The Composition Method Wolfgang Thomas Francqui Lecture, Mons, April 2013
Mastering compositions Wolfgang Thomas
Overview 1. Motivation 2. m -equivalence and the EF-game 3. Applications 4. m -types 5. Monadic types Wolfgang Thomas
Motivation Wolfgang Thomas
Composition and Decomposition General problem: How to know what is true in a composed system if one knows what is true in the components? Essential question in verification. Here we consider two kinds of compositions: Ordered sums (e.g., concatentation of word models) Products Wolfgang Thomas
Recall Automata Given a DFA A = ( Q , Σ , q 0 , δ , F ) “If we know the behaviour of A on u and on v then we know the behaviour on uv .” We capture “behaviour” by the state transformations over Q realized by words. For u ∈ Σ ∗ define u A : Q → Q by u A ( q ) = δ ( q , u ) The set { u A | u ∈ Σ ∗ } of state transformations forms a finite monoid with composition and identity ε A . ( uv ) A = u A ◦ v A Wolfgang Thomas
Composition on Level of Automata A accepts uv iff ( u A ( q 0 ) = p ∧ v A ( p ) ∈ F ) � p ∈ Q Nondeterministic version: Use relation � u � A = { ( p , q ) | A : p u → q } A accepts uv iff � u � A ( q 0 , p ) ∧ � v � A ( p , r ) � p ∈ Q , r ∈ F Wolfgang Thomas
m -Equivalence and the EF-Game Wolfgang Thomas
Composition in Logic How to obtain information whether uv | = ϕ from knowledge about u and v ? Solution: When dealing with a formula ϕ do not look into ϕ but consider all formulas of the same quantifier complexity. More precisely: The quantifier-depth qd ( ϕ ) of a formula ϕ is the maximal number of nested quantifiers in ϕ . The quantifier alternation depth qad ( ϕ ) of ϕ is the number of blocks of existential resp. universal quantifiers in the prenex normal form of ϕ . Wolfgang Thomas
Format of Models Fix a signature with unary relation symbols Q 1 , . . . , Q k and binary relation symbols R 1 , . . . , R ℓ . Obtain relational structures S = ( S , Q S 1 , . . . , Q S k , R S 1 , . . . , R S l ) Satisfaction relation: ( S , s ) | = ϕ ( x ) Special case: Word models ovder Σ = { a , b } : w = ( dom ( w ) , < , Suc, Q a , Q b ) Example w = aaba : dom ( w ) = { 1, 2, 3, 4 } , Q a = { 1, 2, 4 } , Q b = { 3 } . Wolfgang Thomas
Ordered Sums For i ∈ I we are given relational structures M i = ( M i , R i 1 , . . . ) of the same signature First focus on I = { 1, 2 } . M 1 + M 2 is the structure M = ( M 1 ∪ M 2 , R 1 1 ∪ R 2 1 , . . . ) If M i is ordered by < i , then the ordered sum has the following ordering < : a < b iff a , b belong to the same M i and a < i b , or a ∈ M 1 , b ∈ M 2 Similarly for arbitrary orderings ( I , < I ) . Wolfgang Thomas
m -Equivalence Two structures ( S , s ) and ( T , t ) are m-equivalent (short: ( S , s ) ≡ m ( T , t ) ) if ( S , s ) | = ϕ ( x ) ⇐ ⇒ ( T , t ) | = ϕ ( x ) for all formulas ϕ ( x ) of quantifier-depth ≤ m . The m -equivalence classes are also called m -types. Plan: 1. Find a way to show that two structures are m -equivalent. 2. Present some applications 3. Introduce descriptions of the m -types Wolfgang Thomas
A. Ehrenfeucht R. Fra¨ ıss´ e Wolfgang Thomas
Ehrenfeucht-Fra¨ ıss´ e game allows to verify ( S , s ) ≡ m ( T , t ) . A game position is a partial isomorphism: a finite relation { ( s 1 , t 1 ) , . . . , ( s n , t n ) } ⊆ S × T denoted by s �→ t , which is injective and preserves all relations Q S , R S under consideration: s ∈ Q S ⇐ ⇒ p ( s ) ∈ Q T and ( s , s ′ ) ∈ R S ⇐ ⇒ ( p ( s ) , p ( s ′ )) ∈ R T Wolfgang Thomas
Game G m (( S , s ) , ( T , t )) played between two players called Spoiler and Duplicator ( S , s ) and ( T , t ) . There are m rounds. The initial configuration is s �→ t . Given a configuration r , a round is composed of two moves: first Spoiler picks an element s from S or t from T , and then Duplicator reacts by choosing an element in the other structure, i.e. by choosing some t from T , resp. some s from S . The new configuration is r ∪ { ( s , t ) } . After m rounds, Duplicator has won if the final configuration is a partial isomorphism (otherwise Spoiler has won). Wolfgang Thomas
Example 1 Let u = aabaacaa and v = aacaabaa Consider G 2 ( u , v ) (including the linear ordering < ) Duplicator looses: Spoiler can pick the u -positions with the letters b and c , whence Duplicator can only respond by picking the positions with b and c in v , in order to preserve the relations Q b and Q c ; but then the order between the positions < is not preserved. Consider ∃ x ∃ y ( x < y ∧ Q b ( x ) ∧ Q c ( y )) Wolfgang Thomas
