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Finite Model Theory
Lecture 16
Lw1w Summary
and 0/1 Laws
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Outline
• Summary on Lw1w
– All you need to know in 5 slides !
• Start 0/1 Laws: Fagin’s theorem
– Will continue next time
New paper:
Infinitary Logics and 0-1 Laws, Kolaitis&Vardi, 1992
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Summary on
w
L 1w
Notation Comes from in classical logic
• Lab = formulas where:
– Conjunctions/disjunctions of ordinal < a
Çi 2 g fi, Æi 2 g, where g < a
– Quantifier chains of ordinal < b
9i 2 g xi. f, where g < b
• Hence, L1w = [a Law
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Summary on
w
L 1w
Motivation
• Any algorithmic computation that applies FO formulas
is expressible in Lw1w
• Relational machines
• While-programs with statements R := f
• Fixpoint logics: LFP, IFP, PFP, etc, etc
Consequence: cannot express EVEN, HAMILTONEAN
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Summary on
w
L 1w
Canonical Structure
Any algorithmic computation on A can be decomposed
• Compute the ¼k equivalence relation on k-tuples, and
order the equivalence classes ) in LFP
[how do we choose k ???]
• Then compute on ordered structure ) any complexity
Consequence: PTIME=PSPACE iff IFP=PFP
But note that DTC  TC yet L ? NL [ why ?]
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Summary on
w
L 1w
Pebble Games: with k pebbles
• Notation: A 1wk B if duplicator wins
Theorem 1. For any two structures A, B:
• A, B are Lk1w equivalent iff
• A 1wk B
Theorem 2. If A, B are finite:
• A, B are FOk equivalent iff
• A, B are Lk1w equivalent iff
• A 1wk B
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Summary on
w
L 1w
Definability of FOk types
• FOk types are the same as Lk1w types [ why ?]
Theorem [Dawar, Lindell, Weinstein] The type of A
(or of (A, a)) can be expressed by some f 2 FOk
B ² f[b] iff Tpk(A,a) = Tpk(B,b)
Difficult result: was unknown to Kolaitis&Vardi
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0/1 Laws in Logic
Motivation: random graphs
• 0/1 law for FO proven by Glebskii et al., then
rediscovered by Fagin (and with nicer proof)
– Only for constant probability distribution
• Later extended to other logics, and other
probability distributions
Why we care: applications in degrees of belief,
probabilistic databases, etc.
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Definitions
• Let s = a vocabulary
• Let n ¸ 0, and An µ STRUCT[s] be all
models over domain {0, 1, …, n-1}
• Uniform probability distribution on An
• Given sentence f, denote mn(f) its
probability
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Definition
• Denote m(f) = limn ! 1 mn(f) if it exists
Definition A logic L has a convergence law if
for every sentence f, m(f) exists
Definition A logic L has a 0/1 law if for every
sentence f, m(f) exists and is 0 or 1
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Theorems
• Suppose s has no constants
Theorem [Fagin 76, Glebskii et al. 69]
FO admits a 0/1 law
Theorem [Kolaitis and Vardi 92]
Lw1w admits a 0/1 law
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Application
• What does this tell us for database query
processing ?
• Don’t bother evaluating a query: it’s either
true or false, with high probability 
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Examples [ in class ]
• Compute mn(f), then m(f):
R(0,1) /* I’m using constants here */
R(0,1) Æ R(0,3) Æ : R(1,3)
9 x.R(2,x)
: (9 x.9 y.R(x,y))
8 x.8 y.(9 z.R(x,z) Æ R(z,y))
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Types
• We only need rank-0 types (i.e. no
quantifiers)
• Recall the definition
Definition A type t(x) over variables (x1, …,
xm) is conjunction of a maximally consistent
set of atomic formulas over x1, …, xm
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Types
The type t(x) says:
• For each i, j whether xi = xj or xi  xj
• For each R and each xi1, …, xip whether
R(xi1, …, xip) or : R(xi1, …, xip)
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Extension Axioms
Definition Type s(x, z) extends the type t(x) if
{s, t} is consistent;
Equivalently: every conjunct in t occurs in s
Definition The extension axiom for types t, s
is the formula
tt,s = 8 x1…8 xk (t(x) ) 9 z.s(x, z))
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Example of Extension Axiom
t(x1, x2, x3) =
x1  x2 Æ x2  x3 Æ x1  x3 Æ
R(x1,x2) Æ R(x2,x3) Æ R(x2,x2) Æ
: R(x1, x1) Æ : R(x2, x1) Æ …
s(x1, x2, x3, z) =
t(x1, x2, x3) Æ
z  x1 Æ z  x2 Æ z  x3 Æ
R(z,x1) Æ R(x3,z) Æ R(z,z) Æ
: R(x1, z) Æ : (z, x2) Æ …
x1
z
x2
x3
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Example of Extension Axiom
tt,s =
8 x1.8 x2.8 x3. (t(x1, x2, x3) ) 9 z. s(x1, x2, x3, z))
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The Theory T
• Let T be the set of all extension axioms
– Studied by Gaifman
• Is T consistent ?
