Algebraic Structure of Combined Traces –
Appendix
Łukasz Mikulski1
Faculty of Mathematics and Computer Science
Nicolaus Copernicus University
Toruń, Chopina 12/18, Poland
[email protected]
ˆ
Theorem 4. For a given comtrace 𝜏 , its Foata normal form is the ≤-greatest,
ˆ
and lexicographical normal form is the ≤-least, step sequence contained in 𝜏 .
ˆ
Proof. The lexicographical normal form is the ≤-least
step sequence contained
in 𝜏 directly from the definition. We need to prove that Foata canonical form is
greater than any other step sequence contained in 𝜏 .
Let 𝑢 = 𝐴1 . . . 𝐴𝑛 , 𝑣 = 𝐵1 . . . 𝐵𝑚 , 𝑢 ∕= 𝑣, 𝑢 ≡𝛩 𝑣, and 𝑢 be in Foata canonical
form. Moreover, let 𝑖 = 𝑚𝑖𝑛{𝑘∣𝑘 ≤ 𝑛 ∧ 𝐴𝑘 ∕= 𝐵𝑘 }. Note that such a number 𝑖
exists, since 𝑢 ∕= 𝑣 and 𝑢 ≡𝛩 𝑣 so one sequence cannot be a prefix of another. We
have 𝐴1 . . . 𝐴𝑘−1 = 𝐵1 . . . 𝐵𝑘−1 , so directly form the definition of Foata canonical
ˆ 𝑘 , we have 𝐵𝑘 . . . 𝐵𝑚 ≤𝐴
ˆ 𝑘 . . . 𝐴𝑛 , and 𝑣 ≤𝑢.
ˆ
form 𝐵𝑘 ∕= 𝐴𝑘 ∧𝐵𝑘 ⊆ 𝐴𝑘 . Since 𝐵𝑘 ≤𝐴
⊔
⊓
Lemma 5. For every 𝐴 ∈ 𝕊, ≡𝐴 is an equivalence relation.
Proof. Reflexivity an symmetry follow directly from definition. Let 𝑎 ≡𝐴 𝑏 and
𝑏 ≡𝐴 𝑐 which implies that {(𝑎, 𝑏), (𝑏, 𝑎), (𝑏, 𝑐), (𝑐, 𝑏)} ⊆ sin ∗ . Since the star operation is transitive, we have (𝑎, 𝑐) ∈ sin ∗ and (𝑐, 𝑎) ∈ sin ∗ . Hence 𝑎 ≡𝐴 𝑐, and
the proof is complete.
⊔
⊓
Lemma 7. Let 𝐴 ∈ 𝕊 ∖ ˆ
𝕊 be a step that is not indivisible. Then there exist two
steps 𝐵 and 𝐶 such that 𝐴 ∼𝛩 𝐵𝐶. Moreover, 𝐴/≡𝐴 = 𝐵/≡𝐵 ∪ 𝐶/≡𝐶 .
Proof. Since 𝐴 is not indivisible, the relation ≡𝐴 divides 𝐴 into at least two
equivalence classes.
Let 𝐵 be the set of all actions 𝑏 ∈ 𝐴 such that, for all 𝑎 ∈ 𝐴 (𝑏, 𝑎) ∈ sin ∗ ⇒
𝑎 ∈ [𝑏]≡𝐴 . Suppose that 𝐵 is empty. Let us take any 𝑏1 ∈ 𝐴. Then, by 𝐵 = ∅,
there exists 𝑏2 ∈ 𝐴 such that 𝑏2 ∈
/ [𝑏1 ]≡𝐴 and (𝑏1 , 𝑏2 ) ∈ sin ∗ . Continuing in this
way, we can construct an infinite sequence of actions 𝑏𝑖 ∈ 𝐴 such that, for all 𝑖,
𝑏𝑖+1 ∈
/ [𝑏1 ]≡𝐴 ∧ (𝑏𝑖 , 𝑏𝑖+1 ) ∈ sin ∗ .
