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(
,
) =
(
,
)
(
,
) =
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 ˃ } ˃,
or backwards,
{ ˃ ˃ } μ .
The first type of inference corresponds to search for the solutions of
μ · ˃) ˃
,
that is,for J and T 0 such that,
( )
T 0 (μ(
),
(μ(
), ˃ (
)) ˃(
) ;∀
,
.
x
J
x
y
y
x
y
X
˃ · ˃) 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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