数理解析研究所講究録 1222 巻 2001 年 34-39

34

On structures of weak interlaced bilattices
近藤通朗
島根大学総合理工学部

Michiro KONDO
Department of Mathematics and Computer Science, Shimane UniversityAbstract
We sutdy fundamental properties of weak interlaced bilattices and show that
1. For any bounded lattice , there exists an interlaced bilattice such that
$B$

$L$

.

$\mathcal{K}(L)\underline{\simeq}$$\mathrm{C}\mathrm{o}\mathrm{n}s(\mathrm{B})_{1}$2. For any interlaced bilattice$\mathcal{K}(L)\underline{\simeq}$$B$with negation, there exists alattice$L$such thatCons{B).

1Introduction
It is well-known the Kleene’s 3- alued logic in the field of multiple- alued logics. The logic
has three values Jalse, true, and 1[unknown) as truth values. These values have two
informal orderings concerning “amount of knowledge” and “degree of truth”. For example,
truth value,
if we think of acertain proposition such as Riemann’s conjecture assigned
then it is possible that we can conclude the truth value of the proposition as true or false
smaller than true andwith increasing knowledge. Thus in the ordering of knowledge,false. Asentence with 1is between false and true in the ordering of degree of truth.
In this way it can be considered that the three valued logic has two orderings. Belnap([2]), Ginsberg([5]), and others proposed concept of abilattice which has two orderings andproved some fundamental results ([1, 3, 4]). It is shown by Fitting ([3]) that bilattices cangive auniform semantics for many lanuages of logic programming. Since then the theory ofbilattices is ahot reserach field.
On the other hand, as in Fuzzy logics, atruth value can be taken as aclosed interval$[a,b]$ . Let
be the set of all closed intervals of . In this case we also
be alattice and, if $[a,b]\subseteq[c,d]$ then the knowledge in $[a,b]$ isdefine two orderings. For $[a,b]$ ,greater than that in $[c,d]$ . Thus we set $[a,b]\subseteq_{k}[c,d]$ if $[a,b]\subseteq[c,d]$ . Likewise we also define
if $a\leq c$ and $b\leq d$, because $[c,d]$ is greater than $[a,b]$ in the ordering degreeprecise definition is given below hasof truth. The structure
the property of weak interlaced bilattice.
and proved
In $[3, 4]$ , Fitting, Font and Moussavi have investigated the strucutre ofsome results :$[perp] \mathrm{a}\mathrm{s}$$[perp] \mathrm{i}\mathrm{s}$

$L$

$L$

$\mathcal{K}(L)$$[c,d]\in \mathcal{K}(L)$$[a,b]\subseteq_{t}[c, d]$$\mathcal{K}(L)=<\mathcal{K}(L),$$\subseteq_{t},$$\subseteq_{k}>\mathrm{w}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{h}$$\mathcal{K}(L)$

1. If$L$

is abounded lattice, then$\mathcal{K}(L)$is aweak interlaced bilattice ([4]) ;2. If is acomplete lattice with an involution, then $\mathcal{K}(L)\cong Cons(B1$ , where Cons(B)is the set of all consistent elements of an interlaced bilattice with negation and$L$$B$conflation ([3]) ;35
3. If is distributive bilattice with commutative negation and conflation, then Cons(B)for some complete distributive lattice $L([3])$ .
$B$$\mathcal{K}(L)$Now it is natural to ask the following questions:
Ql Is there alattice$\mathcal{W}$$L$such that$\mathcal{W}\cong \mathcal{K}(L)$for every weak interlaced bilattice?

Q2 Is there an interlaced bilatticebounded lattice ?

$B$such that$\mathcal{K}(L)\underline{\simeq}$Cons(B) for every$L$

Q3 Is there alatticewith negation ?

$L$such that$\mathcal{K}(L)\underline{\simeq}Cons(B)$for every interlaced bilattice$B$

In the following, we study properties of

2Definition of$\mathcal{K}(L)$and answer the questions above.

