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Ordered Structures, Gaifman’s Theorem |
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Order-invariant queries |
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Gurevitch’s theroem |
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Gaifman’s theorem (time permitting) |
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Idea: |
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Use some innocuous additional information to
express a query |
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“Forget” the extra information |
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Need to check that the query is invariant w.r.t.
to the extra info |
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Recall: EVEN(A) is not expressible in FO |
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Let s<,+ = {<,+} |
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Consider A 2 STRUCT[s<,+] s.t. |
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< is interpreted as a linear order a1
< … < an |
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(ai, aj, ak) 2
+ iff i+j=k |
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Then, we can write a sentence f that checks
EVEN(A) [ HOW ?? ] |
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f = 9 x. 9 y.(x+x=y Æ : 9 z(y < z)) |
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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) |
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Let s, s’ be two disjoint vocabularies, and C a
class of s’-structures (C µ STRUCT[s’]) |
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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) |
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(A,A’)
² f iff (A,A’’) ² f |
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Notation
f 2 (FO+C)inv |
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We have seen that EVEN 2 (FO + (<,+))inv |
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Main question for today:
Is (FO + <)inv
more powerful than FO ? |
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Theorem (Gurevich) There are properties in (FO +
<)inv that are not in FO. |
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FO &
(FO + <)inv |
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Idea: recall that we can express EVEN in
MSO(<).
Let S(x,y) = :(9
z.x<z Æ z<y) Æ x < y |
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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. |
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3 Steps |
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Design s, fB s.t. A ² fB
iff A is an atomic boolean algebra, A = (2X, µ)
Let EVENB
= (EVEN(|X|) Æ fB |
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Show that EVENB 2 (FO + <)inv |
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Prove (using EF-games) that EVENB Ï
FO |
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Background: two alternative definitions |
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Definition 1.
A boolean algebra is A = (A, µ) s.t.: |
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µ is a
partial order |
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8 x, y 2
A 9 inf(x,y); will denote it x Å y |
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8 x,y 2
A 9 sup(x,y); will denote it x [ y |
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There are minimal, maximal elements: ?, > 2 A |
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8 x,
there exists x 2 A s.t. x Å x = ?, x [ x = >. |
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Definition 2.
A BA is an algebra
(A,[,
Å, ?, >, ) s.t. |
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[, Å =
commutative, associative, idempotent, distributive |
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x Å ? = ?,
x [ ? = x,
x Å > =
x, x [ > = > |
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x Å x = ?, x [ x = > |
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Proposition. The two definitions are equivalent |
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Proof: |
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From µ to Å, [:
define x Å
y = inf(x,y), x [ y = sup(x,y) |
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From Å, [ to µ:
define x µ
y to mean x Å y = x |
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Define:
atom(x) = (x ¹ ?) Æ 8
y. (y µ x ) y=? Ç y = x) |
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Proposition. Let A be finite. Let X = {x | x 2
A, atom(x)}. Then (A,µ) is
isomorphic to (2X, µ) |
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[why ?] |
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Define fB to consists of the axioms
of a boolean algebra (using Definition 1) |
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Show that EVENB 2 (FO + <)inv |
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Now we prove that EVENB Ï FO |
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Proposition.
Let |X|, |Y| ¸ 2k.
Then:
(2X, µ) ºk
(2Y, µ) |
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We prove it using two lemmas. |
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Lemma. Assume (2X, µ) ºk
(2Y, µ). Then the
duplicator has a winning strategy where he responds to ; with ;, to Y with
X, to X with Y. |
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[Why ?] |
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Lemma.
Assume X1 Å X2 = ;, Y1 Å Y2
= ; and that (2X1, µ) ºk (2Y1,
µ), (2X2, µ) ºk (2Y2, µ). Then:
(2X1
[ X2, X1, X2, µ)
ºk (2Y1 [
Y2, Y1, Y2, µ) |
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[Why ?] |
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[How do prove the proposition ?] |
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Having < seems to add lots of extra power |
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Will see this later, for fixpoint logics |
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In FO(<) it is even more surprising |
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The extra structure (boolean algebra) seems
quite artificial. Can we get rid of
it ? |
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Open problem: does < add extra power over
strings ? |
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Let A 2 STRUCT[s] |
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Recall the r-ball (also called r-sphere):
Br(x)
= {y | d(x,y) · r} |
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Br(x) can be expressed in FO
(obviously) |
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Definition A formula f(x) is r-local if all its
quantifiers are bounded to Br(x): |
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9 z.y è 9 z. (z 2 Br(x)) Æ y |
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8 z.y è 8 z.(z 2
Br(x)) ) y) |
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We write f(r) to emphasize that f is
r-local. |
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Theorem [Gaifman] Let s be relational. Then every FO sentence over s is
equivalent to a Boolean combination of sentences of the form: |
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