Since 𝐴 is finite, the elements contained in this sequence have to repeat. Let
𝑏𝑛 = 𝑏𝑚 and 𝑛 < 𝑚. Since sin ∗ is transitive we have (𝑏𝑛+1 , 𝑏𝑛 ) ∈ sin ∗ and
(𝑏𝑚 , 𝑏𝑛+1 ) ∈ sin ∗ , so 𝑏𝑛+1 ∈ [𝑏𝑛 ]≡𝐴 which contradicts the assumption. Hence 𝐵
is not empty.
We consider two cases.
Case 1: 𝐴 ∕= 𝐵.
Let 𝐴 ∕= 𝐵, 𝑏 ∈ 𝐵 and 𝑎 ∈ 𝐴 ∖ 𝑏. From the definition of 𝐵 we have that
(𝑏, 𝑎) ∈
/ sin ∗ , so (𝑏, 𝑎) ∈
/ wdp and (𝑏, 𝑎) ∈
/ ssm. We also have (𝑎, 𝑏) ∈
/ dep since 𝑎
and 𝑏 are both contained in 𝐴. This gives
∀𝑎∈𝐴∖𝐵 ∀𝑏∈𝐵 (𝑎, 𝑏) ∈ wdp ∨ (𝑎, 𝑏) ∈ ind .
Hence
∀𝑎∈𝐴∖𝐵 ∀𝑏∈𝐵 (𝑏, 𝑎) ∈ ser
and finally 𝐴 ∼𝛩 𝐵(𝐴 ∖ 𝐵).
Case 2:𝐴 = 𝐵.
We take an arbitrary 𝑏 ∈ 𝐴. Since 𝐴 is not indivisible, we have [𝑏]≡𝐴 ∕= 𝐴.
Using similar reasoning as in the first case we get
∀𝑎∈𝐴∖[𝑏]≡𝐴 ∀𝑏′ ∈[𝑏]≡𝐴 (𝑏′ , 𝑎) ∈ ser
and finally 𝐴 ∼𝛩 [𝑏]≡𝐴 (𝐴 ∖ [𝑏]).
It remains to prove that indivisible parts of steps are truly indivisible. According to the definition of the comtrace equivalence, 𝐴 ∼𝛩 𝐵𝐶 implies that
𝐵 × 𝐶 ⊆ ser . It means that for every pair of actions 𝑏 ∈ 𝐵 and 𝑐 ∈ 𝐶 we have
(𝑏, 𝑐) ∈
/ sin. Hence for every 𝑎, 𝑏 ∈ 𝐴 we have 𝑎 ≡𝐴 𝑏 ∧ 𝑎 ∈ 𝐵 ⇒ 𝑏 ∈ 𝐵 and
𝑎 ≡𝐴 𝑏∧𝑎 ∈ 𝐶 ⇒ 𝑏 ∈ 𝐶. By the sin∣𝐴 we denote the graph of relation sin limited
to the step 𝐴, namely the vertices induced subgraph of (𝛴, sin). It also means
that graphs of the relation sin limited to steps 𝐵 and 𝐶 not only are vertices
induced parts of the graph of relation sin limited to the step 𝐴, but also are the
division of this graph. Therefore the union of strongly connected components of
graphs sin∣𝐵 and sin∣𝐶 is equal to the set of strongly connected components of
sin∣𝐴 , which end the proof.
⊔
⊓
Proposition 8. All steps contained in the lexicographical normal form of a comtrace are indivisible (𝑚𝑖𝑛𝑙𝑒𝑥(𝜏 ) ∈ 𝑖𝑛𝑑𝑖𝑣(𝜏 )).
