Textural formal context

International Journal of Approximate Reasoning 114 (2019) 182–203

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International Journal of Approximate Reasoning www.elsevier.com/locate/ijar

Textural formal context Murat Diker a,∗ , Aysegül ¸ Altay U˘gur b , Sadık Bayhan c a b c

Hacettepe University, Department of Mathematics, 06532 Beytepe, Ankara, Turkey Hacettepe University, Department of Mathematics and Science Education, 06532 Beytepe, Ankara, Turkey Burdur Mehmet Akif Ersoy University, Department of Mathematics, 15030, I˙ stiklal Campus, Burdur, Turkey

a r t i c l e

i n f o

Article history: Received 7 February 2019 Received in revised form 31 May 2019 Accepted 27 August 2019 Available online 2 September 2019 Keywords: Direlation Dual formal concept Formal concept Formal context Sufficiency operator Texture space

a b s t r a c t In this paper, we discuss formal context in textures. A texturing is a family of subsets of a domain of discourse satisfying certain conditions. Considering a t-formal context, we formulate the notions of extent and intent in terms of p-sets and q-sets. Then we define t-formal concept and t-formal co-concept. For complemented direlations, we prove that they give two different concept lattices which are dually isomorphic to each other. These lattices correspond to a concept lattice in the sense of R. Wille and a dual concept lattice given by Düntsh and Gediga, respectively. In particular, we observe that a texturing, as an imperfect collection of sets of a domain of discourse, provides a remarkable setting which still ensures information with respect to given system. Finally, we prove the main theorem of formal concept analysis for textures. © 2019 Elsevier Inc. All rights reserved.

1. Introduction Formal concept analysis (FCA) was introduced by the German scientist R. Wille in 1982 as a mathematical modeling for data analysis [22]. FCA contributes to many disciplines such as information retrieval, knowledge discovery, and artificial intelligence in information sciences. This leads the researcher to consider the successful counterparts and applications in fuzzy set theory and rough set theory (see e.g., [1,5,6,16–19,23–26]). The main idea of FCA is to organize hierarchically a collection of formal concepts in a context with respect to an order. A formal concept, as a primary tool of formal concept analysis, is a pair of sets where the first is a set of objects and the second is a set of properties which are called extent and intent, respectively. The family of all formal concepts of a formal context is a complete lattice which is called a concept lattice. According to the main theorem of formal concept analysis, every concept lattice is isomorphic to a lattice under certain conditions and vice versa. It is known that a texture is a convenient point-set based setting for fuzzy sets [2]. Recent studies show that there are also remarkable connections between textures and rough sets ([2–4,7–12,21]). A texturing is a family of sets satisfying certain conditions subjected to a power set of a domain of discourse. Essentially, a texturing may not be closed under ordinary set-theoretical complementation. This provides a complement-free environment for mathematical structures. For instance, since a p-set and q-set of a texture impose a powerful duality in textures, a formal concept and a dual formal concept can be simultaneously defined. This paper is a starting point for a future research related to fuzzy concepts, object-oriented concepts, property-oriented concepts in the framework of textures. In a textural case, in spite of the absence of some subsets of a domain of discourse, there may exist some considerable information embedded in data. In other words, as a different matter from the methods of incomplete systems, the missing information may be considered

*

Corresponding author. E-mail addresses: [email protected] (M. Diker), [email protected] (A. Altay U˘gur), [email protected] (S. Bayhan).

https://doi.org/10.1016/j.ijar.2019.08.011 0888-613X/© 2019 Elsevier Inc. All rights reserved.

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and evaluated using a relative comparison between the texturing and the family of all subsets of the domain of discourse. This may be accepted as an answer to question “If we ignore some subsets of the properties or objects of the domain of discourses in the given information system, then how can we observe the total effect to the system using a new (textural) setting in the framework of formal concept analysis?”. In this work, Section 2 is devoted to the basic notions and results related to textures which will be used in the sequel. In Section 3, we discuss the concept lattices in the framework of textures. We formulate the notions of intent and extent in terms of p-set and q-set. Then we define t-sufficiency and t-co-sufficiency operators for textures. We show that these operators constitute two Galois connections. In this case, a direlation (r , R ) from a texture U = (U , U ) to a texture V = ( V , V ) provides two complete lattices Br (U, V) and B R (U, V). If (r , R ) is complemented, then we prove that Br (U, V) and B R (U, V) are dually isomorphic. In case of discrete textures, they correspond to a concept lattice in the sense of Wille [22] and to a dual concept lattice given by Düntsh and Gediga [13] under the notation SC M , respectively. Further, Chen and Yao also considered the dual formal concept lattices in [5]. Finally, in the last section, we prove the main theorem of formal concept analysis for textural concept lattices. The reader may find the dual forms of the proofs in the appendix. 2. Preliminaries In this work, we use the basic concepts and results on textures. For more details about the textures we refer the reader to [2–4,8]. Basic concepts and results Let U be a set. Then U ⊆ P (U ) is called a texturing of U , and (U , U ) is called a texture space, or in brief, a texture, if (i) (U , ⊆) is a complete lattice containing U and ∅, which has the property that arbitrary meets coincide with intersections, and finite joins coincide with unions, that is, for all index set K ,



k∈ K

Ak =



k∈ K

Ak

and for all finite index set K



k∈ K

Ak =



k∈ K

Ak

where { A k | k ∈ K } ⊆ U . j (ii) U is completely distributive, that is, for all index set K , and for all k ∈ K , if J k is an index set and if A k ∈ U , then we have

  k∈ K j ∈ J k

A kj =



A kγ (k) ,

γ ∈C k∈ K



where C is the set of all choice functions γ : K → k∈ K J k such that γ (k) ∈ J k ⊆ K . (iii) U separates the points of U . That is, given u = v in U there exists A ∈ U such that u ∈ A , v ∈ / A, or u ∈ / A , v ∈ A. A mapping c U : U → U is called a complementation on a texture (U , U ) if

∀ A ∈ U , c U (c U ( A )) = A , and ∀ A , B ∈ U , A ⊆ B =⇒ c U ( B ) ⊆ c U ( A ). Then the triple (U , U , c U ) is called a complemented texture space. For u ∈ U , the p-set and q-set are defined by

Pu =



{ A ∈ U | u ∈ A }, and  Qu = {A ∈ U | u ∈ / A },

respectively. The notions of p-set and q-set are primary tools in textures. Most of the results with respect to textures can be proved under the condition of complete distributivity of textures. However, the usage of this form of complete distributivity mostly is not so practical to give the proofs of the results with respect to textures, and hence, we use a characterisation of complete distributivity in terms of the p-set and q-set. Now let us recall the following theorem of Raney: Theorem 2.1. [20] A complete lattice ( L , ) is completely distributive if and only if (i) ∀u , v ∈ L , u  v =⇒ ∃x, y ∈ L, u  x and y  v, and (ii) ∀ z ∈ L, ∃ p , q ∈ L, either z p or q z. Corollary 2.2. The following statements are equivalent:

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(i) (U , ⊆) is completely distributive. (ii) ∀ A , B ∈ U , A ⊆ B =⇒ ∃u ∈ U , A ⊆ Q u and P u ⊆ B. Proof. (i) =⇒ (ii) is given in Theorem 1.2 in [4]. For the implication (ii) =⇒ (i), suppose that A , B ∈ U and A ⊆ B. Since for some u ∈ U , A ⊆ Q u and P u ⊆ B, for P u , Q u ∈ U , the condition (i) of Theorem 2.1 holds. Let A ∈ U . For the condition (ii) of Theorem 2.1, we show that there exists a u ∈ U such that A ⊆ Q u or P u ⊆ A. For every u ∈ U we have two cases: A ⊆ Q u or A ⊆ Q u . If A ⊆ Q u , then the proof is complete. If A ⊆ Q u , then by definition of Q u , we get P u ⊆ A, that is, the condition (ii) of Theorem 2.1 also holds. As a result, we obtain that (U , ⊆) is completely distributive. 2 Discrete texture and fuzzy texture are two primary examples of textures: (i) For the universe U , the family P (U ) = { A | A ⊆ U } is a texturing on U . The (U , P (U )) is called a discrete texture. For u ∈ U , we clearly have P u = {u } and Q u = U \ {u } and the mapping c U : P (U ) → P (U ) is the ordinary complementation on (U , P (U )) defined by c U ( A ) = U \ A for all A ∈ P (U ). (ii) The family M = {(0, r ] | r ∈ [0, 1]} is a complemented texture on M = (0, 1] which is called the fuzzy texture. Here, we have P r = Q r = (0, r ] for all r ∈ (0, 1]. The complementation c M : M → M is defined by ∀r ∈ (0, 1], c M (0, r ] = (0, 1 − r ]. Products of textures The product of a family of textures and its basic properties were given in [3]. For the sake of simplicity, we consider here the product of two textures. Now let (U 1 , U1 ) and (U 2 , U2 ) be any two textures and let us take the family A = { A × U 2 |  A ∈ U1 } {U 1 × B | B ∈ U2 } and define

B={



E j | { E j } j ∈ J ⊆ A and J is an index set}.

j∈ J

Now let K be any index set. Then the family

U1 ⊗ U2 = {



D k | { D k }k∈ K ⊆ B and K is an index set}

k∈ K

is a texture on U 1 × U 2 which is called a product of (U 1 , U1 ) and (U 2 , U2 ). The p-set and q-set of U1 ⊗ U2 are as follows:

P (u 1 ,u 2 ) = P u 1 × P u 2 . Q (u 1 ,u 2 ) = (U 1 × Q u 2 ) ∪ ( Q u 1 × U 2 ). Direlations, relation and corelation For the product texture P (U ) ⊗ V , we denote the p-set by P (u , v ) and the q-set by Q (u , v ) . Let (U , U ), ( V , V ) be texture spaces. Then (i) r ∈ P (U ) ⊗ V is called a relation from (U , U ) to ( V , V ) if R1 r  Q (u , v ) , P u   Q u =⇒ r  Q (u  , v ) , and R2 r  Q (u , v ) =⇒ ∃u  ∈ U such that P u  Q u  and r  Q (u  , v ) . (ii) R ∈ P (U ) ⊗ V is called a corelation from (U , U ) to ( V , V ) if CR1 P (u , v )  R , P u  Q u  =⇒ P (u  , v )  R, and CR2 P (u , v )  R =⇒ ∃u  ∈ U such that P u   Q u and P (u  , v )  R. A pair (r , R ), where r is a relation and R is a corelation from (U , U ) to ( V , V ) is called a direlation from (U , U ) to ( V , V ). Note that if (r , R ) is a direlation from (U , P (U )) to ( V , P ( V )), then r and R are ordinary relations from U to V , that is, r , R ⊆ U × V since P (U ) ⊗ P ( V ) = P (U × V ). Identity and reflexive direlation The identity direlation (i , I ) on (U , U ) is defined by

i=



{ P (u ,u ) | u ∈ U } and I =



{ Q (u ,u ) | u ∈ U  }

where U  = {u | U ⊆ Q u }. Recall that if (r , R ) is a direlation on (U , U ), then r is reflexive if i ⊆ r and R is reflexive if R ⊆ I . Then we say that (r , R ) is reflexive if r and R are reflexive. For discrete textures, we have i = {(u , u ) | u ∈ U } and I = (U × U ) \ i. Now let c U and c V be the complementations on (U , U ) and ( V , V ), respectively. The complement r  of the relation r is the corelation

r =



{ Q (u , v ) | ∃ w , z with r ⊆ Q ( w ,z) , c U ( Q u ) ⊆ Q w and P z ⊆ c V ( P v )}.

