A PSpace Algorithm for Acyclic Epistemic DL $\mathcal {ALCS}5_m$

Abstract

We study the description language $\mathcal {ALCS}5_m$, a variant of $\mathcal {ALCK}_m$ and $\mathcal {ALCS}4_m$. It augments $\mathcal {ALC}$ by allowing multi-modal epistemic operators over concept and role expressions. The epistemic operators are interpreted in modal logic $\mathbf {S}5_m$. By examining design issues of the tableau algorithm specific for $\mathcal {ALCS}5_m$, different from those for $\mathcal {ALCK}_m$ and $\mathcal {ALCS}4_m$, we provide a sound and complete tableau algorithm for deciding the satisfiability of an $\mathcal {ALCS}5_m$ knowledge base with an acyclic TBox, and further show how it can be implemented in PSpace.

Appendices

A Proof of Theorem 1

Theorem 1

Let $\mathbb {M} =\langle S$, $\pi$, $\mathcal{E}_1,\ldots$, $\mathcal{E}_m \rangle$ be an $\mathbf {S}5_m$-structure and $\mathcal {T}$ an acyclic TBox such that $\mathbb {M} \vDash \mathcal {T}$. Let $\mathbb {G}$ be a constraint graph obtained using $\mathcal {T}$, $\alpha$ an L-, G-, or T-rule and $\mathbb {G}_\alpha$ a constraint graph obtained by applying $\alpha$ to $\mathbb {G}$. If $\mathbb {M} \Vdash \mathbb {G}$ via $\sigma$, then there exists a semantic extension $\mathbb {M}_\alpha$ of $\mathbb {M}$ with respect to $N_\varSigma \cup \mathcal {O}_\mathbb {G}$ such that $\mathbb {M}_\alpha \Vdash \mathbb {G}_\alpha$ via $\sigma '$ (extension of $\sigma$) and $\mathbb {M}_\alpha \vDash \mathcal {T}$. Furthermore, $\mathbb {M}_\alpha \Vdash \mathbb {G}$.

Proof

Assume the hypotheses.

• If $\alpha$ is a $\sqcap$-rule, then there is a constraint $a: C_1 \sqcap C_2 \in \mathbb {L}(n)$ in $\mathbb {G}$ and $\{a:C_1, a:C_2\} \nsubseteq \mathbb {L}(n)$. After applying $\sqcap$-rule, $\mathbb {L}(n) = \mathbb {L}(n) \cup \{a:C_1, a:C_2\}$. Since $\mathbb {M} \Vdash \mathbb {G}$ and $a: C_1 \sqcap C_2 \in \mathbb {L}(n)$, by Definition 2, we have $(\mathbb {M}, \sigma (n)) \vDash C_1 \sqcap C_2 (a)$. It follows that $a^{\pi (\sigma (n))} \in (C_1 \sqcap C_2)^{\pi (\sigma (n))}$, which means that $a^{\pi (\sigma (n))} \in C_1^{\pi (\sigma (n))}$ and $a^{\pi (\sigma (n))} \in C_2^{\pi (\sigma (n))}$. Hence, $(\mathbb {M}, \sigma (n)) \vDash C_1(a)$ and $(\mathbb {M}, \sigma (n)) \vDash C_2(a)$. Thus, $\mathbb {G}_\alpha$ obtained by application of $\sqcap$-rule from $\mathbb {G}$ is satisfied by $\mathbb {M}$ via $\sigma$.

