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In-Depth Information
B
and the indiscernibility, sim-
ilarity and inclusion relations obtained from the information system
obtained corresponding to the modal operators
S
ʣ
.Next,
we capture this aspect, which will also lead us to the Truth Lemma.
We recall the following definition. For,
B
ↆA
,and
k
∈{
1
,
2
,
3
}
, we define
R
B
ↆ
W
ʣ
×
W
ʣ
such that
R
B
if and only if
k
(
ʓ,ʔ
)
∈
B
ʱ
∈
ʓ
implies
ʱ
∈
ʔ.
(1)
, we again simply write
R
k
{a}
as
R
a
.Fora
ʓ
W
ʣ
,and
a
For
a
∈A
∈
∈A
,let
ʓ
a
denote the set
{
(
a, v
):(
a, v
)
∈
ʓ
}
. Then we have the following.
Proposition 4.
1. R
B
ↆ R
C
for C ↆ B, k ∈{
1
,
2
,
3
}.
2. ʓ
a
=
ʔ
a
if and only if
(
ʓ,ʔ
)
R
a
.
∈
3. ʓ
a
∩
ʔ
a
∅
if and only if
(
ʓ,ʔ
)
∈
R
a
.
=
R
a
.
4. ʓ
a
ↆ
ʔ
a
if and only if
(
ʓ,ʔ
)
∈
R
B
and ʓ
a
=
ʔ
a
,then
(
ʓ,ʔ
)
R
B∪{a}
5. If
(
ʓ,ʔ
)
∈
∈
.
R
B
and ʓ
a
∩
R
B∪{a}
6. If
(
ʓ,ʔ
)
∈
ʔ
a
=
∅
,then
(
ʓ,ʔ
)
∈
.
R
B
and ʓ
a
ↆ
R
B∪{a}
7. If
(
ʓ,ʔ
)
∈
ʔ
a
,then
(
ʓ,ʔ
)
∈
.
8. R
B
=
a∈B
R
a
, k
∈{
1
,
2
,
3
}
.
Proof.
We provide the proofs of Items 2 and 3.
(2):
First suppose
ʓ
a
=
ʔ
a
, and let
a
ʱ
∈
ʓ
. We need to show
ʱ
∈
ʔ
.From
a
ʱ
axiom 16, we obtain
i
∈
ʓ
for some
i
∈
ʘ
. Therefore,
i
∧
∈
ʓ
. Now using
axiom 11 for
B
=
∅
,weobtain
(
a, v
)
(
a, v
))
ʱ
1
∅
1
∅
ₔ
(
i
→
→
∈
ʓ.
v∈V
a
R
1
∅
Since (
ʓ,ʔ
)
∈
,weobtain
(
a, v
)
(
a, v
))
1
∅
ₔ
(
i
→
→
ʱ
∈
ʔ.
v∈V
a
If possible, let
ʱ∈ ʔ
. Then there exists a
v ∈V
a
such that either (
a, v
)
→
1
∅
1
∅
(
i
→
(
a, v
))
/
∈
ʔ
,or
(
i
→
(
a, v
))
→
(
a, v
)
/
∈
ʔ
. First suppose, (
a, v
)
→
1
∅
1
∅
(
i
→
(
a, v
))
/
∈
ʔ
.Then(
a, v
)
∈
ʔ
,and
(
i
→
(
a, v
))
/
∈
ʔ
. Now using the
fact that
ʓ
a
=
ʔ
a
,weobtain(
a, v
)
∈
ʓ
, and hence
i
∧
(
a, v
)
∈
ʓ
. Therefore, from
1
∅
R
1
∅
axiom 14, we obtain
(
i
→
(
a, v
))
∈
ʓ
. Again using the fact that (
ʓ,ʔ
)
∈
,
1
∅
we obtain
i
→
(
a, v
)
∈
ʔ
. Hence by axiom 14, we have
(
i
→
(
a, v
))
∈
ʔ
,
1
∅
a contradiction. Similarly,
(
i
→
(
a, v
))
→
(
a, v
)
/
∈
ʔ
will also lead us to a
contradiction.
Conversely suppose (
ʓ,ʔ
)
R
a
. We need to show (
a, v
)
∈
∈
ʓ
if and only if
a
(
a, v
)
(
a, v
)
∈
ʔ
. First let (
a, v
)
∈
ʓ
. Then from axiom 8, we obtain
∈
ʓ
,and
hence (
a, v
)
∈
ʔ
. Now suppose (
a, v
)
∈
ʔ
.If(
a, v
)
/
∈
ʓ
, then using axiom 9, we
a
(
obtain
¬
(
a, v
))
∈
ʓ
, and hence
¬
(
a, v
)
∈
ʔ
, a contradiction.
R
a
.Let(
a, v
)
(3):
First suppose
ʓ
a
∩
ʔ
a
=
∅
and we show (
ʓ,ʔ
)
∈
∈
ʓ
a
∩
ʔ
a
.
a
ʱ
a
ʱ
Let
∈
ʓ
. We need to show
ʱ
∈
ʔ
.Wehave(
a, v
)
∧
∈
ʓ
, and hence by