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The complement R  of the corelation R is the relation

R =



{ P (u , v ) | ∃ w , z with P ( w ,z) ⊆ R , P w ⊆ c U ( P u ) and c V ( Q v ) ⊆ Q z }.

The complement (r , R ) of the direlation (r , R ) is the direlation (r , R ) = ( R  , r  ). A direlation (r , R ) is called complemented if r = R  and R = r  . Inverse and symmetric direlation Let (r , R ) be a direlation from (U , U ) to ( V , V ) where (U , U ) and ( V , V ) are any two texture spaces. The direlation (r , R )← = ( R ← , r ← ) from ( V , V ) to (U , U ) is called the inverse of the direlation (r , R ) where

r← =



{ Q ( v ,u ) | r  Q (u , v ) } and R ← =



{ P ( v ,u ) | P (u , v )  R }.

Further, (r , R ) is called symmetric if r = R ← and R = r ← . If (r , R ) is a complemented direlation from (U , P (U )) to ( V , P ( V )), then we have r ← = (U × U ) \ r −1 = R where r −1 = {( v , u ) | (u , v ) ∈ r }. Composition and transitive direlations Let ( p , P ) and (q, Q ) be direlations from (U , U ) to ( V , V ) and from ( V , V ) to ( W , W ), respectively. The composition of ( p , P ) and (q, Q ) is the direlation (q, Q ) ◦ ( p , P ) = (q ◦ p , Q ◦ P ) where

q◦p=



{ P (u , w ) | ∃ v ∈ V with p  Q (u , v ) and q  Q ( v , w ) }

and

Q ◦P=



{ Q (u , w ) | ∃ v ∈ V with P (u , v )  P and P ( v , w )  Q }.

Further, r is transitive if r ◦ r ⊆ r and R is transitive if R ⊆ R ◦ R. Then (r , R ) is called transitive if r and R are transitive. Finally, if (r , R ) is reflexive, symmetric and transitive, then it is called an equivalence direlation. Sections and presections The A-sections and the B-presections with respect to relation and corelation are given as

r→ A =



{ Q v | ∀u , r  Q (u , v ) ⇒ A ⊆ Q u },  R A= { P v | ∀u , P (u , v )  R ⇒ P u ⊆ A },  ← r B= { P u | ∀ v , r  Q (u , v ) ⇒ P v ⊆ B }, and  R← B = { Q u | ∀ v , P (u , v )  R ⇒ B ⊆ Q v }, →

for all A ∈ U and B ∈ V , respectively. 3. Textural formal context Let U be a set of objects and V be a set of properties. Further, let r be a relation from U to V , that is, r ⊆ U × V . Then the triple (U , V , r ) is called a formal context. Recall that for all A ⊆ U and B ⊆ V , the sufficiency operators ∗ : P (U ) → P ( V ) and ∗ : P ( V ) → P (U ) are defined by

A ∗ = { v ∈ V | ∀u ∈ U , u ∈ A =⇒ (u , v ) ∈ r } and B ∗ = {u ∈ U | ∀ v ∈ V , v ∈ B =⇒ (u , v ) ∈ r }, respectively [14,22]. Sufficiency operators were used to define the notion of formal concept by Wille in [22]. For A ⊆ U and B ⊆ V , the pair ( A , B ) is called a formal concept in the formal context (U , V , r ) if

A ∗ = B and B ∗ = A . According to this, in a formal concept, objects in extent share all attributes in intent, and only attributes in intent are possessed by all objects in extent. Further, the dual sufficiency operators  : P (U ) → P ( V ) and  : P ( V ) → P (U ) were given by Düntsch and Gediga in [13] as follows: for all A ⊆ U , B ⊆ V ,

A  = { v ∈ V | ∃u ∈ U , u ∈ U \ A ∧ (u , v ) ∈ / r }, B  = {u ∈ U | ∃ v ∈ V , v ∈ V \ B ∧ (u , v ) ∈ / r }. Recall that ( A , B ) is called a dual formal concept in the formal context (U , V , r ) [5] if

A  = B and B  = A . For dual concept lattice, we refer to [5,13,15]. Now we may give the following definition.

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Definition 3.1. Let U = (U , U ) and V = ( V , V ) be texture spaces and (r , R ) be a direlation from (U , U ) to ( V , V ). Then the quadruple (U, V, r , R ) is called a textural formal context or in brief, a t-formal context. Definition 3.2. Let (U, V, r , R ) be a t-formal context. The mappings : U → V and : V → U defined by

∀ A ∈ U , A =



{ Q v | ∃u ∈ U , A ⊆ Q u ∧ r ⊆ Q (u , v ) }

and

∀B ∈ V, B =

 { Q u | ∃ v ∈ V , B ⊆ Q v ∧ r ⊆ Q (u , v ) }

are called t-sufficiency operator. Further, the mappings  : U → V and  : V → U defined by

∀ A ∈ U , A =



{ P v | ∃u ∈ U  , P u ⊆ A ∧ P (u , v ) ⊆ R }

and

∀B ∈ V, B =

 { P u | ∃ v ∈ V , P v ⊆ B ∧ P (u , v ) ⊆ R ∧ u ∈ U  }

are called t-co-sufficiency operator. If the textures are discrete, that is, U = P (U ) and V = P ( V ), and if (r , R ) is a direlation from U = P (U ) to V = P ( V ), then r and R can be regarded as ordinary relations from U to V . In this case, we may consider two different formal contexts in the sense of Wille. The first is (U , V , r ) and the second is (U , V , R ). Then the following result shows that in the context (U , V , r ), every t-sufficiency operator is also a sufficiency operator and vice versa. However, if (r , R ) is complemented, then for the context (U , V , r ), a t-co-sufficiency operator corresponds to a dual sufficiency operator. Theorem 3.3. Let A ∈ P (U ) = U and B ∈ P ( V ) = V and (r , R ) be a direlation from (U , P (U )) to ( V , P ( V )). Then the following statements hold. (i) For the context (U , V , r ), we have

A = A∗,

B = B∗.

(ii) Let (r , R ) be complemented. Then for the context (U , V , r ), we have

A = A

and

B = B.

Proof. We only prove that A  = A ∗ and A  = A  . The other equalities can be similarly proved. To see the equalities, it is enough to observe that

A =

=

 

{ Q v | ∃u ∈ U , A ⊆ Q u ∧ r ⊆ Q (u , v ) } / r} { V \ { v }| ∃u ∈ U , u ∈ A ∧ (u , v ) ∈

= V \ { v | ∃u ∈ U , u ∈ A ∧ (u , v ) ∈ / r} / A ∨ (u , v ) ∈ r } = { v | ∀u ∈ U , u ∈ = { v | ∀u ∈ U , u ∈ A =⇒ (u , v ) ∈ r } = A ∗ and

A =

=

 

{ P v | ∃u ∈ U  , P u ⊆ A ∧ P (u , v ) ⊆ R } / A ∧ (u , v ) ∈ R } {{ v }| ∃u ∈ U , u ∈

/ A ∧ (u , v ) ∈ / (U × V ) \ R } = { v | ∃u ∈ U , u ∈ = { v | ∃u ∈ U , u ∈ U \ A ∧ (u , v ) ∈ / r} = A.

2

Definition 3.4. Let (U, V, r , R ) be a t-formal context. A pair ( A , B ) where A ∈ U and B ∈ V is called (i) a t-formal concept if A  = B and B  = A, (ii) a t-formal co-concept if A  = B and B  = A.

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Corollary 3.5. Let (U, V, r , R ) be a t-formal context where U = (U , P (U )) and V = ( V , P ( V )). Then we have the following. (i) ( A , B ) is a t-formal concept in (U, V, r , R ) if and only if ( A , B ) is a formal concept in (U , V , r ). (ii) Let (r , R ) be a complemented direlation. Then ( A , B ) is a t-formal co-concept in (U, V, r , R ) if and only if ( A , B ) is a dual formal concept in (U , V , r ). By the above result, in case of discrete textures, every t-formal concept corresponds to a formal concept in the sense of Wille. Further, every t-formal co-concept corresponds to dual formal concept given in [5,13]. Now let us give the following example. Example 3.6. Consider the textures (U , U ) and ( V , V ) where

U = {a, b, c , d}, U = {U , ∅, {a}, {b}, {a, b}, {a, d}, {b, c }, {a, b, c }, {a, b, d}} and

V = {1, 2, 3}, V = { V , ∅, {2}, {3}, {1, 2}, {2, 3}}. The p-sets and q-sets of the texturings U and V are

P a = {a}, P b = {b}, P c = {b, c }, P d = {a, d}, Q a = {b, c }, Q b = {a, d}, Q c = {a, b, d}, Q d = {a, b, c } and

P 1 = {1, 2}, P 2 = {2}, P 3 = {3}, Q 1 = {2, 3}, Q 2 = {3}, Q 3 = {1, 2}, respectively. It is easy to see that the pair (r , R ) given in Table 1 is a direlation from (U , U ) to ( V , V ). We have the following inclusions with respect to the p-sets and q-sets of the product P (U ) ⊗ V and the direlation (r , R ):

r ⊆ Q (a,1) , r ⊆ Q (a,2) , r ⊆ Q (b,1) , r ⊆ Q (b,3) , r ⊆ Q (c ,3) , r ⊆ Q (d,1) , P (a,2) ⊆ R , P (b,3) ⊆ R , P (c ,2) ⊆ R , P (c ,3) ⊆ R , P (d,1) ⊆ R , P (d,2) ⊆ R where

P (a,2) = {(a, 2)}, P (b,3) = {(b, 3)}, P (c ,2) = {(c , 2)}, P (c ,3) = {(c , 3)}, P (d,1) = {(d, 1), (d, 2)}, P (d,2) = {(d, 2)} and

Q (a,1) = {(b, 1), (b, 2), (b, 3), (c , 1), (c , 2), (c , 3), (d, 1), (d, 2), (d, 3), (a, 2), (a, 3)}, Q (a,2) = {(b, 1), (b, 2), (b, 3), (c , 1), (c , 2), (c , 3), (d, 1), (d, 2), (d, 3), (a, 3)}, Q (b,1) = {(a, 1), (a, 2), (a, 3), (c , 1), (c , 2), (c , 3), (d, 1), (d, 2), (d, 3), (b, 2), (b, 3)}, Q (b,3) = {(a, 1), (a, 2), (a, 3), (c , 1), (c , 2), (c , 3), (d, 1), (d, 2), (d, 3), (b, 1), (b, 2)}, Q (c ,3) = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3), (d, 1), (d, 2), (d, 3), (c , 1), (c , 2)}, Q (d,1) = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3), (c , 1), (c , 2), (c , 3), (d, 2), (d, 3)}. Then we may determine the following sets:

{a} = {a, d} = {3},

{b, c } = {2} = {b}

U  = {a, b} = {a, b, c } = {a, b, d} = ∅,

∅ = V , {2} = {b, c }, {3} = {a, d}, ∅ = U , {1, 2} = {2, 3} = V  = ∅, {a, b} = {a} = {1, 2, 3},

{b, c } = {1, 2}, {b} = V ,

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Table 1 The relation r and corelation R. r

1

2

a b c d

×

× × ×

3

R

×

a

1

b c

×

d

2

3

×

×

× ×

× ×

Table 2 Comparison of concepts. t-Formal concepts

t-Formal co-concepts

Formal concepts

Dual formal concepts

({a, d}, {3}) ({b, c }, {2}) (U , ∅) (∅, V )

({b, c }, {1, 2}) ({a, d}, {2, 3}) (∅, V ) (U , ∅)

({a, d}, {3}) ({b, c , d}, {2}) ({d}, {2, 3}) ({c }, {1, 2}) (U , ∅) (∅, V )

