Appendix A. Formal Proofs

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1 Distributed Reasoning for Multiagent Simple Temporal Problems Appendix A Formal Proofs Throughout this paper, we provided proof sketches to convey the gist of the proof when presenting the full proof would break flow of the prose In this Appendix, we provide the formal proofs for all such Theorems A1 Preliminaries In this section, we provide present definitions and lemmas that will be useful in our proofs of correctness for the D DPC, and D PPC algorithms We begin by defining a key relation about elimination orderings Once we have defined this relationship, we will prove some properties about this relationship that will be useful for proving that our algorithms correctly establish PPC Definition 1 Given a graph G = V, E and a total ordering o over V that is v x, v y V such that x y implies (v x o v y v y o v x ) let Go = V, E Eo be the graph that results from triangulating graph G by eliminating vertices in order o and adding fill edges Eo Definition 2 Given a graph G = V, E and (total) elimination order o, we define the precedence relation o, where v x o v y if and only if, at the time of v x s elimination, v y shares an edge with v x and has not been eliminated That is, v x o v y (v x o v y ) (e, e yx E Eo ) Next we label the two key algorithmic operations of the DPC and PPC algorithms, eliminate and reinstate respectively Definition 3 We label lines 3-11 of the DPC algorithm as the eliminate procedure This procedure eliminates a timepoint v k after first using edges e ik and e kj to tighten (and when necessary, to add) each edge e ij for every pair of non-eliminated, neighboring timepoints, v i and v j Definition 4 We label lines 4-8 of the PPC algorithm as the reinstate procedure This procedure reinstates a timepoint v k by, for every pair of previously-reinstated, neighboring timepoints v i and v j, tightening edge e ki and edge e jk with respect to e ij Lemma 1 Let o and o be two distinct total orderings of the vertices, V for some graph G = V, E If o is consistent with the precedence relation o, then G o = G o Proof Assume G o G o Since Go = V, E Eo and G o = V, E E o, if G o G o then E o E o Eo E o implies that there ests at least one edge e such that, either e Eo and e / E o, or e / Eo and e E o WLOG, let e Eo be the first edge that is added to Eo under elimination order o that is not added to E o under o In order for edge e to be added under elimination order o, there must be some vertex v z such that it is eliminated prior to v x and v y and shares an edge with both v x and v y (e xz and e yz respectively) at the time of its elimination By 1

2 Boerkoel & Durfee definition, this implies v z o v x and v z o v y So, under the assumption that o respects the precedence relation o, o eliminates v z prior to v x and v y Since v z is eliminated prior to v x and v y but no fill edge e is added, at least one of e xz or e is absent at the time of v z s elimination under o WLOG, assume e xz is missing If e xz is missing at the time of v z s elimination it cannot be part of the original specification of G, which implies it is a member of E o However, once v z is eliminated, no new edge e xz can ever be constructed, since fill edges are only ever added between non-eliminated vertices Thus, either e is not the first edge that is added to E o under elimination order o that is not added to E o under o, or o does not respect the precedence relation o, but both cases violate the assumptions Therefore, since every time elimination order o adds a fill edge e, it induces a new o relation, any other elimination order o that also satisfies the relation o will also add the fill edge, implying E o E o Next we prove that E o E o, which mirrors the proof that Eo E o WLOG, let e E o be the first edge that is added to E o under elimination order o that is not added to Eo under o In order for edge e to be added under elimination order o, there must be some vertex v z such that it is eliminated prior to v x and v y and shares an edge with both v x and v y (e xz and e yz respectively) at the time of v z s elimination Since e is the first edge added to E o under elimination order o that is not added to E o under o, and also since