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The extended Heyting algebra H ( Υ ) + =
0 ,
Lemma 25
(H( Υ ),
,
,
h ,
λ ,
0
0
ρ ) , where
_) , is an algebraic dual of
the S 4 extended bimodal logic ( by modal logical operators
λ =⊥
( Υ
\
λ_) and
ρ =⊥
ρ( Υ
\
R 1
and
R
, respec-
tively ), so called IM-logic ( Intuitionistic Modal logic [ 26 ]).
Proof It is easy to show that
ρ is
a multiplicative algebraic modal operator. In fact, for any two S 1 ,S 2 H( Υ ) (we
recall that
λ is an additive algebraic modal operator and
0
={⊥}
is an empty database with the empty relation
), we have:
={⊥}∪ Υ
S 2 ) ={⊥}∪ Υ
λS 2 )
λ (S 1
S 2 )
\
λ(S 1
\
(λS 1
={⊥}∪
Υ
\
(λS 1 )
Υ
\
(λS 2 )
= λ (S 1 )
λ (S 2 )
;
λ
0 ={⊥}∪ Υ
λ {⊥} ={⊥}∪
0
\
( Υ
\
Υ )
={⊥}∪∅=⊥
;
and hence
λ is an additive modal algebraic operator, while,
ρ Υ
S 2 ) =⊥
ρ Υ
(S 2 )
0
0
ρ (S 1
S 2 )
=⊥
\
(S 1
\
(S 1 )
Υ
\
ρ Υ
(S 1 )
ρ Υ
(S 2 ) = ρ (S 1 )
0
=⊥
\
\
ρ (S 1 )
;
0
0
0
ρ ( Υ )
=⊥
ρ( Υ
\
Υ )
=⊥
ρ(
)
=⊥
Υ
=
Υ ,
and hence
ρ is a multiplicative modal algebraic operator. Thus, these two algebraic
operators correspond to the logical modal operators
R 1
and
R
.
Consequently, WMTL can be regarded just as a fragment of classical bimodal
logics containing S4, whose dual is the Heyting algebra
= F
0 ,
L alg =
F ( Υ )
( Υ ),
,ρλ
,ρλ
h ,
where ρλ = Γ = T , ρλ ∪=
and, from Corollary 34 , (ρλ h ) = ⇒
.
F ( Υ ) can be embedded
into a complete database Boolean algebra extended by three multiplicative algebraic
modal operators.
Let us show that this database Heyting algebra L alg =
0 , Υ ) is a complete Boolean
algebra with negation operator defined for every A = TA
=
Corollary 36
The algebra BA
(Ob DB sk ,
,
,
,
Ob DB sk
by
A
0
( Υ
\
A) , such that there is a homomorphism :
(i) _ .
BA Υ ,
into the complete Boolean algebra BA Υ =
:
BA
, Υ ) ( Proposition 70 ).
( Υ ,
v ,
v ,
¬ v ,
Consequently , the database Heyting algebra L alg =
F ( Υ ) can be embedded
into the N-modal algebra BA + =
0 , Υ ,
1 ,
2 ,
3 ) , where
(Ob DB sk ,
,
,
,
1 =
is the S 4 algebraic universal modal operator ( Example 42 ) and ,
from Lemma 25 ,
λ R
2 =− λ −=−
λ and
3 = ρ =
ρ
, such that for every
A
=
TA
Ob DB sk
and B
=
TB
Ob DB sk :
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