# On power idealization filter topologies of lattice implication algebras.

1. Introduction and PreliminariesBy generalizing Boolean algebras and Lukasiewicz implication algebras [1], Xu [2] defined the concept of lattice implication algebra which is regarded as an efficient approach to deal with lattice valued logical systems. Later, Xu and Qin [3] defined the concept of the filer topology of a lattice implication algebra which takes the set of all filters of the lattice implication algebra as a base. Based on these definitions and some results in [4], we introduce and study power idealization topologies with respect to filter topologies and power ideals of lattice implication algebras.

Now we recall some definitions and notions of lattice implication algebras and topological spaces.

Let (L, [conjunction], [disjunction], 0,1) be a bounded lattice with the greatest 1 and the smallest 0. A system (L, [conjunction], [disjunction], ', [right arrow], 0,1) is called a quasi-lattice implication algebra if ' : L [right arrow] L is an order-reserving involution and [right arrow]: L x L [right arrow] L is a map (called an implication operator) satisfying the following conditions for any x, y, z [member of] L:

(1) x [right arrow] (y [right arrow] z) = y [right arrow] (x [right arrow] z),

(2) x [right arrow] x = 1,

(3) x [right arrow] y = y' [right arrow] x',

(4) x [right arrow] y = y [right arrow] x = 1 implies x = y,

(5) (x [right arrow] y) [right arrow] y = (y [right arrow] x) [right arrow] x.

A quasi-lattice implication algebra (L, [conjunction], [disjunction], ', [right arrow], 0,1) is called a lattice implication algebra if the implication operator [right arrow] further fulfils the following conditions:

(6) (x [disjunction] y) [right arrow] z = (x [right arrow] z) [conjunction] (y [right arrow] z),

(7) (x [conjunction] y) [right arrow] z = (x [right arrow] z) [disjunction] (y [right arrow] z).

A lattice implication algebra (L, [conjunction], [disjunction], ', [right arrow], 0, 1) will be simply denoted by L.

Let L be a lattice implication algebra and let [phi] be a subset of [2.sup.L]. We use [[phi].sup.c] to denote the complement {L \ A : A [member of] [phi]}, where L \ A={x [member of] L: x [not member of] A}. A subset [tau] [subset or equal to] [2.sup.L] is called a topology on L, if r satisfies the following:

(1) [empty set], L [member of] [tau],

(2) A, B [member of] [tau] implies A [intersection] B [member of] [tau],

(3) {[A.sub.t] [member of] [tau] : t [member of] T} [subset or equal to] [tau] implies [[intersection].sub.t[member of]T] [A.sub.t] [member of] [tau].

Elements of [tau] are called [tau]-open sets and the complements of them are called [tau]-closed. The pair (L, [tau]) is called a topological space. A subset B of [tau] is called a base of [tau], if for each A [member of] [tau] and each x [member of] A, there exists B [member of] B such that x [member of] B [subset or equal to] A.

Let L be an implication algebra. A subset F of L is called a filer, if F satisfies the following: (1) 1 [member of] F;(2) x, x [right arrow] y [member of] F implies y [member of] F. The collection of all filters in L is denoted by F(L), or F briefly. Clearly, F consists a base of some topology [T.sub.F(L)], briefly [T.sub.F]. Usually, [T.sub.F] is called the filter topology generated by F. And the pair (L, [T.sub.F]) is called the filter topological space. A subset U [subset or equal to] L is called [T.sub.F]- neighborhood of x [member of] L, or neighborhood of x in [T.sub.F] if x [member of] U [member of] [T.sub.F]. The set of all [T.sub.F]-neighborhoods of x is denoted by [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since F [subset or equal to] [T.sub.F] and [x) = [intersection]|F : x [member of] F [member of] F} [member of] F, [x) is the smallest element of NT (x).

The closure operator and interior operator of [T.sub.F] are denoted by c and i. Clearly, for every A [subset or equal to] L, c(A) = [intersection]{L\[x) : x [member of] L, [x) [intersection] A = [empty set]} and i(A) = [union]|[x) : x [member of] L,[x) [subset or equal to] A}. The following proposition describes c(A).

Proposition 1. Let (L, [T.sub.F]) be the filer topology generated by F(L). Then for A [subset or equal to] L, c(A) = {x [member of] L : [x) [intersection] A [not equal to] [empty set]}.

Proof. The proof is trivial since [x) is the smallest [T.sub.F]-neighborhood of x.

Let L be a lattice implication algebra and let [2.sup.L] be the power set of L. A nonempty subset J of [2.sup.L] is called a power ideal of L if J satisfies the following: (1) A, B [member of] [2.sup.L] and A [subset or equal to] B [member of] J imply A [member of] J; (2)A, B [member of] J implies A [union] B [member of] J. The collection of all power ideals in [2.sup.L] is denoted by J(L), or briefly J. Note that [J.sub.[empty set]] = {[empty set]} is the smallest power ideal and [J.sub.L] = [2.sup.L] is the greatest power ideal. Moreover, if J, J [member of] J, then (1) J [intersection] J [member of] I;(2) J [disjunction] I = {I [union] J: I [member of] J, J [member of] I} [member of] J.

2. Local Functions and Power Idealization Filter Topologies

Let L be a lattice implication algebra, let [T.sub.F] be the filter topology, and let J be a power ideal. An operator * on [2.sup.L] is defined as follows:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1)

for every A [subset or equal to] L.

The operator * is called the local function with respect to [T.sub.F] and I. [A.sup.*] is called local function of A. We usually write [A.sup.*] (J) or [A.sup.*] instead of [A.sup.*] (I, [T.sub.F]).

Clearly, x [member of] [A.sup.*] if and only if [x) [intersection] A [not member of] I. Thus [A.sup.*] = {x [member of] L : [x) [intersection] A [not member of] I}. The following proposition gives some further details of [A.sup.*].

