s
i=1
{
i
, mi 2( 2
i
+ pri )} χi KmL so σ1 = σ3 + σ4. We calculate
NL/F (σ3) =
s
i=1
(
{℘(ri) + b, mi
2( 2
i
+ pri )} χi
)
=
s
i=1
(
−{a/(℘(ri) + b), mi
2( 2
i
+ pri )} χi
)
+
s
i=1
(
{a, mi
2( 2
i
+ pri )} χi
)
=
s
i=1
(
{pri , mi
2( 2
i
+ pri )} χi
)
+ {a}
s
i=1
(
{mi
2( 2
i
+ pri )} χi
)
Using the K2-identity {z, w} = {z/w, z + w} we see that {pri , mi 2( 2
i
+ pri )} =
{pri ,
m2}
+
2
i
+ pri ),
2}
i
= 2({pri , mi} + {pri /(
2
i
+ pri ),
i
}). Using the fact
that
∑is
i=1
({pri/(
{mi 2( 2
i
+ pri )} χi
)
= 2χ∞ we obtain
NL/F (σ3) = 2
s
i=1
{
pri
2
i
+ pri
, i} + {pri , mi} χi + {a} χ∞ .
Applying Lemma 2.5 we can express
σ3 iL/F
s
i=1
({
pri
2
i
+ pri
, i} + {pri , mi}) χi + {a} χ∞
NL/F (σ5)) (mod 2KmL)
for some σ5 KmL.
As σ1 = σ3 + σ4 we find
σ1
s
i=1
({
pri
2
i
+ pri
,
i
} + {pri , mi}) χi + {a} χ∞
−NL/F (σ5) +
s
i=1
(
{
i
,
2
i
+ pri } χi
)
(mod 2KmL)

s
i=1
({pri , imi} χi) + {a} χ∞ NL/F (σ5) (mod 2KmL)
As pri , a aNL/F L, we see {pri ,
i
/mi} χi, {a} χ∞, NL/F (σ5) Tm and the
lemma follows.
This final lemma gives part (iii) of Theorem 2.1 and completes its proof.
Lemma 2.7. For σ KmF (X) as in Theorem 2.1, σ = iF
(X)/F
) + for
some θ Tm KmF and σ KmF (X).
Proof. Setting σ1 :=
∑s
i=1
(
{
i
(ri + 1 + β), mi( 2
i
+ pri )} χi
)
KmL then ac-
cording to Lemma 2.4, iL(X)/F
(X)
(σ) = iL(X)/L(σ1) + iL(X)/F (σ6) + 2σ7, where
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