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Therefore, we can put all this together with
⊛
⊝
k
,
l
|
k
+
l
=
i
⊞
∞
⊠
,
ʔ
kl
=
ʔ
kl
ʛ
=
ʛ
k
=
(6.9)
k
=
0
k
,
l
≥
0
i
≥
0
where the last step is a straightforward diagonalisation.
Now from the decompositions above we recompose an alternative trajectory of
ʔ
i
's to take
ʔ
via
⃒
to
ʛ
directly. Define
⊛
⊝
k
,
l
|
k
+
l
=
i
⊞
ʔ
i
ʔ
i
ʔ
kl
,
⊠
+
ʔ
i
=
ʔ
i
ʔ
i
ʔ
kl
ʔ
i
+
,
=
=
,
k
,
l
|
k
+
l
=
i
(6.10)
so that from (
6.9
) we have immediately that
ʔ
i
ʛ
=
.
(6.11)
i
≥
0
We now show that
ʔ
0
,
(i)
ʔ
=
˄
−ₒ
(ii)
ʔ
i
ʔ
i
+
1,
from which, with (
6.11
), we will have
ʔ
⃒
ʛ
as required. For (i) we observe that
ʔ
=
ʔ
0
(
6.7
)
ʔ
0
+
ʔ
0
=
(
6.7
)
ʔ
0
=
ʔ
00
+
(
6.8
)
ʔ
00
+
ʔ
00
+
ʔ
0
=
(
6.8
)
(
k
,
l
|
k
+
l
=
0
ʔ
kl
)
(
k
,
l
|
k
+
l
=
0
ʔ
kl
)
ʔ
0
=
+
+
index arithmetic
ʔ
0
ʔ
0
=
+
(
6.10
)
ʔ
0
.
=
(
6.10
)
For (ii) we observe that
ʔ
i
(
k
,
l
|
k
+
l
=
i
ʔ
kl
)
ʔ
i
=
+
(
6.10
)
(
k
,
l
|
k
+
l
=
i
ʔ
k
,
l
+
1
)
˄
−ₒ
+
ʔ
i
+
1
(
6.7
), (
6.8
), additivity
(
k
,
l
|
k
+
l
=
i
(
ʔ
k
,
l
+
1
+
ʔ
k
,
l
+
1
))
ʔ
i
+
1
+
ʔ
i
+
1
=
+
(
6.7
), (
6.8
)
(
k
,
l
|
k
+
l
=
i
ʔ
k
,
l
+
1
)
(
k
,
l
|
k
+
l
=
i
ʔ
k
,
l
+
1
)
ʔ
i
+
1
+
ʔ
i
+
1
=
+
+
rearrange
(
k
,
l
|
k
+
l
=
i
ʔ
k
,
l
+
1
)
+
ʔ
i
+
1,0
+
(
k
,
l
|
k
+
l
=
i
ʔ
k
,
l
+
1
)
+
ʔ
i
+
1
=
(
6.8
)
(
k
,
l
|
k
+
l
=
i
ʔ
k
,
l
+
1
)
+
ʔ
i
+
1,0
+
ʔ
i
+
1,0
+
(
k
,
l
|
k
+
l
=
i
ʔ
k
,
l
+
1
)
+
ʔ
i
+
1
=
(
6.8
)
(
k
,
l
|
k
+
l
=
i
+
1
ʔ
kl
)
+
(
k
,
l
|
k
+
l
=
i
+
1
ʔ
kl
)
+
ʔ
i
+
1
=
index arithmetic
ʔ
i
+
1
+
ʔ
i
+
1
=
(
6.10
)
=
ʔ
i
+
1
,
(
6.10
)
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