no edges can be added after the elimination of one of its endpoints, e xz must already est at the time of both v x s and v z s elimination under o However, since elimination order o does not add e, at least one of v x or v y is eliminated before v z under o WLOG assume v x is eliminated prior to v z However, since at the time of v x elimination, v x and v z share edge e xz and so by definition v x o v z This contradicts the assumption that o respects o Therefore, since the order that vertices that share edges are specified as part of o by definition, if elimination order o respects o, E o E o Since E o E o and E o E o, E o = E o which violates the assumption that G o G o since they only can differ in fill edges Therefore, if o is consistent with the precedence relation o, then G o = G o Lemma 2 Let o be a total elimination order used to triangulate STN G, resulting in graph G o and precedence relation o Any application of DPC that eliminates nodes with respect to the precedence relation o will have the same output as DPC(o, G o ) Proof By contradiction: Let G be the output of DPC and G DP C be the output of DPC Assume G G DP C By Lemma 1, G and G DP C will contain the same edges This implies for at least one edge e, w w DP C Part 1: Suppose after applying both DPC and DPC, there was an edge e, where w DP C < w and, WLOG, this was the first edge that DPC tightened further than DPC This implies that for at least one vertex v z, DPC performs the update w DP C min(w DP C, wxz DP C + wzy DP C ) and either DPC does not, or if it does, min(w DP C, wxz DP C + wzy DP C ) < min(w, wxz + wzy ) However, since DPC only performs the update w DP C min(w DP C, wxz DP C + wzy DP C ) if and only if edges ests between v x, v y, and v z and v z is eliminated before v x and v y, by definition, v z o v x and v z o v y 2

3 Distributed Reasoning for Multiagent Simple Temporal Problems Since DPC eliminates nodes with respect to the precedence relation o, DPC must eliminate v z before eliminating v x and v y, resulting in the update: w min(w, wxz + wzy ), so unless the assumption that DPC respects o is violated, DPC correctly applies the update Since DPC correctly applies the update w min(w, wxz + wzy ), the only way that w DP C w DP C, or wyz assumption that w DP C < w holds true after the update is if either w < < wyz DP C at the time the update is performed But this violates the < w is the first update performed by DPC that is never correctly performed by DPC Thus DPC will never perform an update to the bound w DP C of any edge e that will not also be applied by DPC, thus w DP C w Part 2: Suppose after applying both DPC and DPC, there ests an edge e, where < w DP C, that was the first edge that DPC tightens further than DPC This implies there must be some vertex v z such that DPC eliminates it prior to v x and v y and that shares edges with both v x and v y with tightened values wxz and wzy respectively Further, at the time of v z s elimination, DPC tightens the bound w w using the rule w min(w, Bxz + wzy ) is the first bound that DPC tightens further than DPC using elimination Since w order o and also since DPC does not tighten the bounds of any edge after it eliminates one of its endpoints, DPC will have already tightened wxz DP C by the time it eliminates either v x or v z and DPC will have already tightened wzy DP C by the time it eliminates either v y or v z Thus, if v z appears before v x and v y, DPC would apply the update w DP C min(w DP C, Bxz DP C + wzy DP C ) with wxz DP C = wxz the same update as DPC Thus, w < w DP C and wzy DP C, which is exactly, DPC must never apply the update, = w zy implying v z must appear after either v x or v y in o WLOG, assume v x appears before v z in o However, as shown in Lemma 1, at the time of the elimination of v x or v z, edge e xz must already est, since edges are never added between eliminated vertices Since v x and v z share an edge and v x appears before v z in o, v z This contradicts the assumption that the order DPC eliminates w DP C by definition v x o vertices respects o Therefore, w Conclusion: Since both w DP C w and w w DP C, then w = w DP C However this contradicts the assumption that G G DP C Therefore, the output, G, of an application of DPC will be the same as the output, G DP C, of DPC(o, G o ) if DPC eliminates nodes with respect to the precedence relation o Lemma 3 Let o be a total elimination order used to triangulate STP G, resulting in graph Go and precedence relation o Also let G = V, E be the output of DPC(o, Go ) Then the output, G P P C, of any application of the second phase of the PPC algorithm that reinstates vertices in reverse o order will be the same as the output, G P 3C, of applying P 3 C (o, (o, G )) Proof Note: When a vertex v x is reinstated, both PPC and P 3 C apply the following updates: 3

