Complete Axiomatization for the Bisimilarity Distance on MCs - - PowerPoint PPT Presentation
Complete Axiomatization for the Bisimilarity Distance on MCs Giorgio Bacci , Giovanni Bacci, Kim G. Larsen, and Radu Mardare Dept. of Computer Science, Aalborg University, DK CONCUR 2016 Introduction Kleenes Theorem : fundamental
Complete Axiomatization for the Bisimilarity Distance on MCs
Giorgio Bacci, Giovanni Bacci, Kim G. Larsen, and Radu Mardare
CONCUR 2016
between regular expressions and DFAs
for proving equivalence of regular expressions
behaviors and LTSs
2/25
Expressions: t,s := X | a.t | t +e s | rec X.t
3/25
Expressions: t,s := X | a.t | t +e s | rec X.t
names X∈𝕐
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Expressions: t,s := X | a.t | t +e s | rec X.t
action-prefix a∈𝔹 names X∈𝕐
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Expressions: t,s := X | a.t | t +e s | rec X.t
action-prefix a∈𝔹 probabilistic choice names X∈𝕐
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Expressions: t,s := X | a.t | t +e s | rec X.t
action-prefix a∈𝔹 probabilistic choice names X∈𝕐 recursion
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Expressions: t,s := X | a.t | t +e s | rec X.t
action-prefix a∈𝔹 probabilistic choice names X∈𝕐 recursion t a,1/3 b,2/3
Kleene’s theorem for MCs t = rec X.(a.X +1/3 b.s) s = rec Y.(c.Y)
s c,1
3/25
Expressions: t,s := X | a.t | t +e s | rec X.t
action-prefix a∈𝔹 probabilistic choice names X∈𝕐 recursion t a,1/3 b,2/3
Kleene’s theorem for MCs t = rec X.(a.X +1/3 b.s) s = rec Y.(c.Y)
s c,1 finite MCs
3/25
(B1) ⊢ t +1 s = t (B2) ⊢ t +e t = t (SC) ⊢ t +e s = s +1-e t (SA) ⊢ (t +e s) +e’ u = t +ee’ (s +e’-ee’ u) — for e,e’∈[0,1) (Unfold) ⊢ rec X.t = t[rec X.t / X] (Fix) {t = s[t / X]} ⊢ t = rec X.s — for X guarded in t (Unguard) ⊢ rec X.(t +e X) = rec X.t
1-ee’
4/25
(B1) ⊢ t +1 s = t (B2) ⊢ t +e t = t (SC) ⊢ t +e s = s +1-e t (SA) ⊢ (t +e s) +e’ u = t +ee’ (s +e’-ee’ u) — for e,e’∈[0,1) (Unfold) ⊢ rec X.t = t[rec X.t / X] (Fix) {t = s[t / X]} ⊢ t = rec X.s — for X guarded in t (Unguard) ⊢ rec X.(t +e X) = rec X.t
1-ee’
Stark-Smolka axiomatization
4/25
(B1) ⊢ t +1 s = t (B2) ⊢ t +e t = t (SC) ⊢ t +e s = s +1-e t (SA) ⊢ (t +e s) +e’ u = t +ee’ (s +e’-ee’ u) — for e,e’∈[0,1) (Unfold) ⊢ rec X.t = t[rec X.t / X] (Fix) {t = s[t / X]} ⊢ t = rec X.s — for X guarded in t (Unguard) ⊢ rec X.(t +e X) = rec X.t
1-ee’
Stone’s barycentric axioms
Stark-Smolka axiomatization
4/25
(B1) ⊢ t +1 s = t (B2) ⊢ t +e t = t (SC) ⊢ t +e s = s +1-e t (SA) ⊢ (t +e s) +e’ u = t +ee’ (s +e’-ee’ u) — for e,e’∈[0,1) (Unfold) ⊢ rec X.t = t[rec X.t / X] (Fix) {t = s[t / X]} ⊢ t = rec X.s — for X guarded in t (Unguard) ⊢ rec X.(t +e X) = rec X.t
1-ee’
Milner’s recursion axioms Stone’s barycentric axioms
Stark-Smolka axiomatization
4/25
Baeten-Bergstra-Smolka‘95 & Stark-Smolka‘00
Bandini-Segala‘01
Mislove-Ouaknine-Worrell‘04 (strong-bisimulation) Deng-Palamidessi‘07 (weak-bisimulation & behavioral eq.)