$\mathcal{K}(L)$for any lattice
We define astructure
set of all closed intervals of , that is,$\mathcal{K}(L)$$L$. Let $L=(L, \leq)$ be alattice and $K(L)$ be the$L$$K(L)=\{[a, b]|a\leq b,a, b\in L\}$$[a,b]=\{x|a\leq x\leq b\}$.

For any $[a, b]$ , $[c, d]\in K(L)$ , we define two orderings Ct,$\subseteq_{k}$on $K(L)$ as follows :, $b\leq d$$[a, b]\subseteq_{k}[c,d]\Leftrightarrow a\leq c$ , $b\geq d$$[a,b]\subseteq_{t}[c,d]\Leftrightarrow a\leq c$. It is obvious from definition that $[0,0]$ ([1, 1]) is the
We set $\mathrm{K}(\mathrm{L})=<K(L),$. On the other hand, while $[0, 1]$ is theminimum (maximum) element with respect to. This
minimum element, there is no maximum element with respect to the ordering. Fourand is asemi-lattice conceringmeans thatis alattice with respect tooperators , , ,are defined by$\subseteq_{t},$$\subseteq_{k}>$$\subseteq_{t}$$\subseteq_{k}$$\mathcal{K}(L)$$\bigcap_{t}$$\mathrm{u}_{t}$$\subseteq_{k}$$\subseteq_{t}$$\mathrm{n}_{k}$$\mathrm{u}_{k}$$\inf\subseteq_{t}\{a, b\}=a\bigcap_{t}b$$\sup\subseteq_{t}\{a, b\}=a\mathrm{u}_{t}b$$\inf\subseteq_{k}\{a, b\}=a\bigcap_{k}b$$\sup\subseteq k$

Arelational system $<B,$
1.
$2$

$B$

$\{a, b\}$ $=a \bigcap_{k}b$$\leq_{t},$$\leq_{k}>\mathrm{i}\mathrm{s}$(if it is defined)

called an interlaced bilattice if it satisfies

is anon-empty set. $<B,$$\leq_{t}>,$$<B,$$\leq_{k}>\mathrm{a}\mathrm{r}\mathrm{e}$bounded lattices and satisfy(a) $x\leq_{t}y\Rightarrow x\otimes z\leq_{t}y\otimes z$, $x\oplus z\leq_{t}y\oplus z$(b) $x\leq_{k}y\Rightarrow x\wedge z\leq_{k}y\wedge z$ , $x\vee z\leq_{k}y\vee z$

By $0(1)$ , we mean the minimum (maximum) element with respect to the ordering
the minimum (maximum) element concering to .
We also denote by
Any interlaced bilattice is called distributive when it satisfies$[perp](\mathrm{T})$$\leq_{k}$$\leq_{t}$.

36$x\circ(y\bullet z)=(x\circ z)\bullet(y\mathrm{o}z)$

for ,$\circ$$\bullet\in\{\wedge, \vee, \otimes, \oplus\}$.

. This means twelve equations such as$x\wedge(y\vee z)=(x\wedge y)\vee(x\wedge z)$$x\oplus(y\wedge z)=(x\oplus y)\wedge(x\oplus z)$

Amap

from

$\neg$$B$

into itself is called anegation if$x\leq_{t}y\Rightarrow\neg y\leq_{t}\neg x$$x\leq_{k}y\Rightarrow\neg x\leq_{k}\neg y$$\neg-.x=x$.

For lattices $L_{1}=<L_{1}$ , ,: For $(a,b)$ ,on the product$\bigwedge_{1}$$\mathrm{v}_{1}>\mathrm{a}\mathrm{n}\mathrm{d}L_{2}=<L_{2}$$L_{1}\mathrm{x}L_{2}$$(c, d)\in L_{1}\mathrm{x}$,$\bigwedge_{2}$$L_{2}$,$\mathrm{v}_{2}>$,, we define operations$\wedge,$$\vee,$$\otimes,$$\oplus$$(a,b) \wedge(c, d)=(a\bigwedge_{1}c, b\bigvee_{2}d)$$(a, b) \vee(c,d)=(a\bigvee_{1}c, b\bigwedge_{2}d)$$( \mathrm{a},6)\otimes(c, d)=(a\bigwedge_{1}c, b\bigwedge_{2}d)$$(a, b)\oplus(c,d)=$ ( $a \bigvee_{1}c$, V2 ).

