulfills [MATH] As [MATH] is added to [MATH] , there are [MATH] and [MATH]... |
As there are at most [MATH] while-iterations and as checking the while-condition takes [MATH] time, Algorithm runs in [MATH] time. |
Having a way to reliably produce [MATH] -regions for the net classes of interest, we argue that they are versatile enough to solve all (E)SSP atoms. Otherwise, the computed regions [MATH] would not suffice to synthesize the net [MATH] |
Lemma 5.7 If [MATH] with [MATH] and [MATH] is a TS then [MATH] is [MATH] -inhibitable at [MATH] where [MATH] if and only if (1) [MATH] and the region [MATH] returned by Algorithm on input [MATH] satisfies [MATH] and [MATH] , or |
(2) [MATH] and the region [MATH] returned by Algorithm on input [MATH] satisfies [MATH] and [MATH] , or (3) [MATH] and the region [MATH] returned by Algorithm on input [MATH] satisfies [MATH] |
Two states [MATH] are [MATH] -separable if and only if the region [MATH] returned by Algorithm fulfills [MATH] for [MATH] or [MATH] for [MATH] |
Proof 5.8 The if-direction for an ESSP atom [MATH] is trivial, as [MATH] is inhibited at [MATH] by the pair [MATH] from Algorithm , a [MATH] -region according to Lemma 5.5 Reversely, if [MATH] is inhibited at [MATH] by a [MATH] -region [MATH] then interaction [MATH] is not defined on [MATH] Hence, [MATH] Let [MATH] [MA... |
The if-direction for the SSP atom [MATH] is trivial, again, as the [MATH] -region of Algorithm separates [MATH] Reversely, let [MATH] be separated by a [MATH] -region [MATH] where, without loss of generality, [MATH] and [MATH] The result [MATH] of Algorithm on [MATH] is a [MATH] -region by Lemma 5.5 that fulfills [MATH... |
By the required versatility of the regions from Algorithm , we can now prove Lemma 5.4 Proof 5.9 (Proof of Lemma 5.4 Let [MATH] and [MATH] The idea is to firstly produce a region set [MATH] that solves all (E)SSP atoms of [MATH] If we cannot find [MATH] , then we reject [MATH] There are [MATH] ESSP atoms [MATH] Dependi... |
Next, there are [MATH] SSP atoms [MATH] By Lemma 5.7 , we have to call Algorithm with [MATH] and [MATH] to decide the separability of [MATH] After [MATH] time, either [MATH] solves all SSP atoms or we can reject [MATH] |
Hence, using [MATH] time in total, we decide the feasibility of [MATH] and, in the positive case, get [MATH] Computing [MATH] consumes [MATH] time, which is dominated by the previous costs. |
If [MATH] with [MATH] our approach is to synthesize a net [MATH] for the isomorphic type [MATH] that replacing set with res out with inp used with free , and free with used In order to obtain a [MATH] -net [MATH] , we simply revert the interaction replacement in the flow function [MATH] Obviously, [MATH] is isomorphic ... |
5.2. Net Synthesis for Relatives of Flip-Flop-Nets Last step in proving Theorem 5.1 is to cover item , the relatives of flip-flop nets: |
Lemma 5.10 If [MATH] with [MATH] then a given TS [MATH] can be synthesized into a [MATH] -net [MATH] with [MATH] isomorphic to [MATH] , respectively rejected if [MATH] does not exist, in polynomial time. |
This works simply by modifying Schmitt’s algorithm which is based on the ability to efficiently solve equations over the boolean field [MATH] As flip-flop nets have already been covered there and as types of nets [MATH] with [MATH] are isomorphic to [MATH] , where [MATH] mirrors [MATH] simply by replacing inp with out ... |
(1) [MATH] with [MATH] and (2) [MATH] with [MATH] but not [MATH] The following roughly summarize the main steps of Schmitts algorithm and afterwards we show how this can be modified for our cases. If we are given a TS [MATH] , we firstly interpret it as a directed graph on nodes [MATH] and with (labeled) directed arcs ... |
With the previous definition, our goal for every (E)SSP atom [MATH] is to define a system [MATH] of equations over the boolean field [MATH] using the events [MATH] as variables. If [MATH] has a solution [MATH] assigning either [MATH] or [MATH] to every event, it leads to a region solving the original atom. Otherwise, t... |
Any solution [MATH] to [MATH] is called abstract region with the following meaning: As long as [MATH] contains nop and swap , we can derive real [MATH] -regions [MATH] from [MATH] by defining the support [MATH] for all [MATH] The chord equations make sure that the event parity on any other path to [MATH] is equal to [M... |