4 Boerkoel & Durfee w min(w, w + w ji ) w min(w, w + w ij ) w ix min(w ix, w ij + w jx ) w jx min(w jx, w ji + w ix ) i, j such that e, e E, where v x appears before v i and v j in o By Lemma 2, the call to DPC in line 2 will produce the same output as the call to DPC by the P 3 C algorithm By contradiction: Assume that applying P 3 C (o, (o, G )) achieves a different output than an application of PPC to G that reinstates vertices in reverse o order does Then there ests at least one pair of vertices, v x and v i, where, WLOG, v x appears before v i in o, such that w P 3 C w P P C So either w P 3 C < w P P C or w P 3 C > w P P C WLOG, let w P 3 C w P P C be the first such difference between G P 3C and G P P C Part 1: Assume that after both P 3 C and PPC are applied, w P 3 C < w P P C Thus P 3 C applies some update w P 3 C does not apply or applies when w P 3 C min(w P 3 C, w P 3 C < w P P C Notice that the only time a bound w P 3 C ij is being considered Thus, any updates to w P 3 C + w P 3 C ji ) that PPC either or w P 3 C ij ij is updated during P 3 C is when either v i or v j < w P P C, w P 3 C, or w P 3 C ij must have occurred when processing either v i or v j, both of which appear later than v x in o If e and e ij est in G, and v x appears before v i and v j in o, then by definition, v x will appear before v i and v j in o, thus PPC will also apply this update Since we assumed this was the first time P 3 C and PPC differed, neither w P 3 C < w P P C nor w P 3 C ij < w P P C ij Thus, there is a contradiction, and so w P 3 C w P P C Part 2: Assume that after both P 3 C and PPC are applied, w P P C can be true < w P 3 C Since w is the first place that P 3 C applies a different update than PPC, the difference < w P 3 C < w P 3 C ij at the time of the update w min(w, w + w ji ) Thus, PPC must apply an update that P 3 C does not apply, which can only occur in two cases Case 1: There ests some v k that appears later than v x in o such that v x and v k share an edge during PPC s execution but not P 3 C However, this violates Lemma 2 Case 2: PPC reinstates some v k that shares an edge with v x before reinstating v x, but that appears earlier than v x in o However, if v k shares an edge with v x and appears cannot occur as a result of a tighter bound w P P C before v x in o, then v x o in reverse o Therefore w P 3 C w P 3 C w P P C or w P P C ij v k, which violates the assumption that PPC reinstates vertices Conclusion: Since, for the outputs of P 3 C (o, G ) and PPC, w P 3 C w P P C, w P 3 C must equal w P P C and PPC are identical w P P C and for all x, i Therefore the outputs of P 3 C (o, G ) So far, we have defined a key precedence relation of graph triangulations, o We have shown that any elimination order o that respects this precedence relation will result in the same triangulated graph Further, we have shown that any application of the PPC 4

5 Distributed Reasoning for Multiagent Simple Temporal Problems ( DPC) algorithm that respects the precedence relation o as it eliminates and reinstates vertices and tightens bounds will calculate exactly the same PPC (DPC) STN as applying the P 3 C (DPC) algorithm using o Notice that since we proved this for each phase of the P 3 C algorithm independently, as long both phases of PPC respect o, the total order in which it reinstates vertices in the two phases can be different We must now prove that both our D PPC and our D PPC algorithms correctly apply DPC and PPC respectively to calculate a PPC STP instance A2 The D DPC Algorithm Proof of Correctness This theorem proves the correctness of the D DPC algorithm (Algorithm 7 on page 119) Note, this proof builds on definitions and