Special filters in bounded lattices

1 Introduction

M. Sambasiva Rao recently studied various types of filters and ideals in distributive lattices with pseudocomplements, see [4] and [5]. Some of these findings require the lattice under consideration to be either Stone or weakly Stone, as indicated in the referenced papers. The question is whether the distributivity assumption can be relaxed and, more specifically, if the pseudocomplementation assumption can be eliminated. This approach, in which we consider instead of the pseudocomplement a certain subset (in fact an antichain) of elements behaving similar to the pseudocomplement, has already been introduced by I. Chajda and H. Lānger in [2]. In this paper the authors show that this construction can be successfully used in order to introduce the implication and negation connectives in an intuitionistic-like logic based on any bounded lattice that satisfies the Ascending Chain Condition.

It is well known that some of non-classical logics are formalized by means of lattice structures, i.e., by lattices endowed with an additional unary operation. Concerning classical propositional logic, it is formalized by a Boolean algebra, and intuitionistic logic is usually formalized by relatively pseudocomplemented semilattices or lattices, see, e.g., [2] and [3]. However, other non-classical logics use logical connectives that do not necessarily have a sharp meaning, i.e., the result of negation or implication for given entries need not be an element of the lattice in question, but may be a subset of mutually incomparable elements, see [2] for details. In order to axiomatize such logics bounded lattices are used where pseudocomplementation is replaced by the so-called annihilator, i.e., the subset of maximal elements being disjoint to a given one. Using such an operator, we can introduce sharp and dense elements and a certain generalization of a Stone lattice.

It is generally accepted that the structure of a lattice can be revealed by investigating its filters. This motivated us to study some special filters on bounded lattices where the annihilator serves as an operator replacing pseudocomplementation. For distributive and pseudocomplemented lattices such a study was performed by M. Sambasiva Rao, see [4] and [5]. We obtain similar results for lattices that are neither distributive nor pseudocomplemented. The study of these special filters deepens our understanding of how lattice structure influences logical connectives.

2 Preliminaries

In this paper, we will deal with bounded lattices L = (L, ∨, ∧, 0, 1) satisfying the Ascending Chain Condition (ACC). A poset is said to satisfy the ACC if it has no infinite ascending chains. We identify singletons with their unique element, i.e., we will write a instead of {a}. For a ∈ L and for subsets A,B of L, we define

Here and in the following, Max A denotes the set of all maximal elements of A which obviously is an antichain. Observe that 0⁰ = 1 and 1⁰ = 0 and that A⁰ ≠ ∅. The set a⁰ is in fact a generalization of the pseudocomplement of a introduced by O. Frink [3] or a modification of the annihilator since the set

is in fact the annihilator of the set A as known in lattice theory. Recall that the pseudocomplement a^∗ of a is the greatest element of {x ∈ L | x ∧ a = 0}. The advantage of our approach is that a⁰ can be defined in any lattice with 0. However, a⁰ need not be an element of L, but may be a subset of L, namely an antichain of L.

It is easy to see that in a bounded completely distributive lattice every element has a pseudo-complement. We will indicate the fact that a bounded lattice is pseudocomplemented by denoting it in the form L = (L, ∨, ∧, ^∗, 0, 1).

The element a is called dense if a⁰ = 0 and sharp if a⁰⁰ = a. Let D and S denote the set of all dense and sharp elements of L, respectively. Clearly the following holds:

Moreover, we define the following binary relations and unary operators on 2^L:

The set A is called closed if A‾‾=A. Observe that 0̅ = D and 1̅ = L.

Proof. If a ∈ L, then 1 ∈ a⁰⁰ ∨ x⁰⁰ for all x ∈ D and hence a ∈ D̅. This shows L ⊆ D̅, i.e., D̅ = L. Now let a ∈ L̅. Then 1 ∈ a⁰⁰ ∨ x⁰⁰ for all x ∈ L, especially 1 ∈ a⁰⁰ ∨ 0⁰⁰ = a⁰⁰. Since a⁰⁰ is an antichain, we have a⁰⁰ = 1 whence a ∈ D. This shows L̅ ⊆ D. Conversely, if a ∈ D, then a⁰⁰ = 1 and hence 1 ∈ a⁰⁰ ∨ x⁰⁰ for all x ∈ L, i.e., a ∈ L̅. This shows D ⊆ L̅. Altogether, we obtain L̅ = D.