Proposition 2. Let (L, [T.sub.F]) be the filter topological space and I, J [member of] I. Then

(1) [A.sup.*]([I.sub.[empty set]]) = c(A) and [A.sup.*]([I.sub.L]) = 0;

(2) if A [subset or equal to] B, then [A.sup.*] (J) [subset or equal to] [B.sup.*](J);

(3) if I [subset or equal to] J, then [A.sup.*] (I) [subset or equal to] [A.sup.*](J);

(4) [A.sup.*] (J) = [subset or equal to] ([A.sup.*](J)) [subset or equal to] c(A);

(5) ([A.sup.*])*(1) [subset or equal to] [A.sup.*](J);

(6) if A [subset or equal to] 1, then [A.sup.*] (J) = [empty set];

(7) if A [member of] [T.sup.c.sub.F], then [A.sup.*] (J) [subset or equal to] A;

(8) if B [member of] I, then (A [union] B) * (J) = [A.sup.*](J) = (A\B)*(J);

(9) (A [union] [A.sup.*](J))*(J) = [A.sup.*](J);

(10) (A [union] B)*(J) = [A.sup.*](J) [union] [B.sup.*](J);

(11) [A.sup.*](J)\[B.sup.*](J) = (A\B)*(J)\[B.sup.*](J) c (A\B)* (J);

(12) if {1} [not member of] I, then [x)*(J) = L for each x [member of] L;

(13) if {1} [not member of] I and 1 [member of] A [subset or equal to] L, then [A)*(J) = [[A.sup.*](J)) = L.

Proof. (1) By Proposition 1, x [member of] [A.sup.*] ([I.sub.[empty set]]) if and only if [x) [intersection] A [not equal to] [empty set] if and only if x [member of] c(A). Thus [A.sup.*]([I.sub.[empty set]]) = c(A). Since [x) [intersection] A [member of] [2.sup.L] = [I.sub.L] for each x [member of] L, [A.sup.*] (L) = [empty set].

(2) Let A [subset or equal to] B and x [member of] [A.sup.*] (J). Then [x) [intersection] A [not member of] I. Since I is a power ideal and [x) [intersection] A [subset or equal to] [x) [intersection] B, [x) [intersection] B [not member of] I and so x [member of] [B.sup.*] (J). Thus [A.sup.*] (J) [subset or equal to] [B.sup.*] (J).

(3) Let J [subset or equal to] I and x [member of] [A.sup.*] (J). Then [x) [intersection] A [not member of] J. It follows that [x) [intersection] A [not member of] J and so x [member of] [A.sup.*](J). Thus [A.sup.*](I) [subset or equal to] [A.sup.*] (J).

(4) If x [not member of] c(A), then x [member of] L \ c(A) [member of] [T.sub.F] and so [x) [subset or equal to] L \ c(A). Thus [x) [intersection] A [subset or equal to] (L\ c(A)) [intersection] A = [empty set] [member of] I. This implies x [not member of] [A.sup.*](J) and so [A.sup.*](J) [subset or equal to] c(A). Then c([A.sup.*](J)) [subset or equal to] c(c(A)) = c(A).

It is clear that [A.sup.*](J) [subset or equal to] c([A.sup.*](J)). Next, we prove c([A.sup.*](J)) [subset or equal to] [A.sup.*] (J).

Let x [member of] c([A.sup.*](J)). By Proposition 1, [x) [intersection] [A.sup.*](J) [not equal to] [empty set]. Then there exists y [member of][x) [intersection] [A.sup.*] (J). By y [member of] [A.sup.*] (J), [y) [intersection] A [not member of] I. By y [member of] [x), [y) [subset or equal to] [x). Thus [x)[intersection]A [not member of] I and so x [member of] [A.sup.*] (J). Therefore c([A.sup.*](J)) [subset or equal to] [A.sup.*] (J).

(5) By (4), ([A.sup.*](J))*(J) [subset or equal to] c([A.sup.*](J)) = [A.sup.*](J).

(6) Since [x) [intersection] A [subset or equal to] A [member of] I for each x [member of] L, [A.sup.*] (J) = [empty set].

(7) Suppose that x [member of] [A.sup.*] (J) \ A. Then x [member of] L \ A [member of] [T.sub.F]. Thus [x) [subset or equal to] L \ A and so [x) [intersection] A [subset or equal to] (L\A) = [empty set] [member of] 1. Hence x [not member of] [A.sup.*] (J) which is a contradiction. Therefore [A.sup.*] (J) [subset or equal to] A.

(8) By (2), (A\B)*(J) [subset or equal to] [A.sup.*](J) [subset or equal to] (A [union] B)*(J). Next, we prove the inverse inclusions.

If x [not member of] (A\B)*(J), then ([x)[intersection]A)\B= [x)[intersection](A\B) [member of] I. Thus [x) [intersection] A [subset or equal to] J [union] B [member of] J which follows from I is a power ideal. This implies x [not member of] [A.sup.*](J). Thus [A.sup.*](J) [subset or equal to] (A \ B)*(J) and so [A.sup.*](J) = (A\B)*(J).

If x [not member of] [A.sup.*] (1), then [x)[intersection]A[member of] J. Since B [member of] J,

[x) [intersection](A [union] B) [subset or equal to] ([x) [intersection] A) [union] [beta] [member of] I. (2)

Thus x [member of] (A [union] B)*(J). This implies (A [union] B)*(J) [subset or equal to] [A.sup.*](J) and so (A [union] B)*(J) = [A.sup.*](J).

(9) Clearly, [A.sup.*](J) [subset or equal to] (A [union] [A.sup.*](J))*(J). Conversely, if x [not member of] [A.sup.*](J), then [x) [intersection] A [member of] I. Let [x) [intersection] A = I. Then A [subset or equal to] I U (L \ [x)). By (2), (7), (8), and [x) [member of] [T.sub.F],

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (3)

Thus A [union] [A.sup.*](J) [subset or equal to] (L\[x)) [union] A and so

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (4)

This implies x [not member of] (A [union] [A.sup.*] (J))* (J) and so (A [union] [A.sup.*](J))*(J) [subset or equal to] [A.sup.*](J).

