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P μ Q )(
,
) =
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 P (
x
),
J
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
}
˃)(
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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