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3.5
Probabilistic Bisimulation
With a solid base of the lifting operation, we can proceed to define a probabilistic
version of bisimulation.
Let
s
and
t
be two states in a pLTS. We say
t
can simulate the behaviour of
s
if
whenever the latter can exhibit action
a
and lead to distribution
ʔ
then the former can
also perform
a
and lead to a distribution, say
ʘ
, which then in turn can mimic
ʔ
in
successor states. We are interested in a relation between two states, but it is expressed
by invoking a relation between two distributions. To formalise the mimicking of one
distribution by the other, we make use of the lifting operation investigated in Sect.
3.3
.
Definition 3.5
A relation
R
ↆ
S
×
S
is a
probabilistic simulation
if
s
R
t
implies
a
−ₒ
a
−ₒ
• f
s
ʔ
then there exists some
ʘ
such that
t
ʘ
and
ʔ
R
†
ʘ
.
R
−
1
If both
R
and
are probabilistic simulations, then
R
is a
probabilistic bisimu-
lation
. The largest probabilistic bisimulation, denoted by
∼
, is called
probabilistic
bisimilarity
.
As in the nonprobabilistic setting, probabilistic bisimilarity can be approximated
by a family of inductively defined relations.
Definition 3.6
Let
S
be the state set of a pLTS. We define:
•
∼
0
:
=
S
×
S
•
s
∼
n
+
1
t
, for
n
≥
0, if
a
−ₒ
a
−ₒ
∼
n
)
†
ʘ
;
1. whenever
s
ʔ
, there exists some
ʘ
such that
t
ʘ
and
ʔ
(
a
−ₒ
a
−ₒ
∼
n
)
†
ʘ
.
2. whenever
t
ʘ
, there exists some
ʔ
such that
s
ʔ
and
ʔ
(
=
n
≥
0
∼
n
In general,
•
∼
ˉ
:
∼
ˉ
. However, the two relations coin-
cide when limited to
image-finite
pLTSs where for any state
s
and action
a
, the set
{
∼
is a strictly finer relation than
a
−ₒ
ʔ
∈
D
(
S
)
|
s
ʔ
}
is finite.
Proposition 3.8
On image-finite pLTSs,
∼
ˉ
coincides with
∼
.
Proof
It is trivial to show by induction that
s
∼
t
implies
s
∼
n
t
for all
n
≥
0, and
thus that
s
∼
ˉ
t
.
Now we show that
a
−ₒ
∼
ˉ
is a bisimulation. Suppose
s
∼
ˉ
t
and
s
ʔ
.By
a
−ₒ
∼
n
)
†
ʘ
n
.
Since we are considering image-finite pLTSs, there are only finitely many different
ʘ
n
's. Then for at least one of them, say
ʘ
,wehave
ʔ
(
≥
assumption for all
n
0 there exists some
ʘ
n
with
t
ʘ
n
and
ʔ
(
∼
n
)
†
ʘ
for infinitely many
different
n
's. By a straightforward induction we can show that
s
∼
n
t
implies
s
∼
m
t
for all
m
,
n
≥
∼
n
)
†
ʘ
for all
n
≥
0 with
n>m
. It follows that
ʔ
(
0. We now claim
that
∼
ˉ
)
†
ʘ
(3.12)
To see this, let
A
be any subset of the whole state space that may be countable. For any
n
≥
ʔ
(
∼
n
)
†
ʘ
we know from Theorem
3.5
(1) that
ʔ
(
A
)
0, since
ʔ
(
≤
ʘ
(
∼
n
(
A
)) .
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