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=
⊥
1
◦
⊥
1
, id
B
, id
B
◦
⊥
1
, id
B
,
id
A
,
◦
id
A
,
1
,
1
◦
id
A
,
1
⊥
⊥
⊥
=
⊥
1
,
⊥
1
, id
B
,
id
A
,
1
,
1
:
⊥
⊥
A
B
→
A
B.
Thus,
i
2
◦
i
1
id
A
id
B
={
id
A
, id
B
}=
, and hence we obtain an identity
i
2
◦
i
1
=
id
A
id
B
=
id
A
B
:
→
A
B
A
B.
1
, id
B
]
1
Consequently,
i
1
=[⊥
[
⊥
]:
→
,
id
A
,
A
B
B
A
is an isomorphism, that
is,
A
A
.
Claim 3. (Associativity) It holds that
id
A
[
B
B
1
1
, id
C
]:
⊥
]
[⊥
id
B
,
,
A
(B
C)
C
is an isomorphism.
It is enough to show that
i
3
=[
→
A
B
1
1
, id
C
]:
id
B
,
⊥
]
,
[⊥
B
C
→
B
C
is an
id
B
C
i
3
isomorphism, as follows:
={
id
B
, id
C
}=
(from the fact that
id
B
C
=
id
B
C
and
from the fact that each identity morphism is an isomorphism, it holds that
i
3
is an
isomorphism.
Claim 4.
T(C
id
C
), thus (by point 3 of Definition
23
)
i
3
=
id
B
C
:
B
C
→
B
0
)
0
0
TC
, and hence we cannot
use Proposition
8
in order to demonstrate the isomorphism
C
⊥
=
TC
T
⊥
=
TC
⊥
=
0
C
,butthe
standard categorial method for verification of the isomorphisms. For a given schema
C
⊥
α
∗
(
with the interpretation
α
such that
C
=
C
)
, we have the identity morphism
α
∗
(
M
CC
)
α
∗
(
id
C
=
:
C
→
C
with
M
CC
=
{
1
r
|
r
∈
C
}∪{
1
r
∅
}
)
.
α
∗
(
M
CC
),α
∗
(
0
Let us define the morphisms
i
1
=
{
1
r
∅
}
)
:
C
→
C
⊥
and
i
2
=
α
∗
(
M
CC
),α
∗
(
0
[
{
1
r
∅
}
)
]:
C
⊥
→
C
. Thus,
i
1
=
α
∗
(
M
CC
),α
∗
{
1
r
∅
}
◦
α
∗
(
M
CC
),α
∗
{
1
r
∅
}
i
2
◦
=
id
C
,
1
◦
id
C
,
1
⊥
⊥
=
id
C
◦
1
=
id
C
,
⊥
1
1
id
C
,
⊥
◦⊥
(
from point 2
.
4 of Definition
20
)
=
id
C
:
C
→
C,
i.e., the identity morphism for
C
. While,
i
1
◦
i
2
=
α
∗
(
M
CC
),α
∗
{
1
r
∅
}
◦
α
∗
(
M
CC
),α
∗
{
1
r
∅
}
=
id
C
,
1
◦
id
C
,
1
⊥
⊥
=
id
c
,
1
,
⊥
1
.
1
,
⊥
⊥
id
C
⊥
0
={
i
1
◦
i
2
1
) and hence, from point 3
Thus,
=
id
C
}
(from
id
C
⊥
=
id
C
⊥
0
of Definition
23
,
i
1
◦
i
2
=
id
C
⊥
0
. Consequently, this morphism is the identity mor-
0
. Therefore, both morphisms
i
1
◦
phism of the object
C
⊥
i
2
and
i
2
◦
i
1
are the
1
0
identity morphisms and so
i
1
=
id
C
,
⊥
:
C
→
C
⊥
is an isomorphism and
0
and, analogously, by commutativity,
0
0
hence, (a)
C
C
⊥
⊥
C
C
⊥
C
.
