1
6.3. A Refutation.
In Proposition A, can we replace ELG by the simpler and
more basic SD? We refute this in a strong way. In
particular, we refute Proposition C with ELG removed.
PROPOSITION
α
. For all f,g
∈
SD
∩
BAF there exist A,B,C
∈
INF such that
A
∪
. fA
⊆
C
∪
. gB
A
∪
. fB
⊆
C
∪
. gC.
We will even refute the following weaker Proposition.
PROPOSITION
β
. Let f,g
∈
SD
∩
BAF. There exist A,B,C
⊆
N,
|A|
≥
4, such that
A
∪
. fA
⊆
C
∪
. gB
A
∪
. fB
⊆
C
∪
. gC.
We assume Proposition
β
, and derive a contradiction.
We begin with a modification of Lemmas 5.1.6 and 5.1.7.
Basically, these go through without any change in the
proof, but we provide some additional details.
LEMMA 5.1.6'. Let f,g
∈
SD
∩
BAF. There exist f',g'
∈
SD
∩
BAF such that the following holds.
i) g'S = g(S*)
∪
6S+2;
ii) f'S = f(S*)
∪
g'S
∪
6f(S*)+2
∪
2S*+1
∪
3S*+1.
Proof: In the proof of Lemma 5.1.6, f',g' are constructed
explicitly from f,g. It is obvious that if f,g
∈
SD
∩
BAF,
then f',g'
∈
SD
∩
BAF. The verification goes through
without change. QED
LEMMA 5.1.7'. Let f,g
∈
SD
∩
BAF and rng(g)
⊆
6N. There
exist A
⊆
B
⊆
C
⊆
N\{0}, |A|
≥
3, such that
i) fA
∩
6N
⊆
B
∪
gB;
ii) fB
∩
6N
⊆
C
∪
gC;
iii) fA
∩
2N+1
⊆
B;
iv) fA
∩
3N+1
⊆
B;
v) fB
∩
2N+1
⊆
C;
vi) fB
∩
3N+1
⊆
C;
vii) C
∩
gC =
∅
;
viii) A
∩
fB =
∅
;
Proof: In the proof Lemma 5.1.6, f',g' are constructed
explicitly from f,g. Then A,B,C are used from Proposition