– In a model of T the duplicator always wins [ why ? ]
• Does it have finite models ?
• Does it have infinite models ?
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The Theory T
• Let qk be the conjunction of all extension axioms
for types with up to k variables
• There exists a finite model for qk [why ?]
• Hence any finite subset of T has a model
• Hence T has a model. [can it be finite ?]
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The Model(s) of T
• T has no finite models, hence it must have
some infinite model
• By Lowenheim-Skolem, it has a countable
model
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The Theory T
Theorem T is w-categorical
Proof: let A, B be two countable model.
Idea: use a back-and-forth argument to find an
isomorphism f : A ! B
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The Theory T
Theorem T is w-categorical
Proof: (cont’d)
A = {a1, a2, a3, ….} B = {b1, b2, b3, ….}
Build partial isomorphisms f1 µ f2 µ f3 µ …
such that:
8 n.9 m. an 2 dom(fm)
and
8 n.9 m. bn 2 rng(fm)
[in class]
Then f = ([m ¸ 1 fm) : A ! B is an isomorphism
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The Theory T
Corollary T has a unique countable model R
• R = the Rado graph
= the “random” graph
Corollary The theory Th(T) is complete
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0/1 Law for FO
Lemma
For every extension axiom t, m(t) = limn mn(t) = 1
Proof: later
Corollary For any m extension axioms t1, …, tm:
m(t1 Æ … Æ tm) = 1
Proof
mn(:(t1 Æ … Æ tm))
= mn(: t1 Ç … Ç : tm)
· mn(: t1) + … + mn(: tm) ! 0
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Fagin’s 0/1 Law for FO
Theorem
For every f 2 FO, either m(f) = 0 or m(f) = 1.
Proof.
Case 1: R ² f. Then there exists m extension
axioms s.t. t1, …, tm ² f. Then mn(f) ¸ mn(t1 Æ
… Æ tm) ! 1
Case 2: R 2 f. Then R ² : f, hence m(: f) = 1, and
m(f) = 0
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Proof for the Extension Axioms
• Let t = 8 x. t(x) ) 9 z.s(x, z)
• Assume wlog that t asserts xi  xj forall i  j.
Denote (x) the formula Æi < j xi  xj
– Hence t(x) = (x) Æ t’(x)
• Similarly, s asserts z  xi forall i.
Denote (x, z) = Æi xi  z
– Hence s(x, z) = t(x) Æ (x, z) Æ s’(x, z)
where all atomic predicates in s’(x, z) contain z
• Hence:
t = 8 x.((x) Æ t’(x) ) 9 z. (x,z) Æ s’(x, z))
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Proof for the Extension Axioms
: t = 9 x.((x) Æ t’(x) Æ 8 z.((x, z) ) : s’(x, z)))
mn(: t) · mn(9 x.((x) Æ 8 z.((x, z) ) : s’(x, z))))
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Proof for the Extension Axioms
mn(: t) · mn(9 x.((x) Æ 8 z.((x, z) ) :s’(x, z))))
· a1, ... , ak 2 {1, …, n} mn(8 z. ((x, z) ) :s’(a1, …, ak, z)))
= n(n-1)…(n-k+1) mn(8 z. (x, z) ) :s’(1, 2, …, k, z))
· nk mn(8 z. (x, z) ) :s’(1, 2, …, k, z)) =
= nk z=k+1, n : s’(1,2,…,k,z)
/* by independence !! */
= nk ( 1 - 1 / 22k+1 )n-k /* since s’ is about 2k+1 edges */
!0
when n ! 1
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Complexity
Theorem [Grandjean] The problem whether
m(f) = 0 or 1 is PSPACE complete
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Discussion
• Old way to think about formulas and
models: finite satsfiability/ validity
FO
f valid
f unsatisfiable
Undecidable
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Discussion
• New way to think about formulas and
models: probability
m(f)=0
m(f)=1
FO
f valid
f unsatisfiable
PSPACE
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