(10) [A.sup.*](J) [union] [B.sup.*](J) [subset or equal to] (A [union] B)*(J) is clear. Conversely, if x [not member of] [A.sup.*] (J) [union] [B.sup.*](J), then [x) [intersection] A,[x) [intersection] B [member of] 1. Thus [x) [intersection] (A [union] B) = ([x) [intersection] A) [union] ([x) [union] B) [member of] I. This implies x [not member of] (A [union] B)*(J). Therefore (A [union] B)*(J) [subset or equal to] [A.sup.*] (J) [union] [B.sup.*] (J).

(11) We firstly prove [A.sup.*](J) \ [B.sup.*](J) [subset or equal to] (A \ B)*(J). Assume that x [member of] ([A.sup.*](J) \ [B.sup.*](J)) \ (A \ B)*(J). Then [x) [intersection] (A \ B) [member of] I and [x) [intersection] B [member of] 1. Thus

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (5)

This implies x [not member of] [A.sup.*] (J) which is a contradiction. Thus [A.sup.*] (J)\[B.sup.*] (J) [subset or equal to] (A\B)* (J) and so [A.sup.*](J)\[B.sup.*] (J) [subset or equal to] (A B)*(J)\[B.sup.*](J). Finally, (A\B)*(J)\[B.sup.*](J) [subset or equal to] [A.sup.*](J)\[B.sup.*](J) follows from (2). Therefore [A.sup.*](J) [B.sup.*](J) = (A \ B)*(J) \ [B.sup.*] (J).

(12) Since {1} [not member of] 1, 1 [not member of] I for each I [member of] I. Assume that there exists x [member of] L such that [x) [not member of] L. Let y [not member of] L\ [x)*(J). Thus [y) [intersection] [x) [member of] I. Since [x [disjunction] y) = [y) [intersection] [x), 1 [not member of] [x [disjunction] y) which is a contradiction. Therefore [x)*(J) = L for each x [member of] L.

(13) Assume that there exists y [member of] L \ [A.sup.*] (J). Then 1 [member of] [y) [intersection] A [member of] J which is a contradiction. Thus [A.sup.*] (J) = L and [[A.sup.*](J)) = L. Since 1 [member of] A, L = [1)*(J) [subset or equal to] [A)*(J) follows from (12). Therefore [A)*(J) = L.

Proposition 3. Let (L, [T.sub.F]) be the filter topological space and J [member of] J. The operator [c.sup.*.sub.J] (briefly, [c.sup.*]) on [2.sup.L], defined by [c.sup.*] (A) = A [union] [A.sup.*] for A [subset or equal to] L, satisfies the following statements:

(1) [c.sup.*]([empty set]) = [empty set]; [c.sup.*](L) = L,

(2) [c.sup.*]([c.sup.*](A)) = [c.sup.*](A) [subset or equal to] c(A),

(3) ([c.sup.*](A))* = [c.sup.*]([A.sup.*]) = [A.sup.*],

(4) [c.sup.*](A [union] B) = [c.sup.*](A) [union] [c.sup.*](B),

(5) A [member of] [T.sup.c.sub.F] or A [member of] J implies [c.sup.*](A) = A.

Proof. (1) [c.sup.*]([empty set]) = [empty set] follows from [empty set]* = [empty set]. [c.sup.*](L) = L is clear.

(2) By (4) and (9) of Proposition 2,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (6)

(3) By (5) and (9) of Proposition 2,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (7)

and [c.sup.*]([A.sup.*]) = [A.sup.*] [union] ([A.sup.*])* = [A.sup.*].

(4) By (10) of Proposition 2,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (8)

(5) The result follows from (6) and (7) of Proposition 2.

Theorem 4. Let (L, [T.sub.F]) be the filter topological space and J [member of] J. The operator [c.sup.*] stated in Proposition 3 is the closure operator of a new topology which is finer than TF and the topology generated by [J.sup.c] (note that [J.sup.c] is not a topology since [empty set] [not member of] [J.sup.c] in general case). Such a topology is called a power idealization filter topology and often denoted by [T.sup.*.sub.F] (J, [T.sub.F]), [T.sup.*.sub.F] (1), or [T.sup.*.sub.F].

Proof. Let [T.sup.*.sub.F] = {A [subset or equal to] L : [c.sup.*] (L \ A) = L \ A}. We prove that [T.sup.*.sub.F] is a topology on L.

(1) [empty set], L [member of] [T.sup.*.sub.F] follows from (1) of Proposition 3.

(2) If A, B [member of] [T.sup.*.sub.F], then [c.sup.*](L\A) = L\A and [c.sup.*](L\B) = L\B. Thus

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (9)

This implies A [intersection] B [member of] [T.sup.*.sub.F].

(3) Let [A.sub.t] [member of] [T.sup.*.sub.F] for t [member of] T, where T is an index set. Then [c.sup.*](L\[A.sub.t]) = L \ [A.sub.t] and

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (10)

Therefore [U.sub.t[member of]T][A.sub.t] [member of] [T.sup.*.sub.F].

Finally, by (6) and (7) of Proposition 3, [T.sub.F], [J.sup.c] [T.sup.*.sub.F].