Proof. Suppose, to the contrary, that 𝑚𝑖𝑛𝑙𝑒𝑥(𝜏 ) = 𝑢𝐴𝑣 contains a non-indivisible
step 𝐴. We conclude from Lemma 7 that for two disjoint steps 𝐵 and 𝐶 we have
a step sequence 𝑢𝐵𝐶𝑣 ∈ 𝜏 which is different from the step sequence 𝑚𝑖𝑛𝑙𝑒𝑥(𝜏 ).
ˆ 𝑢𝐴𝑣 so we found a step sequence conSince 𝐵 ⊆ 𝐴 and 𝐴 ∕= 𝐵 we have 𝑢𝐵𝐶𝑣 ≤
tained in 𝜏 that is lexicographically smaller than 𝑚𝑖𝑛𝑙𝑒𝑥(𝜏 ), which contradicts
our assumption. Hence all steps contained in 𝑚𝑖𝑛𝑙𝑒𝑥(𝜏 ) are indivisible.
⊔
⊓
Theorem 9. Let 𝜏 be a comtrace. The set 𝑖𝑛𝑑𝑖𝑣(𝜏 ) is a trace over the concurrent
ˆ
alphabet (ˆ
𝕊, dep).
Proof. To prove the statement of the theorem it is sufficient to show two facts.
ˆ is symmetric and irreflexive. Secondly,
Firstly, we need to prove that relation ind
we need to argue that by the transposing of two subsequent and independent
actions (in fact indivisible steps) we can reach any of other elements of the set
𝑖𝑛𝑑𝑖𝑣(𝜏 ) and cannot go beyond this set.
We start from the first statement. By the definition of the relation ind =
ser ∪ ser −1 is symmetric and irreflexive. Since two indivisible steps 𝐴 and 𝐵 are
ˆ if all pairs of actions (𝑎, 𝑏) ∈ 𝐴 × 𝐵 are independent, we conclude
in relation ind
ˆ is also symmetric and irreflexive.
that the relation ind
ˆ. By the
Let 𝑤 = 𝑢𝐴𝐵𝑣 be a step sequence from 𝑖𝑛𝑑𝑖𝑣𝜏 and (𝐴, 𝐵) ∈ ind
ˆ relation we have 𝐴𝐵 ∼𝛩 𝐶 ∼𝛩 𝐵𝐴. Therefore 𝑢𝐴𝐵𝑣 ≡𝛩
definition of the ind
𝑎𝐵𝐴𝑣 and the set 𝑖𝑛𝑑𝑖𝑣𝜏 is equal to its own trace closure. The last sufficient
statement follows from Lemma 7 (about indivisibility of indivisible steps).
Let us suppose that there are two comtrace equivalent step sequences 𝑢 and 𝑣
belonging to 𝑖𝑛𝑑𝑖𝑣(𝜏 ) that are not trace equivalent. Hence they differ in at least
one projection to binary dependent subalphabet so there are two occurrences
of indivisible steps 𝐴 and 𝐵 that appear in the two different orders and are
ˆ For definiteness, let 𝐴 precedes 𝐵 in the step sequence
dependent ((𝐴, 𝐵) ∈ dep).
𝑢, while 𝐵 precedes 𝐴 in the step sequence 𝑣. From the definition of comtrace
equivalence there exits a sequence of equivalent step sequences (𝑤𝑖 )𝑖=1...𝑛 such
that 𝑢 = 𝑤1 , 𝑤𝑖 ∼𝛩 𝑤𝑖+1 , and 𝑤𝑛 = 𝑣. In this sequence there have to exist
an element 𝑤𝑖 where the considered occurrences of indivisible steps was the last
′
′′
time in the same order as in 𝑢 (𝑤𝑖 = 𝑤𝑖′ 𝑋𝑖 𝑌𝑖 𝑤𝑖′′ and 𝑤𝑖+1 = 𝑤𝑖+1
𝑍𝑖 𝑤𝑖+1
and
𝐴 ⊆ 𝑋𝑖 and 𝐵 ⊆ 𝑌𝑖 ). Hence 𝐴 × 𝐵 ⊆ ser . Moreover, there exists an element 𝑤𝑗
where the considered occurrences first time after 𝑤𝑖 are in the same order as in
′
′′
𝑣 (𝑤𝑗 = 𝑤𝑗′ 𝑍𝑗 𝑤𝑗′′ and 𝑤𝑗+1 = 𝑤𝑗+1
𝑋𝑗 𝑌𝑗 𝑤𝑗+1
and 𝐴 ⊆ 𝑋𝑗 and 𝐵 ⊆ 𝑌𝑗 ). Hence
ˆ, which is impossible and completes
also 𝐵 × 𝐴 ⊆ ser . Therefore (𝐴, 𝐵) ∈ ind
the proof.
