Information Technology Reference
In-Depth Information
xissmall
→
x
+
eissmall
sissmall
s
+
eissmall
x
+
eissmall
→
x
+
2
eissmall
s
+
eissmall
s
+
2
eissmall
···
s
+
te is small
It should be pointed out that this successive application of 'Modus Ponens' is only
formally correct provided the arrow
, representing the conditional statements
is a 'conditional', that is, satisfies the Modus Ponens' inequality a
→
·
(
a
→
b
)
≤
b,
a
+
for all a and b. It happens for instance either with a
→
b
=
b in the case all
statements can be represented in a Boolean algebra, or a
b, if they are
representable in just a lattice (at the end, the negation is not in the problem). In
the current case. 'If x is small, then x
→
b
=
a
·
+
e is small' either can be identified with
'Not(x is small) or (x
+
e is small)', or with '(x is small) and (x
+
e is small)'. Of
a
+
a
+
course, in a Boolean algebra it is a
·
(
b
)=
a
·
a
·
b
=
0
+
a
·
b
=
a
·
b
≤
b,
and in any lattice it is a
·
(
a
·
b
)=(
a
·
a
)
·
b
=
a
·
b
≤
b.
15.3
About the Black's Separation Point with 'small' in [0,10]
The philosopher Max Black asserted, in a more general setting, that a separation
point
B
between the numbers that are small and those that are not, does exist, but
is impossible to find. That is, that the before mentioned number
p
such that
h
)
is not a heap does exist but is not determinable. We will see in the following that
this statement is not always correct, but that a different kind of separation points
s for 'small' can always be found by employing something related with a partial
contradiction of small with itself [8].
(
p
15.3.1
Accepting that 'small' is a gradable predicate in [0,10], and by using 'small' under
the former four rules, it is possible to define which functions
μ
S
:
[
0
,
10
]
→
[
0
,
1
]
can
represent
μ
S
(
x
)=
degree up to which 'x is small'.
Under those rules, those functions do verify:
a)
μ
S
(
0
)=
1
b)
μ
S
(
10
)=
0
c) If
x
≤
y
,then
μ
S
(
y
)
≤
μ
S
(
y
)
d) If
μ
S
(
x
)
>
0, then there is no any
y
∈
(
x
−
e
,
x
+
c
)
such that
μ
S
(
y
)=
0.
