Lecture Notes in Computer Science 3472

264 Verena Wolf
(8) (⊑BC , ⊑CH ):Weshowherethat
M1 ⊑BC M2 =⇒ φem(M1) ⊑CH φem(M2)
holds which has not been considered by Bernardo and Cleaveland [BC00]
or Christoff [Chr90]. Assume M1 ⊑BC M2. We have to show that for each
T ∈T np,re , α ∈ Act ∗
Pr trace
trace
φem (M1)�T (α) � Prφem (M2)�T (α).
For a given T we set T ′ =(ST , →, sT ) ∈T pa
τ
t a −→T t ′ iff t (a,0)
−−−→ t ′ for a ∈ Act.
where → is such that
For a given trace α in T there is only one single prefix β1 of a path β =
β1β2 ∈ Path(T )withtrace(β1) =α. Itiseasytoverifythat
Pr trace
M1�T ′(α) =Pr trace
φem(M1)�T (α).
So if the last state of the sequence β1 is the only success state in T ′ ,wecan
derive
Pr trace
M1�T ′(α) =W �x
M1�T ′ � W �x
M2�T ′ = Pr trace
M2�T ′(α).
Hence we have φem(M1) ⊑CH φem(M2).
⊑BC is strictly finer than ⊑CH which can be seen by a simple counterexample
similar to example 9.7.2 because we can apply test processes with
τ-transitions.
(9) (⊑BC , ⊑CL) : To relate the Markovian testing relation ⊑BC and the fully
probabilistic testing relation ⊑CL, we have to consider the sets of applied test
processes. For ⊑BC some kind of ”passive” test processes are used because
external actions must have rate zero. Only τ-transitions have nonzero rates.
For ⊑CL all transitions in test processes are equipped with probabilities.
Since probabilistic test processes have more distinguishing power than nonprobabilistic
test processes (see relationship (6)), we have
M1 ⊑BC M2 �=⇒ φem(M1) ⊑CL φem(M2)
in general. It is easy to construct a counterexample (similar to Example 6).
Of course, the converse of the statement is also wrong, because of the additional
requirement on the expected duration of a successful path for ⊑BC .
So ⊑BC and ⊑CL are incomparable. Bernardo and Cleaveland defined ⊑CL
in a restrictive way such that the statement holds [BC00].
Figure 9.14 shows a diagram of the relationships discussed above.