1
2.5. EBRT in A,B,fA,fB,
⊆
on (ELG,INF).
In this section, we use the tree methodology described in
section 2.1 to analyze EBRT in A,B,fA,fB,
⊆
on (ELG,INF) and
(EVSD,INF). We handle both BRT settings at once, as they
behave the same way for EBRT in A,B,fA,fB,
⊆
. In particular,
we show that they are RCA
0
secure (see Definition 1.1.43).
Some of this treatment is the same as for EBRT in
A,B,fA,fB,
⊆
on (SD,INF) given in section 2.4. However, many
new features appear that makes this section considerably
more involved than section 2.4.
A key difference between EBRT in A,B,fA,fB,
⊆
on (SD,INF)
and on (ELG,INF) is that the Compelmentation Theorem holds
on (SD,INF), yet fails on (ELG,INF). E.g., it fails for
f(x) = 2x.
Let f:N
k
→
N be partial. Define the following series of
sets by induction i
≥
1.
S
1
= N.
S
i+1
= N\fS
i
.
LEMMA 2.5.1. S
2
⊆
S
4
⊆
S
6
⊆
...
⊆
...
⊆
S
5
⊆
S
3
⊆
S
1
. I.e.,
for all i
≥
1, S
2i
⊆
S
2i+2
⊆
S
2i+1
⊆
S
2i1
.
Proof: We argue by induction on i
≥
1. The basis case is
S
2
⊆
S
4
⊆
S
3
⊆
S
1
.
To see this, clearly
S
3
⊆
S
1
.
N\S
1
⊆
N\S
3
.
S
2
⊆
S
4
.
S
2
⊆
S
1
.
N\S
1
⊆
N\S
2
.
S
2
⊆
S
3
.
fS
2
⊆
fS
3
.
N\fS
3
⊆
N\fS
2
.
S
4
⊆
S
3
.
Now assume the induction hypothesis
S
2i
⊆
S
2i+2
⊆
S
2i+1
⊆
S
2i1
.
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Then
fS
2i
⊆
fS
2i+2
⊆
fS
2i+1
⊆
fS
2i1
.
N\fS
2i1
⊆
N\fS
2i+1
⊆
N\fS
2i+2
⊆
N\fS
2i
.
S
2i
⊆
S
2i+2
⊆
S
2i+3
⊆
S
2i+1
.
fS
2i
⊆
fS
2i+2
⊆
fS
2i+3
⊆
fS
2i+1
.
N\fS
2i+1
⊆
N\fS
2i+3
⊆
N\fS
2i+2
⊆
N\fS
2i
.
S
2i+2
⊆
S
2i+4
⊆
S
2i+3
⊆
S
2i+1
.
QED
LEMMA 2.5.2. Let f:N
k
→
N be partial, where each f
1
(n) is
finite. Let A = S
2
∪
S
4
∪
..., and B = S
1
∩
S
3
∩
... . Then
A
⊆
B, A = N\fB, B = N\fA.
Proof: Let A,B be as given. By Lemma 2.5.1, A
⊆
B.
Fix i
≥
1. S2i = N\fS2i1, S2i
∩
fS2i1 =
∅
, S2i
∩
fB =
∅
.
Since i
≥
1 is arbitrary, A
∩
fB =
∅
. I.e., A
⊆
N\fB.
Since S
2i+1
= N\fS
2i
, we see that for all j
≥
i, S
2i+1
∩
fS
2j
=
∅
. Hence S
2i+1
∩
fA =
∅
. Since i
≥
1 is arbitrary, B
∩
fA =
∅
. I.e., B
⊆
N\fA.
Now let n
∈
N\fB. We claim that for some j
≥
0, n
∉
fS
2j+1
.
Suppose that for all j
≥
0, n
∈
fS
2j+1
. Since f
1
(n) is
finite, there exists x
∈
f
1
(n) which lies in infinitely
many S
2j+1
. Hence there exists x
∈
f
1
(n) such that x
∈
B.
Therefore n
∈
fB. This establishes the claim. Fix j
≥
0
such that n
∉
fS
2j+1
. Then n
∈
S
2j+2
, and so n
∈
A. This
establishes that A = N\fB.
Finally, let n
∈
N\fA. Then for all i, n
∉
fS
2i
. Hence for
all j, n
∈
S
2j+1
. Therefore n
∈
B. This estabslihes that B =
N\fA. QED
LEMMA 2.5.3. Let f:[0,n]
k
→
[0,n] be partial, n
≥
0. There
exist A
⊆
B
⊆
[0,n] such that A = [0,n]\fB and B =
[0,n]\fA.
Proof: Let n,f be as given. Obviously f:Nk
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 Fall '08
 JOSHUA
 Math, fb, a,b, EBRT

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