Silva-Bonchi-Bonsangue-Rutten‘11 (coagebraic bisim.)
…for probabilistic systems
5/25
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bisimilarity distance of Desharnais et al.
6/25
bisimilarity distance of Desharnais et al.
for generative Markov chains
6/25
bisimilarity distance of Desharnais et al.
for generative Markov chains
By using Quantitative Equational Theories* of Mardare-Panangaden-Plotkin (LICS’16)
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bisimilarity distance of Desharnais et al.
for generative Markov chains
By using Quantitative Equational Theories* of Mardare-Panangaden-Plotkin (LICS’16)
s = t s =ε t
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bisimilarity distance of Desharnais et al.
for generative Markov chains
By using Quantitative Equational Theories* of Mardare-Panangaden-Plotkin (LICS’16)
s = t s =ε t
c
p l e t e n e s s a l m
t f
f r e e !
6/25
{ti = si | i ∈ I} ⊢ t = s
inference
7/25
{ti = si | i ∈ I} ⊢ t = s
(Refl) ⊢ t = t (Symm) {t = s} ⊢ s = t (Trans) {t = u, u = s} ⊢ t = s (Cong) {t1 = s1,…,tn = sn} ⊢ f(t1,…,tn) = f(s1,…sn) — for f∈Σ
inference
7/25
quantitative inference
{ti =ε si | i ∈ I} ⊢ t =ε s
εi
(Refl) ⊢ t =0 t (Symm) {t =ε s} ⊢ s =ε t (Triang) {t =ε u, u =δ s} ⊢ t =ε+δ s (NExp) {t1 =ε s1,…,tn =ε sn} ⊢ f(t1,…,tn) =ε f(s1,…sn) — for f∈Σ (Max) {t =ε s} ⊢ t =ε+δ s — for δ>0 (Arch) {t =δ s | δ>ε } ⊢ t =ε s
Mardare-Panangaden-Plotkin (LICS’16)
8/25
𝓑 = (A,ΣA,dA) Quantitative Algebra (A,ΣA) — Universal algebra (A,dA) — metric space
𝓑 ⊨ {ti =ε si | i ∈ I} ⊢ t =ε s
εi
( )
iff for all i∈I. dA(⟦ti⟧,⟦si⟧) ≤ εi implies dA(⟦t⟧,⟦s⟧) ≤ ε Satisfiability
9/25
𝓑 ⊨ ⊢ t =ε s
( )
⊢ t =ε s ∈ 𝓥
( )
soundness completeness
quantitative algebra quantitative theory
10/25
𝓑 ⊨ ⊢ t =ε s
( )
⊢ t =ε s ∈ 𝓥
( )
soundness completeness
𝓑MC 𝓥MC
quantitative algebra quantitative theory
10/25
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Signature: X : 0 | a.- : 1 | +e : 2 | rec X : 1
12/25
Signature: X : 0 | a.- : 1 | +e : 2 | rec X : 1
X
(X)MC =
12/25
Signature: X : 0 | a.- : 1 | +e : 2 | rec X : 1
X
(X)MC = (a. )MC =
a.m
ℳ
m
a,1 ℳ
m
12/25
Signature: X : 0 | a.- : 1 | +e : 2 | rec X : 1
X
(X)MC = (a. )MC =
a.m
ℳ
m
a,1 ℳ
m
𝜈 ℳ
m
( +e )MC =
𝜉 𝒪
n
m+en
ℳ 𝒪
+
e𝜈+(1-e)𝜉
12/25
Signature: X : 0 | a.- : 1 | +e : 2 | rec X : 1
X
(X)MC = (a. )MC =
