Information Technology Reference
In-Depth Information
.
6
.
3
Look that
N
1
(
.
)
=
,
N
1
(
.
)
=
,
N
1
(
.
)
=
.
,
N
1
(
.
)
=
.
,
3
2
0
3
5
0
3
4
0
43
3
6
0
25
.
1, but
N
0
(
N
1
(
3
.
8
)
=
0
3
.
2
)
=
0
.
8
,
N
0
(
3
.
5
)
=
0
.
5
,
N
0
(
3
.
6
)
=
0
.
4
,
N
0
(
3
.
4
)
=
0
.
6
,,
N
0
(
3
.
6
)
=
0
.
4
,
N
0
(
3
.
8
)
=
0
.
2.
Example 2.2.52
The negation is linear, and the standard algebra must verify the
x
10
μ
=
μ
·
˃
+
μ
·
˃
. Determine the triplet
law
(
T
,
S
,
N
)
, and with
μ(
x
)
=
(in
X
=[
0
,
10
]
) and
⊧
⊨
1
,
if
x
∈[
0
,
5
]
7
−
x
˃(
x
)
=
,
if
x
∈[
5
,
7
]
⊩
2
0
,
if
x
∈[
7
,
10
]
μ
·
˃
+
μ
·
˃
=
μ
μ
·
˃
+
μ
·
˃
=
˃
(μ
·
˃
)
=
˃
+
μ
·
˃
.
check that
,
, and
Solution
. From N linear,
N
=
N
0
, it follows that we can take
˕
=
id, and
μ
·
˃
+
μ
·
˃
W
˕
from
=
μ
it follows
T
=
prod
,
S
=
, and
N
=
N
. Hence
˕
˕
W
∗
,
(
T
,
S
,
N
)
=
(
prod
,
N
)
.
⊧
⊨
0
x
−
5
μ
(
x
˃
(
Since
x
)
=
1
−
10
, and
x
)
=
,itfollows
2
⊩
1
⊧
⊨
⊧
⊨
⊧
⊨
x
10
x
x
10
0
x
−
1
(
7
−
x
)
, μ˃
(
(
x
−
5
)
, μ
˃(
7
−
x
x
μ˃(
x
)
=
x
)
=
x
)
=
(
1
−
10
)
.
⊩
20
⊩
20
⊩
2
x
10
0
0
Hence,
⊧
⊨
⊫
⊬
x
x
10
W
∗
(
10
,
0
)
=
x
10
=
μ(
(μ
·
˃
+
μ
·
˃
)(
x
(
7
−
x
)
x
(
x
−
5
)
W
∗
(
x
10
x
)
=
=
x
),
,
)
=
⊩
20
20
⊭
W
∗
(
x
x
10
,
10
)
=
0
⊧
⊨
⊫
⊬
W
∗
(
x
x
10
,
1
−
10
)
=
1
(˃
·
μ
+
˃
·
μ
)(
x
(
7
−
x
)
W
∗
(
7
−
x
x
7
−
x
x
)
=
=
˃(
x
).
,
(
1
−
10
)
=
20
2
2
⊩
⊭
W
∗
(
0
,
0
)
=
0
Finally, since,
Search WWH ::
Custom Search