({b, c }, {1, 2}) ({a}, {1, 3}) ({a, b, c }, {1}) ({a, b, d}, {3}) (∅, V ) (U , ∅)

{a, b, c } = {1, 2}, {a, d} = {2, 3}, {a, b, d} = {2, 3}, U  = ∅, ∅ = V ,

{2} = U

{3} = U ,

{1, 2} = {b, c },

{2, 3} = {a, d}, V  = ∅, ∅ = V . For {b, c } ∈ U , let us take the sets {b, c } and {b, c } . Observe that we have {b, c } ⊆ Q b , {b, c } ⊆ Q c and the inclusions r ⊆ Q (b,1) , r ⊆ Q (b,3) and P (c ,3) ⊆ R yield {b, c } = Q 1 ∩ Q 3 = {2}. For the set {b, c } , note that P a ⊆ {b, c } and P d ⊆ {b, c } and since P (a,2) ⊆ R and P (d,2) ⊆ R, we also find {b, c } = P 2 = {2}. Now let us determine the sets {1, 2} and {1, 2} . We clearly see that {1, 2} ⊆ Q 1 , {1, 2} ⊆ Q 2 and r ⊆ Q (a,1) , r ⊆ Q (b,1) , r ⊆ Q (a,2) , r ⊆ Q (d,1) . Hence, we conclude that {1, 2} = Q a ∩ Q b ∩ Q d = ∅. Since P 3 ⊆ {1, 2}, P (c ,3) ⊆ R and P (b,3) ⊆ R, we obtain that {1, 2} = P b ∪ P c = {b, c }. According to these evaluations, we obtain the t-formal concepts and t-formal co-concepts given in Table 2. On the other hand, in case of discrete textures, we also have to take the following subsets of U and V into account, respectively:

{d}, {c }, {a, c }, {b, d}, {b, c , d}, {a, c , d} and

{1}, {1, 3}. If we take r as an ordinary relation, then we get the formal concepts and dual formal concepts as in Table 2. In this case, a formal context certainly provides more information. Because all subsets of the universe are used in determining all formal concepts of a formal context. Since the given texture does not contain the sets {c } and {d}, the pairs ({c }, {1, 2}) and ({d}, {2, 3}) can not be t-formal concepts. Further, the pair ({b, c }, {2}) is not a formal concept in the sense of Wille. However, ({b, c }, {2}) is a t-formal concept and it still contains correct information about the system in textural hierarchy. This shows that a texturing, as a relatively imperfect collection of sets of the universe in the above sense, provides a remarkable setting which still gives an information with respect to given system. As we easily see in Example 3.21, the direlation is complemented, that is, we have R = r  . Therefore, the families of formal concepts and t-formal concepts are different since the texture does not contain all subsets of the universe. Moreover, the families of dual formal concepts and t-formal co-concepts are also different since the textural complement is weaker than the set theoretical complement. Now we may give the basic properties of the mappings  and  : Theorem 3.7. Let (U, V, r , R ) be a textural formal context. Then (1) (i) (ii) (2) (i) (ii)

 ∀ A 1 , A 2 ∈ U , A 1 ⊆ A 2 =⇒ A  2 ⊆ A1 . ∀ A ∈ U , A ⊆ A  .  ∀ B 1 , B 2 ∈ V , B 1 ⊆ B 2 =⇒ B  2 ⊆ B1 . ∀ B ∈ V , B ⊆ B  .

   Proof. (1) (i) Let A 1 ⊆ A 2 . Suppose that A  2 ⊆ A 1 . Take v ∈ V such that A 2 ⊆ Q v and P v ⊆ A 1 . Then from the first statement, we have

∀u ∈ U , A 2 ⊆ Q u =⇒ r ⊆ Q (u , v ) .

(1)

M. Diker et al. / International Journal of Approximate Reasoning 114 (2019) 182–203

189

   Since P v ⊆ A  1 , there exists u ∈ U such that A 1 ⊆ Q u and r ⊆ Q (u , v ) for some v ∈ V with P v ⊆ Q v . By the assumption, we also have A 2 ⊆ Q u and hence by (1), we conclude that r ⊆ Q (u , v ) . However, P v ⊆ Q v  implies that r ⊆ Q (u , v  ) and this gives a contradiction. (ii) Suppose that A ⊆ A  . Let us choose u ∈ U such that A ⊆ Q u and P u ⊆ A  . Then we have A  ⊆ Q v and r ⊆ Q (u  , v ) for some v ∈ V and u  ∈ U with P u ⊆ Q u  . Now A  ⊆ Q v gives that

∀u ∈ U , A ⊆ Q u =⇒ r ⊆ Q (u , v ) . Since we (2) (i) for some

(2)

also have A ⊆ Q u  , by (2), we find r ⊆ Q (u  , v ) which is an immediate contradiction.    Suppose that B 1 ⊆ B 2 and B  2 ⊆ B 1 . Choose u ∈ U such that B 2 ⊆ Q u and P u ⊆ B 1 . Then by the second statement, u  ∈ U with P u ⊆ Q u  , we have

∃ v ∈ V , B 1 ⊆ Q v and r ⊆ Q (u  , v ) .

(3)

 Clearly, B  2 ⊆ Q u and so

∀ v ∈ V , B 2 ⊆ Q v =⇒ r ⊆ Q (u  , v ) .

(4)

However, we also have B 2 ⊆ Q v and hence, by (4), we obtain r ⊆ Q (u  , v ) . But this is an immediate contradiction by (3). (ii) Let us assume that B ⊆ B  for some B ∈ V . Then for some v  ∈ V with P v ⊆ Q v  , there exists u ∈ U such that B  ⊆ Q u ∧ r ⊆ Q (u , v  ) . Further, B  ⊆ Q u implies that B ⊆ Q v =⇒ r ⊆ Q (u , v ) . On the other hand, B ⊆ Q v and P v ⊆ Q v  gives that B ⊆ Q v  and so r ⊆ Q (u , v  ) is a contradiction. 2 Theorem 3.8. Let (U, V, r , R ) be a textural formal context. Then (1) (i) (ii) (2) (i) (ii)

 ∀ A 1 , A 2 ∈ U , A 1 ⊆ A 2 =⇒ A  2 ⊆ A1 ∀ A ∈ U , A  ⊆ A.  ∀ B 1 , B 2 ∈ V , B 1 ⊆ B 2 =⇒ B  2 ⊆ B1 ∀ B ∈ V , B  ⊆ B.

The operations “” “” are idempotent and monotonic: Corollary 3.9. (i) ∀ A ∈ U , A  = A  .

∀ A 1 , A 2 ∈ U , A 1 ⊆ A 2 =⇒ A  ⊆ A  1 2 . (ii) ∀ A ∈ U , A  = A  .

∀ A 1 , A 2 ∈ U , A 1 ⊆ A 2 =⇒ A  ⊆ A  1 2 . (iii) ∀ B ∈ B , B  = B  .

∀ B 1 , B 2 ∈ V , B 1 ⊆ B 2 =⇒ B  ⊆ B  1 2 . (iv) ∀ B ∈ V , B  = B  .

∀ B 1 , B 2 ∈ V , B 1 ⊆ B 2 =⇒ B  ⊆ B  1 2 . Note that by Theorem 3.7, t-sufficiency and co-t-sufficiency operators constitute Galois connections. Therefore, we immediately have the following theorem. Corollary 3.10. Let { A j | j ∈ J } ⊆ U and { B j | j ∈ J } ⊆ V . Then we have the following equalities:

  j∈ J (ii) ( j ∈ J  (iii) ( j ∈ J  (iv) ( j ∈ J (i) (

  j∈ J A j .    B j ) = j∈ J B j .  A j ) = j ∈ J A  j .  B j ) = j ∈ J B  . j A j ) =

Theorem 3.11. [4] Let (U , U ) be a texture. Then

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(i) A = { P u | A  Q u } for all A ∈ U . (ii) A = { Q u | P u  A } for all A ∈ U . Corollary 3.12. (i) If ( A , B ) is a t-formal concept, then



A=

P v



B=

and

B ⊆ Q v

P u

A ⊆ Q u

(ii) If ( A , B ) is a t-formal co-concept, then

A=



Q v



B=

and

P v ⊆ B

Q u .

P u ⊆ A

Proof.  (i) We prove the first equality leaving the proof of the second to the reader. By Theorem 3.11 (i), we have B = P v . Then by Corollary 3.10 (ii), B ⊆ Q v



A = B = (

P v ) =

B ⊆ Q v



P v .

B ⊆ Q v



(ii) By Theorem 3.11 (ii), we have B =

Q v . Then by Corollary 3.10 (iv),

P v ⊆ A



A = B = (



Q v ) =

2

.

P v ⊆ B Q v

P v ⊆ B

Corollary 3.13. (i) If ( A , B ) is a t-formal concept, then



A=

P u



B=

and

Q v .

B ⊆ Q v

A ⊆ Q u

(ii) If ( A , B ) is a t-formal co-concept, then

A=



Q u



B=

and

Q v .

P v ⊆ B

P u ⊆ A

Proof. (i) By Corollary 3.12 (i),

A=





Pu ⊆

A Q u



Pu

A Q u

=(



⊆(

A Q u



 

) =(

Pu

A Q u

and so we get A =



A Q u



P u )



 P u ) = B  = A

A Q u



P u . The proof of the second equality is similar.

(ii) Since ( A , B ) is a t-formal co-concept, we have A ∇ = B and B ∇ = A. Then, by Corollary 3.10 (iii),

B = A∇ = (



Q u )∇ =

PuA



Q u∇

PuA

Moreover,

A = B∇ = (



Q u∇ )∇ = (

PuA

Hence, by Theorem 3.8 (1)(ii),

 PuA

Q u∇∇∇ )∇ = (

 PuA

Q u∇∇ )∇∇ .

M. Diker et al. / International Journal of Approximate Reasoning 114 (2019) 182–203

A=(





Q u∇∇ )∇∇ ⊆

PuA

and so we get A =

Q u∇∇ ⊆

PuA



PuA



191

Qu = A

PuA

Q u∇∇ . The proof of other equality is similar.

2

Corollary 3.14. Let ( A 1 , B 1 ) and ( A 2 , B 2 ) be any two t-formal concepts or formal co-concepts in (U, V, r , R ). Then

A 1 ⊆ A 2 ⇐⇒ B 2 ⊆ B 1 . Proof. By Theorem 3.7 and 3.8, the proof is immediate.

2

Definition 3.15. We denote the families of all textural formal concepts and all textural formal co-concepts in (U, V, r , R ) by Br (U, V) and B R (U, V), respectively. That is, let

Br (U, V) = {( A , B ) ∈ U × V | A  = B and A = B  } and

B R (U, V) = {( A , B ) ∈ U × V | A  = B and A = B  }. The pairs (Br (U, V), r ) and (B R (U, V), R ) are partially ordered sets where the relations r and R are defined by

( A 1 , B 1 ) r ( A 2 , B 2 ) ⇐⇒ A 1 ⊆ A 2 (or B 2 ⊆ B 1 ), and

( A 1 , B 1 ) R ( A 2 , B 2 ) ⇐⇒ A 1 ⊆ A 2 (or B 2 ⊆ B 1 ), respectively. Theorem 3.16. (i) (Br (U, V), ) is a complete lattice with the infimum and supremum given by



(A j, B j) = (

j∈ J



A j, (

j∈ J



B j ) ),

j∈ J



( A j , B j ) = ((

j∈ J



A j ) ,



j∈ J

B j ).

j∈ J

(ii) (B R (U, V), ) is a complete lattice with the infimum and supremum given by



( A j , B j ) = ((

j∈ J



A j ) ,

j∈ J



B j ),

j∈ J



(A j, B j) = (

j∈ J



A j, (

j∈ J



B j ) ).

j∈ J

Proof. (i) For all j ∈ J , the pair ( A j , B j ) ∈ Br (U, V) is a formal concept and hence A j = B  . Therefore, by Corollary 3.10 (i), j we have

(



A j, (

j∈ J



B j ) ) = (

j∈ J



B j ,(

j∈ J



B j ) ) = ((

j∈ J



B j ) , (

j∈ J



B j ) )

j∈ J

  and so by Theorem 3.7, ( j ∈ J A j , ( j ∈ J B j ) ) is a formal concept. Further, for all j ∈ J , we clearly have

(



A j, (

j∈ J



B j ) ) ( A j , B j ),

j∈ J

  that is, ( j ∈ J A j , ( j ∈ J B j ) ) is a lower bound of the family {( A j , B j ) | j ∈ J }. If ( A , B ) is another lower bound of    {( A j , B j ) | j ∈ J }, then we clearly have A ⊆ j ∈ J A j and hence, ( j ∈ J A j , ( j ∈ J B j ) ) is indeed the desired infimum.

