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Ehrenfeucht-Fraisse Games |
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Everything can be defined “in the finite” |
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f has a
finite model |
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f is
valid in the finite if for every finite A, A ² f; notation ²fin f |
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f is a
consequence in the finite of G if for every finite A, A ² G implies A ² f;
notation G ²fin f |
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“f has a finite model” is r.e. [why] |
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Trachtenbrot’s theorem: “f has a finite model”
is undecidable |
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Hence there is no proof system ` s.t.
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f iff ²fin f [why ?] |
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There exists an infinite set of formulas G s.t.
for every finite subset G0 ½ G, G0 has a finite
model, but G does not have a finite model |
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[why ?] |
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Today’s lecture: |
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Ehrenfeucht-Fraisse games |
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Applications |
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Let A, B 2 STRUCT[s] |
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They are elementary equivalent, A º B,
if 8
f, A ² f iff B ² f |
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They are isomorphic, A @ B,
if there exists a isomorphism f : A !
B |
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If A @ B, then A º B
[why ?] |
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There are A, B s.t. A º B and not A @ B [give
examples in class] |
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If A is finite and A º B then A @ B [why ?] |
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Let c = (c1, …, cn) be the
constants in s |
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Let A, B be two structures in STRUCT[s] |
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Definition A partial isomorphism is given by a =
(a1, …, am) and b = (b1, …, bm)
s.t. the structures (a, c) and (b, c)
are isomorphic. |
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Given two structures A, B |
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Two players: spoiler and duplicator play k
rounds |
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Round i
(i =1, …, k): |
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Spoiler picks a structure A or B |
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Spoiler picks an element in that structure ai
2 A or bi 2 B |
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Duplicator responds by picking an element in the
other structure, bi 2 B or ai 2 A |
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Duplicator wins if a, b is a partial isomorphism |
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If duplicator has a winning strategy in k rounds
then we write A ºk B |
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Note: if A ºn B and k < n then A ºk
B |
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The quantifier rank of a formula f is defined
inductively:
qr(t1 = t2) = qr(R(t1,…,tn))
= 0
qr(f1 Æ f2) = max(qr(f1), qr(f2))
qr(9
x.f) =1+qr(f)
etc |
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FO[k] = all formula f s.t. qr(f) · k |
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Theorem The following two are equivalent: |
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A and B agree on FO[k] |
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A ºk B |
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We will prove this next time. |
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For now, let’s start playing the game ! |
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Let s = ; (i.e. no relation names, no
constants), and assume |A| ¸ k, |B| ¸ k. |
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Theorem A ºk B. [why ?] |
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Corollary EVEN is not expressible in FO when s =
; |
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Theorem Let k > 0 and L1, L2
be linear orders of length ¸ 2k. Then L1 ºk L2. |
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Define: |
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a-1 = min(L1), a0 = max(L1) |
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b-1 = min(L2), b0 = max(L2) |
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Let a = (a-1, a0, a1,
…, ai), b = (b-1,
b0, b1, …, bi) |
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Duplicator plays such that 8 j, l: |
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If d(aj, al) < 2k-i
then d(aj, al) = d(bj, bl) |
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If d(aj, al) ¸ 2k-i
then d(bj, bl) ¸ 2k-i |
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aj · al iff bj ·
bl |
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Remark: if
L1 ºk L2 then the
duplicator has a winning strategy where it responds with min(L2)
to min(L1), and with max(L2) to max(L1),
and vice versa. [why ?] |
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Lemma If L1· a ºk
L2 · b and L1¸
a ºk L2 ¸ b
then (L1
,a) ºk (L2 ,b) |
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[how does this help us prove the theorem ?] |
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Corollary EVEN is not expressible in FO(<) |
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[why ?] |
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Corollary Finite graph connectivity (CONN) is
not expressible in FO |
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[proof in class] |
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