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2
(( a, v )
ʱ )
ʓ .Thisgives
1
(( a, v )
ʱ )
ʓ (due to axiom 6),
axiom 12,
and therefore we obtain ( a, v )
ʱ
ʔ as ( ʓ,ʔ )
R 1
. This gives us ʱ
ʔ as
( a, v )
ʔ .
Conversely, suppose ( ʓ,ʔ )
R a , and we prove ʓ a
ʔ a
=
. Axiom 16
guarantees the existence of an i
ʘ such that i
ʓ . Therefore, by axiom 10, we
a v∈V a (( a, v )
( a, v )))
ʓ and hence v∈V a (( a, v )
1
1
obtain
( i
( i
( a, v )))
ʔ . Therefore, for some v
∈V a ,( a, v )
1
( i
( a, v ))
ʔ .Since i
ʓ ,
we obtain ( a, v ) ∈ ʓ ∩ ʔ and hence ʓ a ∩ ʔ a = .
The following proposition relates the canonical relations corresponding to the
modal operators
k
B and the indiscernibility, similarity and inclusion relations
obtained from the information system
ʣ . Let us use the notations Ind B , Sim B
and In B to denote the indiscernibility, similarity and inclusion relations relative
to the attribute set B induced from the NIS
S
ʣ .
S
Proposition 5. 1. R B = Ind B .
2. R B = Sim B .
3. R B = In B .
Proof. We provide the proof for the indiscernibility relation. Since Ind B =
a∈B Ind a and R B = a∈B R a (by Item 8 of Proposition 4), it is enough
to prove the result for singleton B .Solet B =
Ind b .
Then, we obtain F ʣ ( ʓ,b )= F ʣ ( ʔ, b ). This implies ʓ b = ʔ b , and hence by
Item (2) of Proposition 4, we obtain ( ʓ,ʔ )
{
b
}
. Suppose ( ʓ,ʔ )
R b , as desired. Conversely, let
R b . Then by Item (2) of Proposition 4, we obtain ʓ b = ʔ b .Thisgives
F ʣ ( ʓ,b )= F ʣ ( ʔ, b ) and hence ( ʓ,ʔ )
( ʓ,ʔ )
Ind b .
Once we have the Proposition 5, giving the same argument as in normal modal
logic, we obtain
W ʣ ,
Proposition 6 (Truth Lemma). For any wff ʲ and ʓ
ʣ ,ʓ | = ʲ.
ʲ ∈ ʓ if and only if M
Using the Truth Lemma and Proposition 3 we have
Proposition 7. If ʱ is consistent then there exists a maximal consistent set ʣ
containing ʱ such that
ʣ
M
|
= ʱ.
This gives the desired completeness theorem.
Theorem 2 (Completeness). For any wff ʱ,if
|
= ʱ,then
ʱ.
4 Decidability
In this section we shall prove the following decidability result.
Theorem 3. We can decide for a given wff ʱ,whether
|
= ʱ.
 
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