Example 5. Let L = {0,a,b,c,d,1}, 0' = 1, a' = c, b' = d, c = a, d' = b, 1 = 0, and the implication operator [right arrow] be defined by a [right arrow] b = a V b for a, b [member of] L. Then (L, [conjunction], [disjunction], ', [right arrow]) is the Hasse lattice implication algebra (Figure 1 and Table 1). Then

F(L) = {{1},{a,1},{b,c,1},{a,b,c,d,1},L}, [T.sub.F(L)] = {0,{1},{a,1},{b,c,1},{a,b,c,1},{a,b,c,d,1},L}. (11)

Let J = {[empty set], {0}, {a}, {0, a}}. Then I is a power ideal. It is easy to check that

[T.sup.*.sub.F] = {0, {1}, {a, 1}, {b,c, 1}, {a,b, c, 1}, {b,c,d, 1}, {0, b, c, d, 1}, {a, b, c, d, 1}, L}. (12)

Clearly, [T.sub.F] [subset or equal to] [T.sup.*.sub.F].

Proposition 6. Let (L, [T.sub.F]) be the filter topological space and J, I [member of] I. Then

(1) [T.sup.*.sub.F] ([J.sup.[empty set]]) = [T.sub.F] and [T.sup.*.sub.F] ([I.sub.L]) = [2.sup.L],

(2) if J [subset or equal to] I, then [T.sup.*.sub.F] (J) [subset or equal to] [T.sup.*.sub.F] (I).

Proof. (1) By (1) of Proposition 2, [A.sup.*]([J.sub.[empty set]]) = c(A) and [A.sup.*] ([J.sub.L]) = 0. Thus [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] if and only if [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Similarly, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for each Ad. Therefore (1) holds.

(2) Let J [subset or equal to] I. By (3) of Proposition 2, [c.sup.*.sub.I](A) [subset or equal to] [c.sup.*.sub.J](A) for A [subset or equal to] L. Thus if A [member of] [T.sup.*.sub.F](J), then A [member of] [T.sup.*.sub.F](I). Therefore [T.sup.*.sub.F] (J) [subset or equal to] [T.sup.*.sub.F] (I).

Clearly, if J [member of] J satisfies [T.sup.c.sub.F] [subset or equal to] I, then L [member of] J and so I = [2.sup.L] = [J.sub.L]. Thus by (1) of Proposition 3, [T.sup.*.sub.F] = [J.sup.c] = [2.sup.L]. If [T.sup.c.sub.F] \ {L} [subset or equal to] J, we have the following proposition.

Proposition 7. Let (L, [T.sub.F]) be the filter topological space. If J [member of] J satisfies [T.sup.c.sub.F] \ {L} [subset or equal to] J, then [T.sup.*.sub.F] = [J.sup.c] [union] {[empty set]}.

Proof. [J.sup.c] [union] {[empty set]} [subset or equal to] [T.sup.*.sub.F] follows from Theorem 4. Conversely, suppose that [T.sup.*.sub.F] [not subset or equal to] [J.sup.c] [union] {[empty set]}. Then there exists B [member of] [T.sup.*.sub.F] \ ([J.sup.c] [union] {[empty set]}) such that (L\B)* [subset or equal to] L\B [not equal to] L. Let y [member of] (L \ B) \ (L \ B)*. Then y [not member of] (L\B)*. Thus [y)[intersection](L\B) [member of] J. Put [y)[intersection](L\B) = I. We have L\B [subset or equal to] I [union](L\[y)). Since [T.sup.c.sub.F] \ {L} [member of] I, L\ [y) [member of] I. Thus L\B [subset or equal to] I [union] (L\B) [member of] I and so L \ B [member of] I. Hence B [member of] [J.sup.c]. It is a contradiction. Therefore [T.sup.*.sub.F] [subset or equal to] [J.sup.c] [union] {[empty set]}.

Lemma 8. Let (L, [T.sub.F]) be the filter topological space and J [member of] J. If A [subset or equal to] L satisfies A [intersection] I = [empty set] for each I [member of] J, then [c.sup.*] (A) = c(A).

Proof. [c.sup.*](A) [subset or equal to] c(A) is clear. Conversely, if x [not member of] [c.sup.*](A), then x [not member of] A and x [not member of] [A.sup.*]. Thus I = [x)[intersection] A [member of] I and A [subset or equal to] I [union] (L\ [x)). Since A [intersection] I = [empty set], A [subset or equal to] L \ [x). Observe that x [not member of] L \ [x) and L \ [x) is [T.sub.F]-closed. We have x [not member of] c(A). Thus c(A) [subset or equal to] [c.sup.*](A). Therefore [c.sup.*](A) = c(A).

Proposition 9. If I [member of] [T.sup.c.sub.F], then [T.sup.*.sub.F] = [T.sub.F].

Proof. It is clear that [T.sub.F] [subset or equal to] [T.sup.*.sub.F]. Conversely, let [I.sub.M] be the greatest element of J and A [subset or equal to] L. Thus (A \ [I.sub.m]) [intersection] I = [empty set] for each I [member of] J. By Lemma 8, [c.sup.*](A [I.sub.M]) = c(A \ [I.sub.M]). By (8) of Proposition 2, (A \ [I.sub.M])* = [A.sup.*]. Now, notice that A [intersection] [I.sub.M] [member of] I [subset or equal to] [T.sup.c.sub.F] and thus c(A [intersection] [I.sub.M]) = A [intersection] [I.sub.M]. We have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (13)

This implies [c.sup.*] = c. Therefore [T.sup.*.sub.F] = [T.sub.F].

Lemma 10. Let (L, [T.sub.F]) be the filter topological space and I [member of] J. If J [member of] I and U [member of] [T.sub.F], then U\I [member of] [T.sup.*.sub.F].

Proof. Let P = L \ U. Then P [member of] [T.sup.c.sub.F]. By (7) and (8) of Proposition 2,

[c.sup.*] (P[union]t) = (P[union]t) [union] (P [union] I)* =(P [union] I) [union] P* = P [union] I. (14)

Thus L\(P [union] I)=U[intersection](L\I) = U\I [member of] [T.sup.*.sub.F].

