Enhancements of the bisimulation proof method Davide Sangiorgi - PowerPoint PPT Presentation

Enhancements of the bisimulation proof method Davide Sangiorgi Focus Lab., INRIA (France) and University of Bologna (Italy) Email: Davide.Sangiorgi@cs.unibo.it http://www.cs.unibo.it/˜sangio/ BASICS 2009, Shanghai

CONTENTS ☞ • Introduction [1] • Part I: Examples [3] • Part II: Counterexamples [40] • Part III: Towards an algebra of enhancements [48] • Part IV: Weak bisimilarity [58]

Equality on processes, coinductively Bisimulation: Q A relation R s.t. P R α α Q ′ P ′ R Bisimilarity ( ∼ ) : ∪ {R : R is a bisimulation } Hence: x R y R is a bisimulation (bisimulation proof method) x ∼ y page 1

This talk Enhancements of the bisimulation proof method – Motivations – Results and Examples – Open problems page 2

CONTENTS ✓ • Introduction [1] ☞ • Part I: Examples [3] • Part II: Counterexamples [40] • Part III: Towards an algebra of enhancements [48] • Part IV: Weak bisimilarity [58]

Examples, in CCS-like calculus page 3

Pieces of syntax and transitions a → P ′ P | ! P − a inp : a . P − → P rep : a ! P − → P ′ a a → P ′ → Q ′ P − Q − sumL : sumR : a a P + Q − → P ′ P + Q − → Q ′ a a → P ′ → Q ′ P − Q − parL : parR : a a → P ′ | Q P | Q − P | Q − → P | Q ′ Intuitively: ! P = P | P | . . . | P (indeed P | ! P ∼ ! P , as the transitions are the same) � Process abbreviations: a a . 0 P n � P | . . . | P ( n times) page 4

Enhancements of the bisimulation method: an example A property of replication !( a + b ) ∼ ! a | ! b Proof: Let’s find a bisimulation... page 5

Is this a bisimulation? R � { (!( a + b ) , ! a | ! b ) } page 6

Is this a bisimulation? R � { (!( a + b ) , ! a | ! b ) } No! a ( a + b ) n | 0 | ( a + b ) m | !( a + b ) !( a + b ) − → � R R a a n | 0 | a m | ! a | ! b ! a | ! b − → a ( a + b ) n | 0 | ( a + b ) m | !( a + b ) NB: ∀ n, m , !( a + b ) − → a a n | 0 | a m | ! a | ! b ! a | ! b − → Try again... page 7

Is this a bisimulation? R � ∪ n,m (( a + b ) n | 0 | ( a + b ) m | !( a + b ) , a n | 0 | a m | ! a | ! b ) , { (( a + b ) n | 0 | ( a + b ) m | !( a + b ) , ! a | b n | 0 | b m | ! b ) } page 8

Is this a bisimulation? R � ∪ n,m (( a + b ) n | 0 | ( a + b ) m | !( a + b ) , a n | 0 | a m | ! a | ! b ) , { (( a + b ) n | 0 | ( a + b ) m | !( a + b ) , ! a | b n | 0 | b m | ! b ) } No! b ( a + b ) n | 0 | ( a + b ) m ( a + b ) n − 1 | 0 2 | ( a + b ) m − → | !( a + b ) | !( a + b ) � R R b a n | 0 | a m | ! a | ! b a n | 0 | a m | ! a | b | ! b − → Try again... page 9

It is possible to write the full bisimulation, but one has to be careful We started with the singleton relation { (!( a + b ) , ! a | ! b ) } The added pairs: redundant ? (derivable, laws of ∼ ) Can we work with relations smaller than bisimulations? Advantages: – fewer and simpler bisimulation diagrams – easier to find the relation to work with page 10

Redundant pairs What we would like to have: Q implies R ⊆ ∼ P R α α Q ′ R ∪ { redundant pairs } P ′ R : less work, simpler to find Notation R ֌ S � Q P R α α Q ′ P ′ S page 11

Up-to techniques: example – Rules for transitivity of ∼ (up-to ∼ ) R ֌ ∼ R ∼ implies R ⊆ ∼ Q diagram : P R α α P ′′ R Q ′′ Q ′ ∼ ∼ P ′ page 12

Now we can establish !( a + b ) ∼ ! a | ! b using the singleton relation R � { !( a + b ) , ! a | ! b } and proving that it is a bisimulation up-to ∼ a ( a + b ) n | 0 | ( a + b ) m | !( a + b ) !( a + b ) − → ∼ !( a + b ) R R ! a | ! b ∼ a a n | 0 | a m | ! a | ! b ! a | ! b − → (laws P | ! P ∼ ! P and 0 | P ∼ P , congruence of ∼ ) page 13

A more interesting example !( a . P + b . Q ) ∼ ! a . P | ! b . Q Proof: Let’s find a bisimulation... page 14

Is this a bisimulation up-to ∼ ? R � { (!( a . P + b . Q ) , ! a . P | ! b . Q ) } page 15

Is this a bisimulation up-to ∼ ? R � { (!( a . P + b . Q ) , ! a . P | ! b . Q ) } No! a !( a . P + b . Q ) − → ∼ P | !( a . P + b . Q ) � R R a ! a . P | ! b . Q − → ∼ P | ! a . P | ! b . Q page 16

c → P ′ : – Note also, if P − → a a → c P ′ | P | !( a . P + b . Q ) !( a . P + b . Q ) − − − → → a a → c P ′ | P | ! a . P | ! b . Q ! a . P | ! b . Q − − − → so a bisimulation up-to ∼ should include also such pairs of derivates – Again, these added pairs may be considered redundant (for instance, !( a . P + b . Q ) ∼ ! a . P | ! b . Q implies P ′ | P | !( a . P + b . Q ) ∼ P ′ | P | ! a . P | ! b . Q ) – We can avoid these additional pairs using a different form of up-to page 17

