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Lecture Notes in Computer Science 3472

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262 Verena Wolf<br />

Now let Q m<strong>in</strong> be the fully probabilistic process with the smallest success<br />

probability, i.e. W Q m<strong>in</strong> =m<strong>in</strong>Q ′{WQ ′ | Q ′ ∈ fully {0,1} ( ˆ Q�T )}. LetP ′′ be<br />

the process such that the equation above holds with Q ′ = Q m<strong>in</strong> .Thenwe<br />

have<br />

m<strong>in</strong>P ′ WP ′ � WP ′′ � WQm<strong>in</strong> .<br />

So ˆ P ⊑ must<br />

JY<br />

andomittedhere.<br />

ˆQ. The proof for deriv<strong>in</strong>g ˆ P ⊑ must<br />

JY<br />

ˆQ from ˆ P � must<br />

SE<br />

ˆQ is similar<br />

(3) (⊑JY , ⊑DH ) : Jonsson and Yi def<strong>in</strong>ed a test<strong>in</strong>g relation which boils down to<br />

ord<strong>in</strong>ary simulation but not to the ord<strong>in</strong>ary test<strong>in</strong>g relation, i.e.<br />

ˆP ⊑JY ˆ Q �=⇒ φnp( ˆ P) ⊑DH φnp( ˆ Q).<br />

It is easy to construct a counterexample where ˆ P ⊑JY ˆ Q and φnp( ˆ P) �⊑ may<br />

DH<br />

φnp( ˆ Q). But ⊑ must<br />

JY<br />

have relationship (1) and (2).<br />

(4) (⊑CH , ⊑DH ):Weshowthat<br />

boils down to ⊑ must<br />

DH<br />

P ⊑CH Q =⇒ φnp(P) ⊑DH φnp(Q).<br />

which is not surpris<strong>in</strong>g s<strong>in</strong>ce we<br />

Recall that for ⊑CH test processes T ∈T np,re are applied (compare Def<strong>in</strong>ition<br />

9.14) and with Theorem 9.23 also for ⊑DH . Assume P ⊑CH Q and<br />

φnp(P) may T , i.e. there exists a path α <strong>in</strong> φnp(P)�T end<strong>in</strong>g up <strong>in</strong> a success<br />

state w. S<strong>in</strong>ceT ∈T np,re ,wehavethattrace(α) =β corresponds to a s<strong>in</strong>gle<br />

path <strong>in</strong> T , i.e. there is only a s<strong>in</strong>gle path αT <strong>in</strong> T with trace(αT )=β. Fur-<br />

thermore, all paths α ′ <strong>in</strong> Q�T with trace(α ′ )=β lead to w. Together with<br />

P ⊑CH Q we derive 0 < Pr trace<br />

P�T<br />

(β) � Pr trace<br />

Q�T<br />

(β). So there is at least one<br />

successful path <strong>in</strong> Q�T with a nonzero probability and φnp(P) ⊑ may<br />

DH φnp(Q)<br />

follows. With a similar argument we can derive that φnp(P) ⊑ must<br />

DH φnp(Q).<br />

(5) (⊑CL, ⊑DH ):Itholds[CSZ92]<br />

P ⊑CL Q =⇒ φnp(P) ⊑DH φnp(Q).<br />

First recall that φnp(P) isnotequaltoφnp(φpp(P)) <strong>in</strong> general. So we cannot<br />

derive this statement from (1) and (6). We show that this statement is true<br />

as follows. First observe that P ⊑CL Q iff<br />

WP�T � p =⇒ WQ�T � p, ∀ T ∈Tfp τ , ∀ p ∈ [0, 1].<br />

Furthermore, we have that φnp(P�T ) is isomorphic to φnp(P) � φnp(T )and<br />

T np,re ⊂{T ′ ∈T np<br />

τ<br />

|∃T ∈T fp<br />

τ : φnp(T )=T ′ }.<br />

Moreover, we have that for each T ′ ∈T np<br />

τ there exists at least one test<br />

process T ∈Tfp τ with φnp(T )=T ′ . Hence, if P ⊑CL Q and φnp(P) may T<br />

then there exists a successful path <strong>in</strong> φnp(P�T )andWP�T (x )=p with p ∈

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