process algebra (2if45) extending the basic process algebra

Process Algebra (2IF45)

Extending the Basic Process Algebra

Dr. Suzana Andova

2 Process Algebra (2IF45)

Guarded recursive specification

What do we still miss?

Consider three specifications:E = {X = a.Y, Y = a.X} and F = {Z = a.a.Z} G = {Un = a.Un+1, n 0}

X

a

Y

a

Z

a

a.Z

a

U0 U1 U2

a a a…

Can we derive

BPA(A), E, F, G ├ X = Z or BPA(A), E, F, G ├ X = U0 or

BPA(A), E, F, G ├ Z = U0


Deriving equalities of recursive variables

Consider three specifications:E = {X = a.Y, Y = a.X} and F = {Z = a.a.Z} G = {Un = a.Un+1, n 0}

1. BPA(A), E, F, G ├ X = a.a.X which is exactly the form of Z 2. BPA(A), E, F, G ├ X = U0 cannot be derived directly. We introduce new variables X0, X1, … which are defined using X, as:

X0 = X, X1= a.X, X2 = a.a.X, …, Xn = an X, ….,

Now, we want to show that Xn can be rewritten in the form of Un for any n 0!BPA(A), E, F, G ├ X0 = X = a.a.X = a.X1

BPA(A), E, F, G ├ X1 = a.X = a.a.a.X = a.X2

and in general BPA(A), E, F, G ├ Xn = an X = an a.a.X = a.Xn+1

We conclude that Xn is in the same form as Un for any n 0

4

Recursive principles for Deriving equalities of recursive variables

1. Every guarded recursive specification has a solution, this is called Restricted Recursive Definition Principle (RDP-)

2. Every guarded recursive specification has a unique solution, this is called Recursive Specification Principle (RSP)

1. BPA(A), E, F, G ├ X = a.a.X which is exactly the form of Z. Directly from the RSP and RDP- we can conclude

BPA(A), E, F, G, RDP-, RSP ├ X = Z 2. BPA(A), E, F, G ├ X = U0 cannot be derived directly. We introduce new

variables X0, X1, … which are defined using X, as:X0 = X, X1= a.X, X2 = a.a.X, …, Xn = an X, ….,

Now, we want to show that Xn can be rewritten in the form of Un for any n 0!BPA(A), E, F, G ├ X0 = X = a.a.X = a.X1

BPA(A), E, F, G ├ X1 = a.X = a.a.a.X = a.X2

and in general BPA(A), E, F, G ├ Xn = an X = an a.a.X = a.Xn+1

We conclude that Xn is in the same form as Un for any n 0. Directly from the RSP and RDP- we can conclude

BPA(A), E, F, G, RDP-, RSP ├ X0 = U0

and also BPA(A), E, F, G, RDP-, RSP ├ X = X0 = U0

5

Specifying a Stack

Consider a finite set of data elements D = {d1, d2, …, dn} for some n natural number. Define a recursive specification that describes a stack with unlimited capacity. Elements from D can be added or removed from the stack from the top of the stack.

Steps: 1. First, define the set of atomic actions.

2. As a short hand notation you can use universal sum dD

3. Reason how many recursive variables you may need. 4. Specify the behaviour of the stack process.

6

Specifying a Stack


Steps: 1. First, define the set of atomic actions.

2. As a short hand notation you can use universal sum dD

3. Reason how many recursive variables you may need. 4. Specify the behaviour of the stack process.

Stack1 = S

S = dD push(d).Sd, for any d D

Sd = eD push(e).Sed + dD pop(d).S, for any d D and D*

7

Specifying a Stack again


Steps: 1. First, define the set of atomic actions. 2. The equational theory is extended with operator (general sequential

composition) 3. Can we do it with finitely many equations now?

Stack2 = T

T = dD push(d). ( ( ….) . pop(d).1)

8

Specifying a Stack again


Steps: 1. First, define the set of atomic actions. 2. The equational theory is extended with operator3. Can we do it with finitely many equations now?

Stack2 = T

T = dD push(d). ( ( U ) . pop(d).1)U = 1 + T U

Stack

U pop(d)

pop(d)

push(d)

(U pop(e)) (U pop(d))

pop(e)

push(e)


Extending BPA(A) with sequential composition

Question: Can you find a set of axioms for the new operator?

(A1) x+ y = y+x

(A2) (x+y) + z = x+ (y + z)

(A3) x + x = x

(A4) x+ 0 = x

BPA(A)

Signature: constants 0,1

action prefix a._ (unary operator)

non-deterministic choice _+_ (bin. op.)

sequential composition __ (bin.op.)


