Fixed-point Characterization of Compositionality Properties of Probabilistic Processes Combinators
Daniel Gebler1, Simone Tini2
1VU University Amsterdam, The Netherlands 2University of Insubria, Italy
EXPRESS/SOS workshop September 1, 2014
1 / 18
Fixed-point Characterization of Compositionality Properties of - - PowerPoint PPT Presentation
Fixed-point Characterization of Compositionality Properties of Probabilistic Processes Combinators EXPRESS/SOS workshop September 1, 2014 1 / 18 Daniel Gebler 1 , Simone Tini 2 1 VU University Amsterdam, The Netherlands 2 University of Insubria,
1VU University Amsterdam, The Netherlands 2University of Insubria, Italy
1 / 18
1
n
1, . . . , t′ n)
n
f n
f.
2 / 18
1
n
1, . . . , t′ n)
1) ≤ ϵ1
n) ≤ ϵn
1, . . . , t′ n)) ≤ ωf(ϵ1, . . . , ϵn)
2 / 18
1
n
1, . . . , t′ n)
1) ≤ ϵ1
n) ≤ ϵn
1, . . . , t′ n)) ≤ ωf(ϵ1, . . . , ϵn)
2 / 18
1
n
1, . . . , t′ n)
1) ≤ ϵ1
n) ≤ ϵn
1, . . . , t′ n)) ≤ ωf(ϵ1, . . . , ϵn)
2 / 18
3 / 18
a
0.6
0.4
b
c
1.0
1.0
a
0.6 − ϵ
0.4 + ϵ
b
c
1.0
1.0 4 / 18
a
0.6
0.4
b
c
1.0
1.0
a
0.6 − ϵ
0.4 + ϵ
b
c
1.0
1.0 4 / 18
a
0.6
0.4
b
c
1.0
1.0
a
0.6 − ϵ
0.4 + ϵ
b
c
1.0
1.0 4 / 18
5 / 18
5 / 18
3
a
1/3
2/3
b
0.6
0.4
c
d
1.0
1.0
3
a
1/3
2/3
b
0.6 − ϵ
0.4 + ϵ
c
d
1.0
1.0
3, q′ 3) = 0
3) = 1
3, q3) = 1
6 / 18
3
a
1/3
2/3
b
0.6
0.4
c
d
1.0
1.0
3
a
1/3
2/3
b
0.6 − ϵ
0.4 + ϵ
c
d
1.0
1.0
3, q′ 3) = 0
3) = 1
3, q3) = 1
6 / 18
3
a
1/3
2/3
b
0.6
0.4
c
d
1.0
1.0
3
a
1/3
2/3
b
0.6 − ϵ
0.4 + ϵ
c
d
1.0
1.0
3, q′ 3) = 0
3) = 1
3, q3) = 1
6 / 18
1
2
i for i
7 / 18
1
2
i for i
7 / 18
1
2
i for i
7 / 18
1
2
i) ≤ δi for i = 1, 2
1, t′ 2)) ≤ ϵ
7 / 18
ai,m
bj,n
a
i I pi i if i are distribution terms and pi
i I pi
r f
i are distribution terms and f
r f
i, that is for closed substitution distribution
r f state term
r f i i
8 / 18
ai,m
bj,n
a
i∈I piθi if θi are distribution terms and pi ∈ (0, 1] with ∑ i∈I pi = 1
r f
i, that is for closed substitution distribution
r f state term
r f i i
8 / 18
ai,m
bj,n
a
i∈I piθi if θi are distribution terms and pi ∈ (0, 1] with ∑ i∈I pi = 1
r f
i, that is for closed substitution distribution
r f state term
r f i i
8 / 18
ai,m
bj,n
a
i∈I piθi if θi are distribution terms and pi ∈ (0, 1] with ∑ i∈I pi = 1
distribution
state term
r(f)
i=1
8 / 18
a
0.4
0.6
a
0.3
0.7
a
a
a
a
0.12
0.28
0.18
0.42 9 / 18
a
0.4
0.6
a
0.3
0.7
a
a
a
a
0.12
0.28
0.18
0.42 9 / 18
a
0.4
0.6
a
0.3
0.7
a
a
a
a
0.12
0.28
0.18
0.42 9 / 18
n
i=1 ϵi
i=1
i
n
i=1
n
i=1
10 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
a
1.0
b
1.0
a
(1.0 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
b
1.0
11 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
a
1.0
b
1.0
a
(1.0 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
b
1.0
11 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
a
1.0
b
1.0
a
(1.0 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
b
1.0
11 / 18
1
2
a
r
1 − r
b
c
√
√
a
a
a
a
12 / 18
1
2
a
r
1 − r
b
c
√
√
a
a
a
a
12 / 18
1
2
a
r
1 − r
b
c
√
√
a
a
a
a
12 / 18
1
2
a
r
1 − r
b
c
√
√
a
a
a
a
12 / 18
1
2
a
r
1 − r
b
c
√
√
a
a
a
a
12 / 18
13 / 18
13 / 18
13 / 18
a
a
a
a
a
a
f
a
a
a
f
a
a
a
f
a
a
a
f
14 / 18
a
a
a
a
a
a
a
a
a
f
a
a
a
f
a
a
a
f
14 / 18
a
a
a
a
a
a
f
a
a
a
a
a
a
f
a
a
a
f
14 / 18
a
a
a
a
a
a
f
a
a
a
f
a
a
a
a
a
a
f
14 / 18
a
a
a
a
a
a
f
a
a
a
f
a
a
a
f
a
a
a
14 / 18
x∈V
m∈M
p∈P
15 / 18
x∈V
m∈M
p∈P
15 / 18
x∈V
m∈M
p∈P
15 / 18
x∈V
m∈M
p∈P
15 / 18
x∈V
m∈M
p∈P
15 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
a
a
1.0
a
1.0
b
1.0
a
1.0
a
(1.0 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
b
1.0
d(g(P),g(Q))=1−(1−ϵ)2
d(f(P),f(Q))=1−(1−ϵ)2
16 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
a
a
1.0
a
1.0
b
1.0
a
1.0
a
(1.0 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
b
1.0