The proof of the supremum is similar. (ii) Let us prove the second equality, leaving the proof of the infimum to the interested reader. Let ( A j , B j ) ∈ B R (U, V) where j ∈ J . Then for all j, we have A j = B  and hence, by Corollary 3.10 (iv), we obtain j

(

 j∈ J

A j, (

 j∈ J

B j ) ) = (

 j∈ J

B j ,(

 j∈ J

B j ) ) = ((

 j∈ J

B j ) , (

 j∈ J

B j ) ).

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Therefore, the pair ( j ∈ J A j , ( j ∈ J B j ) ) is a formal co-concept. Moreover, it is an upper bound of the family {( A j , B j ) |  j ∈ J } since A j ⊆ j ∈ J A j for all j ∈ J . If a co-concept ( A , B ) is another upper bound, then for all j ∈ J we have A j ⊆ A     ) is the desired supremum. 2 and therefore, j ∈ J A j ⊆ A, that is ( j∈ J A j , ( j∈ J B j ) We call (Br (U, V), ) a textural concept lattice or in brief, a t-concept lattice. We call (B R (U, V), ) a t-co-concept lattice of the t-formal context (U, V, r , R ). Lemma 3.17. [4] (U , U , c U ) be a complemented texture. Then for all A , B ∈ U ,

A  B =⇒ ∃u ∈ U

with

A  cU ( P u )

and c U ( Q u )  B .

We need the following lemma in the sequel: Lemma 3.18. (i) Let (U , U ) be a texture. Then

(∀u ∈ U )(∃u  ∈ U )( P u ⊆ Q u  ). (ii) Let c U be a complementation on (U , U ). Then

(∀u ∈ U )(∃u  ∈ U )( P u ⊆ c U P u  ). Proof. (i) Suppose that there exists some u ∈ U such that P u ⊆ Q u  for all u  ∈ U . Since P u = that P u = ∅. But this is impossible since u ∈ P u . (ii) The proof is immediate by Lemma 3.17. 2



{ P u  | P u ⊆ Q u  }, we get

The following result shows that if the direlation (r , R ) is complemented, then the operations  and  are complementary dual. Theorem 3.19. Let (U , U , c U ) and ( V , V , c V ) be any two complemented textures and (U, V, r , R ) be a textural formal context. If (r , R ) is a complemented direlation from (U , U ) to ( V , V ), then we have the following equalities. (i) ∀ A ∈ U , A  = c V ((c U ( A )) ). (ii) ∀ B ∈ V , B  = c U ((c V ( B )) ). Proof. (i) Suppose that A  ⊆ c V (c U ( A ) ). Let us choose v  ∈ V such that A  ⊆ Q v  and P v  ⊆ c V (c U ( A )) . Then

∀u ∈ U , A ⊆ Q u =⇒ r ⊆ Q (u , v  ) .

(1)

∈ U such that P v  ⊆ Q v  and P v  ⊆ c V (c U ( A ) ). Then we have (c U ( A )) ⊆ c V ( P v  ). Therefore, for some   v ⊆ c V ( P v  ), we have P u ⊆ c U ( A ) and P (u , v  ) ⊆ R for some u ∈ U . Let us choose v 1 ∈ V such that P v  ⊆ Q v 1 c V ( P v  ). Since u ∈ U  , we get u  ∈ U where P u  ⊆ Q u . Then by Lemma 3.17, there exists u 1 ∈ U such that

Let us take v  v  ∈ V with P

and P v 1 ⊆ P u  ⊆ c U ( P u 1 ) and c U ( Q u 1 ) ⊆ Q u . Now we have A ⊆ Q u 1 and by (1), we have r ⊆ Q (u 1 , v  ) . Further, P v  ⊆ Q v  implies that r ⊆ Q (u 1 , v  ) . We clearly have r  ⊆ Q (u , v 1 ) . Since (r , R ) is complemented, we obtain P (u , v  ) ⊆ R = r  ⊆ Q (u , v 1 ) . However, this is a contradiction, since P v  ⊆ Q v 1 . Now let us show that c V ((c U ( A )) ) ⊆ A  . Suppose that c V ((c U ( A )) ) ⊆ A  and let us choose v ∈ V such that c V ((c U ( A )) ) ⊆ Q v and P v ⊆ A  . Then c V ( Q v ) ⊆ (c U ( A )) and so we may choose v  ∈ U such that c V ( Q v ) ⊆ Q v  and P v  ⊆ (c U ( A )) . By the second statement, we have the implication

P u ⊆ c U ( A ) =⇒ P (u , v  ) ⊆ R . A  implies that for some v 

(2)

Further, P v ⊆ ∈ U with P v ⊆ Q v  , we have A ⊆ Q u and r ⊆ Q (u , v  ) for some u ∈ U . By Lemma 3.18, there exists u 1 ∈ U such that P u ⊆ c U ( P u 1 ). Hence, A ⊆ c U ( P u 1 ) and so P u 1 ⊆ c U ( A ). By (2), we have P (u 1 , v  ) ⊆ R. Since (r , R ) is complemented, P (u , v ) ⊆ R  = r ⊆ Q (u , v  ) and this is a contradiction, since P v ⊆ Q v  . (ii) Suppose that (c V ( B )) ⊆ c U ( B  ). Choose u  ∈ U such that (c V ( B )) ⊆ Q u  and P u  ⊆ c U ( B  ). Further, take u 1 ∈ U such that (c V ( B )) ⊆ c U ( P u 1 ) and c U ( Q u 1 ) ⊆ Q u  . Now for some u  ∈ U  , P u  ⊆ c U ( P u 1 ) such that ∃ v ∈ V , P v ⊆ c V ( B ) and P u  , v ) ⊆ R. Note that we have P u 1 ⊆ c U ( P u  ) and hence, c U ( P u  ) ⊆ Q u 1 . This implies that c U ( Q u 1 ) ⊆ P u  , that is, P u  ⊆ Q u  . Since u  ∈ U  , for some u 2 ∈ U , we have c U ( Q u 2 ) ⊆ Q u  and so c U ( Q u 2 ) ⊆ Q u  whence c U ( Q u 2 ) ⊆ c U ( B  ). This gives that B  ⊆ Q u 2 . Therefore, we have that

M. Diker et al. / International Journal of Approximate Reasoning 114 (2019) 182–203

∀ v ∈ V , B ⊆ Q v =⇒ r ⊆ Q (u 2 , v ) .

193

(3)

Let us take v ∗

∈ V such that P v ⊆ Q v ∗ and P v ∗ ⊆ c V ( B ). Further, let us choose v 1 , v 2 ∈ V such that P v ⊆ Q v 2 , P v 2 ⊆ c V ( P v 1 ) and c V ( Q v 1 ) ⊆ Q v ∗ . Now c V ( Q v 1 ) ⊆ Q v ∗ implies that B ⊆ Q v 1 and hence, by (3), we conclude that r ⊆ Q (u 2 , v 1 ) . As a result, we obtain that P (u  , v ) ⊆ R = r  ⊆ Q (u  , v 2 ) . However, this is a contradiction. For the reverse inclusion, assume that c U ((c V ( B )) ) ⊆ B  . Let us choose u ∈ U such that c U ((c V ( B )) ) ⊆ Q u and P u ⊆  B . The second statement implies that for some u  ∈ U , P u ⊆ Q u  and ∃ v ∈ V , B ⊆ Q v and r ⊆ Q (u  , v ) . By Lemma 3.18, we have u ∗ ∈ U such that P u  ⊆ c U ( P u ∗ ). Then we have P u ⊆ c U ( P u ∗ ) and hence, c U ((c V ( B )) ) ⊆ c U ( P u ∗ ), that is, P u ∗ ⊆ (c V ( B )) . Therefore, we have P v ⊆ c V ( B ) =⇒ P (u ∗ , v ) ⊆ R. Let us take v ∗ , v 2 ∈ V such that B ⊆ c V ( P v 2 ), c V ( Q v 2 ) ⊆ Q v ∗ and P v ∗ ⊆ Q v . Since P v 2 ⊆ c V ( B ), we have P (u ∗ , v 2 ) ⊆ R. Then we obtain that P (u  , v ∗ ) ⊆ R  = r ⊆ Q (u  , v ) which is an immediate contradiction. 2 Corollary 3.20. Let (U, V, r , R ) be a textural formal context. If (r , R ) is a complemented direlation from (U , U ) to ( V , V ), then ( A , B ) is a t-formal concept if and only if (c U ( A ), c V ( B )) is a t-formal co-concept. Proof. By Theorem 3.19, it is enough to observe that we have

c V ( A  ) = (c U ( A )) = c V ( B )

and c U ( B  ) = (c V ( B )) = c U ( A ).

2

Example 3.21. Consider the complemented textures (U , U ) and ( V , V ) given in Example 3.6. It is easy to see that the mappings c U : U → U and c V : V → V defined by

c U ({a}) = {a, b, c }, c U ({a, b, c }) = {a}, c U ({b}) = {a, b, d}, c U ({a, b, d}) = {b}, c U ({a, d}) = {b, c }, c U ({b, c }) = {a, d}, c U ({a, b}) = {a, b}, c U (∅) = U , c U (U ) = ∅, and

c V ({2}) = {2, 3}, c V ({2, 3}) = {2}, c V ({3}) = {1, 2}, c V ({1, 2}) = {3}, c V ( V ) = ∅, c V (∅) = V are complementations on the textures U and V , respectively. Clearly, these complementations are different from ordinary set theoretical complementations. Note that the family

Br (U, V) = {({a, d}, {3}), ({b, c }, {2}), (U , ∅), (∅, V )} is the concept lattice of (U, V, r , R ). Then we may determine the formal co-concept lattice B R (U, V) of (U, V, r , R ). Indeed, by Corollary 3.20, we have

(c U ({a, d}), c U ({3})) = ({b, c }, {1, 2}), (c U ({b, c }), c U ({2}) = ({a, d}, {2, 3}), (c U (U ), c U (∅)) = (∅, V ), (c U (∅), c U ( V ))} = (U , ∅) and hence, the formal co-concept lattice of (U, V, r , R ) is

B R (U, V) = {({b, c }, {1, 2}), ({a, d}, {2, 3}), (∅, V ), (U , ∅)}. Definition 3.22. Let ( L , ≤) and ( L  , ≤ ) be lattices. A mapping h : L → L  is called dually isomorphic if it satisfies the following conditions: (i) h is a bijection. (ii) ∀a, b ∈ L, a ≤ b ⇐⇒ h(b) ≤ h(a). Corollary 3.23. If (r , R ) is complemented, then the lattices Br (U, V) and B R (U, V) are dually isomorphic.

