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Process Algebra (2IF45) Basic Process Algebra (Soundness proof) Dr. Suzana Andova.

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Presentation on theme: "Process Algebra (2IF45) Basic Process Algebra (Soundness proof) Dr. Suzana Andova."— Presentation transcript:

1 Process Algebra (2IF45) Basic Process Algebra (Soundness proof) Dr. Suzana Andova

2 1 Outline of today lecture Soundness property of BPA(A) Example: cooking your own process algebra Process Algebra (2IF45)

3 2 Process Algebra – ingredients Process Algebra (2IF45) Language (signature) Set of constants and operators term  term 1  term 2 label Deduction rules Terms (Expressions) built from constants, operators and variables Equivalence relation on LTSs Set of Axioms (basic equalities) term 1 = term 2 Derivation gives more (derived) equalities Axiom ├ term 1 = term 2

4 3 Process Algebra (2IF45) BPA(A) Process Algebra – soundness property Language: BPA(A) Signature: 0, 1, (a._ ) a  A, + Language terms T(BPA(A)) Axioms of BPA(A): (A1) x+ y = y+x (A2) (x+y) + z = x+ (y + z) (A3) x + x = x (A4) x+ 0 = x Deduction rules for BPA(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 Bisimilarity of LTSsEquality of terms

5 4 Process Algebra (2IF45) Towards axiomatization. Consistency in mind! Summary: We have (so far) defined the language of BPA(A) We have (so far) defined the SOS semantics of BPA(A) process terms We have proved that bisimilarity is congruence relation on T(BPS(A)) NEXT: We aim to define a set of axioms for the language of BPA(A) which is consistent with the semantics, for any process terms s and t if ├ s = t then s t (soundness property) if s t then ├ s = t (completeness property)

6 5 Soundness property of BPA(A) wrt to Process Algebra (2IF45) Soundness property: If BPA(A) ├ t = r then t r, for any terms t and r in T(BPA(A)). Proof: All we need to show is that each axiom is sound, that is, 1.t + r r + t, for any terms t,r  T(BPA(A)) 2.(t + r) + s (t + r) + s, for any terms t, r, s  T(BPA(A)) 3.t + t t, for any term t  T(BPA(A)) 4.t + 0 t, for any term t  T(BPA(A)). WHY IS THIS (considering only axioms) SUFFICIENT?

7 6 Towards the proof that it is indeed “good” Soundness property of axioms wrt to Process Algebra (2IF45) Soundness property: If BPA(A) ├ t = r then t r, for any terms t and r in T(BPA(A)). Proof: All we need to show is that each axiom is sound, that is, 1.t + r r + t, for any terms t,r  T(BPA(A)) 2.(t + r) + s (t + r) + s, for any terms t, r, s  T(BPA(A)) 3.t + t t, for any term t  T(BPA(A)) 4.t + 0 t, for any term t  T(BPA(A)). WHY IS THIS SUFFICIENT? Soundness of A4. We define a relation R = {(p+0, p) | p  T(BPA(A))}. Is it bisimulation relation? Check all 4 transfer conditions for Def. of bisimulation. 1.Assume that p +0 makes a transition, p+0 –a-> p’, for some a  A and q  T(BPA(A)). It can be only (from the rules) that p -a-> p’. We have it almost, but is (p’, p’)  R?! Opsss! R is (again) not defined good. We extend it to R = {(p+0,p) | p  T(BPA(A))}  {(p, p) | p  T(BPA(A))} With this improved definition of R we have it exactly that the target process terms p’ and p’ are R related. NOTE: It is not always the case that the target process terms are the same. In opposite, this is a trivial case.

8 7 Towards the proof that it is indeed “good” Soundness property of axioms wrt to Process Algebra (2IF45) Soundness property: If BPA(A) ├ t = r then t r, for any terms t and r in T(BPA(A)). Soundness of A4 (cont). R = {(p+0,p) | p  T(BPA(A))}  {(p, p) | p  T(BPA(A))} 2. Checking the second transfer condition. Assume that p makes a transition, p –a-> p’, for some a  A and q  T(BPA(A)). Directly from the deduction rule  it follows that p +0 -a-> p’, and the target process terms are (trivially) related, (p’, p’)  R. 3. Checking the third transfer condition. Assume that (p+0) . Either rule  or ⑥ could have been applied to obtain this, meaning that either p  or 0 . But since is NOT 0  then it must be p . This proves that, (p+0)  implies p . 4. Checking the fourth transfer condition. Assume that p . Directly from rule  we obtain that (p +0) . This proves that, p  implies (p+0) . We have to make similar proofs for other three axioms.

9 8 Process Algebra (2IF45) BPA(A) Process Algebra – soundness property Language: BPA(A) Signature: 0, 1, (a._ ) a  A, + Language terms T(BPA(A)) Axioms of BPA(A): (A1) x+ y = y+x (A2) (x+y) + z = x+ (y + z) (A3) x + x = x (A4) x+ 0 = x Deduction rules for BPA(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 Bisimilarity of LTSsEquality of terms

10 9 Example: making a process algebra Our first own BPA(A) extension We extend BPA(A) basic process algebra with an unary operator D(_). D(x) “doubles” every action that x can execute Examples: D(a.b.0) = a.a.b.b.0 D(a.b.1 + c.0) = a.a.b.b.1 + c.c.0 … Construct the BPAD(A) process algebra! Process Algebra (2IF45)

11 10 Example: recipe Follow the steps: Define axioms for the new operator Define SOS rules for the new operator Mix the ingredients: check whether they all fit well! −Congruence of with respect to the new operator −Soundness of the new axioms −Completeness q.e.d. Process Algebra (2IF45)

12 11 Example: Our first own BPA(A) extension Process Algebra (2IF45) Solution: Axioms of BPAD(A) are those of BPA(A) and: D(0) = 0 D(1) = 1 D(a.x) = a.a.D(x) D(x + y) = D(x) + D(y) SOS rules for the term-deduction system of BPAD(A) includes the SOS rules of BPA(A) and the following two rules: For the proofs of congruence of with respect to the new D operator and the soundness proof for the new axioms you should use the proof strategies seen already. x  x’ D(x)  a.D( x’) a a x  D(x) 


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