3 Filters and D-filters

It is evident that if a given bounded lattice L = (L, ∨, ∧, 0, 1) is pseudocomplemented, then for each element x ∈ L, we have x⁰ = x^∗. We start this section with emphasizing certain differences between the properties of our concept x⁰ and pseudocomplements.

It is well known that for distributive pseudocomplemented lattices (L, ∨, ∧, ^∗, 0) with bottom element 0 and a, b ∈ L the following holds:

1. a ≤ b implies b^∗ ≤ a^∗,

2. a ≤ a^∗∗ and a^∗∗∗ = a^∗,

3. (a ∨ b)^∗ = a^∗ ∧ b^∗,

4. (a ∧ b)^∗∗ = a^∗∗ ∧ b^∗∗.

The following example shows that not all statements valid for pseudocomplements hold for x⁰.

Example 1

Consider the non-distributive lattice L depicted in Figure 1.

Figure 1

Non-distributive non-pseudocomplemented lattice

We have

x	0	a	b	c	d	e	f	g	1
x 0	1	fg	eg	g	c	c	a	c	0
x 00	0	a	b	c	g	g	fg	g	1
x̅	F 1	Fd	Fd	Fd	abcf1	abcf1	abcdefg1	abcf1	L

D = F₁ and S := {0, a, b, c, g, 1}. (Here and in the following we often write fg instead of {f, g}, and so on.) Hence L is not pseudocomplemented.

In L we have:

1. a ≤ c and c⁰ = g ≰ {f, g} = a⁰, but c⁰ ≤₁ a⁰.

2. f ≰ {f, g} = f⁰⁰, but f ≤₁ f⁰⁰.

3. (a ∨ b)⁰ = c⁰ = g ≠ {d, g} = {f, g} ∧ {e, g} = a⁰ ∧ b⁰, but (a ∨ b)⁰ ≤₁ a⁰ ∧ b⁰.

4. (d ∧ f)⁰⁰ = d⁰⁰ = g ≠ {d, g} = g ∧ {f, g} = d⁰⁰ ∧ f⁰⁰, but (d ∧ f)⁰⁰ =₁ d⁰⁰ ∧ f⁰⁰.

On the other hand, some of the properties of pseudocomplements are also preserved here, see the following result.

Proof. (i) If b ∈ A ∧ A⁰, then there exists some c ∈ A and some d ∈ A⁰ with c ∧ d = b. Since x ∧ d = 0 for all x ∈ A, we have b = c ∧ d = 0.

(ii) If A ∧ B = 0 and b ∈ A, then b ∧ x = 0 for all x ∈ B and hence there exists some c ∈ B⁰ with b ≤ c.

(iii) This follows from (i) and (ii).

(iv) If A ≤₁ B, then

and hence B⁰ ≤₁ A⁰.

(v) If A ≤₁ B⁰, then B ≤₁ B⁰⁰ ≤₁ A⁰ by (iii) and (iv). Analogously, B ≤₁ A⁰ implies A ≤₁ B⁰.

(vi) Let b ∈ B⁰. Because of B⁰ ≤₁ a, we have b ≤ a, and because of a ≤₁ B⁰ there exists some c ∈ B⁰ with a ≤ c. Together we obtain b ≤ a ≤ c and hence b ≤ c. Since both b and c belong to the antichain B⁰, we conclude b = c and hence b = a showing B⁰ = a.

(vii) Assume b ∈ A⁰. Because of A⁰ ≤₁ B⁰ there exists some c ∈ B⁰ with b ≤ c, and because of B⁰ ≤₁ A⁰ there exists some d ∈ A⁰ with c ≤ d. Together we obtain b ≤ c ≤ d and hence b ≤ d. Since both b and d belong to the antichain A⁰, we conclude b = d and hence b = c ∈ B⁰. This shows A⁰ ⊆ B⁰. Interchanging the roles of A⁰ and B⁰ gives B⁰ ⊆ A⁰. Together we obtain A⁰ = B⁰.