• If $\alpha$ is a $\sqcup$-rule, then there is a constraint $a:C_1 \sqcup C_2 \in \mathbb {L}(n)$ in $\mathbb {G}$ and $\{a:C_1, a:C_2\} \cap \mathbb {L}(n) = \varnothing$. Since $\mathbb {M} \Vdash \mathbb {G}$ and $a: C_1 \sqcup C_2 \in \mathbb {L}(n)$, by Definition 2, we have $(\mathbb {M}, \sigma (n)) \vDash C_1 \sqcup C_2(a)$ and therefore $a^{\pi (\sigma (n))} \in (C_1 \sqcup C_2)^{\pi (\sigma (n))}$. This means that $a^{\pi (\sigma (n))} \in C_1^{\pi (\sigma (n))}$ or $a^{\pi (\sigma (n))} \in C_2^{\pi (\sigma (n))}$. Hence, $(\mathbb {M}, \sigma (n))$ satisfies $C_1(a)$ or $C_2(a)$ (or both). It follows that $\sqcup$-rule can be applied in a way such that $\mathbb {G}_\alpha$ is satisfied by $\mathbb {M}$ via $\sigma$.

• If $\alpha$ is the $\exists$-rule, then there is a node n with $a:\exists R.C\in \mathbb {L}(n)$, no L-, T-, or $\Box$-rule except the $\exists$-rule is applicable, and there is no $b\in \mathcal {O}_\mathbb {G}$ such that $\{(a,b):R, b:C\}\subseteq \mathbb {L}(n)$. After applying $\alpha$, $\{(a,c):R, c:C\}\subseteq \mathbb {L}(n)$ where $c=f_\exists (a:\exists R.C, n)$. Since $\mathbb {M} \Vdash \mathbb {G}$, there must exist an element $d\in \varDelta$ such that $(a^{\pi (\sigma (n))}, d)\in R^{\pi (\sigma (n))}$ and $d\in C^{\pi (\sigma (n))}$. Define the interpretation $\pi '$ as $\pi$ except for the individual c: $c^{\pi '(\sigma (n))}=d$. Let $N_1=N_\varSigma \cup \mathcal {O}_\mathbb {G}\cup \{c\}$ and $N_2=N_\varSigma \cup \mathcal {O}_\mathbb {G}$. Then $(\pi '|_{N_1})|_{N_2} = \pi |_{N_2}$. Therefore, the resulting $\mathbb {G}_\alpha$ is satisfied by $\mathbb {M}_\alpha$ via $\sigma$ where $\mathbb {M}_\alpha =\langle S$, $\pi '$, $\mathcal{E}_1, \ldots$, $\mathcal{E}_m \rangle$ is a semantic extension of $\mathbb {M}$ with respect to $N_2$.

• If $\alpha$ is a $\forall$-rule, then there is a node n with $\{a:\forall R.C, (a,b):R \} \subseteq \mathbb {L}(n)$ and $b:C \notin \mathbb {L}(n)$. Since $\mathbb {M} \Vdash \mathbb {G}$ and $a:\forall R.C \in \mathbb {L}(n)$, by Definition 2, we have $(\mathbb {M}, \sigma (n)) \vDash \forall R.C (a)$, which means that for all $d \in \varDelta$, $(a^{\pi (\sigma (n))},d) \in R^{\pi (\sigma (n))}$ implies $d \in C^{\pi (\sigma (n))}$. Moreover, $(a,b):R \in \mathbb {L}(n)$ implies $(\mathbb {M}, \sigma (n)) \vDash R(a,b)$, which means $(a^{\pi (\sigma (n))},b^{\pi (\sigma (n))}) \in R^{\pi (\sigma (n))}$. After applying the $\forall$-rule, b : C is added to $\mathbb {L}(n)$. The resulting $\mathbb {G}_\alpha$ is satisfied by $\mathbb {M}$ via $\sigma$.

• If $\alpha$ is a $\bot$-rule, then there is a node n such that $\{a:C, a:\lnot C \} \subseteq \mathbb {L}(n)$ and $a:\bot \notin \mathbb {L}(n)$. By Definition 2, $a:C \in \mathbb {L}(n)$ implies $(\mathbb {M}, \sigma (n)) \vDash C (a)$, which means that $a^{\pi (\sigma (n))}\in C^{\pi (\sigma (n))}=\varDelta \setminus (\lnot C)^{\pi (\sigma (n))}$. Therefore, $a^{\pi (\sigma (n))}\notin (\lnot C)^{\pi (\sigma (n))}$. However, $a:\lnot C \in \mathbb {L}(n)$ implies $(\mathbb {M}, \sigma (n)) \vDash \lnot C (a)$, which means that $a^{\pi (\sigma (n))}\in (\lnot C)^{\pi (\sigma (n))}$. Hence, $a^{\pi (\sigma (n))}\in \varnothing =\bot ^{\pi (\sigma (n))}$.