Theorem 11. Let (L, [T.sub.F]) be the filter topological space and J [member of] J. Then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (15)

is a base of [T.sup.*.sub.F]. Moreover,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (16)

is a base of [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for each x [member of] L, where [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is the set of all [T.sup.*.sub.F]-neighborhoods of x in (L, [T.sup.*.sub.F]). Clearly, [x) \ [I.sub.x] is the smallest [T.sup.*.sub.F]-neighborhoods of x, where Ix is the greatest element of J satisfying x [not member of] [I.sub.x].

Proof. By Lemma 10, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Let B [subset or equal to] L. Then B [member of] [T.sup.*.sub.F] [??] L \ B is [T.sup.*.sub.F]-closed [??] (L \ B)* [subset or equal to] (L\B) [??] B [subset or equal to] L\(L\B)*. Thus x [member of] B [??] x [not member of] (L\B)* [??] there exists [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] such that (L\B)[intersection]d [member of] J. Let I = (L\B)[intersection]U. Then L\B [subset or equal to] I[union](L\U) and

x [member of] U \ I = U [intersection] (L \ I) = L \ (I [union] (L \ U)) [subset or equal to] B. (17)

Therefore [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a base of [T.sup.*.sub.F].

Clearly, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Let [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and y [member of] A. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a base of [T.sup.*.sub.F], there are U, V [member of] [T.sub.F] and I, J [member of] J such that x [member of] U\I [subset or equal to] A and y [member of] V\J [subset or equal to] A. We can assume y [not member of] I and x [not member of] J (otherwise, I and J can be replaced by I \ {y} and J {x}, resp.,). Then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (18)

Clearly, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and

(U [union] V)\(I [union] J) [subset or equal to] (U\I) [union] (V\J) [subset or equal to] A. (19)

Therefore [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (x) is a base of [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (x).

Clearly, if [I.sub.x] is the greatest element of J satisfying x [not member of] [I.sub.x], then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is the smallest [T.sup.*.sub.F]-neighborhoods of x.

Let (L, [tau]) be a topological space and I [member of] J. The topology that was generated by B = {U\I :U [member of] [tau], I [member of] J} is denoted by [T.sup.*] (J, [tau]) [5]. Clearly, [T.sup.*] (J, [T.sub.F]) = [T.sup.*.sub.F] (J).

Lemma 12. Let [psi] = {[empty set], L} be the indiscrete topology on L and J [member of] J. Then [T.sup.*] (J, [psi]) = {[empty set]} [union] [I.sup.c].

Proof. By [T.sup.c] \ {L} = {[empty set]} [member of] J and Proposition 7, the proof is obvious.

Theorem 13. Let (L, [tau]) be a topological space and J [member of] J. Then [T.sup.*] (J, [tau]) = [tau] [disjunction] [T.sup.*] (J, [psi]), where [tau] [disjunction] [T.sup.*] (J, [psi]) is the topology generated by the base {U [intersection] V: U [member of] [tau], V [member of] [T.sup.*](J, [psi])}.

Proof. Clearly, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a base of [T.sup.*](I, [tau]). Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. is also a base of [tau] [disjunction] [T.sup.*] (I, [psi]). Therefore [T.sup.*] (J, [tau]) = [tau] [disjuntion] [T.sup.*](J, [psi]).

Corollary 14. Let (L, [T.sub.F]) be the filter topological space and J [member of] J. Then [T.sup.*.sub.F] = [T.sub.F] [disjunction] [T.sup.*](J, [psi]).

Corollary 15. Let (L, [T.sub.F]) be the filter topological space and I, I [member of] J. Then

(1) [T.sup.*] (J [disjunction] I, [psi]) = [T.sup.*] (J, [psi]) [disjunction] [T.sup.*] (I, [psi]),

(2) [T.sup.*.sub.F] (J [disjunction] I) = [T.sup.*](J, [T.sup.*.sub.F] (I)) = [T.sup.*](I, [T.sup.*.sub.F] (J)),

(3) [T.sup.*.sub.F] (J [disjunction] I) = [T.sup.*.sub.F] (J) [disjunction] [T.sup.*.sub.F] (I),

(4) [T.sup.*] (J, [T.sup.*.sub.F] (J)) = [T.sup.*.sub.F] (J).

Proof. (1) By (2) of Proposition 6, [T.sup.*] (J [disjunction] I) [subset or equal to] [T.sup.*] (J) [disjunction] [T.sup.*](I). Conversely, let [empty set] [not equal to] A [member of] [T.sup.*](J [disjunction] I). By Theorem 13, there exist I [member of] I and J [member of] J such that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (20)

Thus [T.sup.*] (J [disjunction] J) [subset or equal to] [T.sup.*] (J) [disjunction] [T.sup.*] (I).

(2) By (1), Theorem 13, and Corollary 14,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (21)

Similarly, [T.sup.*.sub.F] (I [disjunction] J) = [T.sup.*]([T.sup.*.sub.F] (I), I).

(3) By (1) and Theorem 13,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (22)

Therefore [T.sup.*.sub.F] (J [disjunction] I) = [T.sup.*.sub.F] (J) [disjunction] [T.sup.*.sub.F] (I).

(4) Let J = I. Then the proof follows from (2).

Theorem 16. Let (L, TF) be the filter topological space, J, I [member of] J and A [subset or equal to] L. Then

(1) [A.sup.*](J [intersection] I, [T.sub.F]) = [A.sup.*] (J, [T.sub.F]) [union] [A.sup.*] (I, [T.sub.F]),

(2) [A.sup.*](J [disjunction] I, [T.sub.F]) = [A.sup.*](J, [T.sup.*.sub.F] (I)) [intersection] [A.sup.*] (I, [T.sup.*.sub.F] (J)).