(viii) Because of (iii), we have A ≤₁ A⁰⁰ from which we conclude A⁰⁰⁰ = (A⁰⁰)⁰ ≤₁ A⁰ by (iv). Again by (iii) we have A⁰ ≤₁ (A⁰)⁰⁰ = A⁰⁰⁰. Altogether, A⁰⁰⁰ =₁ A⁰. Applying (vii) yields A⁰⁰⁰ = A⁰.

(ix) We have A,B ≤₁ A ∨ B and hence (A ∨ B)⁰ ≤₁ A⁰,B⁰ by (iv) which implies (A ∨ B)⁰ ≤₁ A⁰ ∧ B⁰.

(x) We have A ∧ B ≤₁ A⁰⁰ ∧ B⁰⁰ according to (iii) and hence (A ∧ B)⁰⁰ ≤₁ (A⁰⁰ ∧ B⁰⁰)⁰⁰ according to (iv). Using (ii), (i) and (iv), we see that any of the following statements implies the next one:

Together we obtain (A∧B)⁰⁰ =₁ (A⁰⁰ ∧B⁰⁰)⁰⁰ and hence (A∧B)⁰⁰ = (A⁰⁰ ∧B⁰⁰)⁰⁰ according to (vii). Now we have A⁰⁰ ∧ B⁰⁰ ≤₁ A⁰⁰,B⁰⁰ and hence (A⁰⁰ ∧ B⁰⁰)⁰⁰ ≤₁ A⁰⁰,B⁰⁰ by (iv) and (viii) which implies (A⁰⁰ ∧B⁰⁰)⁰⁰ ≤₁ A⁰⁰ ∧B⁰⁰. Together with A⁰⁰ ∧B⁰⁰ ≤₁ (A⁰⁰ ∧B⁰⁰)⁰⁰ that follows from (iii) this yields (A ∧ B)⁰⁰ =₁ (A⁰⁰ ∧ B⁰⁰)⁰⁰ =₁ A⁰⁰ ∧ B⁰⁰.

(xi) According to (iii), (iv), (ix), (viii) and the assumption, we have □

Theorem 3.1 is very important because we will use it many times in the proofs of most of the following statements.

Proof. According to (x) of Theorem 3.1, we have (a ∧ b)⁰⁰ =₁ a⁰⁰ ∧ b⁰⁰ = a ∧ b and hence (a ∧ b)⁰⁰ = a ∧ b by (vi) of Theorem 3.1, i.e., a ∧ b ∈ S.

Proof. Let a, b ∈ D and c ∈ L. Then according to (x) of Theorem 3.1, we conclude (a ∧ b)⁰⁰ =₁ a⁰⁰ ∧ b⁰⁰ = 1∧ 1 = 1 and hence (a ∧ b)⁰⁰ = 1 according to (vi) of Theorem 3.1 proving a ∧ b ∈ D. If a ≤ c, then c⁰ ≤₁ a⁰ = 0 according to (iv) of Theorem 3.1 and hence c⁰ = 0, i.e., c ∈ D.

It is well known that the set 𝓕 of all filters of L forms a complete lattice with respect to inclusion. Using Proposition 3.3 one can recognize that the set of all D-filters of L forms a complete sublattice of 𝓕 with bottom element D.

D-filters were described for distributive pseudocomplemented lattices by M. Sambasiva Rao [5]. However, we are going to show that similar results can be stated also for lattices which are neither pseudocomplemented nor distributive, but satisfy a weaker condition.

Proof. Obviously, D ⊆ A̅. Let a, b ∈ A̅. Then 1 ∈ (a⁰⁰ ∨ z⁰⁰) ∧ (b⁰⁰ ∨ z⁰⁰) for all z ∈ A and hence 1 ∈ (a⁰⁰ ∧ b⁰⁰) ∨ z⁰⁰ =₁ (a ∧ b)⁰⁰ ∨ z⁰⁰ for all z ∈ A according to (x) of Theorem 3.1, i.e., a ∧ b ∈ A̅. Further, if c ∈ L and a ≤ c, then a⁰⁰ ≤₁ c⁰⁰ according (iv) of Theorem 3.1 and hence 1 ∈ a⁰⁰ ∨ z⁰⁰ ≤₁ c⁰⁰ ∨ z⁰⁰ for all z ∈ A̅ showing c ∈ A̅. Altogether, A̅ is a D-filter of L.

