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Process Algebra (2IF45) Abstraction in Process Algebra Suzana Andova 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 1 Process Algebra (2IF45) 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 2 Process Algebra (2IF45) Branching bisimulation – simple examples first a b is branching bisim to b a a “ related states must have the same potential which does not change until an observable action is executed ” 3 Process Algebra (2IF45) Branching bisimulation – simple examples first is branching bisim to b a b a it is not branching bisim to 4 Process Algebra (2IF45) b a Branching bisimilar processes s t t’ a s’ s t t’’ s tt s’ s’ a 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: a 1. for all states s, t, s’ S, whenever (s, t) R and s → s’ for some a A, then there are a t’ and t’ → t’’ and (s, t’), (s’,t’’) R; states t’, t’’ S such that t a 2. vice versa, for all states s, t, s’ S, whenever (s, t) R and t → t’ for some a A, then there a s’ and s’ → are states s’,s’’ S such that s s’’ and (s’, t), (s’’,t’) R; t’ , t’ and (s, t’) R 3. if (s, t) R and s then there is a state t’ such that t 4. 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 5 less 6 power of the observer most powerful 7 Weak bisimulation just a short comparison a d1 b d2 d3 c d4 b a d1 b d2 d3 c d4 b a d1 8 d2 b d3 c d4 Process Algebra (2IF45) Branching bisimulation and composition 9 Branching bisimulation and composition branching bisimilar! b a a + a 10 + b a b branching bisimilar? NO! Branching bisimulation and composition branching bisimilar! b a a + + b a b branching bisimilar? NO! a Painful conclusion: branching bisimilation is not compositional. 11 Branching bisimulation and composition branching bisimilar components! Not branching bisimilar compositions! + a a + b a b a 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! 12 Rooted Branching Bisimilar processes Rooted branching bisimulation is strengthened variant of branching bisimulation strict enough to obtain compositionality s t a a t’ s’ b r a t’ p a q s t s’ a b q s t a t’t s’ p p (aA i.e. can be from A or can be ) R is Rooted BB between state (s, t) R if R is Branching Bisimulation relation (as already defined) and the root condition: a s’ for a A, then there is a state t’ S such that ta→ t’ and (s’, t’) R; 1. if s → 2. if t a→ t’ for a A, then there is a state s’ S such that sa → 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 13 Axiomatizing Rooted Branching Bisimulations Language: BPA(A) Signature: 0, 1, (a._ )aA, , +, • Language terms T(BPA(A,)) Closed terms C(BPA(A)) Deduction rules for BPA(A) (a A): a x x’ a a.x x a x’ x+y 1 x (x + y) a a y y’ a y’ x+y y (x + y) 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) 0x=0 x1=x 1x=x a.x y = a.(x y) x x’ a xy x’ a x y y’ a x y y’ x y (x y) Strong Bisimilarity on LTSs ⑥ Soundness Completeness 14 Process Algebra (2IF45) Equality of terms Axiomatizing Rooted Branching Bisimulations Language: BPA(A) Signature: 0, 1, (a._ )aA, , +, • Language terms T(BPA(A,)) Closed terms C(BPA(A)) Deduction rules for BPA(A) (a A): a x x’ a a.x x a x’ x+y 1 x (x + y) a a y y’ a y’ x+y y (x + y) x x’ a xy x’ y a x y y’ a x y y’ x y (x y) Strong Bisimilarity on LTSs Rooted Branching 15 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) 0x=0 x1=x 1x=x a.x y = a.(x y) ⑥ Soundness Completeness Process Algebra (2IF45) Equality of terms Axiomazing Rooted branching bisimulation bb + + + Turned into equation looks like: .(x+y) + x = 16 x+y Axiomazing Rooted branching bisimulation rb a a bb + + + Turned into equation looks like: B axiom 17 a.(.(x+y) + x) = a.(x+y) Axiomatizing Rooted Branching Bisimulations Language: BPA(A) Signature: 0, 1, (a._ )aA, , +, • Language terms T(BPA(A,)) Closed terms C(BPA(A)) 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) 0x=0 x1=x 1x=x a.x y = a.