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FOk, Lk1,w,Lw1,w,
and Pebble Games |
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Allow infinite conjunctions, disjunctions: |
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Now restrict everything to just k variables: x1,
…, xk |
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Finally: |
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qr(f) is
defined as before: |
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In general it can be an ordinal, e.g. 5, 99, w+3,
4w + 7, etc |
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Over finite structures, we will show that it
suffices to have qr · w |
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Over ordered finite structures, every property
can be expressed in Lm+11,w,
where m = maxi(arity(Ri)), Ri 2 s |
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Proof: |
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First, show that if s = {<} then every
property on finite linear orders is in Lm+11,w |
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Next, generalize to arbitrary s |
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Consider transitive closure, expressed as LFP: |
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tc(u,v) = lfpR[E(x,y) Ç 9 z.(E(x,z) Æ
R(z,y)](u,v) |
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Can “unfold” it: |
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tc0(x,y)
= E(x,y) |
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tcn+1(x,y)
= 9 z.(E(x,z) Æ 9 x.(x=z Æ tcn(x,y))) |
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Hence tc(u,v) = Çn ¸
0 tcn(u,v) 2 Lw1,
w |
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In general: |
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Theorem LFP, IFP, PFP µ Lw1,w |
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Proof
[in class] |
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There are k pairs of pebbles, (a1, b1),
…, (ak, bk) |
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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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Let A, B 2 STRUCT[s], and k ¸ 0. |
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Round j ¸ 1: |
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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) |
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Duplicator needs to respond by placing pebble b
i on B (or a i on A) |
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The game for n rounds: PGnk(A,B) |
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The game forever: PG1k(A,B) |
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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) |
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Notation:
A º1wk,nB and A º1wk B |
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Theorem |
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A, B 2 STRUCT[s] agree on all sentences of Lk1,w
of qr · n iff A º1wk,nB |
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A, B 2 STRUCT[s] agree on all sentences of Lk1,w
iff A º1wk
B |
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Proof next time. For today we will assume the theorem to be true |
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Theorem EVEN is not expressible in Lw1,w |
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Corollary |
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EVEN is not expressible in LFP, IFP, PFP |
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Corollary |
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LFP &
(LFP+<)inv |
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IFP &
(IFP+<)inv |
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PFP &
(PFP+<)inv |
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Assume A, B agree on all sentences in FOk |
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Then, for every n ¸ 0, they agree on all
sentences in Lk1,w of qr · n |
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Why ?
Consider Çi 2 I fi where 8 i. qr(fi) · n.
Then this is in FOk ! |
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Hence, for every n ¸ 0, A º1wk,nB |
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Hence, A º1wk B |
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Why ? Remember Koenig’s Lemma ? |
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Hence A, B agree on all sentences in Lk1,w |
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Theorem The following are equivalent |
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A, B agree on all FOk sentences |
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A, B agree on all Lk1,w
sentences |
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Definition. Let A 2 STRUCT[s], and a 2 A. The FOk type of (A, a)
is:
tpFOk
(A,a) = { f(x) 2 FOk | A ² f(a) } |
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One could define Lw1,w
types as well, but they turn out to be the same as FOk
types [ why ?? ] |
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The number of FOk types is infinite
(since no restriction on quantifier depth) |
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Each FOk type is definable as Çi
2 I fi, where each fi 2 FOk. However: |
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Theorem. Every FOk type t is defined
in FOk. I.e. there
exists ft(x) s.t.
TpFOk(A,
a) = t , A ² ft(a) |
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Proof: next time. For now assume it holds |
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For every structure A there exists a sentence fA
in FOk s.t. 8 B:
B ²
fA iff A º1wk B |
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Every Lk1,w
formula is equivalent to Çi 2 I fi,
where each fi 2 FOk |
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Notes