Theorem 11. Let 𝑤, 𝑢 be step sequences over comtrace alphabet 𝛩 = (𝕊, sim, ser ).
𝑤 ≡𝛩 𝑢 ⇔ ∀(𝑎,𝑏)∈ind
𝛱𝑎,𝑏 (𝑤) = 𝛱𝑎,𝑏 (𝑢).
/
Proof. ⇒:
We first prove that
𝑤 ≡𝛩 𝑢 ⇒ ∀(𝑎,𝑏)∈ind
𝛱𝑎,𝑏 (𝑤) = 𝛱𝑎,𝑏 (𝑢).
/
According to the definition of comtrace equivalence, it is sufficient to prove the
stated statement it the case of equivalent step sequences 𝑤 = 𝐴 and 𝑢 = 𝐵𝐶.
Let 𝑎, 𝑏 ∈ 𝐴. We consider all but one possible relationships of these actions (the
excepted case is naturally independence).
Case 1: (𝑎, 𝑏) ∈ dep.
Since actions 𝑎 and 𝑏 occur simultaneously in the step 𝐴 it is impossible.
Case 2: (𝑎, 𝑏) ∈ ssm.
Since actions 𝑎 and 𝑏 are strongly simultaneous, Lemma 7 shows that they
both have to occur in step 𝐵 or 𝐶. It means that
𝛱𝑎,𝑏 (𝐴) =⊥ 𝜆 = 𝜆 ⊥= 𝛱𝑎,𝑏 (𝐵)𝛱𝑎,𝑏 (𝐶) = 𝛱𝑎,𝑏 (𝐵𝐶).
Case 3: (𝑎, 𝑏) ∈ wdp ∧ 𝑎, 𝑏 ∈ 𝐵.
Since 𝐵 × 𝐶 ⊆ ser , it is impossible that 𝑏 ∈ 𝐶 and 𝑎 ∈ 𝐵. If they both belong
to one step, we have
𝛱𝑎,𝑏 (𝐴) = (𝑏𝑎)𝜆 = 𝜆(𝑏𝑎) = 𝛱𝑎,𝑏 (𝐵)𝛱𝑎,𝑏 (𝐶) = 𝛱𝑎,𝑏 (𝐵𝐶),
while belonging to the different steps (namely 𝑏 ∈ 𝐵 and 𝑎 ∈ 𝐶) gives
𝛱𝑎,𝑏 (𝐴) = 𝑏𝑎 = 𝛱𝑎,𝑏 (𝐵)𝛱𝑎,𝑏 (𝐶) = 𝛱𝑎,𝑏 (𝐵𝐶),
which completes the first part of the proof.
⇐:
Now, let us assume that we have two step sequences 𝑢, 𝑣 ∈ 𝕊∗ and
∀(𝑎,𝑏)∈ind
𝛱𝑎,𝑏 (𝑤) = 𝛱𝑎,𝑏 (𝑢).
