/ department of mathematics and computer science to the transition system above. termination-behaviour. equation specifying for every recursion variable its transition- and 2. associate behaviour to every recursion variable by means of an 1. label states with process names (a.k.a. recursion variables); General method for expressing transition systems: Expressiveness (3) / department of mathematics and computer science 4/30 Answer Process Algebra (2IMF10) Question Expressiveness (2) / department of mathematics and computer science 3/30 Can we express the following transition system Answer Expressiveness (1) Recursion Bas Luttik MF 6.072 s.p.luttik@tue.nl http://www.win.tue.nl/~luttik Lectures 6–7 / department of mathematics and computer science 2/30 Question Can we express the following transition system a c b in the syntax of BSP( A ) . Yes! The transition system generated by a .( b .0 + 1) + c .0 is isomorphic a b c a X Y X = a . Y + c . Z b Y = b . X + 1 c in the syntax of BSP( A ) . Z = 0 Z No! Infjnite behaviour cannot be expressed in the syntax of BSP( A ) .
5/30 Defjnition / department of mathematics and computer science 9/30 Proposition / department of mathematics and computer science Theorem 10/30 / department of mathematics and computer science Equivalence of recursion variables Example Consider the recursive specifjcation Operational rules for recursion Term model (simplifjed) [ Warning: notation slightly simplifjed; we will generalise later! ] 6/30 Recursive specifjcations [ Remark: we shall not discuss a full-fmedged ground-complete axiomatisation of of recursion variables. Conclusion: we need additional methods to reason about the equivalence / department of mathematics and computer science Recursion: operational semantics Let Σ be a signature and let V R be a set of recursion variables A recursive equation over Σ and V R is an equation of the form Let E be a recursive specifjcation over Σ and V R including, for every X ∈ V R , a defjning equation X = t X . X = t , with X ∈ V R and t a term over Σ and V R . We say that the recursive equation X = t defjnes X . a − → t ′ t X t X ↓ X a X ↓ A recursive specifjcation over Σ and V R is a set of recursive equations − → t ′ X X over Σ and V R consisting of precisely one recursive equation defjning X for every X ∈ V R . Let E be a recursive specifjcation over BSP( A ) and V R . The term algebra for BSP( A )+ E is the algebra { X = a . X , } P (BSP( A )+ E ) = ( C (BSP( A )+ E ), +, ( a .) a ∈ A , 0, 1, ( X ) X ∈ V R ) . Y = a . a . Y Then X ↔ Y . But this equation cannot be derived from BSP( A ) + E . (Prove!) Bisimilarity is a congruence on P (BSP( A )+ E ) . The equational theory BSP( A )+ E is a sound axiomatisation of P (BSP( A )+ E ) / ↔ . the algebra P (BSP( A )+ E ) / ↔ , but we’ll come close.] Is BSP( A )+ E also ground-complete for P (BSP( A )+ E ) / ↔ ?
11/30 / department of mathematics and computer science The converse, however, need not hold: that algebra. The reasoning on the previous slide allows us to conclude that every Consider the recursive specifjcation Example: equivalence of rec. vars. / department of mathematics and computer science 15/30 Consider the recursive specifjcation / department of mathematics and computer science / department of mathematics and computer science 14/30 Solutions: examples Example: equivalence of rec. vars. 12/30 Recursion: algebraic approach bisimilarity is an assignment of closed terms to recursion variables such that the recursion equations are true up to bisimilarity. Defjnition Let E be a rec. spec. over signature Σ and set of variables V R . Furthermore, let A be a Σ -algebra and ι the associated interpretation. 1. The recursive specifjcation E 1 = { X = a .1 } has a solution both in P (BSP( A )) / ↔ and in P (BSP( A )+ E 1 ) / ↔ . A solution of E in A is an extension κ of ι with interpretations of the 2. The recursive specifjcation E 2 = { X = a . X } has a solution in recursion variables in V R as elements of A such that A , κ | = X = t X for P (BSP( A )+ E 2 ) / ↔ , but not in P (BSP( A )) / ↔ . every equation X = t X in E . If κ is a solution of E and X ∈ V R , then we shall call κ ( X ) a solution of X in E . 3. The recursive specifjcation E 3 = { X = X } has many solutions, both in P (BSP( A )) / ↔ and in P (BSP( A )+ E 3 ) / ↔ . NB: a solution of E in an algebra of transition systems modulo { X = a . X , { X = a . X , } } . . Y = a . a . Y Y = a . a . Y Note that we can argue that every solution of X is a solution of Y too: solution of X in whatever algebra (!) must also be a solution of Y in X = a . X = a . a . X . Hence, any solution κ of E in some algebra A satisfjes Exercise: construct a model of BSP( A ) in which Y has a solution that is not also a solution of X . κ ( X ) = κ ( a . a . X ) = a . a . κ ( X ) ( a . the interpretation of a . in A ) , so κ ( X ) is a solution of Y in E .