The structure $L_{1}L_{2}=<L_{1}\mathrm{x}L_{2},$some fundamental results about the structure:$\wedge,\vee,$

Proposition 1(Fitting). If$L_{2}=<L_{1}\mathrm{x}L_{2},$$\wedge,$$\vee,$$\otimes,$bilattice with negation , where$\neg$$\otimes,\oplus>\mathrm{i}\mathrm{s}$called aGinsberg product There areare bounded lattices then the Ginsberg product $L_{1}$an interlaced bilattice. Espectially, $LL$ is an interlaced$L_{1}$$\oplus>is$$d$$b$,$\neg$$L_{2}$

isdefined by $\neg(a,b)=(b, a)$ .

It is proved that the converse holds by Avron ([1]).
,Proposition 2(Avron). For any interlaced bilattice , there are bounded lattices
such that $B\cong L_{1}L_{2}$ . In particular, for any interlaced bilattice with negation, there isa bounded lattice such that $B\cong LL$ .
$B$$L_{1}$$L_{2}$$\leq_{k}$

on

$B$

$L$

It is clear from definition that orderings
Ginsberg product $LL$ , respectively :$\subseteq_{t}$$\subseteq_{k}$inin$\mathrm{K}(\mathrm{L})\Leftrightarrow\leq_{t}$$\mathrm{K}(\mathrm{L})\Leftrightarrow\leq_{k}$$\subseteq_{t},$$\subseteq_{k}$on$\mathcal{K}(L)$are the same as$\leq_{t},$in $LL$in $LL$

Next we give adefinition of aweak interlaced bilattice according to Font ([4]). Astructure
called aweak interlaced bilattice if$\mathcal{W}=<W,$ $\leq_{t},$

$1$

$2$

. $<W,$. $<W,$$\leq_{k}>\mathrm{i}\mathrm{s}$$\leq_{t}>$:lattice$\leq_{k}>$:meet semilattice, $c\leq_{k}d\Rightarrow a\wedge c\leq_{k}b\wedge d,a\vee c\leq_{k}b\vee d$3.

$a\leq_{k}b$4.

$a\leq_{t}b,c\leq_{t}d\Rightarrow a\otimes c\leq_{t}b\otimes d$

5.

$a\leq_{t}b$,$c\leq_{t}d\Rightarrow a\oplus c\leq_{t}b\oplus d$,

if$a\oplus c$

and$b\oplus d$exist

37
3Properties of weak interlaced bilattices
For any weak interlaced bilattice$\mathcal{W}$, if we define$L_{1}=\{x\in \mathcal{W}|x\leq_{k}0\}=[[perp], 0]_{k}$$L_{2}=\{x\in \mathcal{W}|x\leq_{k}1\}=[[perp], 1]_{k}$,then we have
Proposition 3.
$L_{1}=[[perp],0]_{k}=[0, [perp]]_{t}$$L_{2}=[[perp], 1]_{k}=[[perp], 1]_{t}$Proof. Let. Since, we have Iby definitionof weak interlaced bilattice. Prom, it follows that $x\vee[perp]=1$ and hence
that. This means.
Conversely, suppose. If we put $u=0\otimes x$ , then it is clear that $u\leq_{k}0$ and$u\leq_{k}x$ . Since $0\leq_{t}x$ , we have $0(\ x\leq_{t}x\otimes x=x$
and hence $u\leq_{t}x$ . It follows from $1\leq_{k}u$$x\wedge[perp]=x$$x\leq_{t}1$that. Since, we also have
. On the other hand, since $u\leq_{t}x$ ,we get $u\wedge x=u$ . Theses imply that $x\leq_{k}u$ and hence that $x=u$ . Thus we have $x\leq_{k}0$ .
Namely, we have.
The second equation can be proved similarly.
$x\in[[perp], 0]_{k}$$1\leq_{k}x\leq_{k}0$$\vee[perp]\leq_{k}x\vee[perp]\leq_{k}0\vee[perp]$$1\vee[perp]=0\vee[perp]=[perp]$$x\leq_{t}[perp]$$[1, 0]k\subseteq[0, [perp]]_{t}$$x\in[0, [perp]]_{t}$$x\wedge[perp]\leq_{k}x\wedge u$$[0, [perp]]_{t}\subseteq[[perp], 0]_{k}$0The result implies that
andwhereandare defined by$L_{1}$$\leq_{1}$$L_{2}$