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Proof. Let us consider the mapping h : Br (U, V) → B R (U, V) defined by

∀( A , B ) ∈ Br (U, V), h(( A , B )) = (c U ( A ), c V ( B )). We show that the mapping h is one-to-one and onto. Let ( A , B ), ( A 1 , B 1 ) ∈ Br (U, V) and h( A , B ) = h( A 1 , B 1 ). Then by definition of h, we have (c U ( A ), c V ( B )) = (c U ( A 1 ), c V ( B 1 )), that is, c U ( A ) = c U ( A 1 ) and c V ( B ) = c V ( B 1 ). Then c U (c U ( A )) = c U (c U ( A 1 )) and c V (c V ( B )) = c V (c V ( B 1 )). By definition of complementation, we obtain A = A 1 and B = B 1 . Hence we have ( A , B ) = ( A 1 , B 1 ). Now let (C , D ) ∈ B R (U, V). Then C  = D and C = D  . We clearly have (c U (c U (C ))) = D and C = (c V (c V ( D ))) . Now By Theorem 3.19, c V (c U (C )) = D and C = c U (c V ( D ) ). This implies that (c U (C )) = c V ( D ) and c U (C ) = (c V ( D )) . Therefore, (c U (C ), c V ( D )) ∈ Br (U, V). Now we observe that h(c U (C ), c V ( D )) = (c U (c U (C )), c V (c V ( D ))) = (C , D ). Hence, we conclude that h is surjective. Further, if ( A 1 , B 1 ), ( A 2 , B 2 ) ∈ Br (U, V) and ( A 1 , B 1 ) r ( A 2 , B 2 ), then by Corollary 3.20, (c U ( A 1 ), c V ( B 1 )), (c U ( A 2 ), c V ( B 2 )) ∈ B R (U, V) and we clearly have

h(( A 2 , B 2 )) = (c U ( A 2 ), c V ( B 2 )) R (c U ( A 1 ), c V ( B 1 )) = h(( A 1 , B 1 )). Conversely, if h(( A 2 , B 2 )) R h(( A 1 , B 1 ), then we may also conclude that ( A 1 , B 1 )) r ( A 2 , B 2 ).

2

Example 3.24. Note that for the mapping h : Br (U, V) → B R (U, V) given in Corollary 3.23, we have

h(({a, d}, {3})) = (c U {a, d}, c V {3}) = ({b, c }, {1, 2}), h(({b, c }, {2})) = (c U {b, c }, c V {2}) = ({a, d}, {2, 3}), h((U , ∅)) = (c U (U ), c U (∅)) = (∅, V ), h((∅, V )) = (c U (∅), c V ( V )) = (U , ∅). Clearly, h is one-to-one and onto. Further, we have

({a, d}, {3}) r (U , ∅) ⇐⇒ (∅, V ) R ({b, c }, {1, 2}), ({b, c }, {2}) r (U , ∅) ⇐⇒ (∅, V ) R ({a, d}, {2, 3}). Hence, Br (U, V) and B R (U, V) are dually isomorphic. 4. Main theorem of t-formal concept lattices Note that for a direlation (r , R ) from a texture (U , U ) to a texture ( V , V ), we have two complete lattices denoted by Br (U, V) and B R (U, V), respectively. Here, under certain conditions, we state and prove the main theorem for t-formal concept lattice Br (U, V). For t-formal co-concept lattice B R (U, V), we give the dual statements of the respected results without proofs. Since the proofs are not so trivial, in any case, the reader may see them in the appendix. Now let (U, V, r , R ) be a textural formal context. For every point u ∈ U and v ∈ V , t-object concept and t-attribute concept are defined by  ( P u , P u ) and ( P  v , P v ),

respectively. For the relation r, let us consider the mappings

γr : U → Br (U, V), ∀u ∈ U , γr (u ) = ( P u , P u ),  μr : V → Br (U, V), ∀ v ∈ V , μr ( v ) = ( P  v , P v ).

For every point u ∈ U and v ∈ V , t-object co-concept and t-attribute co-concept are defined by

( Q u , Q u ) and ( Q v , Q v ), respectively. Further, for the corelation R let us define the mappings

γ R : U → B R (U, V), ∀u ∈ U , γ R (u ) = ( Q u , Q u ) μ R : V → B R (U, V), ∀ v ∈ V , μ R ( v ) = ( Q v , Q v ). We have the following results:

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Lemma 4.1. (i) (ii) (iii) (iv)

∀ v , v  ∈ V , P v ⊆ Q v  =⇒ μr ( v ) ≤ μr ( v  ). ∀u , u  ∈ U , P u ⊆ Q u  =⇒ γr (u  ) ≤ γr (u ). P u ⊆ Q v ⇐⇒ (∃ v  ∈ V )( P v  ⊆ Q v and γr (u ) ≤ μr ( v  )).   P v ⊆ Q u ⇐⇒ (∃u ∈ U )( P u  ⊆ Q u and γr (u ) ≤ μr ( v )).

Proof. (i) Let P v ⊆ Q v  . Then P v  ⊆ P v and by Corollary 3.9 (i), P  ⊆ P  and so we get μr ( v ) ≤ μr ( v  ). v v   ⊆ P (ii) Let P u ⊆ Q u  . Then P u  ⊆ P u and by Corollary 3.9 (i) we have P u  u , that is, γr (u ) ≤ γr (u ). and hence, (iii) Let P v  ⊆ Q v and γr (u ) ≤ μr ( v  ) for some v  ∈ V . Then by definition of γr and μr , we have P u ⊆ P  v P  ⊆ P u . Further, since P v  ⊆ P  , we get P u ⊆ Q v . Now Suppose that P u ⊆ Q v . Take v  ∈ V such that P u ⊆ Q v  and v v P v  ⊆ Q v . In this case, P v  ⊆ P u , that is, P u ⊆ P  and this implies that γr (u ) ≤ μr ( v  ). v   (iv) Let P u  ⊆ Q u and γr (u  ) ≤ μr ( v ). Then we have P u  ⊆ P u  ⊆ P v and hence, we get P v ⊆ Q u . Conversely, suppose        that P v ⊆ Q u . Take u ∈ U such that P v ⊆ Q u and P u ⊆ Q u . In this case, we get P u ⊆ P v , that is, P  ⊆ P u . Then we v  and we obtain γ (u  ) ≤ μ ( v ). 2 have P u ⊆ P  r r v Corollary 4.2. Suppose that the mappings γr and μr given above satisfy the conditions

∀u , u  ∈ U , P u ⊆ Q u  =⇒ γr (u ) ≤ γr (u  ), and ∀ v , v  ∈ V , P v ⊆ Q v  =⇒ μr ( v  ) ≤ μr u ( v ), respectively. Then we have the following statements: (i) (ii) (iii) (iv) (v)

∀u , u  ∈ U , P u ⊆ Q u  =⇒ γr (u ) = γr (u  ). ∀ v , v  ∈ V , P v ⊆ Q v  =⇒ μr ( v  ) = μr ( v ). ∀ v ∈ V  , P u ⊆ Q v ⇐⇒ γr (u ) ≤ μr ( v ). ∀u ∈ U  , P  v ⊆ Q u ⇐⇒ γr (u ) ≤ μr ( v ). ∀u ∈ U  , ∀ v ∈ V  , γr (u ) ≤ μr ( v ) =⇒ r ⊆ Q (u , v ) .

Proof. The proofs of (i) and (ii) are immediate. (iii) Let γr (u ) ≤ μr ( v ). Take v  ∈ V such that P v  ⊆ Q v . By (ii), μr ( v ) = μr ( v  ) and so we get γr (u ) ≤ μr ( v  ). Now by Lemma 4.1 (iii), we obtain P u ⊆ Q v . Conversely, let P u ⊆ Q v . Suppose that γr (u )  μr ( v ). Then we have P u ⊆ P  v . Hence, for some u  ∈ U and v  ∈ V with P u ⊆ Q u  , P v ⊆ Q v  and r ⊆ Q (u  , v  ) . By the assumption we also have P u ⊆ Q v  and then for all u  ∈ U ,

P u ⊆ Q u  =⇒ r ⊆ Q (u  , v  ) . However, this is an immediate contradiction. (iv) Similar to the proof of (iii). (v) Let u ∈ U  , v ∈ V  and γr (u ) ≤ μr ( v ). Let us choose u  ∈ U and v  ∈ V such that P u  ⊆ Q u and P v  ⊆ Q v . Then by (i) and (ii), γr (u  ) ≤ μr ( v  ). By definition of the mappings of γr and μr , P u  ⊆ P  and hence we get P  ⊆ Q u . Therefore, we v v have

∀ v ∗ ∈ V , P v  ⊆ Q v ∗ =⇒ r ⊆ Q (u , v ∗ ) . This implies that r ⊆ Q (u , v ) as desired.

2

The following example shows the existence of the mappings

γr and μr satisfying the conditions of Corollary 4.2:

Example 4.3. Consider the fuzzy texture ( M , M) and the unit texture ( I , I ) where

M = (0, 1], M = {(0, u ] | u ∈ [0, 1]} and I = [0, 1], I = {[0, u ) | u ∈ [0, 1]} ∪ {[0, u ] | u ∈ [0, 1]}, respectively. The pair (r , R ) where

r = {(u , v ) | v < u } and

R = (0, 1] × {0}

is a direlation from ( M , M) to ( I , I ). Then we get

196

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 { Q v | ∃u  ∈ M , P u  Q u  ∧ r ⊆ Q (u  , v ) }  = {[0, v ) | ∃u  ∈ M , (0, u ]  (0, u  ] ∧ r ⊆ ((0, 1] \ {u  } × [0, 1]) ∪ ((0, 1] × [0, v ))}  = {[0, v ) | ∃u  ∈ M , u  < u ∧ ( v  ≥ v =⇒ (u  , v  ) ∈ / r )}  = {[0, v ) | ∃u  ∈ M , u  < u ∧ ( v  ≥ v =⇒ v  ≥ u  )}

Pu =

=∅∈I and 

 { Q u | ∃ v  ∈ I , P v  Q v  ∧ r ⊆ Q (u , v  ) }  = {(0, u ] | ∃ v  ∈ I , [0, v ]  [0, v  ) ∧ r ⊆ ((0, 1] \ {u } × [0, 1]) ∪ ((0, 1] × [0, v  ))}  = {(0, u ] | ∃ v  ∈ I , v  ≤ v ∧ ( v  ≥ v  =⇒ (u  , v  ) ∈ / r )}  = {(0, u ] | ∃ v  ∈ I , v  ≤ v ∧ ( v  ≥ v =⇒ v  ≥ u  )}

Pv =

= ∅ ∈ M. Now, it is easy to see that the following implications hold:

P u  Q u  =⇒

γr (u ) = γr (u  ),

P v  Q v  =⇒

μr ( v ) = μr ( v  ).