/
Without losing any generality we can assume that 𝑢 = 𝐴𝑢′ is in the lexicographical canonical form and 𝑣 consists of indivisible steps only. We claim that
there exist such 𝑣 ′ , 𝑣 ′′ ∈ 𝕊∗ that 𝑣 = 𝑣 ′ 𝐴𝑣 ′′ . Moreover, we claim that no action
occurring in 𝐴 may occur in 𝑣 ′ and 𝐴 × 𝑎𝑙𝑝ℎ(𝑣 ′ ) ⊆ ind .
Directly from the definition of the projection representation we see that all
projections to the subalphabets containing letters from the indivisible step 𝐴
starts with the letters contained in 𝐴. More precisely, if both 𝑎, 𝑏 ∈ 𝐴 then
𝛱𝑎,𝑏 (𝑢) starts with 𝑎𝑏, 𝑏𝑎 or ⊥, depending on the relation between 𝑎 and 𝑏. If
𝑎 ∈ 𝐴 but 𝑏 ∈
/ 𝐴 however, 𝛱𝑎,𝑏 (𝑢) starts with a single letter 𝑎.
Let 𝑣 ′ be the longest prefix of 𝑣 such that 𝑎𝑙𝑝ℎ(𝑣 ′ ) ∩ 𝐴 = ∅ and 𝑣 = 𝑣 ′ 𝐵𝑣 ′′ .
Obviously, all projections to the subalphabets containing letters from the step
𝐴 are equal for 𝑣 and 𝐵𝑣 ′′ . Moreover, from the definition of the indivisible step,
between every two letters 𝑎, 𝑏 contained in 𝐴 there is a sequence of pairwise
different letters 𝑎 = 𝑎1 , . . . , 𝑎𝑛 = 𝑏 contained in 𝐴 such that for every 𝑖 < 𝑛
we have (𝑎𝑖+1 , 𝑎𝑖 ) ∈ sin. It means that for every such a pair of consecutive
letters we have 𝛱𝑎𝑖 ,𝑎𝑖+1 (𝐵) = 𝑎𝑖 𝑎𝑖+1 if (𝑎𝑖+1 , 𝑎𝑖 ) ∈ wdp or 𝛱𝑎𝑖 ,𝑎𝑖+1 (𝐵) =⊥
if (𝑎𝑖+1 , 𝑎𝑖 ) ∈ ssm. Nevertheless, if 𝑎𝑖+1 is in 𝐵 then also 𝑎𝑖 have to be in 𝐵.
Otherwise 𝛱𝑎𝑖 ,𝑎𝑖+1 (𝐵) would start with 𝑎𝑖+1 . This proves that, since 𝐴 ∩ 𝐵 ∕= ∅,
𝐴 ⊆ 𝐵. Using the same arguments, we see that since 𝐵 is indivisible, no other
letter may occur in 𝐵 and 𝐴 = 𝐵.
It remains to show that 𝐴 × 𝑎𝑙𝑝ℎ(𝑣 ′ ) ⊆ ind . Let 𝑎 ∈ 𝐴 and 𝑐 ∈ 𝑎𝑙𝑝ℎ(𝑣 ′ ).
Naturally, 𝑐 ∈
/ 𝐴 from the definition of sequence 𝑣 ′ . In the step sequence 𝑣 the
letter 𝑐 appears before letter 𝑎 so, if the are not independent, 𝛱𝑎,𝑐 (𝑣) = 𝛱𝑎,𝑐 (𝑢)
starts with 𝑐. But 𝑎 ∈ 𝐴 and 𝑐 ∈
/ 𝐴 so 𝛱𝑎,𝑐 (𝑢) starts with 𝑎. This contradicts our
assumption that 𝑎 and 𝑐 may be not independent and proves that 𝑣 ≡𝛩 𝐴𝑣 ′ 𝑣 ′′ .
Repeating above reasoning, we achieve that 𝑢 is the lexicographical canonical
form of 𝑣 which ends the second part of the proof.
⊔
⊓
Proposition 12. The procedure of computing 𝛱𝜏 from a step sequence 𝑤 ∈ 𝜏
has the time and memory complexities of 𝑂(𝑛𝑘).