In the next example, we show a lattice being neither distributive nor pseudocomplemented, but satisfying the condition from Theorem 3.4.

Example 2

Consider the non-distributive lattice L depicted in Figure 2:

Figure 2

Non-distributive non-pseudocomplemented lattice

We have

x	0	a	b	c	d	e	f	g	1
x 0	1	bcd	acd	abd	abc	0	0	0	0
x 00	0	a	b	c	d	1	1	1	1
x̅	Fe	Fe	Fe	Fe	Fe	F 0	F 0	F 0	F 0

D = F_e and S = {0, a, b, c, d, 1}. Hence L is not pseudocomplemented. Further, F_a is a D-filter of L,{a,b}‾=Fe is a D-filter of L in accordance with Theorem 3.4 and {a, b}^D = {c, d, e, f, g, 1}.

4 Stonean and D-Stonean lattices

The concept of a Stone lattice was introduced by R. Balbes and A. Horn [1], see also the paper [6] by T. P. Speed. Recall from [1] that a bounded pseudocomplemented lattice (L, ∨, ∧, ^∗, 0, 1) is called Stone if

The theory of Stone lattices is well developed, see e.g. [1] and [6] and for filters [5]. This motivated us to introduce and study an analogous concept for bounded lattices that are not necessarily pseudocomplemented.

In analogy to the above definition, we define the following two concepts.

Proof. (i) Assume that L is Stonean and a⁰ ≤₂ a̅. Then a⁰ ⊆ a̅ and hence a‾‾⊂a0‾. Now let b∈a0‾ and c ∈ a̅. Since a⁰ ≤₂ ̅ there exists some d ∈ a⁰ with d ≤ c. We conclude 1 ∈ b⁰⁰∨d⁰⁰ ≤₁ b⁰⁰∨c⁰⁰ and hence 1 ∈ b⁰⁰ ∨ c⁰⁰. This shows b ∈ a̅̅ and hence a0‾⊂‾a‾‾. Together we obtain a‾‾=a0‾.

(ii) According to the observation after Definition 1, L is Stonean if and only if x⁰ ⊆ x̅ for all x ∈ L. Now for every x ∈ L the inclusion x⁰ ⊆ x̅ is equivalent to x‾‾⊆x0‾.

The next result is elementary, but we will use it in the sequel.

Proof. If b ∈ a⁰, then 1 ∈ a⁰⁰ ∨ b⁰⁰ according to (4.1) which is equivalent to a ∨ b ∈ D because of (4.2).

Example 3

1. The lattice visualized in Figure 2 satisfies neither (4.1) nor (4.2) since b ∈ a⁰ and a∨b = e ∈ D, but 1 ∉ e = a ∨ b = a⁰⁰ ∨ b⁰⁰.

2. Consider the non-distributive lattice L depicted in Figure 3:

Figure 3

Non-distributive non-pseudocomplemented non-Stonean lattice

We have

x	0	a	b	c	d	e	f	g	1
x 0	1	bcg	ceg	beg	bce	bcg	0	bce	0
x 00	0	e	b	c	g	e	1	g	1
x̅	Ff	bcdfg1	adefg1	adefg1	abcef1	bcdfg1	L	abcef1	F 0

D = F_f and S = {0, b, c, e, g, 1}. Hence L is not pseudocomplemented. Moreover, L satisfies neither (4.1) nor (4.2) since c ∈ b⁰ and b ∨ c = f ∈ D, but 1 ∉ f = b ∨ c = b⁰⁰ ∨ c⁰⁰. Therefore L is not Stonean.

Proof. The lattice visualized in Figure 1 satisfies (4.1), but not (4.2) since e ∨ g = 1 ∈ D, but 1 g = g ∨ g = e⁰⁰ ∨ g⁰⁰. Hence it is Stonean, but not D-Stonean. Now consider the non-distributive lattice L depicted in Figure 4:

Figure 4

Non-distributive non-pseudocomplemented non-Stonean lattice

We have

D = F_i and S = {0, b, d, g, h, j, 1}. Hence L is not pseudocomplemented. The lattice L satisfies (4.2), but not (4.1) since d ∈ b⁰, but 1 h = b ∨ d = b⁰⁰ ∨ d⁰⁰. Therefore it is not Stonean. □

The following theorem shows how the concept of a Stonean lattice is related with its filters.