Proof. (1) By (3) of Proposition 2, [A.sup.*] (I, [T.sub.F]) [union] [A.sup.*] (J, [T.sub.F]) [subset or equal to] [A.sup.*](I [intersection] J, [T.sub.F]). Conversely, x [not member of] [A.sup.*](I, [T.sub.F]) [union] [A.sup.*](J, [T.sub.F]). Then [x) [intersection] A [subset or equal to] I and [x) [intersection] A [member of] J. Let [x) [intersection] A = I and [x) [intersection] A = J. Then A [subset or equal to] I [union](L\[x)) and A [subset or equal to] J [union] (L\[x)). Thus

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (23)

Thus [x) [intersection] A [subset or equal to] I [intersection] J which implies x [not member of] [A.sup.*](J [intersection] I, [T.sub.F]). Therefore [A.sup.*](J [intersection] I, [T.sub.F]) [subset or equal to] [A.sup.*] (J, [T.sub.F]) [union] [A.sup.*](I, [T.sub.F]).

(3) Let x [member of] [A.sup.*](J [disjunction] I, [T.sub.F]). Then [x) [intersection] A [member of] I [disjunction] J. Then there exist I [intersection] J and J [member of] I such that [x) [intersection] A = I [union] J. We can assume I [intersection] J = [empty set], (otherwise, I can be replaced by I\(I [intersection] J)). Thus x [not member of] I or x [not member of] J, (otherwise, x [member of] I [intersection] J which is a contradiction). Now, we take x [member of] I for example. Then

([x)\I) [intersection] A = [x) [intersection] A [intersection] (L\I) = J. (24)

Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Thus x [not member of] [A.sup.*](I, [T.sup.*.sub.F] (J)) and so x [not member of] [A.sup.*](J, [T.sup.*.sub.F] (I)) [intersection] [A.sup.*](J, [T.sup.*.sub.F] (J)). Hence

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (25)

Conversely, let x [member of] [A.sup.*]([T.sup.*.sub.F] (J), I). Then there exists I [member of] J such that ([x) \ I) [intersection] A [member of] I. Let ([x) \ I) [intersection] A = J. Then [x) [intersection] A = I [union] J which implies x [not member of] [A.sup.*] (J [disjunction] I, [T.sub.F]). Similarly, if x [not member of] [A.sup.*]([T.sup.*.sub.F] (I), J), then x [not member of] [A.sup.*](J [disjunction] I, [T.sub.F]). Therefore [A.sup.*] (J [disjunction] I, [T.sub.F])[subset or equal to] [A.sup.*] (I, [T.sup.*.sub.F] (J)) [intersection] [A.sup.*] (J, [T.sup.*.sub.F] (I)).

Corollary 17. Consider [A.sup.*] (J, [T.sub.F]) = [A.sup.*] (J, [T.sup.*.sub.F] (J)).

Proof. Let J = I. The proof follows from (2) of Theorem 16.

Corollary 18. Consider [T.sup.*.sub.F] (I [intersection] J) = [T.sup.*.sub.F] (1) [intersection] [T.sup.*.sub.F] (I).

Proof. By (2) of Proposition 6, [T.sup.*.sub.F] (J [intersection] I) [subset or equal to] (J) [intersection] [T.sup.*.sub.F] (I).

Conversely, if A [not member of] [T.sup.*.sub.F] (I [intersection] J), then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (26)

Thus (L\A)*(J, [T.sub.F]) [not subset or equal to] (L\A) or (L\A)* (I, [T.sub.F]) [not subset or equal to] (L\A). Thus A [member of] [T.sup.*.sub.F] (J) or A [not member of] [T.sup.*.sub.F] (I). Therefore [T.sup.*.sub.F] (J) [intersection] [T.sup.*.sub.F] (I) [subset or equal to] [T.sup.*.sub.F] (J [intersection] I).

3. Power Idealization Filter Topological Quotient Spaces

Let ([L.sub.1], [conjunction], [disjunction], ', [right arrow] 1, [0.sub.1], [1.sub.1]) and ([L.sub.2], [conjunction], [disjunction], ', [[right arrow].sub.2], [0.sub.2], [1.sub.2]) be two lattice implication algebras. A mapping f from [L.sub.1] to [L.sub.2] is called lattice implication homomorphism, if f(x [right arrow] [1.sub.y]) = f(x) [right arrow] [sub.2]f(y) for any x, y [member of] [L.sub.1]. The set of all lattice implication homomorphisms from [L.sub.1] to [L.sub.2] is denoted by hom([L.sub.1], [L.sub.2]).

Let f [member of] hom([L.sub.1], [L.sub.2]). Then, clearly,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (27)

are topologies [4].

Lemma 19. Let [L.sub.1] and [L.sub.2] be two implication algebras and let f [member of] hom([F.sub.1], [f.sub.2]) and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] be power ideals.

(1) If f is injective, then [f.sup.-1](I) = {[f.sup.-1] (J) : J [member of] J} [member of] I([L.sub.1]).

(2) If f is surjective, then f(J) = {f(J) : I [member of] J} [member of] I([L.sub.2]).

Proof. (1) Since [empty set] [not member of] J, then [empty set] = [f.sup.- 1]([empty set]) [member of] [f.sup.-1](I).

If [I.sub.2] [member of] [f.sup.-1] (I) and [I.sub.1] [subset or equal to] [I.sub.2], then there exist [J.sub.2] [member of] I such that [I.sub.2] = [f.sup.-1](h). Thus f([I.sub.1]) [subset or equal to] f([I.sub.2]) = [J.sub.2] and f([I.sub.1]) [member of] I. Since f is injective, [I.sub.1] = [f.sup.-1](f([I.sub.1])) [member of] [f.sup.- 1](I).

If [I.sub.1], [I.sub.2] [member of] [f.sup.-1] (I), then there exist [J.sub.1], [J.sub.2] [member of] I such that [I.sub.1] = [f.sup.-1]([J.sub.1]) and [I.sub.2] = [f.sup.-1] ([J.sub.2]). One has [I.sub.1] [union] [I.sub.2] = [f.sup.-1] ([J.sub.1]) [union] [f.sup.-1] ([J.sub.2]) = [f.sup.-1] ([J.sub.1] [union] [J.sub.2]). Since [J.sub.1] [union] [J.sub.2] [member of] I, [I.sub.1] [union] [I.sub.2] [member of] [f.sup.-1](I). Therefore [f.sup.-1] (I) is a power ideal.

