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Lemma Forall t, s0: m(t t,s0)
= limn mn(t t,s0) = 1 |
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This is if and only if |
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Lemma Forall s0: m(ss0) = 1 |
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Proof:
Let t1, t2, …, tm be all k-types |
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Then: ss0
= tt1,s0 Æ … Æ ttm,s0 |
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Moreover, this is iff |
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Lemma Forall s0: m(ss0 ¹)
= 1 |
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Proof:
Every ss0 is the conjunction of several expressions ss1¹,
where s1 states s0 over fewer variables, by repeating some of them. |
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Now let’s see the proof from last lecture that m[t]
= 1… |
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What is m(f) in each case below ?
(These are from the book) |
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“There are no isolated nodes”:f = 8 x.9 y.R(x,y) |
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“R is connected” |
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“The domain has even cardinality”: EVEN |
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“R has even cardinality”: PARITY |
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Let qk be the conjunction of all
extension axioms with · k variables |
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I.e. tt,s where t is a type over x =
(x1, …, xk) |
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Lemma.
Let A, B be such that A ² qk, B ² qk . Then A
=1wk B |
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Proof: qk says precisely that the
duplicator has a winning strategy with k pebbles. |
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Theorem Lw1w
has a 0/1 law. |
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Proof Let f 2 Lk1w |
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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 |
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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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Proof via a Transfer Theorem |
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[Kolaitis and Vardi] |
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Lemma. 8
k > 0 there is yk 2 FOk such that: |
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R ² yk |
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For f in
Lk1w there is f’ in FOk such
that:
yk ² f $ f’ |
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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)
R
² 8 x.(9 xj.ai $ al)
Let yk
be their conjunction; clearly R ² fk |
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Claim 2: for any f 2 Lk1w
there exists f’ 2 FOk s.t. yk ² f $ f’
This by induction on f. |
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Let f be in Lk1w
and yk, f’ be as in the Lemma |
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Then m(phik) = 1 and one can check
that m(f) = m(f’) |
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How does the “random graph” R look like ? |
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Does second order logic SO have a 0/1 law ? |
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“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. |
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The following are a.e. equivalent: FO, Lw1w,
LFP, IFP, PFP |
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