(x y) Deduction rules for BPA(A) (a A): a x x’ a a.x x a x’ x+y 1 x (x + y) a a y y’ a y’ x+y y (x + y) x x’ a xy x’ y a x y y’ a x y y’ x y (x y) Strong Bisimilarity on LTSs Rooted Branching 18 ⑥ a.(.(x+y) + x) = a.(x+y) Soundness Completeness Process Algebra (2IF45) Equality of terms Home work • Prove soundness of B axiom wrt rooted BB • Read the proof of ground completeness 19 Process Algebra (2IF45) Combining internal step with other operators Language: BPA(A) Signature: 0, 1, (a._ )aA, , +, • Language terms T(BPA(A,)) Closed terms C(BPA(A)) Deduction rules 20 Axioms Process Algebra (2IF45) Combining internal step with other operators: Hiding operator Language: BPA(A) Signature: 0, 1, (a._ )aA, , +, •, I (I A) Language terms T(BPA(A,)) Closed terms C(BPA(A)) Deduction rules for I 21 Axioms for I Process Algebra (2IF45) turns external actions into internal steps Combining internal step with other operators: Encapsulation operator Language with Signature: 0, 1, (a._ )aA, , +, H (H A) 22 Process Algebra (2IF45) blocks actions Combining internal step with other operators: Parallel composition and communication Language: TCP(A) Signature: 0, 1, (a._ )aA, , +, •, I (I A), ||, |, Language terms T(BPA(A, )) Closed terms C(BPA(A, )) Axioms for parallel composition with silent step: x ╙ .y = x ╙ y x |.y = 0 23 Process Algebra (2IF45) ╙, H, Exercises • see distributed copies 24 Process Algebra (2IF45) 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 25 Process Algebra (2IF45) Abstraction and Recursion and Fairness Z X U Y a a 0 26 Observation: 1. they are rooted bb bisimilar 2. implicitly internal loop is left eventually = fairness 0 Process Algebra (2IF45) Abstraction and Recursion and Fairness Z X Observation on LTSs: 1. they are rooted bb bisimilar 2. implicitly internal loop is left eventually = fairness U Y a a 0 0 As recursive specifications: X = .Y Y = .Y + a.0 27 Z = .U U = a.0 RSP+RDP? Process Algebra (2IF45) X=Z Abstraction and Recursion and Fairness Z X Observation on LTSs: 1. they are rooted bb bisimilar 2. implicitly internal loop is left eventually = fairness U Y a a 0 0 As recursive specifications: X = .Y Y = .Y + a.0 Z = .U U = 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 28 Process Algebra (2IF45) Abstraction and Recursion and Fairness: problem 1. dealing with guardedness for some action i to be turned internal “soon” by applying I for I = {i} X’ = i.Y’ Y’ = i.Y’ + a.0 represents X = .Y Y = .Y + a.0 applying {i} X i Y X’ i Y’ a 0 29 a 0 Process Algebra (2IF45) Abstraction and Recursion and Fairness: problem 1. dealing with guardedness Z’ = i.U’ U’ = a.0 for some action i to be turned internal “soon” by applying I for I = {i} X’ = i.Y’ Y’ = i.Y’ + a.0 represents represents Z = .U U = a.0 X = .Y Y = .Y + a.0 applying {i} X i Y X’ 30 i i U’ Y’ a 0 applying {i} Z’ Z U a a a 0 0 Process Algebra (2IF45) 0 Abstraction and Recursion and Fairness: problem 1. dealing with guardedness Z’ = i.U’ U’ = a.0 for some action i to be turned internal “soon” by applying I for I = {i} X’ = i.Y’ Y’ = i.Y’ + a.0 represents represents Z = .U U = a.0 X = .Y Y = .Y + a.0 applying {i} X i Y X’ How to connect them i Y’ a a 0 ? 0 i a a 0 OK! Process Algebra (2IF45) Z’ U’ U 0 OK! 31 applying {i} Z Abstraction and Recursion and Fairness: problem 2. derivation rules We want to derive that I(X’) = I(Z’)! We need new rules for this! X’ = i.Y’ Y’ = i.Y’ + a.0 Something like this shall help: Y’ = i.Y’ + a.0 . I(Y’) = . I(a.0) 32 Process Algebra (2IF45) Abstraction and Recursion and Fairness: Fairness rule KFAR1b a bit more general rule: x1 = i1.x1 + y1, i1 I . I(x1) = . I(y1) 33 Process Algebra (2IF45) Abstraction and Recursion and Fairness: Fairness rule KFARnb 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)) 34 Process Algebra (2IF45) Abstraction and Recursion and Fairness: Example of tossing a coin 35 Process Algebra (2IF45) 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. 36 Process Algebra (2IF45)