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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
(μ
P
(
x
),
1
)
if
y
=
y
0
J
(μ
P
(
x
), μ
y
0
(
y
))
=
J
(μ
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
(μ
P
∗
(
x
),
J
(μ
P
(
x
), μ
y
0
(
y
)))
∈
X
Sup
x
Min
(μ
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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