Process Algebra (2IF45) Abstraction and Recursions in Process Algebra Suzana Andova

1 Outline of the lecture Combining silent steps with recursion Fairness rules Some examples Fairness: is this what it is really about? Process Algebra (2IF45)

2 Axiomatizing Rooted Branching Bisimulations Language: BPA  (A) Signature: 0, 1, (a._ ) a  A, , +, Language terms T(BPA  (A,)) Closed terms C(BPA  (A)) Rooted Branching 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 Deduction rules for BPA  (A) (a  A  ): x  x’ x + y  x’ a a  1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

3 Process Algebra (2IF45) Combining internal step with other operators: Hiding operator Language: BPA  (A) Signature: 0, 1, (a._ ) a  A, , +,,  I (I  A) Language terms T(BPA  (A,)) Closed terms C(BPA  (A)) turns external actions into internal steps

4 Process Algebra (2IF45) Combining internal step with other operators: Encapsulation operator Language with Signature: 0, 1, (a._ ) a  A, , +,  H (H  A) blocks actions

5 Process Algebra (2IF45) Combining internal step with other operators: Parallel composition and communication Language: TCP  (A) Signature : 0, 1, (a._ ) a  A, , +,,  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

6 Exercises Process Algebra (2IF45)

7 Home work You DID prove soundness of B axiom wrt rooted BB! You DID Read the proof of ground completeness Process Algebra (2IF45)

8 Bisimilarity vs. Derivability Process Algebra (2IF45) Rooted branching bisimulation TCP  (A,  ) …. a.( .(x+y) +x) = a.(x+y) … a  a b b Language: TCP  (A) Signature : 0, 1, (a._ ) a  A, , +,,  I (I  A), ||, |, ╙,  H, Language terms T(BPA  (A,  )) Closed terms C(BPA  (A,  ))

9 Process Algebra (2IF45) Rooted branching bisimulation TCP  (A,  ) …. a.( .(x+y) +x) = a.(x+y) … a  a b b  a   a TCP  (A,  ) (+RSP,RDP,…) X = . Y Y = . Y +a.0 Can we derive X = . a.0? Language: TCP  (A) Signature : 0, 1, (a._ ) a  A, , +,,  I (I  A), ||, |, ╙,  H, Language terms T(BPA  (A,  )) Closed terms C(BPA  (A,  )) Bisimilarity vs. Derivability

10 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 (Do you remember what a solution of a rec. spec. is?) Guardedness and hiding operator:  I cannot appear in t X in X = t X X = i.  I (X), where i  I has solutions i.a.1 but also i. b.1 Process Algebra (2IF45) Guardedness with silent step is defined 

11 Abstraction and Recursion and Fairness Process Algebra (2IF45) X Y  a 0  Z U  a 0

12 Abstraction and Recursion and Fairness Process Algebra (2IF45) X Y  a 0  X = .Y Y = .Y + a.0 Z U  a 0 Z = .U U = a.0 HOW? RSP+RDP ? X = Z Observation on LTSs: 1.they are rooted bb bisimilar 2.implicitly internal loop is left eventually = fairness As recursive specifications:

13 Abstraction and Recursion and Fairness Process Algebra (2IF45) X Y  a 0  X = .Y Y = .Y + a.0 Z U  a 0 Z = .U U = a.0 X = Z At least two problems: 1.One is not guarder recursive specification! 2.Even if it is somehow made guarded, (but how?) B axiom is not sufficient to rewrite one spec into another Observation on LTSs: 1.they are rooted bb bisimilar 2.implicitly internal loop is left eventually = fairness As recursive specifications: HOW? RSP+RDP ?

14 Process Algebra (2IF45) X = .Y Y = .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 i applying  {i} (X’) Abstraction and Recursion and Fairness: problem 1. dealing with guardedness

15 Process Algebra (2IF45) Z = .U U = a.0 Z’ = i.U’ U’ = a.0 Abstraction and Recursion and Fairness: problem 1. dealing with guardedness X = .Y Y = .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 i applying  {i} (X’) represents Z’ U’ i a 0 applying  {i} (Z’) Z U  0 a

16 Process Algebra (2IF45) Z = .U U = a.0 Z’ = i.U’ U’ = a.0 Z’ U’ i a 0 applying  {i} (Z’) Z U  0 a Abstraction and Recursion and Fairness: problem 1. dealing with guardedness X = .Y Y = .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 i applying  {i} (X’) represents OK! How to connect them ?

17 Process Algebra (2IF45) X’ = i.Y’ Y’ = i.Y’ + a.0 Something like this shall help: Y’ = i.Y’ + a.0, i  I .  I (Y’) = .  I (a.0) Abstraction and Recursion and Fairness: problem 2. derivation rules We want to derive that  I (X’) = . a.0 =  I (Z’)! We need new rules for this!

18 Process Algebra (2IF45) a bit more general rule: x 1 = i 1.x 1 + y 1, i 1  I, I  A .  I (x 1 ) = .  I (y 1 ) Abstraction and Recursion and Fairness: Fairness rules KFAR b General KFAR n b rule is: x 1 = i 1.x 2 + y 1, x 2 = i 2.x 3 + y 2, … x n = i n.x 1 + y n, {i 1, … i n }  I  {  }, there is i k   .  I (x 1 ) = . (  I (y 1 ) + … +  I (y n ))

19 Process Algebra (2IF45) General KFAR n b rule is: x 1 = i 1.x 2 + y 1, x 2 = i 2.x 3 + y 2, … x n = i n.x 1 + y n, i 1, … i n  I, there is i k   .  I (x 1 ) = . (  I (y 1 ) + … +  I (y n )) Abstraction and Recursion and Fairness: Fairness rule KFAR n b

20 Process Algebra (2IF45) Abstraction and Recursion and Fairness: Example of tossing coins This exercise has be worked out during the lecture. If you didn’t attend the lecture you can find a similar exercise in [1] to show applicability of the KFAR rule

21 Process Algebra (2IF45) Abstraction and Recursion and Fairness: Example of throwing a die This exercise has be worked out during the lecture. If you didn’t attend the lecture you can find a similar exercise in [1] to show applicability of the CFAR rule

22 Process Algebra (2IF45) There will be no lecture on 29 th of May