Finite Model Theory Lecture 4 Ordered Structures, Gaifman’s Theorem

Outline Order-invariant queries Gaifman’s theorem (time permitting) Gurevitch’s theroem Gaifman’s theorem (time permitting)

Invariant Queries Idea: Use some innocuous additional information to express a query “Forget” the extra information Need to check that the query is invariant w.r.t. to the extra info

Example Recall: EVEN(A) is not expressible in FO Let s<,+ = {<,+} Consider A 2 STRUCT[s<,+] s.t. < is interpreted as a linear order a1 < … < an (ai, aj, ak) 2 + iff i+j=k Then, we can write a sentence f that checks EVEN(A) [ HOW ?? ]

Example (cont’d) f = 9 x. 9 y.(x+x=y Æ : 9 z(y < z)) For any vocabulary s and A 2 STRUCT[s], if A’ is any structure over the same universe and vocabulary s<,+ then (A, A’) ² f iff EVEN(A)

C-Invariance Let s, s’ be two disjoint vocabularies, and C a class of s’-structures (C µ STRUCT[s’]) Definition A sentence f over s [ s’ is called C-invariant if for any A 2 STRUCT[s] and A’, A’’ 2 C, s.t. A=A’=A” (i.e. same universes) (A,A’) ² f iff (A,A’’) ² f Notation f 2 (FO+C)inv

C-Invariance (cont’d) We have seen that EVEN 2 (FO + (<,+))inv Main question for today: Is (FO + <)inv more powerful than FO ?

Order-Invariant FO Theorem (Gurevich) There are properties in (FO + <)inv that are not in FO. FO & (FO + <)inv

The Proof Idea: recall that we can express EVEN in MSO(<). Let S(x,y) = :(9 z.x<z Æ z<y) Æ x < y We can’t say 9 P in FO. Instead, we will use a vocabulary s that allows to assert the existence of all 2n sets P. 9 P.( 8 x.y.(S(x,y) ) (P(x)Æ: P(y) Ç : P(x)ÆP(y)) Æ P(min) Æ : P(max)) EVEN=

The Proof 3 Steps Design s, fB s.t. A ² fB iff A is an atomic boolean algebra, A = (2X, µ) Let EVENB = (EVEN(|X|) Æ fB Show that EVENB 2 (FO + <)inv Prove (using EF-games) that EVENB Ï FO

Boolean Algebras Background: two alternative definitions Definition 1. A boolean algebra is A = (A, µ) s.t.: µ is a partial order 8 x, y 2 A 9 inf(x,y); will denote it x Å y 8 x,y 2 A 9 sup(x,y); will denote it x [ y There are minimal, maximal elements: ?, > 2 A 8 x, there exists x 2 A s.t. x Å x = ?, x [ x = >.

Boolean Algebras (con’t) Definition 2. A BA is an algebra (A,[, Å, ?, >, ) s.t. [, Å = commutative, associative, idempotent, distributive x Å ? = ?, x [ ? = x, x Å > = x, x [ > = > x Å x = ?, x [ x = >

Boolean Algebra’s (cont’d) Proposition. The two definitions are equivalent Proof: From µ to Å, [: define x Å y = inf(x,y), x [ y = sup(x,y) From Å, [ to µ: define x µ y to mean x Å y = x

Boolean Algebras (cont’d) Define: atom(x) = (x ¹ ?) Æ 8 y. (y µ x ) y=? Ç y = x) Proposition. Let A be finite. Let X = {x | x 2 A, atom(x)}. Then (A,µ) is isomorphic to (2X, µ) [why ?]

Step 1 Define fB to consists of the axioms of a boolean algebra (using Definition 1)

Step 2 Show that EVENB 2 (FO + <)inv Recall: 9 P.( 8 x.y.(S(x,y) ) (P(x)Æ: P(y) Ç : P(x)ÆP(y)) Æ P(min) Æ : P(max)) EVEN = 9 p.( 8 x.y.(atom(x) Æ atom(y) Æ S(x,y) ) (x µ p Æ: (y µ p) Ç : (x µ p) Æ y µ p) Æ minatom µ p Æ : (maxatom µ p) EVENB =

Step 3 Now we prove that EVENB Ï FO Proposition. Let |X|, |Y| ¸ 2k. Then: (2X, µ) k (2Y, µ) We prove it using two lemmas.

Step 3 Lemma. Assume (2X, µ) k (2Y, µ). Then the duplicator has a winning strategy where he responds to ; with ;, to Y with X, to X with Y. [Why ?] Lemma. Assume X1 Å X2 = ;, Y1 Å Y2 = ; and that (2X1, µ) k (2Y1, µ), (2X2, µ) k (2Y2, µ). Then: (2X1 [ X2, X1, X2, µ) k (2Y1 [ Y2, Y1, Y2, µ) [How do prove the proposition ?]

Summary Having < seems to add lots of extra power Will see this later, for fixpoint logics In FO(<) it is even more surprising The extra structure (boolean algebra) seems quite artificial. Can we get rid of it ? Open problem: does < add extra power over strings ?

Gaifman’s Theorem Let A 2 STRUCT[s] Recall the r-ball (also called r-sphere): Br(x) = {y | d(x,y) · r} Br(x) can be expressed in FO (obviously)

Gaifman’s Theorem Definition A formula f(x) is r-local if all its quantifiers are bounded to Br(x): 9 z.y  9 z. (z 2 Br(x)) Æ y 8 z.y  8 z.(z 2 Br(x)) ) y) We write f(r) to emphasize that f is r-local.

Gaifman’s Theorem Theorem [Gaifman] Let s be relational. Then every FO sentence over s is equivalent to a Boolean combination of sentences of the form: 9 x1 … 9 xs(Æi=1,sa(r)(xi)) Æ (Æ1 · i < j · s d>2r(xi, xj))