index.md (3745B)
1 +++ 2 title = 'Lecture 5' 3 template = 'page-math.html' 4 +++ 5 # Lecture 5 6 If two states are bisimilar, then they are modally equivalent. 7 Prove by induction on definition of formulas. 8 9 For finitely branching models, if two states modally equivalent, then they are bisimilar. 10 Prove by taking modal equivalence as bisimulation. 11 12 Let Z := {(x, x') | for all φ: x ⊨ φ iff x' ⊨ φ}. 13 Z is a bisimulation. 14 - Local harmony is satisfied. 15 - zig: suppose (x, x') ∈ Z and x → y. 16 Because x ⊨ ◇ T also x' ⊨ ◇ T. 17 So x' has finitely many successors y'₁...y'n with n ≥ 1. 18 Suppose for all i: y is not modally equivalent to y'ᵢ. 19 There are φ₁...φn such that y ⊨ φᵢ and y'ᵢ ⊭ φᵢ. 20 x ⊨ ◇(φ₁ ∧ ... ∧ φn) so also x' ⊨ ◇ (φ₁ ∧ ... ∧ φn). 21 Contradiction. 22 23 24 Asymmetry is not modally definable. 25 To deal with only frames, we can use surjective bounded morphisms. 26 27 Function f : W → W' is bounded morphism from (W,R) to (W',R') if: 28 - for all w,v ∈ W: if Rwv then Rf(w)f(v) 29 - for all w ∈ W and for all v' ∈ W': if R'f(w)v' then there exists v ∈ W such that f(v) = v' and Rwv 30 31 A bounded morphism f is surjective if for every w' ∈ W' there exists w ∈ W such that f(w) = w' 32 33 If f: W → W' is surjective bounded morphism from (W,R) to (W',R'), then if (W,R) ⊨ φ then (W', R') ⊨ φ 34 35 ### Bisimulation games for two players 36 Spoiler S claims M,s an N,t to be different. 37 Duplicator D claims they are similar. 38 39 Play consists of sequence of links, starting with link s ~ t. 40 41 At current link m ~ n (with m in M and n in N): 42 - if m an n different in their atoms, then S wins 43 - if not, then S picks a successor x either of m or of n 44 - then D has to find a matching transition to y in the other model 45 - play continues with next link x ~ y (or y ~ x) 46 47 If player cannot make a move, he loses. 48 D wins infinite games. 49 50 Modal depth of formulas. 51 Modal formulas are 'nearsighted': 52 - md(p) = md(⊥) = md(T) = 0 53 - md(¬φ) = md(φ) 54 - md(φ ∨ ψ) = md(φ ∧ ψ) = max{md(φ), md(ψ)} 55 - md(□ φ) = md(◇ φ) = md(φ) + 1 56 57 We need a formula of model depth k to distinguish states x and y. 58 Spoiler can win bisimulation game in k rounds. 59 Every winning strategy for Spoiler corresponds to a distinguishing formula. 60 Games of less than k rounds can be won by Duplicator. 61 Formulas of modal depth less than k cannot distinguish between x and y. 62 63 M,s and N,t satisfy the same formulas up to modal depth k iff duplicator has winning strategy in the k-round game starting in s ~ t. 64 If Spoiler can win in k rounds, then there is a distinguishing formula of modal depth k. 65 66 ### Transforming and constructing models 67 #### Tree unraveling 68 Basically taking the states with possible paths between them, and drawing a tree. 69 70 Example: 71 72 <table> 73 <tr> <th>Model</th> <th>Tree unraveling</th> </tr> 74 75 <tr> 76 <td> 77 78  79 80 <details> 81 <summary>Graphviz code</summary> 82 83 <!-- :Tangle(dot) model-diagram.dot --> 84 ```dot 85 digraph g { 86 1 -> 2 87 1 -> 1 88 } 89 ``` 90 91 </details> 92 </td> 93 <td> 94 95  96 97 <details> 98 <summary>Graphviz code</summary> 99 100 <!-- :Tangle(dot) tree-unraveling.dot --> 101 ```dot 102 digraph g { 103 a [label="(1)"] 104 b [label="(1,2)"]; a -> b 105 c [label="(1,1)"]; a -> c 106 d [label="(1,1,2)"]; c -> d 107 e [label="(1,1,1)"]; c -> e 108 f [label="(1,1,1,2)"]; e -> f 109 g [label="..."]; e -> g 110 } 111 ``` 112 113 </details> 114 115 </td> 116 </tr> 117 </table> 118 119 If two trees have the same structure, they're bisimilar. 120 121 #### Model contraction 122 Basically getting rid of 'unnecessary' states. 123 124  125 126 ### Validity and satisfiability 127 If φ is satisfiable, then it is satisfiable using a model with at most $2^{s(\phi)}$ elements with s(φ) the number of subformulas of φ.