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Finite Model TheoryLecture 13

FOk, Lk1,,L

1,,and Pebble Games

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Infinitary Logic

• Allow infinite conjunctions, disjunctions:

• Now restrict everything to just k variables: x1, …, xk

• Finally:

Çi 2 I i, Æi 2 I iÇi 2 I i, Æi 2 I i L1,

L1,

Lk1,

Lk1,

L1, = [

¸ Lk

1, L

1, = [¸

Lk1,

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Quantifier Rank

• qr() is defined as before:

• In general it can be an ordinal, e.g. 5, 99, +3, 4 + 7, etc

• Over finite structures, we will show that it suffices to have qr ·

qr(Çi 2 I i) = qr(Æi 2 I i) = supi2 I qr(i)

qr(9 x.) = qr(8 x.) = qr() + 1

qr(Çi 2 I i) = qr(Æi 2 I i) = supi2 I qr(i)

qr(9 x.) = qr(8 x.) = qr() + 1

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Example

• Over ordered finite structures, every property can be expressed in Lm+1

1,, where m = maxi(arity(Ri)), Ri 2

• Proof:• First, show that if = {<} then every

property on finite linear orders is in Lm+11,

• Next, generalize to arbitrary

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Fixpoints

Consider transitive closure, expressed as LFP:

• tc(u,v) = lfpR[E(x,y) Ç 9 z.(E(x,z) Æ R(z,y)](u,v)

Can “unfold” it:

• tc0(x,y) = E(x,y)

• tcn+1(x,y) = 9 z.(E(x,z) Æ 9 x.(x=z Æ tcn(x,y)))

Hence tc(u,v) = Çn ¸ 0 tcn(u,v) 2 L1,

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Fixpoints

• In general:

Theorem LFP, IFP, PFP µ L1,

• Proof [in class]

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Pebble Games

• There are k pairs of pebbles, (a1, b1), …, (ak, bk)

• Pebble games are played much like Ehrenfeucht-Fraisse games, HOWEVER now pebbles can be moved, after they have been placed on a structure

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Pebble Games

• Let A, B 2 STRUCT[], and k ¸ 0.

• Round j ¸ 1:– Spoiler picks a pebble, ai (or b i), i = 1, …, k; if

a i (b i) was already placed on a structure, then it removes it; spoiler places the pebble a i on the structure A (or b i on the structure B)

– Duplicator needs to respond by placing pebble b i on B (or a i on A)

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Pebble Games

• The game for n rounds: PGnk(A,B)

• The game forever: PG1k(A,B)

• Duplicator has a winning strategy for PGn(A,B) if 8 j · n, (a1 ! b1, …, ak ! bk) is a partial isomorphism; similarly for PG1(A,B)

• Notation: A 1k,nB and A 1

k B

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Pebble Games

TheoremA, B 2 STRUCT[] agree on all sentences of Lk

1, of qr · n iff A 1k,nB

A, B 2 STRUCT[] agree on all sentences of Lk

1, iff A 1k B

Proof next time. For today we will assume the theorem to be true

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Example

Can you win this with k=2 pebbles ?

A= B=

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Example

Can you win this with k=4 pebbles ?

A= B=

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EVEN

Theorem EVEN is not expressible in L1,

Corollary EVEN is not expressible in LFP, IFP, PFP

Corollary LFP & (LFP+<)inv

IFP & (IFP+<)inv

PFP & (PFP+<)inv

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A Property

• Assume A, B agree on all sentences in FOk

• Then, for every n ¸ 0, they agree on all sentences in Lk

1, of qr · n– Why ? Consider Çi 2 I i where 8 i. qr(i) · n. Then

this is in FOk !

• Hence, for every n ¸ 0, A 1k,nB

• Hence, A 1k B

– Why ? Remember Koenig’s Lemma ?

• Hence A, B agree on all sentences in Lk1,

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A Property

Theorem The following are equivalent

• A, B agree on all FOk sentences

• A, B agree on all Lk1, sentences

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Definability of Types

Definition. Let A 2 STRUCT[], and a 2 A. The FOk type of (A, a) is: tpFOk (A,a) = { (x) 2 FOk | A ² (a) }

One could define L1, types as well, but they

turn out to be the same as FOk types [ why ?? ]

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Definability of Types

• The number of FOk types is infinite (since no restriction on quantifier depth)

• Each FOk type is definable as Çi 2 I i, where each i 2 FOk. However:

Theorem. Every FOk type is defined in FOk. I.e. there exists (x) s.t. TpFOk(A, a) = , A ² (a)

Proof: next time. For now assume it holds

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Applications

• For every structure A there exists a sentence A in FOk s.t. 8 B: B ² A iff A 1

k B

• Every Lk1, formula is equivalent to Çi 2 I i,

where each i 2 FOk