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(μ
P
ₒ
μ
Q
)(
,
)
=
(μ
P
(
), μ
Q
(
)),
x
y
J
x
y
for all
x
,
is to be taken at each case if, of course, it gives a conditional. That is, if from the
premises
∈
X
,
y
∈
Y
. The problem is which function
J
:[
0
,
1
]×[
0
,
1
]ₒ[
0
,
1
]
follows 'y is Q' as a logical consequence.
Formally speaking, it should exist a continous t-norm
T
0
such that
{
'x is P', 'If x is P, then y is Q'
}
T
0
(μ
P
(
x
),
J
(μ
P
(
x
), μ
Q
(
y
)))
μ
Q
(
y
)
for all
x
∈
X
,
y
∈
Y
. This condition, that should hold for any
μ
P
(
x
)
∈[
0
,
1
]
, and
any
μ
Q
(
y
)
∈[
0
,
1
]
, conducts to the inequality
T
0
(
a
,
J
(
a
,
b
))
b
for all
a
,
J
should verify to represent conditional statements. It is called
the
Modus Ponens Inequality
, since it allows the scheme of reasoning
,
b
in
[
0
,
1
]
If x is P
,
then y is Q
xisP
yisQ
called the
scheme of Modus Ponens
.
J
is called a
T
0
-conditional function (shortly,
T
0
-conditional).
There is a theorem, whose proof will be omitted, showing that being
T
0
a contin-
uous t-norm, it is
T
0
(
a
,
J
(
a
,
b
))
b
⃔
J
(
a
,
b
)
J
T
0
(
a
,
b
)
=
sup
{
z
∈[
0
,
1
];
T
0
(
z
,
a
)
b
}
.
Hence, for each
T
0
, the greatest
T
0
-conditional is the function
J
T
0
, since, it verifies
the MP-inequality
T
0
(
a
,
J
T
0
(
a
,
b
))
=
min
(
a
,
b
)
b
.
Remark 3.2.6
For reasons that will be latter on presented,
T
-conditionals
J
T
are
called R-implications (R shorting residuated). They come directly from the Boolean
equation
a
+
b
=
sup
{
z
;
a
·
z
b
}
.
X
,take
If
μ, ˃
∈{
0
,
1
}
(μ
ₒ
˃)(
x
,
y
)
=
J
T
(μ(
x
), ˃ (
y
))
=
sup
{
z
∈
[
0
,
1
];
T
(
z
,μ(
x
))
˃(
y
)
}
.If
μ(
x
)
∈{
0
,
1
}
,˃(
y
)
∈{
0
,
1
}
,itis
⊧
⊨
J
T
(
0
,
0
)
=
1
J
T
(
,
)
=
0
1
1
J
T
(μ(
x
), ˃ (
y
))
=
J
T
(
,
)
=
⊩
1
0
0
J
T
(
1
,
1
)
=
1
,
that coincides with the values of max
(
1
−
μ(
x
), ˃ (
y
))
. That is, all R-implications
μ
+
˃
do coincide with the Boolean material conditional
in the case that
μ
and
˃
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