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2 = ↓
m
(∃
m
.
prime(l)
(
m
))
By using rule (1) twice, we have
2 = ↓
m
(∃
m
,
k
:
N
. (
prime(l)
(
m
) ∧
prime(l)
(
k
) ∧
n
=
m
+
k
))
In a similar procedure, we can determine that 2 is also the lower bound of
k
, i.e.,
2 = ↓
k
(∃
m
,
k
:
N
. (
prime(l)
(
m
) ∧
prime(l)
(
k
) ∧
n
=
m
+
k
))
Now using
n
=
m
+
k
and the lower bound of
k
, we can get the upper bound of
m
, which is
n
-2. Thus
we have
2 ≤
m
≤
n
-2
Again, using
n
=
m
+
k
, we have
k
=
n
-
m
As for variable
i
in
prime(l)
(
l
) ≡ (
l = 2
∨
l
> 2 ∧ ∀
i
:
N
. (1 <
i
<
l
→
l
%
i
≠ 0))
2 ≤
i
is determined by the fact that
2 = ↓
i
(∃
i
.1<
i
<
l
)
Using rule (2), we get
2 = ↓
i
(∀
i
:
N
. (1 <
i
<
l
→
l
%
i
≠ 0))
i
≤
m
-1 is determined by the fact that
m
-1 = ↑
i
(∃
i
.1<
i
<
l
)
Using rule (2'), we get
m
-1 = ↑
i
(∀
i
:
N
. (1 <
i
<
l
→
l
%
i
≠ 0))
Therefore, we have
2 ≤
i
≤
m
-1
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