Further, we have

 { P v | ∃u  ∈ M  , P u   Q u ∧ P (u  , v ) ⊆ R }  = {[0, v ] | ∃u  ∈ M  , (0, u  ]  (0, u ] ∧ ({u  } × [0, v ] ⊆ R )}  = {[0, v ] | ∃u  ∈ M  , u < u  ∧ ( v  ≤ v =⇒ (u  , v  ) ∈ R )}  = {[0, v ] | ∃u  ∈ M  , u < u  ∧ ( v  ≤ v =⇒ (u  , v  ) ∈ (0, 1] × {0})}

Q u∇ =

= [0, 1] and

 { P u | ∃ v  ∈ I , P v   Q v ∧ P (u , v  ) ⊆ R , u ∈ I  }  = {(0, u ] | ∃ v  ∈ I , [0, v  ]  [0, v ) ∧ ({u } × [0, v  ] ⊆ R , u ∈ M  }  = {(0, u ] | ∃ v  ∈ I , v ≤ v  ∧ ( v  ≤ v  =⇒ (u , v  ) ∈ R , u ∈ M  }  = {(0, u ] | ∃ v  ∈ I , v ≤ v  ∧ ( v  ≤ v  =⇒ (u , v  ) ∈ (0, 1] × {0}, u ∈ M  }

Q v∇ =

= (0, 1]. Therefore, we also have

P u  Q u  =⇒

γ R (u ) = γ R (u  ), and

P v  Q v  =⇒

μ R ( v ) = μ R ( v  ).

Theorem 4.4. Let ( L , ≤) be a complete lattice, (U , U ) and ( V , V ) be any two textures. Let implication

∀u , u  ∈ U , P u ⊆ Q u  =⇒ γ (u ) = γ (u  ). Then the pair (r , R ) is a direlation from (U , U ) to ( V , V ) where

r=

  { P (u , v ) | γ (u ) ≤ μ( v )} and R = { Q (u , v ) | u ∈ U  , μ( v ) ≤ γ (u )}.

γ : U → L be a function satisfying the

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Proof. (R1) Let r  Q (u , v ) and P u   Q u . By definition of r, for some v 1 ∈ V we get P (u , v 1 )  Q (u , v ) such that γ (u ) ≤ μ( v 1 ). Clearly, we have P v 1  Q v and so μ( v 1 ) ≤ μ( v ). Since γ satisfies the implication in Corollary 4.2 (i) and P u  ⊆ Q u , we have γ (u  ) = γ (u ). Hence, γ (u  ) ≤ μ( v 1 ) and so P (u , v 1 ) ⊆ r. Since P v 1  Q v , we conclude that r  Q (u , v ) . (R2) Let r  Q (u , v ) . By definition of r, there exists v 1 ∈ V such that P (u , v 1 )  Q (u , v ) and γ (u ) ≤ μ( v 1 ). Let us choose u  ∈ U such that P u  Q u  . Therefore γ (u  ) ≤ γ (u ) and so γ (u  ) ≤ μ( v 1 ). Hence, we get P (u  , v 1 ) ⊆ r. Since P v 1  Q v , we conclude that r  Q (u  , v ) . (CR1) Let P (u , v )  R and P u  Q u  . By definition of R, for some v 1 ∈ V we get P (u , v 1 )  Q (u , v ) such that μ( v 1 ) ≤ γ (u ). Clearly, we have P v 1  Q v . Since P u   Q u , we have γ (u ) ≤ γ (u  ). Hence μ( v 1 ) ≤ γ (u  ) and so R ⊆ Q (u  , v 1 ) . Since P v 1  Q v , we conclude that P (u  , v )  R. (CR2) Let P (u , v )  R. By definition of R, there exists u 1 ∈ U  and v 1 ∈ V such that P (u , v )  Q (u 1 , v 1 ) and μ( v 1 ) ≤ γ (u 1 ). Clearly, we have u = u 1 and P v ⊆ Q v 1 . Since u ∈ U  , we may choose u  ∈ U such that P u  ⊆ Q u . Take u  ∈ U such that P u  ⊆ Q u  and P u  ⊆ Q u . Then γ (u ) ≤ γ (u  ) and so μ( v 1 ) ≤ γ (u  ). Further, we have u  ∈ U  . Hence, we get R ⊆ Q (u  , v 1 ) . Since P v  Q v 1 , we conclude that P (u  , v )  R. 2 Let L be a lattice. Recall that a subset A of L is called join dense if every element of L can be written as a supremum of a subset of A. Dually, a subset B of L is called meet dense if every element of L can be written as an infimum of a subset of B. Now we may give the main theorem for textures as we have promised: Theorem 4.5. (i) Let L be a complete lattice. If L is isomorphic to Br (U, V), then there exist the mappings γ : U → L and μ : V → L such that γ (U  ) is join dense and μ( V  ) is meet dense in L. (ii) Let ( L , ≤) be a complete lattice, (U , U ) and ( V , V ) be any two textures. Suppose that there exist the mappings γ : U → L and μ : V → L satisfying the conditions (a) ∀u , u  ∈ U , P u ⊆ Q u  =⇒ γ (u ) = γ (u  ), and (b) ∀ v , v  ∈ V , P v ⊆ Q v  =⇒ μ( v  ) = μ( v ) where γ (U  ) is join dense and μ( V  ) is meet dense in L. If r is the relation given in Theorem 4.4, then Br (U, V) and ( L , ≤) are isomorphic. Moreover, for all u ∈ U and v ∈ V  the relation r satisfies the following equivalence:

γ (u ) ≤ μ( v ) ⇐⇒ r ⊆ Q (u, v ) . Proof. (i) Let

ϕ : Br (U, V) → L be an isomorphism. Take the mapping γr : U → L defined by

∀u ∈ U , γr (u ) = ϕ ( P u , P u ). Let l ∈ L. Since ϕ is an isomorphism, there is a t-formal concept ( A , B ) in Br (U, V) where ies 3.12 (i) and 3.13 (i), we get

ϕ( A, B ) = ϕ(



P u ,

A ⊆ Q u



P u ) =

A ⊆ Q u



ϕ ( A , B ) = l. Then by Corollar-

ϕ ( P u , P u ).

A ⊆ Q u

However, since {ϕ ( P u , P u ) | A ⊆ Q u } ⊆ γr (U  ), we conclude that μr : V → L defined by

γr (U  ) is join dense. Now let us consider the mapping

 ∀ v ∈ V , μr ( v ) = ϕ ( P  v , P v ).

If l ∈ L, then there is a t-formal concept ( A , B ) ∈ Br (U, V) such that also get

ϕ( A, B ) = ϕ(



B ⊆ Q v

P v ,



B ⊆ Q v

P  v )=



ϕ ( A , B ) = l. Then by Corollaries 3.12 (i) and 3.13 (i), we

 ϕ( P  v , P v ).

B ⊆ Q v

Since     {ϕ ( P  v , P v ) | B ⊆ Q v } ⊆ {ϕ ( P v , P v ) | V ⊆ Q v } ⊆ μr ( V ),

we obtain that μr ( V  ) is meet dense. (ii) Let ( L , ≤) be a complete lattice, and let γ : U → L and μ : V → L be the mappings satisfying the properties given in the theorem. For all ( A , B ) ∈ Br (U, V), let us define the mapping

ϕ : Br (U, V) → L

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ϕ (( A , B )) =





{γ (u ) | u ∈ A  }. Let ( A 1 , B 1 ), ( A 2 , B 2 ) ∈ Br (U, V) and ( A 1 , B 1 ) r ( A 2 , B 2 ). Then A 1 ⊆ A 2 and hence, A 1 ⊆



A 2 . Therefore,

ϕ (( A 1 , B 1 )) =





{γ (u ) | u ∈ A 1 } ≤





{γ (u ) | u ∈ A 2 } = ϕ (( A 2 , B 2 )).

That is, ϕ is order preserving. Let us prove that Br (U, V) by ψ(l) = ( Al , B l ) where

Al = Bl =

ϕ is an isomorphism. First, for all l ∈ L, we define the mapping ψ : L →



{ Q u | γ (u )  l},



{ Q v | l  μ( v )} 





for l ∈ L. We prove that ψ(l) is a formal concept of Br (U, V), that is, A l = B l and B l = Al . Let Al  B l . Let us choose v ∈ V 

such that A l  Q v and P v  B l . Then we have

∀u ∈ U , Al  Q u =⇒ r  Q (u , v ) .

(∗)

Al . Now take u 

Let u ∈ Al . Clearly, we have P u ⊆ ∈ U such that P u  Q u  . In this case, we also have Al  Q u  and by (∗), we get r  Q (u  , v ) . By definition of r, there exists v 1 ∈ V such that P (u  , v 1 )  Q (u  , v ) and γ (u  ) ≤ μ( v 1 ). Since P u  Q u  , γ (u ) = γ (u  ) and thus, γ (u ) ≤ μ( v 1 ). Finally, P v 1  Q v implies that μ( v 1 ) ≤ μ( v ). This gives γ (u ) ≤ μ( v ). 

Now let us show that {u ∈ U  | γ (u ) ≤ l} = Al . Let γ (u ) ≤ l for some u ∈ U  . Then there exists a point u 0 ∈ U such that P u 0  Q u . Suppose that P u 0  Al . Then we have u  ∈ U with P u 0  Q u  and γ (u  )  l. Hence, γ (u ) = γ (u 0 ) = γ (u  )  l. 



But this is a contradiction. Therefore, we obtain P u 0 ⊆ Al . Now clearly, A l  Q u , namely u ∈ A l . Conversely, if u ∈ A l , then

γ (u ) ≤ l. Clearly, we have u ∈ U  and then we obtain Al ⊆ {u ∈ U  | γ (u ) ≤ l}.  On the other hand, since γ (U  ) is join dense in L, for the set {u ∈ U  | γ (u ) ≤ l} = Al , we obtain    l= {γ (u ) | u ∈ U  , γ (u ) ≤ l} = {γ (u ) | u ∈ Al } ≤ μ( v ).

Al ⊆ Q u and hence,

That is, we get l ≤ μ( v ). Since P v  B l , we have a point v 0 ∈ V such that P v  Q v 0 and l  μ( v 0 ). Since P v  Q v 0 , we get

μ( v ) ≤ μ( v 0 ) and so l  μ( v ). But this is a contradiction and hence, Al ⊆ B l . Now, suppose that B l  Al . Let us choose 

v ∈ V such that B l  Q v and P v  Al . From the second statement, there exists a point v  ∈ V with P v  Q v  such that

Al  Q u , r ⊆ Q (u , v  ) for some u ∈ U . Then P (u , v )  r and γ (u ) ≤ l. By definition of r, γ (u )  μ( v ) and so l  μ( v ). However, B l ⊆ Q v is a contradiction.    Now we show that B l = Al . Suppose that B l  Al . Choose u ∈ U such that B l  Q u and P u  Al . By the first statement, we have

∀ v ∈ V , B l  Q v =⇒ r  Q (u , v ) .

(∗∗)

Let v ∈ B l . We clearly have P v ⊆ B l . Let us take a point v  ∈ V with P v  Q v  . Since B l  Q v  , by (∗∗), we have r  Q (u , v  ) . By definition of r, we have a point v  ∈ V such that P (u , v  )  Q (u , v  ) and γ (u ) ≤ μ( v  ). Now it is easy to see that μ( v  ) = μ( v ) = μ( v  ) and so we obtain γ (u ) ≤ μ( v ). We have proved that γ (u ) ≤ {μ( v ) | v ∈ B l }. 