Proof. (4.2) ⇒ (i): This follows from

for all filters F of L.

1. ⇒ (ii): For all x ∈ L, we have x‾=Fx‾=FxD=xD.

2. ⇒ (iii): Let F,G be filters of L. If a ∈ F ∩ G, then a = a ∨ a ∈ F ∨ G. If, conversely, a ∈ F ∨G, then there exists some b ∈ F and some c ∈ G with b∨c = a and hence a ∈ F ∩G. This shows F ∩ G = F ∨ G. Now the following are equivalent:

   F ∩ G ⊆ D , F ∨ G ⊆ D , x ∨ y ∈ D  for all  x ∈ F  and all  y ∈ G , y ∈ x D  for all  x ∈ F  and all  y ∈ G , y ∈ x ‾  for all  x ∈ F  and all  y ∈ G , 1 ∈ x 00 ∨ y 00  for all  x ∈ F  and all  y ∈ G , F ⊆ G ‾ .

3. ⇒ (4.2): For all x, y ∈ L the following are equivalent: x∨y ∈ D; ⊆ D; ∩⊆ D; Fx⊆Fy‾;x∈y‾;1∈x00∨y00. F_x∨y F_x F_y

The results of the previous theorem can be checked in the following example.

Example 4

Consider the non-distributive lattice L depicted in Figure 5.

Figure 5

Non-distributive non-pseudocomplemented D-Stonean lattice

We have

x	0	a	b	c	d	1
x 0	1	bc	cd	bd	bc	0
x 00	0	d	b	c	d	1
F x ‾ = F x D = x ‾ = x D	F 1	bc1	acd1	abd1	bc1	F 0

D = F₁ and S = {0, b, c, d, 1}. Hence L is not pseudocomplemented, but it is D-Stonean. Moreover, for x, y ∈ L both F_x ∩ F_y ⊆ D and Fx⊆Fy‾ are equivalent to 1 ∈ {x, y} or (x, y ∈ {a, b, c, d} and x ≠ y and {x, y} ≠ {a, d}) in accordance with Theorem 4.3.

5 Coherent and closed filters

In the following we define a certain operator c on the lattice of filters of a bounded lattice which shares some properties with a closure operator. It is a natural task to investigate when a filter F coincides with its closure c(F). Such filters will be called coherent. In particular, we describe coherent filters in D-Stonean lattices.

Now let us define the operator c on filters of L as follows:

Proof. Let a ∈ F and b ∈ L. Then 1 ∈ F ∩ a̅ and either b ∈ a̅ and hence b = b ∧ 1 ∈ a̅ ∧ F or b ∈ F and hence b = 1∧ b ∈ a̅ ∧ F. This shows a ∈ c(F) and therefore F ⊆ c(F).

Since x̅ = L for all x ∈ D, we have F ⊆ c(F) for all filters F of L being contained in D.

We can show that c(F) is closed with respect to ∧ under a weak condition.

Proof. (i) If a ∈ D, then a⁰⁰ = 1 and hence a̅ = L whence L = L ∧ 1 ⊆ L ∧ F = a̅ ∧ F ⊆ L, i.e., a̅ ∧ F = L showing a ∈ c(F). Therefore D ⊆ c(F). If b ∈ c(F), c ∈ L and b ≤ c, then b⁰⁰ ≤₁ c⁰⁰ and hence L = b̅ ∧ F ⊆ c̅ ∧ F ⊆ L which implies c̅ ∧ F = L, i.e., c ∈ c(F).

(ii) We have

(iii) If x‾∧y‾⊆x∧y‾ for all x, y ∈ c(F), then for all x, y ∈ c(F), we have

and hence x∧y‾∧F=L, which means nothing else than x ∧ y ∈ c(F).

The condition in (ii) of Proposition 5.1 holds for all proper filters of the lattice from Figure 2. For D-Stonean lattices, we can prove the following result.