(2) By [empty set] [member of] I, [empty set] = f([empty set]) [member of] f(J).

If [J.sub.2] [member of] f(I) and [J.sub.1] [subset or equal to] [J.sub.2], then there exists [I.sub.2] [member of] J such that [J.sub.2] = f([I.sub.2]). Let [I.sub.1] = [f.sup.-1] ([J.sub.1]). Then [I.sub.1] [subset or equal to] [I.sub.2] and [I.sub.1] [member of] J. Since f is surjective, [J.sub.1] = f([I.sub.1]) [member of] f(J).

If [J.sub.1], [J.sub.2] [member of] f(I), then there exist [I.sub.1], [I.sub.2] [member of] J such that [J.sub.1] = f([I.sub.1]) and [J.sub.2] = f([I.sub.2]). Thus [I.sub.1] [union] [I.sub.2] [member of] J. Hence [J.sub.1] [union] [J.sub.2] = f([I.sub.1])[union]f([I.sub.2]) = f([I.sub.1] [union] [I.sub.2]) [member of] f(J). Therefore f(J) is a power ideal.

Lemma 20. Let [L.sub.1] and [L.sub.2] be two implication algebras and f [member of] hom([F.sub.1], [F.sub.2]). Then

(1) if f is injective, then for each [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is the smallest [T.sub.1]-neighborhood of x;

(2) if f is bijective, then for each y [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is the smallest [T.sub.r]-neighborhood of y.

Proof. (1) Clearly, [f(x)) is the smallest [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]-neighborhood of f(x). Then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Let [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and [f(x)) [subset or equal to] f(V). Thus [f.sup.-1] ([f(x))) [subset or equal to] [f.sup.-1] (f(V)) = V. Therefore [f.sup.-1]([f(x))) is the smallest one.

(2) Clearly, [[f.sup.-1](y)) is the smallest [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]-neighborhood of [f.sup.-1](y). Thus [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Now, let [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Thus [[f.sup.-1] (y)) [subset or equal to] [f.sup.-1](U) and so f([[f.sup.-1](y))) [subset or equal to] f([f.sup.-1](U)) = U. Therefore (2) holds.

Lemma 21. Let [L.sub.1] and [L.sub.2] be two implication algebras and f [member of] hom([F.sub.1], [F.sub.2]).

(1) If f is injective and I [member of] J([L.sub.2]), then for each x [member of] [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is the smallest neighborhood of f(x), where [J.sub.M] is the greatest element of J satisfying f(x) [not member of] [J.sub.M].

(2) If f is bijective and J [member of] I([L.sub.1]), then for each [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is the smallest neighborhood of [f.sup.-1](y), where [I.sub.M] is the greatest element of 1 satisfying [f.sup.-1](y) [not member of] [I.sub.M].

Theorem 22. Let [L.sub.1] and [L.sub.2] be two implication algebras and f [member of] hom([F.sub.1], [F.sub.2]).

(1) If f is injective and J [member of] I([L.sub.2]), then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (28)

(2) If f is bijective and I [member of] I([L.sub.1]), then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (29)

Proof. (1) Let [c.sup.*.sub.2] and [c.sub.l] be the closure operators of the left side and the right side of the equation. We only need to prove [c.sup.*.sub.1] = [c.sub.l].

Let A [subset or equal to] [L.sub.1] and x [member of] [c.sup.*.sub.1] (A). Then x [not member of] A and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. By (1) of Lemma 21, [f.sup.-1]([f(x))) [intersection] A [member of] [f.sup.-1](I). Thus there exists J [member of] I such that [f.sup.-1] ([f(x))) [intersection] A = [f.sup.-1] (I). Since x [not member of] A and f is injective, f(x) [not member of] J. Let [J.sub.M] [member of] I be the greatest one satisfying f(x) [not member of] [J.sub.M]. Then [f.sup.-1] ([f(x))) [intersection] A [subset or equal to] [f.sup.-1]([J.sub.M]). Thus

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (30)

By Lemma 21 and Proposition 1, x [not member of] [c.sub.l](A). Therefore [c.sub.l](A) [subset or equal to] [c.sup.*.sub.1] (A).

Conversely, let y [not member of] [c.sub.l](A). By Proposition 1,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (31)

Thus x [not member of] A, [f.sup.-1]([f(x))) [intersection] A [subset or equal to] [f.sup.-1]([J.sub.M]) and [f(x)) [intersection] f(A) [subset or equal to] [J.sub.M]. Then [J.sub.1] = [f(x)) [intersection] f(A) [member of] I. Since f is injective, [f.sup.-1]([f(x))) [intersection] A = [f.sup.-1]([J.sub.1]). By (1) of Lemma 20, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Therefore x [not member of] [c.sup.*.sub.1] (A) and [c.sup.*.sub.1] (A) [subset or equal to] [c.sub.l](A).

(2) Let [c.sup.*.sub.2] and [c.sub.r] be the closure operators of the left side and the right side of the equation. We only need to prove [c.sup.*.sub.2] = [c.sub.r].