• If $\alpha$ is the $\Diamond$-rule, then there are two cases.

  • There is a node n with $a:\Diamond _i C\in \mathbb {L}(n)$, $a:C\notin \mathbb {L}(n)$, no L-, T-, or $\Box$-rule is applicable, and n has no i-neighbor l with $a:C\in \mathbb {L}(l)$. By Definition 2, $a:\Diamond _i C\in \mathbb {L}(n)$ implies $(\mathbb {M}, \sigma (n)) \vDash \Diamond _i C(a)$. This means that there is a state s with $(\sigma (n),s)\in \mathcal {E}_i$ and $a^{\pi (s)}\in C^{\pi (s)}$. After applying the $\Diamond$-rule, a new node $n'$ is added to $\mathbb {G}$ with $\mathbb {L}(n')=\{a:C\}$ where $n'=f_\Diamond (a:\Diamond _i C, n)$ and $\mathbb {L}(n',n'')=\{i\}$ for each i-neighbor $n'$ of n. Extend $\sigma$ to $\sigma '$ such that $\sigma '(n')=s$. Since $\mathbb {M}$ is an $\mathbb {S}5$-structure, $(\sigma '(n'),\sigma ( n''))=(s, \sigma (n''))\in \mathcal {E}_i$ for each i-neighbor $n''$ of n. Therefore, $\mathbb {M}$ satisfies the resulting $\mathbb {G}_\alpha$ via $\sigma '$.

  • There is a node n with $(a,b):\Diamond _i R\in \mathbb {L}(n)$, $(a,b):R\notin \mathbb {L}(n)$, no L-, T-, or $\Box$-rule is applicable, and n has no i-neighbor l with $(a,b):R\in \mathbb {L}(l)$. The proof of this case is similar to that of the previous case and is omitted.

• If $\alpha$ is the $\Box$-rule, then there are two cases.

  • There are two nodes n and $n'$ with $a:\Box _i C\in \mathbb {L}(n)$, n being an i-neighbor of $n'$ and $a: C\notin \mathbb {L}(n')$. After the application, $a: C\in \mathbb {L}(n')$. Since $\mathbb {M}\Vdash \mathbb {G}$ and $a:\Box _i C\in \mathbb {L}(n)$, by Definition 2, we have $(\mathbb {M},\sigma (n))\vDash \Box _i C(a)$. Moreover, $i\in \mathbb {L}(n, n')$ implies that $(\sigma (n),\sigma (n'))\in \mathcal {E}_i$. Therefore, $(\mathbb {M},\sigma (n'))\vDash C(a)$. It follows that $\mathbb {M}$ satisfies the resulting $\mathbb {G}_\alpha$ via $\sigma$.

  • There are two nodes n and $n'$ with $(a,b):\Box _i R\in \mathbb {L}(n)$, n being an i-neighbor of $n'$ and $(a,b): R\notin \mathbb {L}(n')$. The proof of this case is similar to that of the previous case and is omitted.

• If $\alpha$ is a T-rule, then $a: A \in \mathbb {L}(n)$, $A \doteq D \in \mathcal {T}$ and $a:D' \notin \mathbb {L}(n)$ where $D'$ is the NNF of D. After applying $\alpha$, we have $a:D'\in \mathbb {L}(n)$. Since $\mathbb {M} \Vdash \mathbb {G}$ and $\mathbb {M} \vDash \mathcal {T}$, $a^{\pi (\sigma (n))} \in A^{\pi (\sigma (n))} = (D')^{\pi (\sigma (n))}$. Therefore, $(\mathbb {M}, \sigma (n)) \vDash D'(a)$ and hence, $\mathbb {M} \Vdash \mathbb {G}_\alpha$ via $\sigma$.