a.m
ℳ
m
a,1 ℳ
m
𝜈 ℳ
m
( +e )MC =
𝜉 𝒪
n
m+en
ℳ 𝒪
+
e𝜈+(1-e)𝜉
(rec X. )MC =
ℳ 𝜈
m
X e 1-e
ℳ 𝜈 e𝜈
1-e
recX.m12/25
Bisimilarity distance for MCs
(Desharnais et al. TCS’04)
dMC( , ) = min { ∫ Λ(dMC) dω | ω∈Ω(𝜈,𝜉) }
𝜈 ℳ
m
𝜉 𝒪
n
it is the least 1-bounded pseudometric satisfying
13/25
Bisimilarity distance for MCs
(Desharnais et al. TCS’04)
dMC( , ) = min { ∫ Λ(dMC) dω | ω∈Ω(𝜈,𝜉) }
𝜈 ℳ
m
𝜉 𝒪
n
it is the least 1-bounded pseudometric satisfying Kantorovich lifting
13/25
Bisimilarity distance for MCs
(Desharnais et al. TCS’04)
dMC( , ) = min { ∫ Λ(dMC) dω | ω∈Ω(𝜈,𝜉) }
𝜈 ℳ
m
𝜉 𝒪
n
it is the least 1-bounded pseudometric satisfying Kantorovich lifting
13/25
couplings = probabilistic “relations”
Bisimilarity distance for MCs
(Desharnais et al. TCS’04)
dMC( , ) = min { ∫ Λ(dMC) dω | ω∈Ω(𝜈,𝜉) }
𝜈 ℳ
m
𝜉 𝒪
n
it is the least 1-bounded pseudometric satisfying Kantorovich lifting Λ(dMC) — greatest 1-bounded pseudometric on (𝔹×MC)∪𝕐 s.t, for all a∈𝔹, Λ(dMC)((a, ),(a, )) = dMC( , )
ℳ
m
ℳ
m
𝒪
n
𝒪
n
13/25
couplings = probabilistic “relations”
m
a,1/2 1/2
Z
m = rec X. (a.X +1/2 Z) n = rec Y. (a.Y +1/3 Z)
n
a,1/3 2/3
Z
dMC?
14/25
d( , ) = = Λ(d)((a, ),(a, )) + Λ(d)((a, ), ) + Λ(d)( , ) = d( , ) +
m a,1/2
1/2
Z
1 3
n a,1/3
2/3
Z Z
m a,1/2
1/2
Z
n a,1/3
2/3
Z
m a,1/2
1/2
Z Z Z
1 6 1 2 1 3 1 6
m a,1/2
1/2
Z
n a,1/3
2/3
Z
1/3 1/6 1/2
𝜈((a,m))=1/2 𝜈(Z)=1/2 𝜉((a,n)) 1/3 = 𝜉(Z) 2/3 =
ω*
transition probabilities
15/25
d( , ) = = Λ(d)((a, ),(a, )) + Λ(d)((a, ), ) + Λ(d)( , ) = d( , ) +
m a,1/2
1/2
Z
1 3
n a,1/3
2/3
Z Z
m a,1/2
1/2
Z
n a,1/3
2/3
Z
m a,1/2
1/2
Z Z Z
1 6 1 2 1 3 1 6
m a,1/2
1/2
Z
n a,1/3
2/3
Z
1/3 1/6 1/2
𝜈((a,m))=1/2 𝜈(Z)=1/2 𝜉((a,n)) 1/3 = 𝜉(Z) 2/3 =
ω*
transition probabilities
15/25
d( , ) = = Λ(d)((a, ),(a, )) + Λ(d)((a, ), ) + Λ(d)( , ) = d( , ) +
m a,1/2
1/2
Z
1 3
n a,1/3
2/3
Z Z
m a,1/2
1/2
Z
n a,1/3
2/3
Z
m a,1/2
1/2
Z Z Z
1 6 1 2 1 3 1 6
m a,1/2
1/2
Z
n a,1/3
2/3
Z
1/3 1/6 1/2
𝜈((a,m))=1/2 𝜈(Z)=1/2 𝜉((a,n)) 1/3 = 𝜉(Z) 2/3 =
ω*
transition probabilities
= 1
15/25
d( , ) = = Λ(d)((a, ),(a, )) + Λ(d)((a, ), ) + Λ(d)( , ) = d( , ) +
m a,1/2
1/2
Z
1 3
n a,1/3
2/3
Z Z
m a,1/2
1/2
Z
n a,1/3
2/3
Z
m a,1/2
1/2
Z Z Z
1 6 1 2 1 3 1 6
m a,1/2
1/2
Z
n a,1/3
2/3
Z
1/3 1/6 1/2
𝜈((a,m))=1/2 𝜈(Z)=1/2 𝜉((a,n)) 1/3 = 𝜉(Z) 2/3 =
ω*
transition probabilities
= 1 = 0
15/25
