Two new kinds of protoconcepts based on three-way decisions model

Abstract

With the development of formal concept analysis, classical concept lattice cannot solve some problems in practice because of its strict conditions. Protoconcept, which is an extension of the formal concept, provides a new method of data processing. Three-way decisions is more in line with human cognition. Not like the decision of two ways, three-way cognition expands its scope of application. However, up to now, protoconcept is influenced by two-way decisions which leads to the restrictions of its applications. Hence, it is urgent to find a method to generalize protoconcept from the framework of two-way decisions. To give the solution to this urgent problem, the main contributions of this paper are the following: First, using the combination of protoconcept and three-way decisions, two models of three-way protoconcept are obtained. The algorithms for searching three-way protoconcept are built. Second, we investigate the constructions of three-way protoconcepts in double Boolean algebra. Third, the relationships between protoconcepts and three-way protoconcepts are found. Moreover, using examples that come from the real world, we explain the main results obtained above and show some applications. The two models for protoconcepts provided in this paper generalize the fields of thought from two-way decisions to three-way decisions. This follows that they are more in line with human thinking. Hence, the two models will certainly generalize the applications of protoconcept theory in the future.

Appendix

1.Proof of Property 1

According to the definition of operators $\sqcap ,\sqcup ,\wedge ,\vee$, we have

$c\sqcap d=(E_{1} \cap E_{2},(E_{1} \cap E_{2})^{\lessdot })$,

$c\sqcup d=(((M_{1},N_1) \cap (M_{2},N_2))^{ > rdot }, (M_{1},N_1) \cap (M_{2},N_2))$,

$c\wedge d=(((M_{1},N_1)\cup (M_{2},N_2))^{ > rdot },(M_{1},N_1)\cup (M_{2},N_2))$,

$c\vee d=(E_{1} \cup E_{2},(E_{1} \cup E_{2})^{\lessdot })$.

$\bullet$ To prove $c\sqcap d \le c,d \le c \sqcup d$.

• We know $E_1\cap E_2\subseteq E_1 \ \text {and} \ E_1 \cap E_2\subseteq E_2$ ${\mathop {\Longrightarrow }\limits ^{(C6),(C1)}}$ $(E_1\cap E_2)^{\lessdot } \supseteq E_1^{\lessdot }\cup E_2^{\lessdot }=(M_1,N_1)^{ > rdot \lessdot } \cup (M_2,N_2)^{ > rdot \lessdot } \supseteq (M_1,N_1) \cup (M_2,N_2) \supseteq (M_1,N_1)$ and $(E_1\cap E_2)^{\lessdot }\supseteq (M_2,N_2)$. Thus, $c\sqcap d \le c,d \le c \sqcup d$.

$\bullet$ To prove $c\sqcap d \le c \wedge d.$

• By (C6) and (C1), we know $(E_{1} \cap E_{2})^{\lessdot }\supseteq E_{1}^{\lessdot }\cup E_{2}^{\lessdot }=(M_1,N_1)^{ > rdot \lessdot } \cup (M_2,N_2)^{ > rdot \lessdot } \supseteq (M_1,N_1)\cup (M_2,N_2)$.

• According to (C5 and (C1), it holds $((M_{1},N_1) \cup (M_{2},N_2))^{ > rdot }= (M_{1},N_1)^{ > rdot } \cap (M_{2},N_2)^{ > rdot }=E_1^{\lessdot > rdot } \cap E_2^{\lessdot > rdot } \supseteq E_1 \cap E_2$ ${\mathop {\Longrightarrow }\limits ^{ \text {Lem}5}}$ $(E_{1} \cap E_{2},(E_{1} \cap E_{2})^{\lessdot })\le (((M_{1},N_1)\cup (M_{2},N_2))^ { > rdot },(M_{1},N_1)\cup (M_{2},N_2))$ $\Longrightarrow$ $c\sqcap d \le c \wedge d$.

$\bullet$ To prove $c\vee d \le c \sqcup d.$

• According to (C5) and (C1), we obtain $(E_{1} \cup E_{2})^{\lessdot }=E_1^{\lessdot }\cap E_2^{\lessdot }=(M_1,N_1)^{ > rdot \lessdot }\cap (M_2,N_2)^{ > rdot \lessdot }\supseteq (M_1,N_1)\cap (M_2,N_2)$.

• Then, by (C6) and (C1), we have $((M_{1},N_1) \cap (M_{2},N_2))^{ > rdot }\supseteq E_1\cup E_2$ ${\mathop {\Longrightarrow }\limits ^{\text {Lem }5}}$ $(E_{1} \cup E_{2},(E_{1} \cup E_{2})^{\lessdot }) \le (((M_{1},N_1) \cap (M_{2},N_2))^{ > rdot }, (M_{1},N_1) \cap (M_{2},N_2))$ $\Longrightarrow$ $c \vee d \le c\sqcup d$. $\square$

2.Proof of Theorem 2

According to Definition 4, we should verify $({\mathfrak {B}} (T),\sqcap ,\sqcup ,\lnot ,\lrcorner ,\bot ,\top )$ satisfying properties (1a), (1b), ..., (11a), (11b), and (12). We can easily prove these properties according to Definition 9, except (5a) and (12). Hence, the proofs of other properties are omitted. Suppose c, d, z are $(E_{1}, (M_{1},N_1))$, $(E_{2}, (M_2,N_2))$, and $(E_{3}, (M_{3},N_3))$, respectively.