• If $\alpha$ is an N-rule, then $a:\lnot A\in \mathbb {L}(n)$, $A \doteq D \in \mathcal {T}$ and $a:D' \notin \mathbb {L}(n)$ where $D'$ is the NNF of $\lnot D$. After applying the N-rule, $a:D'\in \mathbb {L}(n)$. Since $\mathbb {M} \Vdash \mathbb {G}$ and $\mathbb {M} \vDash \mathcal {T}$, we have $(\mathbb {M}, \sigma (n)) \vDash A \doteq D$. It follows that $a^{\pi (\sigma (n))} \in (\lnot A)^{\pi (\sigma (n))} = (\lnot D)^{\pi (\sigma (n))}=(D')^{\pi (\sigma (n))}$. Thus, $(\mathbb {M}, \sigma (n)) \vDash D'(a)$. Therefore, $\mathbb {M} \Vdash \mathbb {G}_\alpha$ via $\sigma$.

It follows that after the application of every expansion rule, the resulting constraint graph $\mathbb {G}_\alpha$ is satisfied by $\mathbb {M}_\alpha$ which, except after applying an $\exists$-rule, is the same as $\mathbb {M}$. When $\alpha$ is an $\exists$-rule, $\mathbb {M}_\alpha$ differs from $\mathbb {M}$ only in the interpretation of the newly picked individual name. Therefore, $\mathcal {T}$ is valid in $\mathbb {M}_\alpha$. Moreover, since $\mathbb {M}_\alpha$ is a semantic extension of $\mathbb {M}$ restricted to $N_\varSigma \cup \mathcal {O}_\mathbb {G}$, the constraint graph $\mathbb {G}$ is satisfied by $\mathbb {M}_\alpha$. $\square$

B Proof of Lemma 4

Lemma 4

Let $\mathcal {T}$ be an acyclic TBox and $\mathbb {G}$ an open complete constraint graph with respect to local, global and terminological expansion rules. Then for every $A\in N_\mathcal {C}$ and every $a\in \varDelta$, $a:\lnot A\in \mathbb {L}(n)$ implies $(\mathbb {M}_\mathbb {G}, n)\vDash \lnot A(a)$.

Proof

There are two cases, and for both, since $\mathbb {G}$ is open, $a:A \notin \mathbb {L}(n)$.

• (1) When A is primitive, since $\mathbb {G}$ is open, $a: \lnot A \in \mathbb {L}(n)$ implies $a:A \notin \mathbb {L}(n)$. Then, $a \notin A^{\pi (n)}$ by Definition 3. Thus, $a \in (\lnot A)^{\pi (n)}$. Therefore, $(\mathbb {M}_\mathbb {G},n) \vDash \lnot A(a)$ by Theorem 1 and Definition 2.

• (2) If A is not primitive, i.e., there is a definition $A \doteq D \in \mathcal {T}$, we prove by induction on the structure of D. For the base case where the concept names involved in D are primitive, we have the following cases:

  • 1. D is of the form $\lnot B$ where B is primitive. Since $\mathbb {G}$ is complete, $a:B \in \mathbb {L}(n)$. By Definition 3, $a \in B^{\pi (n)}$ if and only if $a \notin (\lnot B)^{\pi (n)}$. Since $\mathbb {G}$ is open, $a: \lnot A \in \mathbb {L}(n)$ implies $a:A \notin \mathbb {L}(n)$. However, $A^{\pi (n)}= \{b \phantom {l}|\phantom {l} b:A \in \mathbb {L}(n) \} \cup (\lnot B)^{\pi (n)}$. This implies $a \notin A^{\pi (n)}$. Thus, $a \in (\lnot A)^{\pi (n)}$. Therefore, $(\mathbb {M}_\mathbb {G},n) \vDash \lnot A(a)$.