properties established in Section A1 Theorem 1 D DPC correctly establishes DPC on the multiagent STP Proof Notice that the semantics of D DPC dictate that each agent i eliminates its private timepoints VP i in some order oi P before eliminating its shared timepoints V S i, which are eliminated in a globally consistent order o S Despite the fact that agents eliminate timepoints concurrently, using a fine enough granularity of time, this implies that globally, all private timepoints are eliminated in some order o D, which respects the partial order o i P i and o S WLOG, let o D = o 1 P o2 P on P o S, where appends two orderings together This proof proceeds to show that D DPC establishes DPC on G by showing that it calculates the same result as DPC(o D, G) by demonstrating that D DPC correctly applies DPC with respect to o D We begin this proof by appealing to Lemma 2 which states that any application of DPC that respects precedence relation o D achieves the same output as DPC(o D, G) We show that, despite its concurrent execution, D DPC eliminates vertices (and so applies DPC) in a way that respects precedence relation o D and therefore achieves the same output as the same out as DPC(o D, G) We do this by considering the elimination of some timepoint vx, i where vx i belongs to agent i Assume that there ests some v y such that v y o D vx i but has not been eliminated by the time agent i eliminates vx i Case 1: vx i and v y belong to the same agent i Notice v y o D vx i implies v y od vx i However, if both v y and vx i belong to agent i, they must both appear in o i P o S (constructed in lines 1 and 7) and therefore, by construction of o D, v y o i vx i This presents a contradiction since v y o i vx i is true if and only if agent i executing D DPC eliminates v y before eliminating vx, i but we assumed agent i will have not eliminated v y by the time it eliminates vx i Therefore, v y must belong to some agent j where i j Case 2: vx i is private and v y belongs to some agent j where i j Since vx i is private, by definition there can be no edge e E Further, since vx i is private, all of its neighbors are local to agent i, and since, by definition of o D, the only vertices that agent i eliminated at this point are also private, no fill edge between vx i and v y could have been added Therefore v y must belong to agent i However, we have already shown that this can never be the case, thus establishing a contradiction Therefore, if vx i is private, at the time that agent i executing D PPC eliminates vx i there can est no v y 5

6 Boerkoel & Durfee such that v y o D vx i but has not been eliminated So for the assumption to hold, vx i must be shared, which brings us to the third and final case Case 3: v i x is shared and v y belongs to some agent j where i j At this point, vx i and v y must be shared, v y must belong to some agent j, where j i, and we assume both that v y o D vx i and agent i eliminates vx i before agent j eliminates v y Because of the assumption that agent i eliminates vx i before agent j eliminates v y, the following sequence of events can never occur agent j eliminates v y agent i synchronizes its view of the MaSTP agent i eliminates vx i However before the elimination of v i x, agent i first has to obtain a lock on the shared elimination order (line 5) Thus, if v y appears before v i x, the agent i would learn of this in line 10 Line 11 would then ensure that agent i waits to receive all pertinent edge updates (wrt v y ) Thus, agent i could never eliminate v i x at the same time as, or prior to v y, if v y appears before it in o D Therefore, whether v i x is private or shared and v y belongs to agent i or some other agent j i, D DPC correctly eliminates timepoints with respect to o D, and so by Lemma 2, calculates the same output as DPC(o D, G) A3 D PPC is Deadlock Free Here we prove Theorem 5, originally stated on page 123, which establishes that the D DPC algorithm (Algorithm 5 on page 123) is deadlock free Theorem 5 D PPC is deadlock free Proof This is a continuation of the proof of Theorem 5 (page 123), which already established that line 1 is deadlock free Notice that each agent reinstates nodes in reverse o S order (line 3) By contradiction, assume line 8, which represents