are lattices with ordering$\leq_{1}$and$\leq_{2}$in , respectively,$B$$\leq_{2}$$\leq_{1}=\leq\iota=\geq_{k}$$\leq_{2}=\leq t=\leq_{k}$Thus we can consider the Ginsberg product $L_{1}L_{2}$ , which becomes an interlaced bilattice.
Moreover we can prove

Proposition 4. Let$\mathcal{W}$be any weak interlaced bilattice. For any x$\in \mathcal{W}$

x $=(x\otimes 0)\oplus(x\otimes 1)=(x\wedge[perp])\vee(x\vee 1)$, we have

Proof. See Avron [1] COr.3.8$\square$Now we investigate arealtion between aweak interlaced bilatticebilattice $L_{1}L_{2}$ constructed by .

$\mathcal{W}$and an interlaced$\mathcal{W}$

Lemma 1. A map
embedding.

$\xi$

:

$\mathcal{W}arrow L_{1}\mathrm{x}L_{2}$defined by$\xi(x)=(x\otimes 1, x\otimes \mathrm{O})=(x\vee[perp], x\wedge[perp])$is anThis means that
Theorem 1. Any weak interlaced bilattice can be embedded into an interlaced bilattice.

As to the question Ql, we can give anegative answer by presentiong acounter example.
Let
be the set $\{0, 1, a, b, [perp], 1\}$ such thatand $1\leq_{k}a\leq_{k}0,$. It is obvious thatis aweak interlaced bilattice. Suppose that there is alatticesuch that. If $|L|\geq 3$ , then there exists an element $a\in L$ such that $0<a<1$ .
For that element we have $[0, 0]$ , $[0, a]$ , $[0, 1]$ , $[a, 1]$ , $[a, a]$ ,and. Since, it must be $|L|\leq 2$ . But, in this case, we have
. This means that there
is no lattice such that.
$\mathcal{W}$$0\leq_{t}a\leq_{t}[perp]\leq_{t}b\leq_{t}1$$b\leq_{k}1$$L$$[perp]\leq_{k}$$\mathcal{W}$$\mathcal{W}\cong \mathcal{K}(L)$$[1, 1]\in \mathcal{K}(L)$$|\mathcal{W}|=5$$|\mathcal{K}(L)|\leq 3$$L$$\mathcal{W}\cong \mathrm{K}\{\mathrm{L})$$|\mathcal{K}(L)|\geq 6$

384Characterization of$\mathcal{K}(L)$for abounded lattice . Let
In this section we consider the properties ofis closed under the operations. It is clear thatby $\eta([a,b])=(a, b)$ , then we canamap:
not closed under . If we define
prove that$L$$\mathcal{K}(L)$$\{(\mathrm{a}, b)|a\leq b\}\subseteq L\mathrm{x}L$$\mathrm{I}(L)$$\vee$

Lemma 2.