d( , ) = = Λ(d)((a, ),(a, )) + Λ(d)((a, ), ) + Λ(d)( , ) = d( , ) +
m a,1/2
1/2
Z
1 3
n a,1/3
2/3
Z Z
m a,1/2
1/2
Z
n a,1/3
2/3
Z
m a,1/2
1/2
Z Z Z
1 6 1 2 1 3 1 6
m a,1/2
1/2
Z
n a,1/3
2/3
Z
1/3 1/6 1/2
𝜈((a,m))=1/2 𝜈(Z)=1/2 𝜉((a,n)) 1/3 = 𝜉(Z) 2/3 =
ω*
transition probabilities
= 1 = 0
15/25
d( , ) = = Λ(d)((a, ),(a, )) + Λ(d)((a, ), ) + Λ(d)( , ) = d( , ) +
m a,1/2
1/2
Z
1 3
n a,1/3
2/3
Z Z
m a,1/2
1/2
Z
n a,1/3
2/3
Z
m a,1/2
1/2
Z Z Z
1 6 1 2 1 3 1 6
m a,1/2
1/2
Z
n a,1/3
2/3
Z
Solution: dMC( , ) =
m a,1/2
1/2
Z
n a,1/3
2/3
Z
1 4
1/3 1/6 1/2
𝜈((a,m))=1/2 𝜈(Z)=1/2 𝜉((a,n)) 1/3 = 𝜉(Z) 2/3 =
ω*
transition probabilities
= 1 = 0
15/25
16/25
Axiomatization (first attempt)
(Unfold) ⊢ rec X.t = t[rec X.t / X] (Fix) {t = s[t / X]} ⊢ t = rec X.s — for X guarded in t (Unguard) ⊢ rec X.(t +e X) = rec X.t (B1) ⊢ t +1 s =0 t (B2) ⊢ t +e t =0 t (SC) ⊢ t +e s =0 s +1-e t (SA) ⊢ (t +e s) +e’ u =0 t +ee’ (s +e’-ee’ u) — for e,e’∈[0,1) (IB) {t =ε s, t’ =ε’ s’} ⊢ t +e t’ =δ s +e s’ — for δ ≤ eε+(1-e)ε’ (Top) ⊢ t =1 s
1-ee’
17/25
Axiomatization (first attempt)
Milner’s recursion axioms
(Unfold) ⊢ rec X.t = t[rec X.t / X] (Fix) {t = s[t / X]} ⊢ t = rec X.s — for X guarded in t (Unguard) ⊢ rec X.(t +e X) = rec X.t (B1) ⊢ t +1 s =0 t (B2) ⊢ t +e t =0 t (SC) ⊢ t +e s =0 s +1-e t (SA) ⊢ (t +e s) +e’ u =0 t +ee’ (s +e’-ee’ u) — for e,e’∈[0,1) (IB) {t =ε s, t’ =ε’ s’} ⊢ t +e t’ =δ s +e s’ — for δ ≤ eε+(1-e)ε’ (Top) ⊢ t =1 s
1-ee’
17/25
Axiomatization (first attempt)
Milner’s recursion axioms
(Unfold) ⊢ rec X.t = t[rec X.t / X] (Fix) {t = s[t / X]} ⊢ t = rec X.s — for X guarded in t (Unguard) ⊢ rec X.(t +e X) = rec X.t (B1) ⊢ t +1 s =0 t (B2) ⊢ t +e t =0 t (SC) ⊢ t +e s =0 s +1-e t (SA) ⊢ (t +e s) +e’ u =0 t +ee’ (s +e’-ee’ u) — for e,e’∈[0,1) (IB) {t =ε s, t’ =ε’ s’} ⊢ t +e t’ =δ s +e s’ — for δ ≤ eε+(1-e)ε’ (Top) ⊢ t =1 s
1-ee’
Interpolative barycentric axioms
(Mardare-Panangaden-Plotkin LICS’16)
17/25
(IB) {t =ε s, t’ =ε’ s’} ⊢ t +e t’ =δ s +e s’ — for δ ≤ eε+(1-e)ε’
m = rec X. (a.X +1/2 Z) n = rec Y. (a.Y +1/3 Z) the terms from the example…
18/25
a.X +1/2 Z =0 (a.X +1/3 a.X) +1/2 Z (B2) =0 a.X +1/6 (a.X +2/5 Z) (SA)
(IB) {t =ε s, t’ =ε’ s’} ⊢ t +e t’ =δ s +e s’ — for δ ≤ eε+(1-e)ε’
m = rec X. (a.X +1/2 Z) n = rec Y. (a.Y +1/3 Z) the terms from the example…
18/25