16/30 is completely guarded. / department of mathematics and computer science 19/30 / department of mathematics and computer science Guardedness Example 1. The recursive specifjcation 2. The recursive specifjcation Exercise 5.5.1 is not completely guarded. 20/30 / department of mathematics and computer science Guardedness Defjnition Example Although the recursive specifjcation Determine whether, in the following terms, the occurrences of the is not completely guarded, it is guarded. occurs in the scope of an action prefjx. / department of mathematics and computer science the previous slide), then we may be able to say more about the denote the same process! 18/30 Defjnition Guardedness Example: equivalence of rec. vars. Consider the recursive specifjcation Perhaps if we exclude some models (e.g., the answers to the exercise on An occurrence of a recursion variable X in a closed term s is guarded if it { X = a . X , } A term s is completely guarded if all occurrences of all recursion variables . Y = a . a . Y in s are guarded. A recursive specifjcation E is completely guarded if all right-hand sides of all equations in E are completely guarded. equivalence of X and Y in the above recursive specifjcation. Note that, for models in which X and Y both have a unique solution, recursion variables X and Y are guarded, unguarded, or both: the reasoning on slide 14 would suffjce to conclude that X and Y indeed a . X , Y + b . X , b .( X + Y ) , a . Y + X . A recursive specifjcation E is guarded if there exists a completely guarded E 1 = { X 1 = a . X 1 , Y 1 = a . X 1 } recursive specifjcation F with V R ( E ) = V R ( F ) and BSP( A )+ E ⊢ X = t for all X = t ∈ F . E 2 = { X 1 = a . X 1 , Y 1 = X 1 } E 2 = { X 1 = a . X 1 , Y 1 = X 1 } But E 1 and E 2 have exactly the same solutions in every model!
21/30 / department of mathematics and computer science Extension with recursion [Not in the book!] / department of mathematics and computer science Example RSP as a proof principle / department of mathematics and computer science 23/30 24/30 Recursive Specifjcation Principle Defjnition 22/30 Determine whether the following recursive specifjcations are guarded or Guardedness Defjnition Exercise 5.5.2 / department of mathematics and computer science unguarded: A recursive specifjcation E is guarded if there exists a completely guarded recursive specifjcation F with V R ( E ) = V R ( F ) and BSP( A )+ E ⊢ X = t for all X = t ∈ F . Let Σ be a signature; we say that Σ -algebra A satisfjes the Recursive Specifjcation Principle (RSP) if every guarded recursive specifjcation E over Σ and some set V R of variables has at most one solution. 1. { X = Y , Y = a . X } ; and 2. { X = a . Y + Z , Y = b . Z + X , Z = c . X + Y } . Consider rec. spec. E consisting of the following equations: Let A be a set of actions, and let T ( A ) be some process theory. X = a . X + b . X , Let Rec be the collection of all recursive specifjcations over T ( A ) . Y = a . Y + b . Z , and Z = a . Z + b . Y . T ( A ) rec : extension of T ( A ) with for every rec. spec. E over T ( A ) and for every ( X = t X ) ∈ E : We can prove that X = Y in the context of E as follows: 1. a constant symbol µ X . E , and Defjne two sequences of terms ˜ t = t X , t Y , t Z and ˜ u = u X , u Y , u Z by 2. an axiom µ X . E = µ t X . E ( Rec ) . t X ≡ X , t Y ≡ Y , t Z ≡ Z , and u X ≡ X , u Y ≡ X , u Z ≡ X . [We write µ t . E for the term obtained from t by replacing all occurrences of a Then both ˜ t and ˜ u denote solutions of E (verify!). recursion variable X by µ X . E (see book for an inductive defjnition).] Since E is guarded, by RSP, ˜ t = ˜ u , so X ≡ u Y = t Y ≡ Y .
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