Example 2 as before, however with successor relation Suc only in word models, besides Q a , Q b , Q c . u = aabaacaa and v = aacaabaa Duplicator wins. If Spoiler picks a position with b or c or a position adjacent to one of them, Duplicator reacts accordingly in the other word; Otherwise Duplicator reacts by corresponding positions. Consider, e.g., ∃ x ∃ y ( Suc ( x , y ) ∧ Q a ( x ) ∧ Q b ( y )) Wolfgang Thomas
Example 3 Word models with order but without successor, singleton alphabet { a } . Format: ( dom ( w ) , < , Q a ) . Duplicator wins G 2 ( aaa , a n ) for any n ≥ 3 : In the first round, Spoiler may pick a first position, a last position, or a non-border position in one of the two words, and Duplicator reacts accordingly. This allows Duplicator also to respond correctly (i.e., order-preserving) in the second round. Wolfgang Thomas
G 3 ( a i , a j ) Here after the first round we get the situation a i = a i 1 aa i 2 and a j = a j 1 aa j 2 Remembering the 2-rounds game, Duplicator will win if i 1 , j 1 are both ≥ 3 or else i 1 = j 1 , and similarly for i 2 , j 2 . Duplicator can reach such a decomposition in the first round if i , j are both ≥ 7 , or if i = j . Wolfgang Thomas
In general, . . . With k rounds ahead, Duplicator ensures that corresponding letter-blocks delimited by chosen positions are of length ≥ 2 k − 1 or are of the same length. Duplicator wins G m ( a i , a j ) for any i , j ≥ 2 m − 1 Duplicator also wins G m ( w i , w j ) for any word w and i , j ≥ 2 m − 1 . Keep in mind: Sentences of qd m can desribe repetitions up to threshold 2 m − 1 but otherwise can just say “many”. Wolfgang Thomas
Describing Winning Strategy When does Duplicator win G m (( S , s ) , ( T , t )) ? Specify, for each k = 0, . . . , m , a set I k of partial isomorphisms (describing game positions) which would Duplicator allow to win with k rounds ahead. There should be nonempty sets I m , . . . , I 0 of partial isomorphisms, each of them extending s �→ t , such that for all k = m , . . . , 1 : (back property) ∀ p ∈ I k ∀ t ∈ T ∃ s ∈ S such that p ∪ { ( s , t ) } ∈ I k − 1 (forth property) ∀ p ∈ I k ∀ s ∈ S ∃ t ∈ T such that p ∪ { ( s , t ) } ∈ I k − 1 . Write ( S , s ) � m ( T , t ) . Wolfgang Thomas
Ehrenfeucht-Fra¨ ıss´ e Theorem For m ≥ 0 , the following are equivalent: 1. ( S , s ) ≡ m ( T , t ) 2. ( S , s ) � m ( T , t ) 3. Duplicator wins G m (( S , s ) , ( T , t )) . Wolfgang Thomas
Applications Wolfgang Thomas
Non-Definability The language { a n | n is even } is not first-order definable. Suppose a defining first-order sentence ϕ exists, with < only, say of quantifier-depth m . We have a 2 m ≡ m a 2 m + 1 We have a 2 m | = ϕ . So also a 2 m + 1 | = ϕ . This model is of odd length, contradiction. Wolfgang Thomas
Finally Composition! In order to know whether a formula ϕ of qd m holds in uv , it suffices to know the m -types of u and v . Composition Lemma If u ≡ m u ′ and v ≡ m v ′ , then u · v ≡ m u ′ · v ′ . Use the Ehrenfeucht-Fra¨ ıss´ e Theorem. Duplicator has winning strategies for the games G m ( u , u ′ ) and G m ( v , v ′ ) . The strategy “on the segments u and u ′ use the first strategy, on the segments v and v ′ use the second strategy” guarantees Duplicator to win also the game G m ( u · v , u ′ · v ′ ) . Wolfgang Thomas
Products The direct product of S 1 , S 2 has S 1 × S 2 as its universe, with relations R S 1 × S 2 ( a 1 , b 1 ) . . . ( a n , b n ) iff R S 1 a 1 . . . a n and R S 2 b 1 . . . b n Special forms: Reduced product, synchronized product. Landmark paper by Feferman and Vaught 1959 Wolfgang Thomas
Composition for Products In order to know whether a formula of qd m holds in S × T , it suffices to know the m -types of S and T . Show: If S ≡ m S ′ and T ≡ m T ′ , then S × T ≡ m S ′ × T ′ . Wolfgang Thomas
m -Types Wolfgang Thomas
How to Describe Types? Hintikka formulas: ψ 0 � � s ( ¯ x ) : = x ∧ ¬ R ¯ R ¯ x S ,¯ s ∈ R S s �∈ R S R , ¯ R ,¯ ψ m + 1 ∃ x ψ m � s ( ¯ x ) : = s , s ( ¯ x , x ) S ,¯ S ,¯ = ψ m s ∈ S , S ,¯ s | s , s ( ¯ x , x ) S ,¯ ψ m � ∧ ∀ x s , s ( ¯ x , x ) S ,¯ = ψ m s ∈ S , S ,¯ s | S ,¯ s , s Wolfgang Thomas
Shorter Notation T 0 ( S , ¯ s ) : = { ψ ( x ) | ψ atomic , ( S , ¯ s ) | = ψ ( x ) } T m + 1 ( S , ¯ s ) : = { T m ( S , ¯ s , s ) | s ∈ S } Wolfgang Thomas
Recommend
More recommend