the only blocking communication in this algorithm, introduces a deadlock This implies that there are two (or more) agents, i and j, where i j such that both agent i and agent j are simultaneously waiting for communication from each other in line 8 Thus, there ests a timepoint vx j VX i V j L for which agent i is waiting to receive updated edges from agent j, while there is also a vy i V j X V L i for which agent j is waiting to receive updated edges from agent i Notice that vk i (the timepoint that agent i is currently considering) must appear before vj y (hence the need for blocking communication), but after vx i in agent i s copy of o S, because otherwise agent i would have already sent agent j all edge updates pertaining to vx i (line 14) in the previous loop iteration in which vx i was reinstated However, for the same reason, v j k (the timepoint that agent j is currently considering) must appear before vx i but after vy j in o S But this is a contradiction, because o S is constructed in a way that consistently and totally orders all shared timepoints This argument extends inductively to three or more agents, and so line 8 can also not be the cause of a deadlock This is a contradiction Therefore the D PPC algorithm is deadlock free 6

7 Distributed Reasoning for Multiagent Simple Temporal Problems A4 The D PPC Algorithm Proof of Correctness Here we prove Theorem 6, originally stated on page 124, which establishes the correctness of the D DPC algorithm (Algorithm 7 on page 119) Note, this proof builds on definitions and properties established in Section A1 and Theorems 5 and 1 Theorem 6 D PPC correctly establishes PPC on the MaSTN Proof Notice that the semantics of D DPC dictate that each agent i eliminates its private timepoints VP i in some order oi P before eliminating its shared timepoints V S i, which are eliminated in a globally consistent order o S Despite the fact that agents eliminate timepoints concurrently, using a fine enough granularity of time, this implies that globally, all private timepoints are eliminated in some order o D, which respects the partial order o i P i and o S WLOG, let o D = o 1 P o2 P on P o S, where appends two orderings together This proof proceeds to show that D PPC establishes PPC G by showing that it calculates the same result as P 3 C (o D, G) by demonstrating that D PPC reinstates vertices (and so correctly applies PPC) with respect to o D We start by acknowledging the proof of Theorem 1, which demonstrates that line 1 correctly establishes DPC By Lemma 3, if the reverse sweep of the PPC algorithm reinstates vx i after it reinstates v y if v x o D v y, it achieves the same out as P 3 C (o, G) We now show that, despite its concurrent execution, the last time D PPC reinstates v x is after it reinstates v y if v x o D v y By contradiction, assume agent i reinstates vx i (ie, applies line 3-14 of the D PPC Algorithm) before v y, despite the fact that vx i o D v y Case 1: vx i and v y belong to the same agent i Notice vx i o D v y implies vx i od v y However, if both v y and vx i belong to agent i, they must both appear in o i (ie, o i P o S ) and therefore, by construction of o D, vx i o i v y However, line 3 explicitly reinstates nodes in reverse o i order Therefore, v y must belong to some agent j where i j Case 2: v y V j P for some agent j i vx i o D v y implies there is an edge between vx i and v y at the time vx i is eliminated, which, by definition, implies v y is not private Further, if v y is private, Theorem 1 states agent j can reason over it independently of agent i Thus, either way we have a contradiction, thus v y cannot be private to some other agent j Case 3: v y VX i, that is v y VL i for some agent j i In this case, vx i cannot be private, since vx i o D v y implies that there ests an edge connecting vx i to a node belonging to another agent Further, if vx i were private, Theorem 1 states agent i can reason over it independently of agent j So, by definition, e belongs to EX i Thus, agent i would be explicitly forced to block in line 8, until receiving edge updates w zy, w yz v z ste xz E i, which can only occur after v y has been reinstated, updates calculated by agent j in lines 9-12, and edge update sent to agent i in line 14 Hence, all three cases present contradictions, implying that it is impossible for agent i to reinstate vx i prior to v y when vx i o D v y, 7