Now we let us show that { v ∈ V  | l ≤ μ( v )} = B l . Let l ≤ μ( v ) for some v ∈ V  . Let us take a v 0 ∈ V with P v 0  Q v . Suppose that P v 0  B l . By definition of B l , there exists v  ∈ V such that P v 0  Q v  and l  μ( v  ). Since P v 0  Q v and P v 0  Q v  , we get μ( v 0 ) = μ( v ) = μ( v  ) and hence, we conclude that l  μ( v ). But this is a contradiction. Hence, P v 0 ⊆ B l , 



that is, B l ⊆ Q v and hence, v ∈ B l . If v ∈ B l , then B l ⊆ Q v and by definition of B l , we obtain l ≤ μ( v ). Since we also have v∈

V  , we obtain that v

γ (u ) ≤





∈ Bl ⊆ {v ∈ 

{μ( v ) | v ∈ B l } =

V |l



≤ μ( v )}. On the other hand, since μ( V  ) is meet dense in L,

{μ( v ) | v ∈ V  , l ≤ μ( v )} = l

γ (u ) ≤ l. Since P u  Al , there exists u 0 ∈ U such that P u  Q u0 and γ (u 0 )  l. Since P u  Q u0 , we have γ (u 0 ) = γ (u ) and so γ (u )  l. But this is a contradiction and this means that B l ⊆ Al . Now let Al  B l . Choose u ∈ U such  that Al  Q u and P u  B l . By the second statement, for some u  ∈ U and v ∈ V , we get P u  Q u  , B l  Q v and γ (u  ) ≤  μ( v ), that is, r ⊆ Q (u , v ) . Let choose a point v  ∈ V with B l  Q v  and P v   Q v . Since Al  Q u and B l  Q v  , γ (u  ) ≤ l and l ≤ μ( v  ) and this implies that γ (u  ) ≤ μ( v  ), that is, P (u  , v  ) ⊆ r. Since P v   Q v , we obtain r  Q (u  , v ) . But this is a and we get



contradiction. Now we have proved that B l ⊆ Al . Further, the map ψ is order preserving since l ≤ t implies that A l ⊆ A t . Moreover,

ϕ (ψ(l)) = ϕ (( Al , B l )) =





{γ (u ) | u ∈ Al } =



{γ ( u ) | u ∈ U  , γ ( u ) ≤ l } = l

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and

 ψ(ϕ ( A , B )) = ψ( {γ (u ) | u ∈ A  }) = ψ(t ) = ( A t , B t ). We show that ( A t , B t ) = ( A , B ). Suppose that A  A t . Choose a point A  Q u and P u  A t for some u ∈ U . A  Q u implies that u ∈ A  and hence, we have γ (u ) ≤ t. Since P u  A t , we have a point u  ∈ U with P u  Q u  and γ (u  )  t. Hence, γ (u )  t. This is a contradiction. Now, let us prove the inclusion A t ⊆ A  = choose u ∗ ∈ U such that A t  Q u ∗ A. Suppose the contrary. Then we may   ∗ and P u ∗  A . By definition of A t , γ (u ) ≤ u ∈ A  γ (u ). On the other hand, P u ∗  A implies that A   Q v and r ⊆  ∗  Q (u  , v ) for some v ∈ V and u ∈ U where P u ∗  Q u  . Take v ∈ V such that A  Q v ∗ and P v ∗  Q v . We show that



γ (u ) ≤ μ( v ∗ ). Let u ∈ A  . Then A  Q u , that is, P u ⊆ A and hence A  ⊆ P u . Therefore, P u  Q v ∗ . Then P v ∗ ⊆ P u    ∗ ∗ ∗ ∗ and so P u ⊆ P v ∗ . We have γ (u ) ≤ μ( v ∗ ), that is, u ∈ A  γ (u ) ≤ μ( v ). This gives that γ (u ) ≤ μ( v ). Since P u  Q u  ,  ∗  ∗ we get γ (u ) ≤ γ (u ) and so γ (u ) ≤ μ( v ). By definition of r, P (u  , v ∗ ) ⊆ r. Since P v ∗  Q v , we obtain r  Q (u  , v ) . But this u∈ A

is a contradiction. Now let us prove the last part of the theorem. For u ∈ U and v ∈ V  , let γ (u ) ≤ μ( v ). Take a v  ∈ V such that P v  ⊆ Q v . Then by the assumption, we get μ( v ) = μ( v  ) and hence, μ( v  ) ≤ γ (u ). Then by definition of r, we have P (u , v  ) ⊆ r. This gives that r ⊆ Q (u , v ) . Now assume that r ⊆ Q (u , v ) . By definition of r, there exist u  ∈ U and v  ∈ V where P (u  , v  ) ⊆ Q (u , v ) and γ (u  ) ≤ μ( v ). Now, we have u = u  and P v  ⊆ Q v . By the assumption, μ( v ) = μ( v  ), that is, γ (u ) ≤ μ( v ). 2 5. Main theorem of t-formal co-concept lattices Lemma 5.1. (i) (ii) (iii) (iv)

∀ v , v  ∈ V , P v  Q v  =⇒ μ R ( v ) ≤ μ R ( v  ). ∀u , u  ∈ U , P u  Q u  =⇒ γ R (u  ) ≤ γ R (u ). P u  Q v∇ ⇐⇒ (∃u  ∈ U ) ( P u  Q u  ∧ μ R ( v ) ≤ γ R (u  )). P v  Q u∇ ⇐⇒ (∃ v  ∈ V ) ( P v  Q v  ∧ μ R ( v  ) ≤ γ R (u )).

Corollary 5.2. Suppose that the mappings γ R and μ R given above satisfy the conditions

∀u , u  ∈ U , P u ⊆ Q u  =⇒ γ R (u ) ≤ γ R (u  ), and ∀ v , v  ∈ V , P v ⊆ Q v  =⇒ μ R ( v  ) ≤ μ R ( v ), respectively. Then we have the following statements: (i) (ii) (iii) (iv) (v)

∀ v , v  ∈ V , P v ⊆ Q v  =⇒ μ R ( v ) = μ R ( v  ). ∀u , u  ∈ U , P u ⊆ Q u  =⇒ γ R (u  ) = γ R (u ). ∀ v ∈ V , P v ⊆ Q u∇ ⇐⇒ μ R ( v ) ≤ γ R (u ). ∀u ∈ U , P u ⊆ Q v∇ ⇐⇒ μ R ( v ) ≤ γ R (u ). ∀u ∈ U  , ∀ v ∈ V , μ R ( v ) ≤ γ R (u ) =⇒ P (u , v ) ⊆ R.

Theorem 5.3. (i) Let L be a complete lattice. If L is isomorphic to B R (U, V), then there exist the mappings γ : U → L and μ : V → L such that γ (U ) is meet dense and μ( V ) is join dense in L. (ii) Let ( L , ≤) be a complete lattice, (U , U ) and ( V , V ) be any two textures. Suppose that there exist the mappings γ : U → L and μ : V → L satisfying the conditions

∀u , u  ∈ U , P u ⊆ Q u  =⇒ γ (u  ) = γ (u ) and ∀ v , v  ∈ V , P v ⊆ Q v  =⇒ μ( v ) = μ( v  ) where γ (U ) is meet dense and μ( V ) is join dense in L. If R is the corelation given in Theorem 4.4, then B R (U, V) and ( L , ≤) are isomorphic. Further, for all u ∈ U  and v ∈ V we have

μ( v ) ≤ γ (u ) ⇐⇒ P (u, v ) ⊆ R .

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6. Conclusion This paper presents a discussion on formal concepts in the framework of textures. Since textures has conspicuous connections with the theories of rough set and fuzzy set, an investigation of formal concepts in the framework of textures may provide a new link for mathematical foundations about the classifications of data. A t-formal concept is a generalization of a formal concept in the sense of Rudolf Wille. No doubt that the lack of an object or attribute of an information system may lead us to different observations with respect to knowledge embedded in data. In such a case, the change of the whole system may be easily evaluated in the context of textures. As it is seen in Example 3.6, the missing information can be relatively compared considering the respected objects and attributes. For this purpose, we have defined the textural counterparts of sufficiency and dual sufficiency operators for a (textural) relation and a corelation between any two textures, respectively. These operators also hold some desirable properties as Galois connectivity. A textural complementation is a mapping on a texture satisfying the involution and reversing the ordinary set inclusion. Although a textural complementation is weaker than the set-theoretical complementation, the textural data operators, t-sufficiency and t-co-sufficiency mappings are (complementary) dual. This shows that t-concept and t-co-concept lattices are dually isomorphic. The notions of a t-object concept and t-attribute concept can be defined in terms of p-set and q-set of textures. For a textural formal context (U, V, r , R ) and the t-concept lattice- Br (U, V), we have stated and proved the main theorem of formal concept analysis. Moreover, if the direlation (r , R ) is complemented, then we have shown that the lattices Br (U, V) and B R (U, V) are dually isomorphic. Therefore, the main theorem can be also stated and proved for the formal co-concept lattice B R (U, V). Since the dual proofs are not so trivial, for the sake of the reader, we have given them in the appendix. Note that Theorem 4.5 is a generalization of the main theorem of formal concept analysis. Indeed, if we get P (U ) = U and P ( V ) = V and if (r , R ) is a direlation from (U , P (U )) to ( V , P ( V )), then by Theorem 3.3, Br (U, V) turns into the concept lattice B (U , V ) of Wille in the formal context (U , V , r ). In this case, the conditions (a) ∀u , u  ∈ U , P u ⊆ Q u  =⇒ (b) ∀ v , v  ∈ V , P v ⊆  Q v  =⇒

γ (u ) = γ (u  ) , μ( v  ) = μ( v )

are nothing but the conditions (a ) ∀u , u  ∈ U , u = u  =⇒ γ (u ) = γ (u  ), (b ) ∀ v , v  ∈ V , v = v  =⇒ μ( v ) = μ( v  ) which are trivial implications since γ and μ are functions. Further, the equivalence γ (u ) ≤ μ( v ) ⇐⇒ r ⊆ Q (u , v ) will be γ (u ) ≤ μ( v ) ⇐⇒ (u , v ) ∈ r. Note that we also have U = U  and V = V  . Likewise, Theorem 5.3 also generalizes Proposition 3 given in [13]. A next step of this research will lead us to a new approach to formal fuzzy concept lattices. 7. Appendix: Proofs of the dual results Proof of Lemma 5.1. (i) Let P v ⊆ Q v  . Then Q v  ⊆ Q v and by Theorem 3.8 (i), Q v∇ ⊆ Q v∇ and so we get μ R ( v ) ≤ μ R ( v  ). ⊆ Q u∇∇ , that is, γ R (u  ) ≤ γ R (u ). (ii) Let P u ⊆ Q u  . Then Q u  ⊆ Q u and by Theorem 3.9 (i) we have Q u∇∇    (iii) Let P u ⊆ Q u  and μ R ( v ) ≤ γ R (u ) for some u ∈ U . Then we have Q v∇ ⊆ Q u∇∇ and hence we get Q v∇ ⊆ Q u  and so  ∇ ∇  ∇ P u  Q v . Conversely, suppose that P u ⊆ Q v . Take u ∈ U such that P u  ⊆ Q v and P u ⊆ Q u  . In this case, we get Q v∇ ⊆ Q u  ,  that is, Q v∇ ⊆ Q u∇∇  . Then we have μ R ( v ) ≤ γ R (u ).  (iv) Let P v ⊆ Q v  and μ R ( v ) ≤ γ R (u ) for some v  ∈ V . Then by definition of γ and μ, we have Q v∇ ⊆ Q u∇∇ and hence, ∇∇ ⊆ Q  , we get Q ∇ ⊆ Q  . Hence, we obtain P  Q ∇ . Now Suppose that P ⊆ Q ∇ . Take ∇ Q u ⊆ Q v∇∇  . Further, since Q v  v v v v u u u  v ∈ V such that P v ⊆ Q v  and P v  ⊆ Q u∇ . In this case, Q u∇ ⊆ Q v  , that is, Q v∇  Q u∇∇ . Hence, we get μ R ( v  ) ≤ γ R (u ). Proof of Corollary 5.2. The proofs of (i) and (ii) are immediate. (iii) Let μ R ( v ) ≤ γ R (u ). Take v  ∈ V such that P v ⊆ Q v  . By (i), μ R ( v ) = μ R ( v  ) and so we get μ R ( v  ) ≤ γ R (u ). Now by Lemma 5.1 (iv), we obtain P v ⊆ Q u∇ . Conversely, let P v ⊆ Q u∇ . Then we have Q u∇ ⊆ Q v . Hence, Q v∇ ⊆ Q u∇∇ and so we obtain μ R ( v ) ≤ γ R (u ). (iv) Let μ R ( v ) ≤ γ R (u ). Take u  ∈ U such that P u ⊆ Q u  . By (i), γ R (u ) = γ R (u  ) and so we get μ R ( v ) ≤ γ R (u  ). Now by Lemma 5.1 (iii), we obtain P u ⊆ Q v∇ . Conversely, let P u ⊆ Q v∇ . Then we have Q v∇ ⊆ Q u . Hence, Q v∇ ⊆ Q u∇∇ and so we obtain μ R ( v ) ≤ γ R (u ). (v) Let u ∈ U  , v ∈ V and μ R ( v ) ≤ γ R (u ). Let us choose u  ∈ U and v  ∈ V such that P u  Q u  and P v  Q v  . Then by (i) and (ii), μ R ( v  ) ≤ γ R (u  ). By definition of mappings of γ R and μ R , Q v∇ ⊆ Q u∇∇ and so we have Q v∇ ⊆ Q u  . Then we obtain  P u  Q v∇ . Therefore,

∀ v ∗ ∈ V , P v ∗  Q v  =⇒ P (u , v ∗ )  R . Since P v  Q v  , we have P (u , v )  R, as required.