d(g(P),g(Q))=1−(1−ϵ)2
d(f(P),f(Q))=1−(1−ϵ)2
16 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
a
a
1.0
a
1.0
b
1.0
a
1.0
a
(1.0 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
b
1.0
d(g(P),g(Q))=1−(1−ϵ)2
d(f(P),f(Q))=1−(1−ϵ)2
16 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
a
a
1.0
a
1.0
b
1.0
a
1.0
a
(1.0 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
b
1.0
d(g(P),g(Q))=1−(1−ϵ)2
d(f(P),f(Q))=1−(1−ϵ)2
16 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
a
a
1.0
a
1.0
b
1.0
a
1.0
a
(1.0 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
b
1.0
d(g(P),g(Q))=1−(1−ϵ)2
d(f(P),f(Q))=1−(1−ϵ)2
16 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
a
a
1.0
a
1.0
b
1.0
a
1.0
a
(1.0 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
b
1.0
d(g(P),g(Q))=1−(1−ϵ)2
d(f(P),f(Q))=1−(1−ϵ)2
16 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
a
a
1.0
a
1.0
b
1.0
a
1.0
a
(1.0 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
b
1.0
d(g(P),g(Q))=1−(1−ϵ)2
d(f(P),f(Q))=1−(1−ϵ)2
16 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
b
b
a
a
1.0
a
1.0
a
(1 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
a
1.0
mg(x1)=mg(x2)=0
d(g(P′,P′),g(Q′,Q′))=0
sup(mg,1{x1,x2})(x1)=1
mf(x)=∑2
i=1 sup(mg,1{x1,x2})(xi)=2
d(f(P),f(Q))=1−(1−ϵ)2
17 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
b
b
a
a
1.0
a
1.0
a
(1 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
a
1.0
mg(x1)=mg(x2)=0
d(g(P′,P′),g(Q′,Q′))=0
sup(mg,1{x1,x2})(x1)=1
mf(x)=∑2
i=1 sup(mg,1{x1,x2})(xi)=2
d(f(P),f(Q))=1−(1−ϵ)2
17 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
b
b
a
a
1.0
a
1.0
a
(1 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
a
1.0
mg(x1)=mg(x2)=0
d(g(P′,P′),g(Q′,Q′))=0
sup(mg,1{x1,x2})(x1)=1
mf(x)=∑2
i=1 sup(mg,1{x1,x2})(xi)=2
d(f(P),f(Q))=1−(1−ϵ)2
17 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
b
b
a
a
1.0
a
1.0
a
(1 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
a
1.0
mg(x1)=mg(x2)=0
d(g(P′,P′),g(Q′,Q′))=0
sup(mg,1{x1,x2})(x1)=1
mf(x)=∑2
i=1 sup(mg,1{x1,x2})(xi)=2
d(f(P),f(Q))=1−(1−ϵ)2
17 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
b
b
a
a
1.0
a
1.0
a
(1 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
a
1.0
mg(x1)=mg(x2)=0
d(g(P′,P′),g(Q′,Q′))=0
sup(mg,1{x1,x2})(x1)=1
mf(x)=∑2
i=1 sup(mg,1{x1,x2})(xi)=2
d(f(P),f(Q))=1−(1−ϵ)2
17 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
b
b
a
a
1.0
a
1.0
a
(1 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
a
1.0
mg(x1)=mg(x2)=0
d(g(P′,P′),g(Q′,Q′))=0
sup(mg,1{x1,x2})(x1)=1
mf(x)=∑2
i=1 sup(mg,1{x1,x2})(xi)=2
d(f(P),f(Q))=1−(1−ϵ)2
17 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
b
b
a
a
1.0
a
1.0
a
(1 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
a
1.0
mg(x1)=mg(x2)=0
d(g(P′,P′),g(Q′,Q′))=0
sup(mg,1{x1,x2})(x1)=1
mf(x)=∑2
i=1 sup(mg,1{x1,x2})(xi)=2
d(f(P),f(Q))=1−(1−ϵ)2
17 / 18
a
1.0
b
1.0
a
1.0 − ϵ
ϵ
b
1.0
d(P,Q)=ϵ
a
a
b
b
a
a
1.0
a
1.0
a
(1 − ϵ)2
ϵ − ϵ2
ϵ − ϵ2
ϵ2
a
1.0
mg(x1)=mg(x2)=0
d(g(P′,P′),g(Q′,Q′))=0
sup(mg,1{x1,x2})(x1)=1
mf(x)=∑2
i=1 sup(mg,1{x1,x2})(xi)=2
d(f(P),f(Q))=1−(1−ϵ)2
17 / 18
▶ derive from a given PGSOS specification the compositionality property of each
▶ derive from a given compositionality property the syntactic requirements for
18 / 18
▶ derive from a given PGSOS specification the compositionality property of each
▶ derive from a given compositionality property the syntactic requirements for
18 / 18
▶ derive from a given PGSOS specification the compositionality property of each
▶ derive from a given compositionality property the syntactic requirements for
18 / 18
▶ derive from a given PGSOS specification the compositionality property of each
▶ derive from a given compositionality property the syntactic requirements for
18 / 18