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In-Depth Information
In conclusion
•
If all the partial outputs
μ
Q
1
,μ
Q
2
, ..., μ
Q
n
, are consequences of the input
μ
P
∗
,
also the final output
μ
Q
∗
is a consequence of
μ
P
∗
.
•
If at least one of the partial outputs
μ
Q
i
(
1
i
n
)
is just a conjecture of the
input
μ
P
∗
, also the final output
μ
Q
∗
is a conjecture of
μ
P
∗
.
Nevertheless, it is not usual that
μ
Q
∗
results to be a consequence of the single input
μ
P
∗
. Let's us introduce a necessary and sufficient condition for it in the particular
case in which there is only one rule represented by
J
(
a
,
b
)
=
ab
(Larsen).
Letitbe“If
x
is
P
, then
y
is
Q
”(
x
,
y
∈
X
), and
(μ
P
ₒ
μ
Q
)(
x
,
y
)
=
μ
P
(
x
)μ
Q
(
y
)
,
with the input
x
=
x
0
. Then,
μ
Q
∗
(
y
)
=
μ
P
∗
(
x
0
)μ
Q
(
y
),
∀
y
∈
X
.
Provided
μ
{
x
0
}
=
μ
0
,tohave
μ
{
x
0
}
(
y
)
μ
Q
∗
(
y
)
=
μ
P
∗
(
x
0
)μ
Q
(
y
)
, it is neces-
sary that, with
y
=
x
0
,1
μ
P
∗
(
x
0
)μ
Q
∗
(
x
0
)
or 1
=
μ
P
∗
(
x
0
)
=
μ
Q
∗
(
x
0
)
. Hence,
μ
Q
∗
∈
Cons
(
{
μ
{
x
0
}
}
)
implies
μ
P
∗
(
x
0
)
=
μ
Q
∗
(
x
0
)
=
1.
Provided
μ
P
∗
(
x
0
)
=
μ
Q
∗
(
x
0
)
=
1, from
μ
Q
∗
(
y
)
=
μ
P
∗
(
x
0
)μ
Q
(
y
)
, follows
μ
Q
∗
(
y
)
=
μ
Q
(
y
)
, for all
y
∈
X
, and,
•
=
If
y
x
0
,
μ
{
x
0
}
(
x
0
)
=
1
=
μ
Q
∗
(
x
0
)
•
=
μ
{
x
0
}
(
)
=
μ
Q
∗
(
)
If
y
x
0
,
y
0
y
,
that is
μ
{
x
0
}
(
y
)
μ
Q
∗
(
y
)
, for all
y
∈
X
.
Hence, in this particular case, the necessary and sufficient condition for being
μ
Q
∗
∈
Cons
(
{
μ
{
x
0
}
}
)
is that
μ
P
∗
(
x
0
)
=
μ
Q
∗
(
x
0
)
=
1. Nevertheless, what happens in
most of the cases is that
μ
Q
∗
∈
Conj
(
{
μ
{
x
0
}
}
)
, with
μ
Q
∗
∈
Sp
(
{
μ
{
x
0
}
}
)
,or
μ
Q
∗
∈
Hyp
(
{
μ
{
x
0
}
}
)
.
3.7 Two Final Examples
Let's show an example in which the output is a speculation of the input and other in
which the output is a hypothesis.
Example
. Rule, “If
x
is small, then
y
is big”, with
X
=
Y
=[
0
,
10
]
and
J
(
a
,
b
)
=
ab
y
10
, and
x
0
=
x
(Larsen), with
μ
S
(
x
)
=
1
−
10
,
μ
B
(
y
)
=
5. It is
5
10
,
y
10
)
=
y
20
,
μ
Q
∗
(
)
=
(
−
y
J
1
and from the graphic
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