8 Boerkoel & Durfee Conclusion: Hence we have shown that D PPC either achieves the same output as applying DPC and PPC with respect to o, and thus establishes PPC on G that is equivalent to P 3 C (o D, G) A5 The MaTD Algorithm Proof of Completeness Here we prove Theorem 10, originally stated on page 135, which establishes the completeness of the MaTD algorithm (Algorithm 9 on page 134) Theorem 10 The MaTD algorithm is complete Proof The basic intuition for this proof is provided by the fact that the MaTD algorithm is simply a more general, distributed version of the basic backtrack-free assignment procedure that can be consistently applied to a DPC distance graph We show that when we choose bounds for new, unary decoupling constraints for v k (effectively in line 12), w zk, w kz are path consistent with respect to all other variables This is because not only is the distance graph DPC, but also the updates in lines guarantee that w zk, w kz are path consistent with respect to v k for all j > k (since each such path from v j to v k will be represented as an edge e jk in the distance graph) So the only proactive edge tightening that occurs, which happens in line 12 and guarantees that w zk + w kz = 0, is done on path-consistent edges and thus will never introduce a negative cycle (or empty domain) Fact 1: After lines 1-2 of the MaTD algorithm (Algorithm 9; page 134), if no decoupling ests, line 2 is guaranteed to terminate the algorithm by returning inconsistent, since, by definition, any consistent MaSTP has at least one solution schedule, which is a de facto temporal decoupling Fact 2: Lines 1-2 of the MaTD algorithm establish DPC, which implies that for every (external) timepoint variable v k, the weights of all edges involving v k (including, in particular, the weights of e zk, w zk and w kz ), are directionally path consistent with respect to all variables v j such that v j appears before v k in o S Now we will show by induction for every external timepoint v k that the decoupling bounds computed in line 12 and constructed in line 14 are path consistent with respect to every other variable v j where j > k in o S (Part 1) and form a non-empty domain (that is b zk + b kz 0) (Part 2) Base case(k = n): The base case is trivial, since when k = n there est no v j such that j > k Thus upon entering line 12, w zk and w kz are path consistent with respect to every variable v j where j k (Fact 2) Also, since line 2 returns inconsistent if the problem instance is, this guarantees that w zk + w kz 0 (Fact 1) In line 12, the incoming weights w zk and w kz are either left unchanged or tightened, but not beyond w zk + w kz 0 Thus the bounds constructed in line 14 are path consistent Inductive case(k < n): Assume that the bounds of all decoupling constraints chosen for all variables v j for j = k + 1,, n are partially path consistent, that is w jz + w zj 0, w jz w jx + w xz, and w zj w zx + w for all x j 8

9 Distributed Reasoning for Multiagent Simple Temporal Problems Part 1: Here we show that the bounds of the decoupling constraints computed in line 12 and constructed in line 14 are as least as tight as the tightest esting path between v k and z By contradiction, assume there ests some timepoint v j where j > k in o S such that WLOG w kz > w kj + w jz Note that since DPC is established in lines 1-2, any path from v j to v k will be represented as an edge e jk with path consistent weights w jk, w kj in the distance graph (Fact 1) Notice also that if v j is local to the agent of v k, then the update in line 12 ensures that w kz w kj + w jz, thus v j must be external to the agent of v k However, then the update in line 10 ensures that w zk w jk w jz Since we inductively assumed that w jz and w zj were chosen to be path consistent, w zj w jz So the update in line 11 implies w kz w kj w zj w kj + w jz Thus there is a contradiction since we have shown that lines 10,12 (and for w zk, lines 11,12) ensure that w kz and w zk represent the tightest path between v k and z coming into line 12, which only further tightens w kz and w zk (if at all) Thus the bounds chosen in line 