Proof. Let a ∈ c(F). Then a̅ ∧ F = L and hence there exists some b ∈ a̅ and some c ∈ F with b ∧ c = a. Now

and hence b = b ∨ (b ∧ c) = b ∨ a ∈ D ⊆ F which implies a = b ∧ c ∈ F. □

Note that the condition of L being D-Stonean is only sufficient but not necessary.

Proof. (i) and (ii) We use Lemma 2.1.

(iii) This follows from (i).

6 Maximal, prime and median filters

Prime filters and maximal filters play an important role in the theory of rings, but also for semirings, especially for bounded distributive lattices. The question is if similar notions can be developed for lattices which need not be distributive. The aim of this section is to answer this question positively. We establish connections between maximal, prime, median and coherent filters as well as D-filters.

First let us recall some well-known classes of lattice filters. We call a filter F of a lattice (L, ∨, ∧)

• proper if F ≠ L,

• maximal if it is a maximal proper filter,

• prime if x, y ∈ L and x ∨ y ∈ F imply x ∈ F or y ∈ F.

It is well known (see, e.g., [7]) that every maximal filter of a distributive lattice is prime. Unfortunately, this does not hold for non-distributive lattices. For example, the filter F_a of the lattice in Figure 5 is maximal, but it is not prime since b ∨ c = 1 ∈ F_a, but b, c F_a. However, we can prove the following result.

Proof. (i) Assume F to be maximal and a ∈ L \ F. Then G := {x ∈ L | there exists some f ∈ F with f ∧ a ≤ x} is a filter of L strictly including F. Since F is maximal, we conclude G = L. Hence 0 ∈ G, i.e., there exists some g ∈ F with g ∧ a ≤ 0, it means g ∧ a = 0. Therefore there exists some b ∈ a⁰ with g ≤ b. Since F is a filter, we conclude b ∈ F and hence b ∈ a⁰ ∩ F whence a⁰ ∩ F ≠ ∅. Conversely, assume x⁰ ∩ F ≠ ∅ for all x ∈ L \ F. Suppose F not to be maximal. Then there exists a proper filter H of L strictly including F. Let c ∈ H \ F. Then c⁰ ∩ F ≠ ∅. Let d ∈ c⁰ ∩ F. Then c, d ∈ H. Since H is a filter of L, we have 0 = c ∧ d ∈ H and hence H = L, a contradiction. This shows that F is maximal.

1. Assume F to be maximal, but not being a D-filter. Then there exists some a ∈ D \ F. According to (i) 0 ∩ F = a⁰ ∩ F ≠ ∅ and hence 0 ∈ F which implies F = L, a contradiction. Therefore F is a D-filter.

2. Assume F to be maximal and (x ∨ y) ∧ z = 0 for all x, y ∈ L \ F and all z ∈ F with x ∧ z = y ∧ z = 0. Suppose F to be not prime. Then there exist some a, b ∈ L \ F with a ∨ b ∈ F. According to (i), we have a⁰ ∩ F, b⁰ ∩ F ≠ ∅. Let f ∈ a⁰ ∩ F and g ∈ b⁰ ∩ F and put h := f ∧ g. Then h ∈ F and a ∧ f = b ∧ g = 0 and hence a ∧ h = b ∧ h = 0. By the above assumption, we conclude 0 = (a ∨ b) ∧ h ∈ F which implies that F is not proper which is a contradiction. Hence F is prime.

Of course, condition (iii) of Theorem 6.1 holds for any distributive lattice. However, the following example shows that it may be satisfied also in a non-distributive lattice.

Example 10

The filter F_a of the non-Stonean lattice (L, ∨, ∧, 0, 1) visualized in Figure 3 is median, closed and coherent since b ∈ L\F_a, 1 ∈ x⁰⁰∨b⁰⁰ for every x ∈ F_a and Fa‾‾={b,c,d,f,g,1}‾= F_a. The filter F_a of the non-Stonean lattice (L, ∨, ∧, 0, 1) depicted in Figure 4 is prime, median, closed and coherent since c ∈ L \ F_a, 1 ∈ x⁰⁰ ∨ c⁰⁰ for all x ∈ F_a and Fa‾‾=Fc‾=Fa. We list all closed filters of the lattices from Figure 1 to Figure 5:

Figure 6

Non-distributive lattice

Fig.	closed filters
1	F 0 , Fd, Ff, F 1
2	F 0 , Fe
3	F 0 , Fa, Fd, Ff
4	F 0 , Fa, Fc, Fi
5	F 0 , Fb, Fc, Fd, F 1

Next, we present several basic properties of proper, prime and median filters.

