Finite Model Theory
Lecture 17
0/1 Laws
1
Extension Axioms Revisited
tt,s = 8 x. (t(x) ) 9 z.s(x, z))
Extension
axiom
ss0 = 8 x.9 z. s0(x,z)
Where s(x, z) = t(x) Æ s0(x, z)
“Loose”
extension
axiom
Lemma Forall t, s0: m(t t,s0) = limn mn(t t,s0) = 1
This is if and only if
Lemma Forall s0: m(ss0) = 1
Proof: Let t1, t2, …, tm be all k-types
Then: ss0 = tt1,s0 Æ … Æ ttm,s0
2
Extension Axioms Revisited
ss0 = 8 x.9 z. s0(x,z)
“Loose”
extension
axiom
ss0 = 8 x.(x))9 z. s0(x,z)
“not-so-loose”
extension
axiom
Moreover, this is iff
Lemma Forall s0: m(ss0 ) = 1
Proof: Every ss0 is the conjunction of several
expressions ss1, where s1 states s0 over fewer
variables, by repeating some of them.
3
Extension Axioms Revisited
• Now let’s see the proof from last lecture
that m[t] = 1…
4
Examples
What is m(f) in each case below ?
(These are from the book)
• “There are no isolated nodes”:f = 8 x.9 y.R(x,y)
• “R is connected”
• “The domain has even cardinality”: EVEN
• “R has even cardinality”: PARITY
5
0/1 Law for
w
L 1w
Let qk be the conjunction of all extension axioms
with · k variables
• I.e. tt,s where t is a type over x = (x1, …, xk)
Lemma. Let A, B be such that A ² qk, B ² qk . Then
A =1wk B
Proof: qk says precisely that the duplicator has a
winning strategy with k pebbles.
6
0/1 Law for
w
L 1w
Theorem Lw1w has a 0/1 law.
Proof Let f 2 Lk1w
Case 1. There exists a model A for both f and qk .
Then, for every model B of qk we have B ² f,
hence m(f) = 1
Case 2. There is no such model. Then : f and qk
have a common model and we prove m(f) = 0.
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0/1 Law for
w
L 1w
Proof via a Transfer Theorem
[Kolaitis and Vardi]
Lemma. 8 k > 0 there is yk 2 FOk such that:
1. R ² yk
2. For f in Lk1w there is f’ in FOk such that:
yk ² f $ f’
8
Proof of the Lemma
1. Claim: there are only finitely many
inequivalent FOk formulas over R
[Proof: in class ?? ]
a1, a2, …, am
Hence:
R ² 8 x(p(x) $ ai(x))
R ² 8 x(: ai $ : aj)
Closure axioms
R ² 8 x.(9 xj.ai $ al)
Let yk be their conjunction; clearly R ² fk
9
Proof of the Lemma
• Claim 2: for any f 2 Lk1w there exists f’ 2
FOk s.t. yk ² f $ f’
This by induction on f.
10
Proof of 0/1 Law for
k
L
1w
• Let f be in Lk1w and yk, f’ be as in the
Lemma
• Then m(phik) = 1 and one can check that
m(f) = m(f’)
11
Discussion
• How does the “random graph” R look like ?
• Does second order logic SO have a 0/1 law ?
• “Almost everywhere equivalence”. Two logics L,
L’ are a.e. equivalent if forall f 2 L 9 f’ in L’ s.t.
m(f $ f’) = 1.
– The following are a.e. equivalent: FO, Lw1w, LFP, IFP,
PFP
12