a.X +1/2 Z =0 (a.X +1/3 a.X) +1/2 Z (B2) =0 a.X +1/6 (a.X +2/5 Z) (SA) a.Y +1/3 Z =0 Z +2/3 a.Y (SC) =0 (Z +1/4 Z) +2/3 a.Y (B2) =0 Z +1/6 (a.Y +2/5 Z) (SA)+(SC)
(IB) {t =ε s, t’ =ε’ s’} ⊢ t +e t’ =δ s +e s’ — for δ ≤ eε+(1-e)ε’
m = rec X. (a.X +1/2 Z) n = rec Y. (a.Y +1/3 Z) the terms from the example…
18/25
1/3 1/6 1/2
(a,X) Z (a,Y) Z
ω* a.X +1/2 Z =0 (a.X +1/3 a.X) +1/2 Z (B2) =0 a.X +1/6 (a.X +2/5 Z) (SA) a.Y +1/3 Z =0 Z +2/3 a.Y (SC) =0 (Z +1/4 Z) +2/3 a.Y (B2) =0 Z +1/6 (a.Y +2/5 Z) (SA)+(SC)
(IB) {t =ε s, t’ =ε’ s’} ⊢ t +e t’ =δ s +e s’ — for δ ≤ eε+(1-e)ε’
m = rec X. (a.X +1/2 Z) n = rec Y. (a.Y +1/3 Z) the terms from the example…
18/25
1/3 1/6 1/2
(a,X) Z (a,Y) Z
ω* a.X +1/2 Z =0 (a.X +1/3 a.X) +1/2 Z (B2) =0 a.X +1/6 (a.X +2/5 Z) (SA) a.Y +1/3 Z =0 Z +2/3 a.Y (SC) =0 (Z +1/4 Z) +2/3 a.Y (B2) =0 Z +1/6 (a.Y +2/5 Z) (SA)+(SC)
(IB) {t =ε s, t’ =ε’ s’} ⊢ t +e t’ =δ s +e s’ — for δ ≤ eε+(1-e)ε’
m = rec X. (a.X +1/2 Z) n = rec Y. (a.Y +1/3 Z) the terms from the example…
18/25
1/3 1/6 1/2
(a,X) Z (a,Y) Z
ω* a.X +1/2 Z =0 (a.X +1/3 a.X) +1/2 Z (B2) =0 a.X +1/6 (a.X +2/5 Z) (SA) a.Y +1/3 Z =0 Z +2/3 a.Y (SC) =0 (Z +1/4 Z) +2/3 a.Y (B2) =0 Z +1/6 (a.Y +2/5 Z) (SA)+(SC)
(IB) {t =ε s, t’ =ε’ s’} ⊢ t +e t’ =δ s +e s’ — for δ ≤ eε+(1-e)ε’
m = rec X. (a.X +1/2 Z) n = rec Y. (a.Y +1/3 Z) the terms from the example…
18/25
The quantitative equational framework
all operators to be non-expansive
(NExp) {t1 =ε s1,…,tn =ε sn} ⊢ f(t1,…,tn) =ε f(s1,…sn) — for f∈Σ
19/25
The quantitative equational framework
all operators to be non-expansive
(NExp) {t1 =ε s1,…,tn =ε sn} ⊢ f(t1,…,tn) =ε f(s1,…sn) — for f∈Σ
… but the NExp axiom is not sound for recursion
𝓑MC ⊨ {t =ε s} ⊢ rec X.t =ε rec X.s
( )
(see Gebler-Larsen-Tini FoSSaCS’15)
19/25
we keep all the axioms of quantitative algebras but the NExp axiom
(Refl) ⊢ t =0 t (Symm) {t =ε s} ⊢ s =ε t (Triang) {t =ε u, u =δ s} ⊢ t =ε+δ s (NExp) {t1 =ε s1,…,tn =ε sn} ⊢ f(t1,…,tn) =ε f(s1,…sn) — for f∈Σ (Max) {t =ε s} ⊢ t =ε+δ s — for δ>0 (Arch) {t =δ s | δ>ε } ⊢ t =ε s
20/25
we keep all the axioms of quantitative algebras but the NExp axiom
(Refl) ⊢ t =0 t (Symm) {t =ε s} ⊢ s =ε t (Triang) {t =ε u, u =δ s} ⊢ t =ε+δ s (NExp) {t1 =ε s1,…,tn =ε sn} ⊢ f(t1,…,tn) =ε f(s1,…sn) — for f∈Σ (Max) {t =ε s} ⊢ t =ε+δ s — for δ>0 (Arch) {t =δ s | δ>ε } ⊢ t =ε s
is NOT the original quantitive equational framework!