Based on the specific type of nets [MATH] and the atom [MATH] , we augment [MATH] with additional equations to obtain [MATH] If, beside nop and swap [MATH] does not contain anything beyond [MATH] then, to solve an SSP atom [MATH] , it is already enough to extend [MATH] with the equation [MATH] to find [MATH] This equat... |
How to solve an ESSP atom [MATH] depends on the availability of inp out used , and free in [MATH] Taking flip-flop nets, this brings along only inp and out and we obtain [MATH] by complementing [MATH] with the equation [MATH] and for all transitions [MATH] with the equations [MATH] This makes sure that all sink states ... |
Compared to the number of transitions in [MATH] , the system [MATH] has at most a linear amount of equations. As solving [MATH] is in polynomial time, solving (E)SSP atoms is tractable for flip-flop nets. Having the polynomial size set [MATH] of regions for all these atoms after polynomial time, we can synthesize the n... |
Proof 5.11 (Proof of Lemma 5.10 As discussed before, we only have to show for every respective [MATH] that the ESSP atoms [MATH] can be expressed as extensions of [MATH] The idea is to create one or more systems [MATH] for every [MATH] that have at least one solution [MATH] if and only if the atom can be solved with a ... |
For [MATH] , we simply use the same system as for flip-flop nets. Having a solution [MATH] , we first define support and signature as before where for [MATH] we let [MATH] if [MATH] for any (that also means all) [MATH] Otherwise, if [MATH] , we switch to the complementary support, which turns [MATH] to [MATH] and thus,... |
To get [MATH] , we take [MATH] and add the equation [MATH] plus for every transition [MATH] the equations [MATH] These equations make sure that [MATH] can be assigned used or free and that states incident to [MATH] behave correctly. If [MATH] is a solution to [MATH] , we again define support and signature like in all p... |
6. Conclusion In this paper we investigate the complexity of boolean net synthesis for the 128 practically more relevant nop -afflicted classes. In total, we prove 84 cases NP-hard and provide polynomial time algorithms for 36 classes. As a side product, this paper introduces a very general reduction scheme that serves... |
For the eight classes ( nop inp , [ used ]), ( nop out , [ free ]), ( nop set res ) and ( nop swap ) extended with at least one of set and res , we leave the complexity of synthesis open. They remain for future work. |
While the first four items of this list are just surprisingly difficult with respect to their very limited interactions set, the really interesting classes are the last four which are built only from nop swap set , and res None of these events can be used to solve ESSP-atoms. Consequently, any input TS [MATH] with at l... |
7. Technical Proofs from Section Proof 7.1 (Proof of Lemma 3.5 ): Firstly, if [MATH] installs a non empty freezer then, by Lemma 3.4 , we have [MATH] |
Let [MATH] For abbreviation we define [MATH] and [MATH] We now argue for [MATH] and see later that the arguments are symmetrically true for [MATH] |
If condition ( ) is satisfied then by [MATH] , we have for all [MATH] that [MATH] which with [MATH] and [MATH] implies that [MATH] and [MATH] Symmetrically, condition ( ) implies that [MATH] |
and [MATH] Let [MATH] and [MATH] and [MATH] Firstly, [MATH] [MATH] ) implies that at least one element of [MATH] has a signature from [MATH] [MATH] ), cf. Figure .1. Secondly, [MATH] is a source of [MATH] and [MATH] is a sink of [MATH] which implies [MATH] [MATH] ), cf. Figure By a similar argument and [MATH] for all [... |
We now argue that there is exactly one variable event of [MATH] with a signature different from nop To do so, we show that if [MATH] then [MATH] |
If [MATH] [MATH] ) and [MATH] [MATH] ) implying that [MATH] [MATH] ) then we can immediately conclude the following fact: By [MATH] [MATH] ) and [MATH] [MATH] ) and Figure we have that [MATH] [MATH] ) which implies [MATH] [MATH] ). Moreover, by [MATH] [MATH] ), we have [MATH] [MATH] ). The inclusion (exclusion) of [MAT... |