Proof. The proof is straightforward. The algorithm is naturally divided into 𝑛
stages grouped by steps of input step sequence. In each stage we process single
action and add it to at most 𝑘 sequences updating at most 𝑘 counters. Hence
each stage can be done in the constant time. Therefore whole procedure has time
complexity of 𝑂(𝑛𝑘).
⊔
⊓
Theorem 13. Testing comtrace equivalence can be done in the time complexity
of 𝑂(𝑛𝑘).
Proof. Notice that output of described procedure has also memory complexity
of 𝑂(𝑛𝑘). Hence for having two step sequences we can compute their projection
representations and compare them sequence by sequence.
⊔
⊓
Theorem 15. Let 𝛱𝜏 be the projection representation of a comtrace 𝜏 , and
𝑀 (𝛱) be a maximal possible step of 𝛱𝜏 . For every allowed set 𝐵 ∈ 𝕊, we have
𝜏 = 𝐵 ∘ 𝜎, where 𝛱𝜎 = 𝑒𝑥𝑡𝑟(𝛱𝜏 , 𝐵).
Proof. By the Theorem 11 it is sufficient to prove that 𝛱𝜏 = 𝛱𝐵∘𝜎 . In other
words we have to show that for all (𝑎, 𝑏) ∈
/ ind we have 𝛱𝜏 (𝑎, 𝑏) = 𝛱𝐵 (𝑎, 𝑏) ∘
𝛱𝜎 (𝑎, 𝑏).
The proof falls naturally into three parts, depending on the type of relation
between the considered actions. Let us examine the projections onto pairs of
fully dependent letters ((𝑎, 𝑏) ∈ dep). We have 𝛱𝐵 (𝑎, 𝑏) equal to the first letter
of 𝛱𝜏 (𝑎, 𝑏) if ∣𝐵 ∩ {𝑎, 𝑏}∣ = 1 and 𝛱𝐵 (𝑎, 𝑏) equal to 𝜆 otherwise. In both cases
𝛱𝐵 (𝑎, 𝑏) ∘ 𝛱𝜎 (𝑎, 𝑏) = 𝛱𝜏 (𝑎, 𝑏).
Almost the same proof works for the remaining two cases, when (𝑎, 𝑏) ∈ wdp
or (𝑎, 𝑏) ∈ ssm.
⊔
⊓
Proposition 16. Projection set 𝛱 is a projection representation of a comtrace
if and only if the described procedure ends with an empty projection set.
Proof. We give the proof only for the case when maximal strategy is used. Note
that the input data is finite and procedure stops when the set 𝑀 (𝛱) is empty
for the remaining set of words. From Theorem 15 we deduce that if the remaining projection set is empty then the input is a projection representation of a
constructed comtrace.
Suppose that we have nonempty projection set 𝛱 that is a projection representation of comtrace 𝜏 and empty set of allowed actions. Let us consider an
arbitrary step sequence 𝑢 = 𝐴1 . . . 𝐴𝑛 that is contained in 𝜏 and arbitrary action 𝑎 contained in 𝐴1 . Then, by the definition of projection representation, the
action 𝑎 have to be possibly allowed. This proves that 𝐴1 ⊆ 𝑐𝑝𝑎. Moreover, since
in any projection before or simultaneously with 𝑎 may occur only other action
from the step 𝐴1 , if existence of action 𝑏 is the necessary condition for presence
of action 𝑎 ((𝑎, 𝑏) ∈ 𝑐𝑛𝑑) then 𝑏 is also an element of 𝐴1 . Therefore none of the
action from step 𝐴1 is impossible, which contradicts the emptiness of the set of
allowed actions and ends the proof.
⊔
⊓
Theorem 17. The procedure of computing canonical forms from a projection
representation of a comtrace has the time complexity of 𝑂(𝑛𝑘 2 ).