Extending BPA(A) with sequential composition

BPA(A)



non-deterministic choice _+_ (bin. op.)

sequential composition __ (bin.op.)

(A1) x+ y = y+x

(A2) (x+y) + z = x+ (y + z)

(A3) x + x = x

(A4) x+ 0 = x

(A5) (x+ y) z = x z+y z

(A6) (x y) z = x (y z)

(A7) 0 x = 0

(A8) x 1 = x

(A9) 1 x = x

(A10) a.x y = a.(x y)


Term deduction system for BPA(A) with sequential composition

• Set of action names (actions): A

• Signature: constants 0,1 (note 0,1 A)


non-deterministic choice _+_ (binary operator)

sequential composition __ (binary operator)• Behaviour expressed by action transitions _ _ for a in A and

termination _ • Behavioural equivalence is bisimilarity

a

• Deduction rules

1

x x’

x + y x’ a.x x

a

a

x (x + y)

a y y’

x + y y’

a

a

y (x + y) ⑥


Term deduction system for BPA(A) with sequential composition

• Deduction rules

1

x x’

x + y x’ a.x x

a

a

x (x + y)

a y y’

x + y y’

a

a

y (x + y) ⑥




non-deterministic choice _+_ (binary operator)

sequential composition __ (binary operator)• Behaviour expressed by action transitions _ _ for a in A and


a

• Deduction rules for sequential com.

x x’

x y x’

a

a

x y (x y)

x y y’

x y y’

a

a


Extending BPA(A) further

BPA(A)


action prefix a._

non-deterministic choice _+_

sequential composition _ _

cut_cakewater

coffee

in_cup

in_plate

||

What are possible scenarios?


Extending BPA(A) further and further

cut_cakewater

coffee

in_cup

in_plate

||

What are possible scenarios now?

sit_sofa

watchTV

sit_sofa

watchTV


Term deduction system for BPA(A) extended to TCP(A)



action prefix a._

non-deterministic choice _+_

sequential composition _ _ • Behaviour expressed by action transitions _ _ for a in A and


• Deduction rules

1

x x’

x + y x’ a.x x

a

a

x (x + y)

a y y’

x + y y’

a

a

y (x + y) ⑥

• Deduction rules for sequential com.

x x’

x y x’

a

a

x y (x y)

x y y’

x y y’

a

a


Term deduction system for BPA(A) extended to TCP(A)



action prefix a._

nondeterminism _+_

sequential compos. __• Behaviour expressed by action transitions _ _ for a in A and


• Deduction rules

x x’

x || y x’ || y

a

a

x yx || y

y y’

x || y x || y’

a

a

x yx | y

x x’ y y’, (a,b) = c

x || y x’ || y’

a

c

b x x’ y y’, (a,b) = c

x | y x’ || y’

a

c

b

communication function (_,_)

parallel composition _ || _

communication composition. _ | _


Extending BPA(A) to TCP(A)BPA(A)


action prefix a._

non-determ. _+_

sequential comp. __

x|| y = x ╙ y + y ╙ x + x | y

x || 1 = x

(x || y) || z = x || (y || z)

0 ╙ x = 0

1 ╙ x = 0

a.x ╙ y = a.(x || y)

(x + y) ╙ z = x ╙ z + y ╙ z

(x ╙ y) ╙ z = x ╙ (y || z)

communication function (_,_)

parallel composition_ || _

communication compos. _ | _

left merge operator _ ╙ _

unary operator ∂H

a.x | b.y = c.(x || y) if (a,b) = c

a.x | b.y = 0 if (a,b) not defined

0| x = 0

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

1 | 1 = 1

a.x | 1 = 0

x | y = y | x

(x | y) | z = x | (y | z)

1 | x + 1 = 1

(x | y) ╙ z = x | (y ╙ z)

18

Example


B

A

C

21

4

3

Specify the communication process:

19

Example


1. Draw the transition system generated by the operational rules for the process X || Y given by the recursive specification X = a.XY = b.Y

There is no communication, so = .

2. Draw the transition system generated by the operational rules for the process X || Y given by the recursive specification X = a.Y + b.Y; Y = a.X + b.X.

There is no communication, so = . 3. Also draw the transition system generated by the operational rules for the

process Z = (a.1 || b.1) Z4. Are the transition systems found in the previous two items (2 and 3) bisimilar? If

yes, show a bisimulation. If no, argue why not.

process algebra (2if45) extending the basic process algebra

Documents

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rsp x0

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