Let y [not member of] [c.sup.*.sub.2](A). Then y [not member of] A and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. By (2) of Lemma 20, f([[f.sup.-1](y))) [intersection] A [member of] f(J). Thus there exists I [member of] J such that f([[f.sup.-1](y))) [intersection] A = f(J). Since f is bijective, [[f.sup.-1] (y)) [intersection] [f.sup.-1] (A) = I. By y [not member of] A, [f.sup.-1](y) [not member of] [f.sup.-1](A) and so [f.sup.-1](y) [not member of] I. Since [I.sub.M] [member of] I is the greatest element of J satisfying [f.sup.-1](y) [not member of] [I.sub.M], [[f.sup.- 1](y))[intersection] [f.sup.-1] (A) [subset or equal to] [I.sub.M] and [[f.sup.-1] (y)) [intersection] ([L.sub.1] \ [I.sub.M]) [intersection] [f.sup.-1] (A) = [empty set]. By f being bijective again, we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (32)

By

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (33)

y [not member of] [c.sub.r](A). Hence [c.sup.*.sub.2] (A) [subset or equal to] [c.sub.r](A).

Conversely, let z [not member of] [c.sub.r](A). Since IM is the greatest element of I satisfying [f.sup.-1](z) [not member of] [I.sub.M], by (2) of Lemma 21, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is the smallest neighborhood of [f.sup.-1](z). Thus f([[f.sup.-1] (z)) \ [I.sub.M]) [intersection] A = [empty set]. Since f is bijective,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (34)

This implies [[f.sup.-1] (z)) [intersection] [f.sup.-1](A) [subset or equal to] [I.sub.M] and [[f.sup.-1] (z)) [intersection] [f.sup.-1](A) [member of] J. Thus f([[f.sup.-1](y))) [intersection] A [member of] f(J) which implies [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since z [not member of] [c.sub.r](A), z [not member of] A. Therefore z [not member of] [c.sup.*.sub.2] (A) and so [c.sup.*.sub.2] (A) [subset or equal to] [c.sub.r](A).

Generally, if f [member of] hom([L.sub.1], [L.sub.2]) is surjective but not bijective, then (2) of Theorem 22 fails.

Example 23. Let [L.sub.1] = {[0.sub.1],a,b,c,d,[1.sub.1]} be the Hasse lattice implication algebra of Example 5. Let [L.sub.2] = {[0.sub.2], e, h, [1.sub.2]} and [0.sup.'.sub.2] = [1.sub.2], e' = h, h' = e, and [1.sup.'.sub.2] = [0.sub.2]. The Hasse diagram and the implication operator of [L.sub.2] are shown by Figure 2 and Table 2. Then it is clear that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (35)

A mapping f from [L.sub.1] to [L.sub.2] is defined as

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (36)

It easy to check f [member of] hom([L.sub.1], [L.sub.2]) and f is surjective. Let J = {[empty set], {c}}. Then J [member of] J([L.sub.1]) and f(J) = {0, {h}} [member of] 1([L.sub.2]).

Since {h} [member of] f(J), by Theorem 4,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (37)

Observe that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. We have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (38)

Moreover, by [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

In fact, we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (39)

Therefore [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Corollary 24. Let [L.sub.1] and [L.sub.2] be two implication algebras and let f [member of] hom([F.sub.1], [f.sub.2]) be bijective.

(1) If I [member of] J([L.sub.2]), then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (40)

(2) If J [member of] J([L.sub.1]), then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (41)

Proof. The proof follows from Theorem 22.

Corollary 25. Let [L.sub.1] and [L.sub.2] be two implication algebras and f [member of] hom([f.sub.1] [f.sub.2]).

(1) If f is injective and [I.sub.1], [I.sub.2] [member of] I([L.sub.2]), then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (42)

(2) If f is bijective and [J.sub.1], [J.sub.2] [member of] I([L.sub.1]), then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (43)

Proof. (1) Clearly, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. By (3) of Corollary 15 and Theorem 22,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (44)

(2) Is similar to (1).

Corollary 26. Let [L.sub.1] and [L.sub.2] be two implication algebras and let f [member of] hom([F.sub.1], [F.sub.2]) be bijective. If I [member of] J([L.sub.2]) and J [member of] I([L.sub.1]), then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (45)

Proof. The proof follows from (4) of Corollary 15 and Theorem 22.

http://dx.doi.org/10.1155/2014/812145

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work is supported by the National Natural Science Foundations of China (no. 11471202) and the Natural Science Foundation of Guangdong Province (no. S2012010008833).

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[4] Y. Xu, D. Ruan, K. Qin, and J. Liu, Lattice-Valued Logic, Springer, 2003.

[5] D. Jankovic and T. R. Hamlett, "New topologies from old via ideals," The American Mathematical Monthly, vol. 97, no. 4, pp. 295-310, 1990.

Shi-Zhong Bai (1,2) and Xiu-Yun Wu (1,3)

(1) School of Mathematics and Computational Science, Xiangtan University, Xiangtan 411105, China

(2) School of Mathematics and Computational Science, Wuyi University, Jiangmen 529020, China

(3) Department of Mathematics and Computational Science, Hunan University of Science and Engineering, Yongzhou 425100, China

Correspondence should be addressed to Xiu-Yun Wu; wuxiuyun2000@126.com

Received 24 June 2014; Accepted 2 August 2014; Published 28 August 2014

Academic Editor: Jianming Zhan

Table 1: The implication operator of L = {0, a, b, c, d, 1}. [right arrow] 0 a b c d 1 0 1 1 i 1 1 1 a c 1 b c b 1 b d a 1 b a 1 c a a 1 1 a 1 d b 1 1 b 1 1 1 0 a b c d 1 Table 2: The implication operator of [L.sub.2] = {[0.sub.2], e, h, [1.sub.2]}. [right arrow] [0.sub.2] e h [1.sub.2] [0.sub.2] 1 [1.sub.2] [1.sub.2] [1.sub.2] e h [1.sub.2] h [1.sub.2] h e e [1.sub.2] [1.sub.2] [1.sub.2] 02 e h [1.sub.2]

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Title Annotation: | Research Article |
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Author: | Bai, Shi-Zhong; Wu, Xiu-Yun |

Publication: | The Scientific World Journal |

Article Type: | Report |

Date: | Jan 1, 2014 |

Words: | 8034 |

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