Proof. (i) Assume x⁰⁰ ∨ y⁰⁰ ≤₁ (x ∨ y)⁰⁰ for all x, y ∈ L and let F be a prime D-filter of L, a ∈ L \ F and b ∈ a. Then 1 ∈ a⁰⁰ ∨ b⁰⁰ ≤₁ (a ∨ b)⁰⁰. Since (a ∨ b)⁰⁰ is an antichain, we have (a ∨ b)⁰⁰ = 1 which implies (a ∨ b)⁰ = (a ∨ b)⁰⁰⁰ = 0, i.e., a ∨ b ∈ D. Since F is a D-filter, we have D ⊆ F and therefore a ∨ b ∈ F. Because a /∈ F and F is prime, we conclude b ∈ F proving a̅ ⊆ F.

(ii) Assume x⁰⁰ ∨ y⁰⁰ ≤₁ (x ∨ y)⁰⁰ for all x, y ∈ L and let F be a median D-filter of L, a ∈ F and b ∈ a. Since F is median there exists some c ∈ L \ F with 1 ∈ a⁰⁰ ∨ c⁰⁰. Hence c∈a‾=a‾‾‾⊆b‾. Since c /∈ F, we have c̅ ⊆ F according to (i) and we obtain b∈b‾‾⊆c‾⊆F proving a‾‾⊆F.

(iii) If we would have a⁰ ⊆ F for some a ∈ F, then 0 = a∧a⁰ ⊆ F and hence F = L contradicting the assumption that F is proper.

(iv) Suppose L to be D-Stonean, F to be a prime D-filter of L and a⁰ ⊈ F. Then there exists some b ∈ a⁰ \ F. Since L is D-Stonean, we have 1 ∈ a⁰⁰ ∨ b⁰⁰ and hence a ∨ b ∈ D ⊆ F according to the assumption that F is a D-filter. Because F is prime and b /∈ F, we obtain a ∈ F. The converse implication follows from (iii).

The next result illuminates the relationship between coherent and median filters.

Proof. Let a ∈ F.

1. Assume F to be coherent. Then a ∈ c(F), i.e., a̅ ∧ F = L and hence there exists some b ∈ a̅ and some c ∈ F with b ∧ c = 0. Because of b ∈ a̅, we have 1 ∈ a⁰⁰ ∨ b⁰⁰. Now b ∈ F would imply 0 = b ∧ c ∈ F and hence F = L contradicting the maximality of F. Hence b /∈ F showing F to be median.

2. Assume F̅ ⊈ F. Then there exists some b ∈ F̅ \ F. Hence 1 ∈ a⁰⁰ ∨ b⁰⁰ and b ∈ L \ F proving F to be median.

3. Since a∈L∖F‾, we have 1 ∈ a⁰⁰ ∨ x⁰⁰ for all x ∈ L \ F. Because F is proper there exists such an x.

M. Sambasiva Rao [5] has found that median prime filters of a distributive pseudocomplemented lattice have an interesting property. The same applies to more general lattices.

Proof. (i) Assume F to be a D-filter of L, a ∈ F and a̅ = b̅. Since F is median there exists some c ∈ L \ F with 1 ∈ a⁰⁰ ∨ c⁰⁰. Hence c ∈ a̅ = b̅, i.e., 1 ∈ b⁰⁰ ∨ c⁰⁰. Since L is D-Stonean and F is a D-filter, we have b ∨ c ∈ D ⊆ F. Because F is prime, we conclude b ∈ F.

(ii) Suppose a ∨ b ∈ F. Since F is prime, we have a ∈ F or b ∈ F. Because F is median, in the first case we see that there exists some c ∈ L \ F with 1 ∈ a⁰⁰ ∨ c⁰⁰. Since L is D-Stonean, we obtain a ∨ c ∈ D. The case b ∈ F can be treated analogously.