1 finite model theory lecture 13 fo k, l k 1, ,l 1, , and pebble games
Post on 20-Dec-2015
216 Views
Preview:
TRANSCRIPT
1
Finite Model TheoryLecture 13
FOk, Lk1,,L
1,,and Pebble Games
2
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,
3
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
4
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
5
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,
6
Fixpoints
• In general:
Theorem LFP, IFP, PFP µ L1,
• Proof [in class]
7
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
8
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)
9
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
10
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
11
Example
Can you win this with k=2 pebbles ?
A= B=
12
Example
Can you win this with k=4 pebbles ?
A= B=
13
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
14
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,
15
A Property
Theorem The following are equivalent
• A, B agree on all FOk sentences
• A, B agree on all Lk1, sentences
16
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 ?? ]
17
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
18
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
top related