$\bullet$ To prove (5a).

• We can acquire $c \sqcap c=(E_1,E_1^{\lessdot })$ and $c \sqcap (c \sqcup d)=(E_1\cap ((M_1,N_1)\cap (M_2,N_2))^{ > rdot },(E_1\cap (M_1,N_1)\cap (M_2,N_2))^{ > rdot })^{\lessdot })$.

• If $(E_1,(M_1,N_1))$ is OE-protoconcept ${\mathop {\Longrightarrow }\limits ^{\text {Def } 8}}$ $(M_1,N_1)^{ > rdot }=E_1^{\lessdot > rdot }$ ${\mathop {\Longrightarrow }\limits ^{(C1)}}$ $((M_1,N_1)\cap (M_2,N_2))^{ > rdot }\supseteq (M_1,N_1)^{ > rdot }=E_1^{\lessdot > rdot }\supseteq E_1$ $\Longrightarrow$ $c \sqcap (c \sqcup d)=(E_1,E_1^{\lessdot })=c \sqcap c$.

$\bullet$ To prove (12).

• $(E_1,(M_1,N_1))$ is OE-protoconcept ${\mathop {\Longrightarrow }\limits ^{\text {Prop }1}}$ $E_1^{\lessdot > rdot }=(M_1,N_1)^{ > rdot }$ and $E_1^{\lessdot }= (M_1,N_1)^{ > rdot \lessdot }$.

• Due to $(c \sqcap c) \sqcup (c \sqcap c){\mathop {=}\limits ^{\text {Def }9}} (E_1^{\lessdot > rdot },E_1^{\lessdot })$ and $(c \sqcup c) \sqcap (c \sqcup c) {\mathop {=}\limits ^{\text {Def }9}} ((M_1,N_1)^{ > rdot },(M_1,N_1)^{ > rdot \lessdot })$ ${\mathop {\Longrightarrow }\limits ^{\text {Prop }1}}$ $(c \sqcap c) \sqcup (c \sqcap c)=(c \sqcup c) \sqcap (c \sqcup c)$. $\square$

3.Proof of Theorem 5

Let (E, M) be a protoconcept. Then $(E,E^{\lessdot })$ is an OE-protoconcept by Definition 8.

$\bullet$ To prove $\varphi$ is $\sqcap$-preserving mapping.

• Let $(E_1,M_1)$ and $(E_2,M_2)$ be any two protoconcepts. Then, by Definition 9, we can get that $\varphi (E_1,M_1)\sqcap \varphi (E_2,M_2)=(E_1,E_1^{\lessdot })\sqcap (E_2,E_2^{\lessdot })=(E_1\cap E_2,(E_1\cap E_2)^{\lessdot })$ and $\varphi ((E_1,M_1) \sqcap (E_2,M_2)) =\varphi ((E_1\cap E_2,(E_1 \cap E_2)^{\lessdot }))=(E_1\cap E_2,(E_1\cap E_2)^{\lessdot })$. Thus, we confirm that $\varphi (E_1,M_1)\sqcap \varphi (E_2,M_2)$ = $\varphi ((E_1,M_1) \sqcap (E_2,M_2))$. Then, we obtain $\varphi$ is a $\sqcap$-preserving mapping.

$\bullet$ To prove $\varphi$ is $\vee$-preserving mapping.

• Let $(E_1,M_1)$ and $(E_2,M_2)$ be any two protoconcepts. Then, using Definition 9, we can know that $\varphi (E_1,M_1)\vee \varphi (E_2,M_2)= (E_1,E_1^{\lessdot }) \vee (E_2,E_2^{\lessdot })=(E_1\cup E_2,(E_1\cup E_2)^{\lessdot })$ and $\varphi ((E_1,M_1) \vee (E_2,M_2)) =\varphi ((E_1\cup E_2,(E_1\cup E_2)^{\lessdot }))=(E_1\cup E_2,(E_1\cup E_2)^{\lessdot })$. Thus, $\varphi (E_1,M_1)\vee \varphi (E_2,M_2)$ = $\varphi ((E_1,M_1) \vee (E_2,M_2))$ holds. we get $\varphi$ is a $\vee$-preserving mapping.

$\bullet$ To prove $\varphi$ is order-preserving mapping.

• $(E_1,M_1)\le (E_2,M_2)$ ${\mathop {\Longrightarrow }\limits ^{\text {Lem }3}}$ $E_1 \subseteq E_2$ ${\mathop {\Longrightarrow }\limits ^{\text {(C2)}}}$ $E_1^{\lessdot } \supseteq E_2^{\lessdot }$ ${\mathop {\Longrightarrow }\limits ^{\text {Lem }3}}$ $(E_1,E_1^{\lessdot })\le (E_2,E_2^{\lessdot })$ ${\mathop {\Longrightarrow }\limits ^{\text {Theo }5}}$ $\varphi (E_1,M_1)\le \varphi (E_2,M_2)$ ${\mathop {\Longrightarrow }\limits ^{\text {Def }11}}$ $\varphi$ is an order-preserving mapping.

$\square$