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In-Depth Information
(
,
)
=
(
,
)
(
,
)
=
Analogously,
J
a
b
T
a
b
are not implication functions, since
J
0
0
(
,
)
=
⃒
(
a
1
,
)
(
a
2
,
)
T
0
0
0, and
a
1
a
2
J
b
J
b
. Also Sasaki operators are
not, for example, 0
.
4
<
0
.
8, but
J
S
(
0
.
4
,
0
.
9
)
=
max
(
1
−
0
.
4
,
min
(
0
.
4
,
0
.
9
))
=
max
(
0
.
6
,
0
.
4
)
=
0
.
6
<
0
.
8
=
J
S
(
0
.
8
,
0
.
9
)
, that is,
J
S
is non-decreasing in the
second variable.
To represent conditional statements as they appear in language, the concept of
implication function is sometimes excessive. What is needed are just T-conditionals,
except in the cases where more properties are necessary to represent the meaning of
the conditional statements.
Remark 3.2.15
There is a question that is not independent of the representation
J
for the conditional statement. The question is: What are we going to do with
J
?
Which is the purpose for using
J
?
We are to make an inference that, in principle, could be forwards
{
μ, μ
ₒ
˃
}
˃,
or backwards,
{
˃
,μ
ₒ
˃
}
μ
.
The first type of inference corresponds to search for the solutions of
μ
·
(μ
ₒ
˃)
˃
,
that is,for
J
and
T
0
such that,
(
∗
)
T
0
(μ(
),
(μ(
), ˃ (
))
˃(
)
;∀
,
∈
.
x
J
x
y
y
x
y
X
˃
·
(μ
ₒ
˃)
μ
, that
The second type corresponds to search for the solutions of
is, for
J
,
T
1
and
N
such that,
(
∗∗
)
T
1
(
N
(˃ (
y
)),
J
(μ(
x
), ˃ (
y
))
N
(μ(
x
))
;∀
x
,
y
∈
X
.
Hence, given
J
, we need to know
T
0
such that
T
0
(
a
,
J
(
a
,
b
))
b
, for forward
inference, and given
J
and
N
, we need to know
T
1
such that
T
1
(
N
(
b
),
J
(
a
,
b
))
, for backwards inference.
Notice that the two t-norms in (
∗
) and (
∗∗
) are not necessarily coincidental. For
example, given
J
N
(
a
)
N
0
, can we do backwards
inference? To answer this question we just need to know if there is a continuous
t-norm
T
1
such that
T
1
(
(
a
,
b
)
=
max
(
1
−
a
,
b
)
, with
N
=
1
−
b
,
max
(
0
,
max
(
1
−
a
,
b
))
1
−
a
, with
a
=
1 it results
T
1
(
1
−
b
,
b
)
=
0, and
T
1
=
W
. Then, since,
W
(
1
−
b
,
max
(
1
−
a
,
b
))
=
max
(
0
,
1
−
b
+
max
(
1
−
a
,
b
)
−
1
)
1
−
a
−
b
,
if
a
+
b
1
=
max
(
0
,
max
(
1
−
a
−
b
,
0
))
=
1
−
a
,
because
0
,
if
a
+
b
>
1
b
0 implies
−
b
0, and 1
−
a
−
b
1
−
a
. Finally, the answer is: Yes, with
T
1
=
W
. It can also be done backwards inference in the following cases,
1
−
b
,
if
a
b
1. With
J
min
, since
W
(
1
−
b
,
J
min
(
a
,
b
))
=
1
−
a
.
(
−
,
)
=
,
>
W
1
b
b
0
if
a
b
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