20/25
we keep all the axioms of quantitative algebras but the NExp axiom
(Refl) ⊢ t =0 t (Symm) {t =ε s} ⊢ s =ε t (Triang) {t =ε u, u =δ s} ⊢ t =ε+δ s (NExp) {t1 =ε s1,…,tn =ε sn} ⊢ f(t1,…,tn) =ε f(s1,…sn) — for f∈Σ (Max) {t =ε s} ⊢ t =ε+δ s — for δ>0 (Arch) {t =δ s | δ>ε } ⊢ t =ε s
the Archimedian axiom will be used to recover completeness is NOT the original quantitive equational framework!
20/25
{m =ε n} ⊢ m =1/3ε+1/6 n
from what we have seen in the example before and (Fix)+(Unfold)+(Top)+(IB) we obtain
21/25
{m =ε n} ⊢ m =1/3ε+1/6 n
from what we have seen in the example before and (Fix)+(Unfold)+(Top)+(IB) we obtain
(Top) ⊢ m =1 n
21/25
{m =ε n} ⊢ m =1/3ε+1/6 n
from what we have seen in the example before and (Fix)+(Unfold)+(Top)+(IB) we obtain
(Top) ⊢ m =1 n
greatest fixed point
21/25
{m =ε n} ⊢ m =1/3ε+1/6 n
from what we have seen in the example before and (Fix)+(Unfold)+(Top)+(IB) we obtain
(Top) ⊢ m =1 n
(Max) {t =ε s} ⊢ t =ε+δ s — for δ>0 (Arch) {t =δ s | δ>ε } ⊢ t =ε s
⊢ m =1/4 n
1 1/4
1/2 1/3
…
(⊢ m =ε n)
greatest fixed point
21/25
Sound & Complete Axiomatization
Milner’s recursion axioms
(Unfold) ⊢ rec X.t = t[rec X.t / X] (Fix) {t = s[t / X]} ⊢ t = rec X.s — for X guarded in t (Unguard) ⊢ rec X.(t +e X) = rec X.t (Cong) {t =0 s} ⊢ rec X.t =0 rec X.s (B1) ⊢ t +1 s =0 t (B2) ⊢ t +e t =0 t (SC) ⊢ t +e s =0 s +1-e t (SA) ⊢ (t +e s) +e’ u =0 t +ee’ (s +e’-ee’ u) — for e,e’∈[0,1) (IB) {t =ε s, t’ =ε’ s’} ⊢ t +e t’ =δ s +e s’ — for δ ≤ eε+(1-e)ε’ (Top) ⊢ t =1 s
1-ee’
Interpolative barycentric axioms
(Mardare-Panangaden-Plotkin LICS’16)
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Sound & Complete Axiomatization
Milner’s recursion axioms
(Unfold) ⊢ rec X.t = t[rec X.t / X] (Fix) {t = s[t / X]} ⊢ t = rec X.s — for X guarded in t (Unguard) ⊢ rec X.(t +e X) = rec X.t (Cong) {t =0 s} ⊢ rec X.t =0 rec X.s (B1) ⊢ t +1 s =0 t (B2) ⊢ t +e t =0 t (SC) ⊢ t +e s =0 s +1-e t (SA) ⊢ (t +e s) +e’ u =0 t +ee’ (s +e’-ee’ u) — for e,e’∈[0,1) (IB) {t =ε s, t’ =ε’ s’} ⊢ t +e t’ =δ s +e s’ — for δ ≤ eε+(1-e)ε’ (Top) ⊢ t =1 s
1-ee’
Interpolative barycentric axioms
(Mardare-Panangaden-Plotkin LICS’16)
22/25
A quantitative Kleene’s theorem
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A quantitative Kleene’s theorem
d⊢([t],[s]) = inf{ ε | ⊢ t =ε s }
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A quantitative Kleene’s theorem
isometric isomorphism
d⊢([t],[s]) = inf{ ε | ⊢ t =ε s }
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future work…
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future work…
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future work…
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future work…
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