Kripke Contexts, Double Boolean Algebras with Operators and Corresponding Modal Systems

Abstract

The notion of a context in formal concept analysis and that of an approximation space in rough set theory are unified in this study to define a Kripke context. For any context (G,M,I), a relation on the set G of objects and a relation on the set M of properties are included, giving a structure of the form ((G,R), (M,S), I). A Kripke context gives rise to complex algebras based on the collections of protoconcepts and semiconcepts of the underlying context. On abstraction, double Boolean algebras (dBas) with operators and topological dBas are defined. Representation results for these algebras are established in terms of the complex algebras of an appropriate Kripke context. As a natural next step, logics corresponding to classes of these algebras are formulated. A sequent calculus is proposed for contextual dBas, modal extensions of which give logics for contextual dBas with operators and topological contextual dBas. The representation theorems for the algebras result in a protoconcept-based semantics for these logics.

Appendix A: Proofs

Proof in Theorem 16, that $(D,\sqcap ,\sqcup ,\lnot ,\top ,\bot )$ is a Boolean algebra: Let $\mathfrak {O}$ be a dBao such that for all $a\in D$, $\lnot a=\lrcorner a$ and $\lnot \lnot a=a$. Let $x,y\in D$ such that $x\sqsubseteq y$ and $y\sqsubseteq x$. By Proposition 4(4), $x\sqcap x=y\sqcap y$ and $x\sqcup x=y\sqcup y$. Using Proposition 5(3), $\lnot \lnot x=\lnot \lnot y$ and so $x=y$. Therefore $(D,\sqsubseteq )$ is a partially ordered set. From Definition 2(2a and 2b) it follows that $\sqcap ,\sqcup$ is commutative, while Definition 2(3a and 3b) gives that $\sqcap ,\sqcup$ is associative. Using Definition 2(5a) and Proposition 5(3), $x\sqcap (x\sqcup y)=x\sqcap x=\lnot \lnot x$. So $x\sqcap (x\sqcup y)= x$. Again using Definition 2(5b) and Proposition 5(3), $x\sqcup (x\sqcap y)= x$. Therefore $(D,\sqcap ,\sqcup ,\lnot ,\top ,\bot )$ is a bounded complemented lattice. To show it is a distributive lattice, we show that for all $x,y, \in D$ $x\sqcap y= x\wedge y$ and $x\vee y= x\sqcup y$. Rest of the proof follows from Definition 2(6a and 6b).

Let $x,y\in D$. Then $x,y\sqsubseteq x\sqcup y$. Proposition 5(2) gives $\lnot (x\sqcup y)\sqsubseteq \lnot x,\lnot y$. Therefore by Proposition 4(6), $\lnot (x\sqcup y)\sqcap \lnot y\sqsubseteq \lnot x\sqcap \lnot y$ and $\lnot (x\sqcup y)\sqcap \lnot (x\sqcup y)\sqsubseteq \lnot (x\sqcup y)\sqcap \lnot y$. So $\lnot (x\sqcup y)\sqcap \lnot (x\sqcup y)\sqsubseteq \lnot x\sqcap \lnot y$. By Proposition 5(1), $\lnot (x\sqcup y)\sqsubseteq \lnot x\sqcap \lnot y,$ and by Proposition 5(2), $\lnot (\lnot x\sqcap \lnot y)\sqsubseteq \lnot \lnot (x\sqcup y)=(x\sqcup y)\sqcap (x\sqcup y)\sqsubseteq x\sqcup y$. Hence $x\vee y\sqsubseteq x\sqcup y$. Using Proposition 4(5) and Proposition 5(2), $\lnot x\sqcap \lnot y\sqsubseteq \lnot x,\lnot y$. So $\lnot \lnot x\sqsubseteq \lnot (\lnot x\sqcap \lnot y)$ and $\lnot \lnot y\sqsubseteq \lnot (\lnot x\sqcap \lnot y)$. Therefore $x\sqsubseteq \lnot (\lnot x\sqcap \lnot y)=x\vee y$ and $y\sqsubseteq \lnot (\lnot x\sqcap \lnot y)=x\vee y$. Proposition 4(6) gives $x\sqcup y \sqsubseteq x\vee y$, as $(x\vee y)\sqcup (x\vee y)=\lrcorner \lrcorner (x\vee y)=\lnot \lnot (x\vee y)=x\vee y$. So $x\sqcup y=x\vee y$. Dually we can show that $x\sqcap y=x\wedge y$.

Proof of Theorem 25

      2a.

Now,

Similarly we can show that $\alpha \sqcap (\beta \sqcap \gamma )\vdash (\alpha \sqcap \beta )\sqcap \gamma$.

3a.

5a.

6a. Proof is identical to that of 5a. $\square$