Process Algebra (2IF45) Abstraction in Process Algebra Suzana Andova.

Process Algebra (2IF45)

Abstraction in Process Algebra

Suzana Andova

2

Outline of the lecture

• Our way of dealing with internal behaviour: branching bisimulation

• How we capture Abstraction in Process Algebra• combining it with other concepts


3

Abstraction

Abstraction is used to • check the correctness of implementation against the system

specification

• reduce and simplify the model to enable better, fasted and cleaner model analysis


Question: How do we chose to relate behaviours with internal steps?

Branching bisimulation

4 Process Algebra (2IF45)

Branching bisimulation – simple examples first

a

b

is branching bisim to a

a b

“ related states must have the same potential which does not change until an observable action is executed ”


Branching bisimulation – simple examples first

a b

is branching bisim to a

b

it is not branching bisim to a

b

6

Branching bisimilar processes

t

t’

a

s

a

s’

t’’

t

s

s’t’’

t s

s’

t

Branching Bisimulation relation: A binary relation R on the set of state S of an LTS is branching bisimulation relation iff the following transfer conditions hold:

1. for all states s, t, s’ S, whenever (s, t) R and s → s’ for some a A, then there are states t’, t’’ S such that t t’ and t’ → t’’ and (s, t’), (s’,t’’) R;2. vice versa, for all states s, t, s’ S, whenever (s, t) R and t → t’ for some a A, then there

are states s’,s’’ S such that s s’ and s’ → s’’ and (s’, t), (s’’,t’) R;3. if (s, t) R and s then there is a state t’ such that t t’ , t’ and (s, t’) R4. whenever (s, t) R and t then there is a state s’ such that s s’ , s’ and (s’, t) R

Two LTSs s and t are branching bisimilar, s b t, iff there is a branching bisimulation relation R

such that (s, t) R

a

a

a

a

7

less

po

we

r o

f th

e o

bs

erve

r

Spectrum of behavioural relations

8

most powerful

9

Weak bisimulation just a short comparison


a b c d1 d2 d3 d4

a b c d1 d2 d3 d4

b

a b c d1 d2 d3 d4

b

10

Branching bisimulation and composition

11


a

a

a

a

b

b b

branching bisimilar!

branching bisimilar? NO!

+ +

12


a

a

a

a

b

b b

branching bisimilar!

branching bisimilar? NO!

+ +

Painful conclusion: branching bisimilation is not compositional.

13


a

a

a

a b b

branching bisimilar components!

++

What to do? Two choices:1. Make the relation weaker and relate the two compositions too!2. Make the relation stronger and do not relate the two components

from the beginning!

Not branching bisimilar compositions!

14

Rooted Branching Bisimilar processes

t’

q

b

s’ b

p

r

t’

s’

pq

t s’

p

t’

t s a a

t s a a

t s a a

R is Rooted BB between state (s, t) R if R is Branching Bisimulation relation (as already defined) and the root condition:

1. if s → s’ for a A, then there is a state t’ S such that t → t’ and (s’, t’) R;2. if t → t’ for a A, then there is a state s’ S such that s → s’ and (s’, t’) R;3. s if and only if t

LTSs s and t are rooted branching bisimilar, s rb t, iff there is a rooted branching bisimulation

relation R such that (s, t) R

a a

a

Rooted branching bisimulation is strengthened variant of branching bisimulation strict enough to obtain compositionality

a

(aA i.e. can be from A or can be )


Axiomatizing Rooted Branching Bisimulations

Language: BPA(A)

Signature: 0, 1, (a._ )aA, , +, •

Language terms T(BPA(A,))

Closed terms C(BPA(A))

Equality of terms

x+ y = y+x

(x+y) + z = x+ (y + z)

x + x = x

x+ 0 = x

(x+ y) z = x z+y z

(x y) z = x (y z)

0 x = 0

x 1 = x

1 x = x

a.x y = a.(x y)

Completeness

Soundness

Deduction rules for BPA(A) (a A): x x’ x + y x’

a

a

1

x (x + y)

a.x x a

y y’ x + y y’

a

a

y (x + y) ⑥

x x’ x y x’

a

a

x y (x y)

x y y’ x y y’

a

a

Strong Bisimilarity on LTSs



Language: BPA(A)

Signature: 0, 1, (a._ )aA, , +, •



Strong Bisimilarity on LTSs Equality of terms

x+ y = y+x

(x+y) + z = x+ (y + z)

x + x = x

x+ 0 = x

(x+ y) z = x z+y z

(x y) z = x (y z)

0 x = 0

x 1 = x

1 x = x

a.x y = a.(x y)

Completeness

Soundness


a

a

1

x (x + y)

a.x x a

y y’ x + y y’

a

a

y (x + y) ⑥

x x’ x y x’ y

a

a

x y (x y)

x y y’ x y y’

a

a

Rooted Branching

17

x y

x

+

+

x y

+

.(x+y) + x = x+y

Turned into equation looks like:

Axiomazing Rooted branching bisimulation

bb

18

a

x y

x

+

+

…

a

x y

+

…

rb

B axiom a.(.(x+y) + x) = a.(x+y)