  • 2. D is of the form $B_1 \sqcap B_2$ where $B_1$ and $B_2$ are primitive. Since $\mathbb {G}$ is complete, $a: \lnot B_1 \sqcup \lnot B_2 \in \mathbb {L}(n)$, and $a: \lnot B_1$ or $a: \lnot B_2$ is in $\mathbb {L}(n)$. W.l.o.g., suppose $a: \lnot B_1\in \mathbb {L}(n)$. Since $\mathbb {G}$ is open, $a: B_1\notin \mathbb {L}(n)$. Because $B_1$ is primitive, by Definition 3, $a \notin B_1^{\pi (n)}$. Thus, $a \in (\lnot B_1)^{\pi (n)}$, which implies that $a \notin (B_1 \sqcap B_2)^{\pi (n)}$. However, $A^{\pi (n)}= \{b \phantom {l}|\phantom {l} b:A \in \mathbb {L}(n) \} \cup (B_1 \sqcap B_2)^{\pi (n)}$ and $a:A \notin \mathbb {L}(n)$. Hence, $a \notin A^{\pi (n)}$. Thus, $a \in (\lnot A)^{\pi (n)}$. Therefore, $(\mathbb {M}_\mathbb {G},n) \vDash \lnot A(a)$.

  • 3. D is of the form $B_1 \sqcup B_2$ where $B_1$ and $B_2$ are primitive. Since $\mathbb {G}$ is complete, $a: \lnot B_1 \sqcap \lnot B_2 \in \mathbb {L}(n)$ and $\{a: \lnot B_1, a: \lnot B_2\} \subseteq \mathbb {L}(n)$. Since $\mathbb {G}$ is open, $a: B_1\notin \mathbb {L}(n)$ and $a: B_2\notin \mathbb {L}(n)$. Because $B_1$ and $B_2$ are primitive, by Definition 3, $a \notin B_1^{\pi (n)}$ and $a \notin B_2^{\pi (n)}$. Thus, $a \notin (B_1 \sqcup B_2)^{\pi (n)}$. However, $A^{\pi (n)}= \{b \phantom {l}|\phantom {l} b:A \in \mathbb {L}(n) \} \cup (B_1 \sqcup B_2)^{\pi (n)}$ and $a:A \notin \mathbb {L}(n)$. Hence, $a \notin A^{\pi (n)}$. Thus, $a \in (\lnot A)^{\pi (n)}$. Therefore, $(\mathbb {M}_\mathbb {G},n) \vDash \lnot A(a)$.

  • 4. D is of the form $\exists R.B$ where B is primitive. Since $\mathbb {G}$ is complete, $a: \forall R. \lnot B \in \mathbb {L}(n)$ and for every b, if $(a,b):R \in \mathbb {L}(n)$, then $b:\lnot B \in \mathbb {L}(n)$. Suppose $(a,b):R \in \mathbb {L}(n)$. Since B is primitive and $\mathbb {G}$ is open, it follows that $b \notin B^{\pi (n)}$. Moreover, $(a,b) \in R^{\pi (n)}$ by Lemma 3 and Definition 1. Hence, for every b, $(a,b) \in R^{\pi (n)}$ implies $b \notin B^{\pi (n)}$. Thus, $a \in (\forall R. \lnot B)^{\pi (n)}$ and therefore, $a \notin (\exists R.B)^{\pi (n)}$. However, $A^{\pi (n)}= \{c \phantom {l}|\phantom {l} c:A \in \mathbb {L}(n) \} \cup (\exists R.B)^{\pi (n)}$ and $a:A \notin \mathbb {L}(n)$. Hence, $a \notin A^{\pi (n)}$, which is equivalent to $a \in (\lnot A)^{\pi (n)}$. Therefore, $(\mathbb {M}_\mathbb {G},n) \vDash \lnot A(a)$.