Proof. The procedure consist of at most 𝑛 parts. In each part we make some operations on at most 𝑘 2 lists and graph of size 𝑘 2 . All graph operations, including
computing the compensation graph and choosing minimal or maximal filter are
linear in the size of graph. This gives overall time complexity of 𝑂(𝑛𝑘 2 ).
⊔
⊓
Proposition 18. Let 𝜏 ∈ 𝕊∗ be a trace-like comtrace and 𝑤 ∈ 𝑖𝑛𝑑𝑖𝑣(𝜏 ). Then
each step of 𝑤 is a singleton.
Proof. The proof is straightforward. Notice that since the relation sin is empty,
every action 𝑎 of every step 𝐴 ∈ 𝕊 forms an indivisible step. Hence all indivisible
steps are singletons, which ends the proof.
⊔
⊓
Theorem 20. Let 𝜏 be a trace-like comtrace and 𝜎 be its trace representation.
Then 𝛱𝜏 = 𝛱𝜎 .
Proof. Let 𝜏 = 𝐴1 . . . 𝐴𝑛 . Notice that the sets of binary and unary dependent
subalphabets are equal in both cases. We claim that only actions from the different steps may appear in the same projection. Indeed, since the pair of action can
only be dependent or independent there is no other possibility. Therefore, the
projection of a step contained in trace-like comtrace is a single action or equals 𝜆.
If an action 𝑎 is contained in the step 𝐴1 , we have 𝛱𝑎,𝑏 (𝐴1 ) = 𝑎 = 𝛱𝑎,𝑏 (𝑙𝑒𝑥(𝐴1 ))
for every 𝑏 dependent with 𝑎. In the remaining situations, when neither 𝑎 nor 𝑏
is contained in 𝐴1 , we have 𝛱𝑎,𝑏 (𝐴1 ) = 𝜆 = 𝛱𝑎,𝑏 (𝑙𝑒𝑥(𝐴1 )), which completes the
proof.
⊔
⊓
Theorem 21. Let 𝜏 ∈ 𝕊∗ be a trace-like comtrace and 𝜎 be its trace representation. Then the linearizations of the Foata and the lexicographical canonical
forms of 𝜏 are equal to the Foata and the lexicographical canonical forms of 𝜎,
respectively.
Proof. Let us start with the Foata canonical forms. Let 𝑢 = 𝐴1 . . . 𝐴𝑛 be the
Foata canonical form of comtrace 𝜏 . Notice that since for every 𝑖 < 𝑛, there
is no ∅ ∕= 𝐴 ⊆ 𝐴𝑖+1 such that 𝐴𝑖 × 𝐴 ⊆ ser and 𝐴 × (𝐴𝑖+1 ∖𝐴) ⊆ ser , for
every action 𝑎 ∈ 𝐴𝑖+1 there exists an action 𝑏 ∈ 𝐴𝑖 such that (𝑎, 𝑏) ∈
/ ser .
Hence for each 𝑖 < 𝑛 and 𝑎 occurring in 𝐴𝑖+1 , there is 𝑏 occurring in 𝐴𝑖 such
that (𝑎, 𝑏) ∈ dep. Moreover, all actions belonging to the same step are pairwise
independent. Noticing that linearization of a step 𝐴 give minimal with respect to
the lexicographical order sequence builded with actions present in 𝐴 completes
the first part of the proof.
Let us now consider the lexicographical canonical form of the comtrace 𝜏 .
ˆ
Since singletons are always ≤-smaller
than any other nonempty step, it is obvious
that every step of the lexicographical canonical form of trace-like comtrace is a
ˆ
singleton. Moreover, in the case singletons (𝑎)≤(𝑏)
if and only if 𝑎 ≤ 𝑏, so the
linearization of the lexicographical canonical form of 𝜏 is the ≤-smaller word
among the linearizations of all representatives of 𝜏 that are builded only with
singletons. This completes the proof.
⊔
⊓