M. Diker et al. / International Journal of Approximate Reasoning 114 (2019) 182–203

Proof of Theorem 5.3. (i) Let defined by

201

ϕ be an isomorphism between B R (U, V) and ( L , ≤). Let us consider the mapping γ R : U → L

γ R (u ) = ϕ ( Q u∇∇ , Q u∇ )

∀u ∈ U ,

Take l ∈ L. Since ϕ is an isomorphism, there is a t-formal co–concept ( A , B ) in B R (U, V) where Corollary 3.13 we get



ϕ( A, B ) = ϕ(

PuA

∇∇

Q u)

PuA



= ϕ(

Q u )∇ )

PuA



= ϕ ((



Q u, (

ϕ ( A , B ) = l. Then, by

,



Q u∇ )

PuA

( Q u∇∇ , Q u∇ ))

PuA



=

ϕ ( Q u∇∇ , Q u∇ ) = l.

PuA

However, since {ϕ ( Q u∇∇ , Q u∇ ) | P u  A } ⊆ γ R (U ), we conclude that Similarly, if we consider the mapping μ R : V → L defined by

γ R (U ) is meet dense.

μ R ( v ) = ϕ ( Q v∇ , Q v∇∇ )

∀v ∈ V ,

we may show that μ R ( V ) is join dense in ( L , ≤). (ii) Let ( L , ≤) be a complete lattice and γ R : U → L and define

μ R : V → L satisfying the properties stated above. Then we

ϕ : B R (U, V) → L by ϕ (( A , B )) = fore,



{μ R ( v ) | v ∈ / B }. Let ( A 1 , B 1 ), ( A 2 , B 2 ) ∈ B R (U, V) and ( A 1 , B 1 ) ≤ ( A 2 , B 2 ). Then A 1 ⊆ A 2 ( B 2 ⊆ B 1 ). There-

ϕ (( A 1 , B 1 )) = Hence,



{μ R ( v ) | v ∈ / B1} ≤



{μ R ( v ) | v ∈ / B 2 } = ϕ (( A 2 , B 2 )).

ϕ is order preserving. Let us prove that ϕ is an isomorphism. First, we define

ψ(l) = ( Al , B l ) where

Al = Bl =

 

{ P u | l  γ R (u )},

{ P v | μ R ( v )  l}

for l ∈ L and we show that ψ(l) is a formal co-concept of B R (U, V). Let us prove that A l∇ = B l and B l∇ = Al . Let Al∇  B l .

Choose v ∈ V such that A l∇  Q v and P v  B l . There exists a v  ∈ V with P v   Q v such that P u  Al and P (u , v  ) ⊆ R for some u ∈ U . Since P u  Al , we have l ≤ γ R (u ). Further, since P v   Q v , we get R  Q (u , v ) . Then μ R ( v )  γ R (u ) and so μ R ( v )  l. Since P v  B l , we obtain μ R ( v ) ≤ l. But this is a contradiction. Now let B l  Al . Take a point v ∈ V such that B l  Q v and P v  Al . Choose a v  ∈ V with P v  Q v  and P v   Al . From the second statement,

∀u ∈ U  , P u  Al =⇒ P (u , v  )  R

(∗)

Let P u  Al . Then, by (∗), P (u , v  )  R. Choose a point v 1 ∈ V with P (u , v  )  Q (u , v 1 ) such that μ R ( v 1 ) ≤ γ R (u ). Since P v  Q v  and P v   Q v 1 , we have P v  Q v 1 and so we get μ R ( v ) ≤ μ R ( v 1 ). Then μ R ( v ) ≤ γ R (u ). Now let us show that {u ∈ U  | l ≤ γ R (u )} = {u | u ∈ / Al }. Let l ≤ γ R (u ) for some u ∈ U  . Then there exists a point u  ∈ U such that P u  Q u  . We want to show that A l ⊆ Q u  . Suppose that A l  Q u  . Then we have a point u  ∈ U with P u   Q u  and l  γ R (u  ). Hence, γ R (u ) = γ R (u  ) = γ R (u  ), and we get l  γ R (u ). But this is a contradiction. Therefore, we obtain Al ⊆ Q u  and so P u  Al , namely u ∈ / Al . Conversely, if u ∈ / Al , then P u  Al and so by definition of Al , we obtain l ≤ γ R (u ). Since γ R (U ) is meet dense in L, for the set {u ∈ U | l ≤ γ R (u )} = {u | u ∈ / Al }, we obtain

μR (v ) ≤



{u | u ∈ / Al } =



{γ R (u ) | l ≤ γ R (u )} = l.

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Then we get μ R ( v ) ≤ l. Since B l  Q v , we have a point v 0 ∈ V such that P v 0  Q v and μ R ( v 0 )  l. But P v 0  Q v and so μ R ( v 0 ) ≤ μ R ( v ). Hence, μ R ( v )  l, which is a contradiction. Now we show that B l = Al . Suppose that B l  Al . Choose u ∈ U such that B l  Q u and P u  Al . By the first statement,

we have a point u  ∈ U  with P u   Q u such that P v  B l and P (u  , v ) ⊆ R for some v ∈ V . Since P u   Al , we have l ≤ γ R (u  ). Take a point v  ∈ V such that P v  Q v  and P v   B l . Since P v   B l , we get μ R ( v  ) ≤ l and so μ R ( v  ) ≤ γ R (u  ). By definition of R, we get R ⊆ Q (u  , v  ) . Since P v  Q v  , we have P (u  , v )  R. But this is a contradiction. Now let A l  B l . Choose u ∈ U such that A l  Q u and P u  B l . Take a point u  ∈ U  such that P u  Q u  , P u   B l . Then

∀ v ∈ V , P v  B l =⇒ P (u  , v )  R

(∗∗)

Let P v  B l . Then, by (∗∗), P (u  , v )  R. Choose a point v 1 ∈ V with P (u  , v )  Q (u  , v 1 ) such that μ R ( v 1 ) ≤ γ R (u  ). Clearly, we have P v  Q v 1 . Since P u  Q u  , we get μ R ( v ) ≤ γ R (u ). Since P v  B l , we have μ R ( v ) ≤ l. Now let us show that { v ∈ V | μ R ( v ) ≤ l} = { v | v ∈ / B l }. Let μ R ( v ) ≤ l for some v ∈ V . Then there exists a point v  ∈ V such that P v  Q v  . We want to show that B l ⊆ Q v  . Suppose that B l  Q v  . Then we have a point v  ∈ V such that P v   Q v  and μ R ( v  )  l. Since P v  Q v  and P v   Q v  , we have μ R ( v ) = μ R ( v  ) and so μ R ( v )  l, which is a contradiction. Then we get B l ⊆ Q v  and so P v  B l . If v ∈ / B l , then P v  B l and so μ R ( v ) ≤ l. / B l }, Since μ R ( V ) is join dense in L, for the set { v ∈ V | μ R ( v ) ≤ l} = { v | v ∈



{v | v ∈ / Bl } =



{μ R ( v ) | μ R ( v ) ≤ l} = l ≤ γ R (u ).

That is l ≤ γ R (u ). Since A l  Q u , we have a u 0 ∈ U with P u 0  Q u and l ≤ γ R (u 0 ). But P u 0  Q u implies that l  γ R (u ), which is a contradiction. The map ψ is order preserving since l ≤ t implies A l ⊆ A t . Moreover,

ϕ (ψ(l)) = ϕ (( Al , B l )) =



{μ R ( v ) | v ∈ / Bl } =



{μ R ( v ) | μ R ( v ) ≤ l } = l

and

 ψ(ϕ ( A , B )) = ψ( {μ R ( v ) | v ∈ / B }) = ψ(t ) = ( A t , B t ). We show that ( A t , B t ) = ( A , B ). Suppose that B t  B. Choose a v ∈ V such that B t  Q v and P v  B. Clearly, P v  B implies that μ R ( v ) ≤ t. Since B t  Q v , there exists a v  ∈ V with P v   Q v and μ R ( v  )  t. Hence, μ R ( v )  t. This is a contradiction. Now, let us prove the inclusion B  = B ⊆ B t . Suppose the contrary. Then we may choose v ∗ ∈ V such that B   Q v ∗  and P v ∗  B t . By definition of B t , μ R ( v ∗ ) ≤ v ∈/ B μ R ( v ). On the other hand, B   Q v ∗ implies that P (u , v 0 ) ⊆ R for  some u ∈ U and v 0 ∈ V where P v 0  Q v ∗ and P u  B  . Take u ∗ ∈ U such that P u  Q u ∗ and P u ∗  B  . We show that  ∗     ∗ ∗ v∈ / B μ R ( v ) ≤ γ R (u ). Let P v  B. Then B ⊆ Q v and  hence Q v ⊆ B . Therefore, P u  Q v , that is, Q v ⊆ Q u and so Q u∗ ⊆ Q v . Thus we obtain μ R ( v ) ≤ γ R (u ∗ ), that is, v ∈/ B μ R ( v ) ≤ γ R (u ∗ ). This gives that μ R ( v ∗ ) ≤ γ R (u ∗ ). Since P u  Q u ∗ , we get γ R (u ) ≤ γ R (u ∗ ) and so μ R ( v ∗ ) ≤ γ R (u ). By definition of R, R ⊆ Q (u , v ∗ ) . Since P v 0  Q v ∗ , we obtain P (u , v 0 )  R. However, this is a contradiction. For the last part of the theorem, let u ∈ U  , v ∈ V and μ( v ) ≤ γ (u ). Then we have a point v  ∈ V such that P v  Q v  . By (ii), we have μ( v  ) ≤ γ (u ). By definition of R, we get R ⊆ P (u , v  ) . Since P v  Q v  , we obtain P (u , v )  R, as desired. Conversely, let P (u , v )  R. Then there exists a point v  ∈ V with P (u , v )  Q (u , v  ) such that μ( v  ) ≤ γ (u ). Hence, P v  Q v  . Since μ( v  ) = μ( v ), we get μ( v ) ≤ γ (u ). Declaration of competing interest There is no conflict of interest. Acknowledgements The authors are grateful to the anonymous referees, whose comments have contributed to the clarity of the proofs and discussions in this paper. References [1] [2] [3] [4] [5] [6]

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