$\mathrm{I}(L)=$$\otimes \mathrm{a}\mathrm{n}\mathrm{d}$$\oplus \mathrm{b}\mathrm{u}\mathrm{t}$$\mathcal{K}(L)arrow \mathrm{I}(L)$$\eta$$\eta:\mathcal{K}(L)arrow \mathrm{I}(L)$$\wedge,$: bijection and$\eta([a, b]\bigcap_{t}[c, d])=\eta([a, b])\otimes\eta([c,d])$$\{(\mathrm{a}, b]\mathrm{U}_{t}[c,d])=\eta([a, b])\oplus\eta([c,d])$$\eta([a,b]\bigcap_{k}[c,d])=\eta([a, b])\wedge\eta([c,d])$$if[a,b]\oplus[c, d]$exists$\eta([a,b]\mathrm{u}_{k}[c,d])=\eta([a,b])\vee\eta([c,d])$We call the map at-k dual isomorphism and identify the isomorphism with the t-k
dual isomorphism, that is,$\eta$$\mathcal{K}(L)\underline{\simeq}\mathrm{I}(L)\subseteq LL$

In any interlaced bilattice $LL$ , the negation$\neg$$\neg(a,b)=(b,a)$

is defined by.

An element $(a, b)$ in $LL$ is called consistent when it satisfies $(a, b)\leq_{t}\neg(a, b)$ , that is,$(a, b)$:$\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\Leftrightarrow a\leq b$

If we denote by Cons(B) the set of all consistent elements of an interlaced bilattice , since, then we haveCons(L $L$ )$B$$=\mathrm{I}(\mathrm{L})$

Theorem 2. For any bounded lattice$L$,$\mathrm{K}(\mathrm{L})\cong \mathrm{I}(L)=Cons(LL)$This means that we can answer the question Q2 as Yes. Moreover, for the structure, we can show$\mathrm{I}(L)$Theorem 3.

Proof. Let

is the weak interlaced bilattice generated by $\mathrm{A}=\{(a,a)|a\in L\}$.

. For every element $(a, b)\in$
be any weak interlaced bilattice such that A$(a,a)\wedge(b,b)=$( A $b,a\vee b$) $=(a,b)\in \mathcal{W}$ .
$(a,a)$, we have,, since$\subseteq \mathcal{W}$$\mathcal{W}$$\mathrm{I}(\mathrm{L})(a\leq b)$

Thus$\mathrm{I}(L)$$a$$(b,b)\in\Delta\subseteq \mathcal{W}$$\mathrm{I}(L)\subseteq \mathcal{W}$.

$\square$

As to the question Q3, let be any interlaced bilattice with negation. Since thereis alattice such that $B\cong LL$ , by identifying with $LL$ , we have Cons(B). This means that
Cons(L (E) )
$B$

$B$

$L$

$L$

$=$

$=\mathrm{I}(L)\cong \mathcal{K}(L)$Theorem 4. For any interlaced bilattice$B$$with$Cons(B)negation, there is a lattice$\underline{\simeq}\mathcal{K}(L)$Therefore we can answer the question Q3 as Yes.

$L$such that39References[1] A.Avron, The structure of interlaced bilattices, Math. Struct, in Comp. Science, vo1.6(1996), 287-299.
[2] N.D.Belnap, Jr., A useful four-valued logic, J.M.Dunn and G.Epstein- (eds) Modern Usesof Multiple-Valued Logic, D.Reidel (1977).

[3] M. Fitting, Kleene \prime s Logic, generalized, Jour.Logic Comput., vol.l (1991), 797-810.
[4] J.M.Font and M.Moussavi, Notes on a six-valued extension of three-valued logic, Jour.
of Applied Non-Classical Logic, vo1.3 (1993), 173-187.

[5] M.Ginsberg, Multivalued logic: a unifom approach to reasoning in artificial intelligence,
Computational Intelligence, vo1.4 (1988), 265-316.

Michiro Kondoe-mail kondo@cis.shimane-u.ac.jp
Department of Mathematics and Computer Science
Shimane University, Matsue, 690-8504
JAPA