Consequently, [MATH] satisfies for all [MATH] the condition [MATH] which makes [MATH] a one-in-three model of [MATH] ): All [MATH] satisfy that if [MATH] in [MATH] then [MATH] in [MATH] Consequently, [MATH] assures state synchronization for [MATH] -labeled transitions: [MATH] For abbreviation we define [MATH] and [MATH... |
Let [MATH] and [MATH] By definition of [MATH] we have [MATH] Firstly, we note that by [MATH] if and only if [MATH] we have [MATH] implying [MATH] for [MATH] Secondly, we show that for [MATH] there is exactly one variable event with a border crossing signature and both of the others are mapped to nop By [MATH] [MATH] an... |
Let [MATH] and [MATH] For [MATH] the signature [MATH] [MATH] ) implies [MATH] [MATH] ), cf. Figure .3. If [MATH] [MATH] ) then, by [MATH] and [MATH] , we easily have [MATH] and [MATH] [MATH] and [MATH] ), c.f. Figure .5. Symmetrically to the argumentation for [MATH] , this implies that either exactly one or all variabl... |
Proof 7.2 (Proof of Lemma 3.7 Let [MATH] and [MATH] ): Firstly, we show that [MATH] for all [MATH] If [MATH] assume that [MATH] [MATH] ). As [MATH] , we have [MATH] [MATH] ), cf. Figure .2, Figure .3 and Figure .4. Hence, we have [MATH] [MATH] ) implying that [MATH] [MATH] ) and, thus, [MATH] is not inhibited at the ke... |
If [MATH] [MATH] ), then [MATH] contains [MATH] [MATH] ) and [MATH] Assume that [MATH] [MATH] ). All states in [MATH] incident to event [MATH] have to be part (outside of) the support. By [MATH] [MATH] ) we get that [MATH] [MATH] ) cannot be in exit enter ) and, thus, [MATH] [MATH] ). Thus, [MATH] [MATH] ) which clearl... |
Secondly, assume that we inhibit [MATH] at the key state [MATH] , which means [MATH] We show that [MATH] and [MATH] Applying Lemma 3.4 to [MATH] , we have [MATH] Let [MATH] By [MATH] [MATH] and [MATH] we have [MATH] and [MATH] If [MATH] then [MATH] implies [MATH] and [MATH] If [MATH] then by Lemma 3.4 we have [MATH] an... |
To prove the existence of an announced key region of [MATH] for [MATH] we, firstly, define the following subsets and operation containers and, secondly, show how they are to composed to a corresponding region: |
(1) [MATH] (2) [MATH] (3) [MATH] (4) [MATH] (5) for [MATH] [MATH] (6) [MATH] [MATH] [MATH] [MATH] For [MATH] the set [MATH] allows the signature [MATH] , where for [MATH] we have that |
[EQUATION] Clearly, the region [MATH] inhibits [MATH] at [MATH] and satisfies the condition of the lemma, cf. Figure Hence, the first claim is proven. |
): Let [MATH] By definition of [MATH] clearly we have either [MATH] [MATH] and [MATH] or [MATH] [MATH] and [MATH] Both cases easily imply [MATH] and, consequently, [MATH] Similarly, we obtain [MATH] By [MATH] , we easily get [MATH] if and only [MATH] which by [MATH] implies [MATH] , too. Now, if [MATH] , then, by [MATH... |
To prove the existence of an announced key region of [MATH] for [MATH] we, firstly, define the following subsets and, secondly, show how they are to composed to a corresponding region: |
(1) [MATH] and [MATH] (2) [MATH] and [MATH] (3) [MATH] and [MATH] (4) [MATH] and [MATH] (5) for [MATH] [MATH] For [MATH] the set [MATH] allows the signature [MATH] , where for [MATH] we have that [MATH] |
[EQUATION] Clearly, the region [MATH] inhibits [MATH] at [MATH] and satisfies the condition of the lemma, cf. Figure Hence, the lemma is proven. |
8. Concluding the ESSP and the SSP from a Key Region In this section we show for [MATH] and [MATH] that the inhibition of the key event at the key state in [MATH] by a [MATH] -region implies the ESSP and the SSP for [MATH] with respect to [MATH] In our reduction, events of the same kind are numbered from [MATH] up to [... |
8.1. Concluding the ESSP and the SSP for [MATH] In this section, we show for [MATH] that [MATH] has the (E)SSP if [MATH] is inhibitable at [MATH] in [MATH] Our approach for the ESSP is as follows: Let [MATH] be a state and [MATH] be an event such that [MATH] We present a subset of states [MATH] such that for [MATH] the... |
(1) [MATH] : Here the set of events are presented which are inhibited at the target states by the current region. (2) Support: Here the set [MATH] is presented. |
(3) Target States: Here a set of states [MATH] is presented, such that for [MATH] all events of [MATH] are inhibited at [MATH] by the appropriate region with the support [MATH] Clearly, [MATH] |
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