14 are guaranteed to be at least as tight as any esting path between v k and z Part 2 Here we show that the bounds of the decoupling constraints constructed in line 14 form a non-empty domain By contradiction, assume that w kz + w zk < 0 Once DPC is established in lines 1-2, (at which point inconsistent is returned for any input distance graphs with negative cycles), w zk and w kz are tightened in lines 9,10-11, and 12 However, notice that line 12 guarantees that w zk + w kz 0 and lines simply recovers path consistency with respect to any local variable v j where j > k in o S, which is guaranteed to be path consistent based on the inductive assumption This implies that line 9 introduces the negative cycle That is, there ests some v x and v y such that w zk = w xk w xz and w kz = w kx w zx and where x, y > k in o S and v x,v y are external to the agent of v k, which together implies w xk w xz + w ky w zy < 0 Then, if x = y, w xk w xz + w ky w zy < 0 (1) w zx + w xk + w kx + w xz < 0 (2) w xk + w kx < 0 (3) where (2) holds by simple replacement (x = y), and (3) holds inductively (since w xz +w zx = 0) However, (3) is a contradiction, since the only time e xk will have been updated is during the DPC, which for this case would have returned inconsistent So x y WLOG, let v x appear before v y in o S Then, w xk w xz + w ky w zy < 0 (4) w zx + w xk + w ky + w yz < 0 (5) w zx + w + w yz < 0 (6) w zx + w xz < 0 (7) where (5) holds inductively (since w xz + w zx = 0; w yz + w zy = 0), (6) holds since DPC is established in lines 1-2 (since w w xk + w ky ), and (7) holds since x < y in o S, and thus line 10 (depending on whether e is external or not) ensures w xz w w zy = w + w yz However, (7) is an obvious contradiction Thus, the decoupling bounds chosen for v k are guaranteed to form a non-empty domain 9

10 Boerkoel & Durfee Therefore we have shown inductively that the decoupling bounds chosen for v k are at least as tight as the tightest possible path between v k and z and always form a non-empty domain Thus, Algorithm 9 always finds a temporal decoupling of a MaSTN, if one ests A6 The MaTDR Proof of Minimal Decoupling Here we prove Theorem 12, originally stated on page 138, which establishes that the constraints that the MaTDR algorithm (Algorithm 10 on page 137) generates form a minimal temporal decoupling Theorem 12 The local constraints calculated by the MaTDR algorithm form a minimal temporal decoupling of S Proof Notice, that the MaTDR subroutine is only called if the input network is consistent (and a valid decoupling has been found) We prove by contradiction that if any bound on an edge in C is relaxed, C may no longer form a temporal decoupling of G Assume there ests a bound of an edge in C that can be relaxed such that C still forms a decoupling of G WLOG, let δ xz be the bound on edge e xz that can be relaxed by some positive value ɛ xz and still form a temporal decoupling of G Notice that during the execution of the MaTDR, δ xz is updated exclusively in line 9, and WLOG, let the loop where j = y be the last time that δ xz is updated, that is, δ xz < w δ zy Then after line 9 is executed, δ xz = w δ zy If v y appears before v x in o, then δ zy will have already been updated (prior to δ xz due to line 8) But this leads to a contradiction, since δ xz +δ zy = w implies that δ xz +ɛ xz +δ zy > w since ɛ xz is positive, and thus a bound of δ xz + ɛ xz would no longer imply that e will be satisfied Thus, v y must appear after v x in o Let δzy IN and δzy OUT be the input and output values of δ zy respectively Then, as already shown δ xz + δzy IN = w and by our assumption δ xz + ɛ xz + δzy OUT w, which implies δzy OUT δzy IN ɛ xz For this to be true, there must est some timepoint v w such that w x, w appears before y in o, and δzy OUT = w wy δ wz Then, w wy δ wz δzy IN ɛ xz However, line 9 would have guaranteed that δ wz w wy δzy IN and so δzy IN w wy δ wz, which leads to the contradiction δzy IN δzy IN ɛ xz Therefore, if any bound on any edge in C is relaxed, C may no longer by a decoupling of G In other words, if we relaxed a bound of some edge in C, the bound on some other edge in C must be tightened to guarantee that C decouples G 10

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