Axiomazing Rooted branching bisimulation

bb

Turned into equation looks like:



Language: BPA(A)

Signature: 0, 1, (a._ )aA, , +, •



Strong Bisimilarity on LTSs Equality of terms

x+ y = y+x

(x+y) + z = x+ (y + z)

x + x = x

x+ 0 = x

(x+ y) z = x z+y z

(x y) z = x (y z)

0 x = 0

x 1 = x

1 x = x

a.x y = a.(x y)

a.(.(x+y) + x) = a.(x+y)

Completeness

Soundness


a

a

1

x (x + y)

a.x x a

y y’ x + y y’

a

a

y (x + y) ⑥

x x’ x y x’ y

a

a

x y (x y)

x y y’ x y y’

a

a

Rooted Branching

20

Home work

• Prove soundness of B axiom wrt rooted BB• Read the proof of ground completeness



Combining internal step with other operators

Language: BPA(A)

Signature: 0, 1, (a._ )aA, , +, •



Axioms Deduction rules


Combining internal step with other operators:Hiding operator

Language: BPA(A)

Signature: 0, 1, (a._ )aA, , +, •, I (I A)



Axioms for I Deduction rules for I

turns external actions into internal steps


Combining internal step with other operators:Encapsulation operator

Language with

Signature: 0, 1, (a._ )aA, , +, H (H A) blocks actions


Combining internal step with other operators:Parallel composition and communication

Language: TCP(A)

Signature: 0, 1, (a._ )aA, , +, •, I (I A), ||, |, ╙, H,

Language terms T(BPA(A, ))

Closed terms C(BPA(A, )) Axioms for parallel composition with silent step:

x ╙ .y = x ╙ y

x |.y = 0

25

Exercises

• see distributed copies


26

Abstraction, silent steps and Recursion

Guardedness and silent steps: cannot be a guard of a variable

X = .X has solutions ..a.1 but also ..b.1

Guardedness and hiding operator: I cannot appear in tX in X = tX

X = i.I(X), where i I has solutions i.i.a.1 but also i.i.b.1


27

Abstraction and Recursion and Fairness


X

Y

a

0

Z

U

a

0

Observation:1. they are rooted bb bisimilar2. implicitly internal loop is left eventually

= fairness

28



X

Y

a

0

X = .YY = .Y + a.0

Z

U

a

0

Z = .UU = a.0

RSP+RDP? X = Z

Observation on LTSs:1. they are rooted bb bisimilar2. implicitly internal loop is left eventually

= fairness

As recursive specifications:

29



X

Y

a

0

X = .YY = .Y + a.0

Z

U

a

0

Z = .UU = a.0

RSP+RDP? X = Z

At least two problems: 1. Those are not guarder recursive specifications!2. Even if they are somehow made guarded, B axiom is not sufficient

to rewrite one spec into another

Observation on LTSs:1. they are rooted bb bisimilar2. implicitly internal loop is left eventually

= fairness

As recursive specifications:


X = .YY = .Y + a.0

X’ = i.Y’Y’ = i.Y’ + a.0

for some action i to be turned internal “soon”

by applying I for I = {i}

represents

X

Y

a

0

X’

Y’

i

a

0

iapplying {i}

Abstraction and Recursion and Fairness: problem 1. dealing with guardedness


Z = .UU = a.0

Z’ = i.U’U’ = a.0

Z’

U’

i

a

0

Z

U

0

applying {i}

a


X = .YY = .Y + a.0

X’ = i.Y’Y’ = i.Y’ + a.0



represents

X

Y

a

0

X’

Y’

i

a

0

iapplying {i}

represents


Z = .UU = a.0

Z’ = i.U’U’ = a.0

Z’

U’

i

a

0

applying {i} Z

U

0

a


X = .YY = .Y + a.0

X’ = i.Y’Y’ = i.Y’ + a.0



represents

X

Y

a

0

X’

Y’

i

a

0

iapplying {i}

represents

OK!OK!

How to connect them

?


X’ = i.Y’Y’ = i.Y’ + a.0

Something like this shall help:

Y’ = i.Y’ + a.0

. I(Y’) = . I(a.0)

Abstraction and Recursion and Fairness: problem 2. derivation rules

We want to derive that I(X’) = I(Z’)! We need new rules for this!


a bit more general rule:

x1 = i1.x1 + y1, i1 I

. I(x1) = . I(y1)

Abstraction and Recursion and Fairness: Fairness rule KFAR1

b


General KFAR rule is:

x1 = i1.x2 + y1,x2 = i2.x3 + y2,…

xn = in.x1 + yn, i1, … in I , there is ik

. I(x1) = . (I(y1) + … + I(yn))

Abstraction and Recursion and Fairness: Fairness rule KFARn

b


Abstraction and Recursion and Fairness:Example of tossing a coin

37

Home Work (part2)

• Study the Coin tossing example• Study the complete proof for ABP, derivation up to abstraction

and derivation by means of fairness derivation rules.


Date post:	14-Dec-2015
Category:	Documents
Upload:	emilie-loud
View:	224 times
Download:	0 times

Process Algebra (2IF45) Abstraction in Process Algebra Suzana Andova.

Documents