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μ Q
Hence, the graphic of the output
is the one shown figure
1
0.5
μ
Q
0
1
0.5
when x 0 =
3
.
5.
Fifth Step: Numerical Consequent
This is the case in which
μ Q is y
=
y 0 or y
∈{
y 0 }
. That is “ y is Q ” corresponds to
y is y 0 ”, and
1f y
=
y 0
μ Q (
) = μ y 0 =
y
0f y
=
y 0
In this case:
J
P (
x
),
1
)
if
y
=
y 0
J
P (
x
), μ y 0 (
y
)) =
J
P (
x
),
0
)
if
y
=
y 0 ,
and the output
μ Q depends on the values of J
,
namely, on J
(
a
,
1
)
and J
(
a
,
0
)
.
Notice that if:
J is an S -implication, J
(
a
,
1
) =
1; J
(
a
,
0
) =
N
(
a
)
.
J is an Q -implication, J
(
a
,
1
) =
S
(
N
(
a
),
a
)
; J
(
a
,
0
) =
N
(
a
)
.
J is an R -implication, J
(
a
,
1
) =
1; J
(
a
,
0
) =
Sup
{
z
∈[
0
,
1
];
T
(
z
,
a
) =
0
}
.
J is an ML -implication, J
(
a
,
1
) =
a ; J
(
a
,
0
) =
0.
When J is an ML -implication, with J
(
a
,
1
) =
a and J
(
a
,
0
) =
0:
μ Q (
y
) =
Sup
x
Min
P (
x
),
J
P (
x
), μ y 0 (
y
)))
X
Sup
x
Min
P (
x
), μ P (
x
)),
if
y
=
y 0
= μ Q =
X
0
,
if
y
=
y 0 ,
Notice that if the input is also numerical, i.e. x
=
x 0 then the output is:
Sup
x X μ P (
x
)
if
y
=
y 0 and x
=
x 0
μ Q =
0
if
y
=
y 0 and x
=
x 0 .
 
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