# A Fast Algorithm for Three-Way Object-Oriented Concept Acquisition
## 2 Preliminaries
By combining the operators \(\square\), \(\lozenge\), \(\square\), and \(\bar{\lozenge}\), a pair of object-induced three-way object-oriented operators (OEO operators), \"\(\triangleright\):\" \(\mathcal{P}(U)\rightarrow\mathcal{DP}(V)\) and \"\(\lhd\):\" \(\mathcal{DP}(V)\rightarrow\mathcal{P}(U)\), can be defined as follows [22; 30]:  
\[X^{\square} =\{y\in V|\forall x\in U(xRy\Rightarrow x\in X)\}\]2 \[X^{\square} =\{y\in V|\forall x\in U(xRy\Rightarrow x\in X)\}\]3  
where \(X\subseteq U\) and \(Y,Z\subseteq V\). If and only if \(X=(Y,Z)^{\triangle}\) and \(X^{\rhd}=(Y,Z)\), the pair \((X,(Y,Z))\) is called an object-induced three-way object-oriented concept, i.e., an OEO-concept [22; 30]. \(X\) and \((Y,Z)\) are called the extent and the intent of the OEO-concept \((X,(Y,Z))\), respectively. The set of all OEO-concepts of \((U,V,R)\) is denoted by \(OEOL(U,V,R)\). For any \((X,(Y,Z))\) and \((A,(B,C))\in OEOL(U,V,R)\),  
\[X^{\square} =\{y\in V|\forall x\in U(xRy\Rightarrow x\in X)\}\]4  
The relation \(\leq\) defined above is a partial-order relation on \(OEOL(U,V,R)\)[22; 30]. For any \((X_{1},(Y_{1},Z_{1})),(X_{2},(Y_{2},Z_{2}))\in OEOL(U,V,R)\),  
\[X^{\square} =\{y\in V|\forall x\in U(xRy\Rightarrow x\in X)\}\]5 (15) \[X^{\square} =\{y\in V|\forall x\in U(xRy\Rightarrow x\in X)\}\]6  
\[X^{\square} =\{y\in V|\forall x\in U(xRy\Rightarrow x\in X)\}\]7  
\[X^{\square} =\{y\in V|\forall x\in U(xRy\Rightarrow x\in X)\}\]8  
_where \((X,a_{i})\in C_{s}\cup C_{s}^{c}\)._  
In order to prove that \(\lambda((X,a_{i}))\) is an OEO-concept, Propositions 1 and 2 are presented separately. **Proposition 1**. : _For a formal context \((U,V,R)\) and its complement context \((U,V,R^{c})\), if \(X\subseteq U\), then \((X^{\square\Diamond}\cup X^{\overline{\Diamond}},(X^{\square},X^{\overline{ \Box}}))\) is an OEO-concept._  
Proof. : According to the definition of the OEO-concept [22; 30], we need to only prove that \(\forall X\subseteq U\), \((X^{\square\Diamond}\cup X^{\overline{\Diamond}})^{\rhd}=(X^{\square},X^{ \overline{\Box}})\). First, we can obtain \(X^{\square\Diamond}\subseteq(X^{\square\Diamond}\cup X^{\overline{\Diamond}})\). It follows from the property (Equation (7)) that \(X^{\square\Diamond}\subseteq(X^{\square\Diamond}\cup X^{\overline{\Diamond}})\square\). Further, we can deduce from the property in Equation (10) that \(X^{\square}\subseteq(X^{\square\Diamond}\cup X^{\overline{\Diamond}})\square\). Second, we know from the property in Equation (9) that \(X^{\square\Diamond}\subseteq X\) and \(X^{\overline{\Diamond}\Diamond}\subseteq X\). Consequently, we obtain \((X^{\square\Diamond}\cup X^{\overline{\Diamond}\overline{\Diamond}})\subseteq X\cup X\). This implies \((X^{\square\Diamond}\cup X^{\overline{\Diamond}\overline{\Diamond}})\subseteq X\). Thus, we can conclude that \((X^{\square\Diamond}\cup X^{\overline{\Diamond}\overline{\Diamond}})^{ \square}=X^{\square}\). Similarly, \((X^{\square\Diamond}\cup X^{\overline{\Diamond}\overline{\Diamond}})^{ \square}=X^{\overline{\Box}}\). Hence, based on the definition of the OEO concept, \((X^{\square\Diamond}\cup X^{\overline{\Diamond}\overline{\Diamond}},(X^{ \overline{\Box}},X^{\overline{\Box}}))\) is an OEO-concept. For any given object set \(X\in U\), an OEO-concept can be obtained based on Proposition 1. It can subsequently be used in the proof of Proposition 2. **Proposition 2**. : _For a formal context \((U,V,R)\) and its complement context \((U,V,R^{c})\), any \((X,a_{i})\in C_{s}\cup C_{s}^{c}\), \(\lambda((X,a_{i}))\) is an OEO-concept._  
Proof. : Based on the function \(\lambda\) (Definition 3), let \(\lambda((X,a_{i}))=(X,(Y,Z))\). We need to prove that \((X,(Y,Z))\) satisfies such conditions that \(Y=X^{\square},Z=X^{\overline{\Box}}\) and \(X=Y^{\square}\cup Z^{\overline{\Diamond}}\). According to Proposition 1, it is an OEO-concept. (1) First, we prove that \(Y=X^{\square}\). According to Definition 3, we can obtain \(Y=\cup a_{j}\), where \(\forall a_{j}\in V,g(a_{j})\subseteq X\).