  • 5. D is of the form $\forall R.B$ where B is primitive. Since $\mathbb {G}$ is complete, $a: \exists R. \lnot B \in \mathbb {L}(n)$ and there exists b such that $(a,b):R \in \mathbb {L}(n)$ and $b:\lnot B \in \mathbb {L}(n)$. Since B is primitive and $\mathbb {G}$ is open, we have $b \notin B^{\pi (n)}$. By Lemma 3 and Definition 1, we have $(a,b) \in R^{\pi (n)}$. Therefore, there exists b such that $(a,b) \in R^{\pi (n)} \wedge b \notin B^{\pi (n)}$. Thus, $a \in (\exists R. \lnot B)^{\pi (n)}$ and hence, $a \notin (\forall R.B)^{\pi (n)}$. However, $A^{\pi (n)}= \{c \phantom {l}|\phantom {l} c:A \in \mathbb {L}(n) \} \cup (\forall R.B)^{\pi (n)}$ and $a:A \notin \mathbb {L}(n)$. Thus, $a \notin A^{\pi (n)}$, which is equivalent to $a \in (\lnot A)^{\pi (n)}$. Therefore, $(\mathbb {M}_\mathbb {G},n) \vDash \lnot A(a)$.

  • 6. D is of the form $\Diamond _i B$ where B is primitive. Since $\mathbb {G}$ is complete, $a: \Box _i \lnot B \in \mathbb {L}(n)$ and $a:\lnot B \in \mathbb {L}(n')$ for each $n'$ with $i \in \mathbb {L}(n,n')$. Note that B is primitive and $\mathbb {G}$ is open. Hence, $a \notin B^{\pi (n')}$ whenever $i\in \mathbb {L}(n,n')$. Thus, $a \in \bigcap _{n' \in \mathcal {E}_i(n)} (\lnot B)^{\pi (n')}$. Then, $a \in (\Box _i \lnot B)^{\pi (n)}$. Therefore, $a \notin (\Diamond _i B)^{\pi (n)}$. However, $A^{\pi (n)}= \{b \phantom {l}|\phantom {l} b:A \in \mathbb {L}(n) \} \cup (\Diamond _i B)^{\pi (n)}$ and $a:A \notin \mathbb {L}(n)$. Hence, $a \notin A^{\pi (n)}$. It follows that $(\mathbb {M}_\mathbb {G},n) \vDash \lnot A(a)$.

  • 7. D is of the form $\Box _i B$ where B is primitive. Since $\mathbb {G}$ is complete, we have $a: \Diamond _i \lnot B \in \mathbb {L}(n)$ and there exists $n'$ such that $i\in \mathbb {L}(n,n')$ and $a:\lnot B \in \mathbb {L}(n')$. Note that B is primitive and $\mathbb {G}$ is open. Thus, $a \notin B^{\pi (n')}$. Hence, $a \in \bigcup _{n' \in \mathcal {E}_i(n)}(\lnot B)^{\pi (n')}$. Then, $a \in (\Diamond _i \lnot B)^{\pi (n)}$. Therefore, $a \notin (\Box _i B)^{\pi (n)}$. However, $A^{\pi (n)}= \{a \phantom {l}|\phantom {l} a:A \in \mathbb {L}(n) \} \cup (\Box _i B)^{\pi (n)}$ and $a:A \notin \mathbb {L}(n)$. Hence, $a \notin A^{\pi (n)}$. It follows that $(\mathbb {M}_\mathbb {G},n) \vDash \lnot A(a)$.

  Note that for the first five cases, the correctness of the implication that if $a: \lnot A \in \mathbb {L}(n)$, then $(\mathbb {M}_\mathbb {G},n) \vDash \lnot A(a)$ depends on the fact that the constraint graph $\mathbb {G}$ has no applicable local or terminological expansion rules. For the last two cases, the correctness of the implication depends on the fact that $\mathbb {G}$ has no applicable global or terminological expansion rules. The induction step is similar to the corresponding base case, except that in the general case, in order to show that $a \notin D^{\pi (n)}$, we use the induction hypothesis rather than relying on the fact that a concept name is primitive when the concept name occurring in D is not primitive.

$\square$