Up-to techniques: example – Rules for transitivity of ∼ (up-to ∼ ) – rules for substitutivity of ∼ (up-to context) C ( R ) � { ( C [ P ] , C [ Q ]) : P R Q } R ֌ C ( R ) implies R ⊆ ∼ Q diagram : P R α α � [ P ′ ] � [ Q ′ ] C C R page 18

Example of composition of techniques We can put together up-to ∼ and up-to context R ֌ ∼ C [ R ] ∼ implies R ⊆ ∼ Q diagram : P R α α � [ Q ′′ ] � [ P ′′ ] R C ∼ ∼ C Q ′ P ′ page 19

Back to our proof: R � { !( a . P + b . Q ) , ! a . P | ! b . Q } is a bisimulation up-to ∼ and up-to context a ( a . P + b . Q ) n | P | ( a . P + b . Q ) m !( a . P + b . Q ) − → | !( a . P + b . Q ) R a ( a . P ) n | P | ( a . P ) m | ! a . P | ! b . Q ! a . P | ! b . Q − → page 20

Back to our proof: R � { !( a . P + b . Q ) , ! a . P | ! b . Q } is a bisimulation up-to ∼ and up-to context a ( a . P + b . Q ) n | P | ( a . P + b . Q ) m !( a . P + b . Q ) − → | !( a . P + b . Q ) ∼ P | !( a . P + b . Q ) R P | ! a . P | ! b . Q ∼ a ( a . P ) n | P | ( a . P ) m | ! a . P | ! b . Q ! a . P | ! b . Q − → page 21

Back to our proof: R � { !( a . P + b . Q ) , ! a . P | ! b . Q } is a bisimulation up-to ∼ and up-to context a ( a . P + b . Q ) n | P | ( a . P + b . Q ) m !( a . P + b . Q ) − → | !( a . P + b . Q ) ∼ � P | !( a . P + b . Q ) R � P | ! a . P | ! b . Q ∼ a ( a . P ) n | P | ( a . P ) m | ! a . P | ! b . Q ! a . P | ! b . Q − → page 22

Back to our proof: R � { !( a . P + b . Q ) , ! a . P | ! b . Q } is a bisimulation up-to ∼ and up-to context a ( a . P + b . Q ) n | P | ( a . P + b . Q ) m !( a . P + b . Q ) − → | !( a . P + b . Q ) ∼ � P | !( a . P + b . Q ) R R � P | ! a . P | ! b . Q ∼ a ( a . P ) n | P | ( a . P ) m | ! a . P | ! b . Q ! a . P | ! b . Q − → page 23

More up-to techniques: example – Rules for transitivity of ∼ (up-to ∼ ) – rules for substitutivity of ∼ (up-to context) – rules for invariance of ∼ under injective substitutions (up-to injective substitutions) Inj ( R ) � { ( P σ, Qσ ) : P R Q , σ injective on names } R ֌ Inj ( R ) implies R ⊆ ∼ Q P R α α diagram : implies R ⊆ ∼ � � P ′ σ Q ′ σ R σ : an injective function page 24

More composition of techniques R ֌ ∼ C [ Inj ( R )] ∼ implies R ⊆ ∼ Q diagram : P R α α � [ Q ′′ σ � [ P ′′ σ � ] R C � ] ∼ C ∼ Q ′ P ′ More sophistication ⇒ – more powerful technique – harder soundness proof for the technique page 25

More examples, in a higher-order calculus (the Ambient calculus) page 26

Ambients: syntax Processes P ::= n � P � ambient | in n . P in action | out n . P out action | open n . P open action | P | P parallel | νn P restriction | . . . page 27

The in movement m n m n − → | in m . P Q | Q P The out movement m n m n − → | P 1 | P 2 Q out n . P 1 | P 2 | Q page 28

Example property The perfect-firewall equation in Ambients P : a process with n not free in it νn n � P � ∼ 0 Proof: Let’s find a bisimulation... page 29

Is this a bisimulation? R � { ( νn n � P � , 0) } page 30

Is this a bisimulation? R � { ( νn n � P � , 0) } enter k � Q � Suppose n � P � − − − − − − − − − → n � P � No! (the loop: simplifies the example, not necessary) νn n � P � 0 R enter k � Q � enter k � Q � � R k � Q | νn n � P � � k � Q � | 0 Try again... page 31

Is this a bisimulation? R � { ( νn n � P � , 0) } ∪ k,Q { ( k � Q | νn n � P � � , k � Q � | 0) } page 32

Is this a bisimulation? R � { ( νn n � P � , 0) } ∪ k,Q { ( k � Q | νn n � P � � , k � Q � | 0) } No! Suppose Q = h � out k . R � | Q ′ k � Q | νn n � P � � k � Q � | 0 R k � Q ′ | νn n � P � � | h � R � � R k � Q ′ � | h � R � | 0 Try again... page 33

Is this a bisimulation? R � { ( νn n � P � , 0) } ∪ k,Q { ( k � Q | νn n � P � � , k � Q � | 0) } Also: Suppose Q = in h . Q ′ k � Q | νn n � P � � k � Q � | 0 R enter h � R � enter h � R � h � R | k � Q ′ | νn n � P � � � h � R | k � Q ′ � � | 0 � R Try again... page 34