Reasoning with fuzzy extensions of OWL and OWL 2

Abstract

Fuzzy Description Logics (f-DLs) have been proposed as logical formalisms capable of representing and reasoning with vague/fuzzy information. They are envisioned to be helpful for many applications that need to cope with such type of information such as multimedia processing, decision making, automatic negotiation and more. Recent results have provided with many tableaux algorithms for supporting reasoning over quite expressive f-DLs. However, no (direct) tableaux algorithm for reasoning with fuzzy extensions of DLs such as $\mathcal{SHOIQ }$ and $\mathcal{SROIQ }$ exists today. $\mathcal{SHOIQ }$ and $\mathcal{SROIQ }$ are particularly interesting formalisms as they constitute the logical underpinnings of the Web ontology languages OWL DL and OWL 2 DL. In the current paper, we present an algorithm for reasoning with the fuzzy DLs f-$\mathcal{SHOIQ }$ and f-$\mathcal{SROIQ }$. In addition, we also provide a tableaux algorithm for fuzzy nominals, thus providing reasoning support for the fuzzy DL language (we call) f-$\mathcal{SHO }_\mathtt{{f}}\mathcal{IQ }$.

Appendix: Omitted proofs

Lemma 1

An ${\text{ f }_{KD}\text{- }\mathcal{SHOIQ }}$ knowledge base ${\Sigma }$ is satisfiable iff there exists a fuzzy tableau $T$ for ${\Sigma }$.

Proof

The proof is similar to the one for ${\text{ f }_{KD}\text{- }\mathcal{SHIN }}$ [50], with the addition of nominals, qualified number restrictions and GCIs. Thus, in the following, we only illustrate the different cases.

For the ‘if’ direction, let $T=(\mathbf S ,\mathcal{L },\mathcal{E },\mathcal{V })$ be a fuzzy tableau for ${\Sigma }$ ; then, we can construct a model $\mathcal{I } =(\Delta ,\cdot ^\mathcal{I })$ of ${\Sigma }$ in a similar way as in [50] by setting $\Delta ^\mathcal{I } =\mathbf S , a^\mathcal{I } =\mathcal{V } (a)$, for each $a\in {\mathbf{I}} _{\Sigma }, \top ^\mathcal{I } (s)=\mathcal{L } (s,\top )$ and $\bot ^\mathcal{I } (s)=\mathcal{L } (s,\bot )$ for all $s\in \mathbf S$, while for concepts and roles we have the following:

where $R_{\mathcal{E }}(s,t)=\mathcal{E } (R,{\langle } s,t{\rangle })$, for all ${\langle } s,t{\rangle } \in \mathbf S \times \mathbf S$, and $R^+_{\mathcal{E }}$ represents its $\sup -\min$ transitive closure of $R_{\mathcal{E }}$ [33].

As with ${\text{ f }_{KD}\text{- }\mathcal{SHIN }}$ [50], Properties 1 and 2 ensure the correct interpretations of the top and bottom concepts, Properties 9 and 10 the correct interpretation of inverse roles and role hierarchies, respectively, Property 16 (due to the results in [35, 48]) that $\mathcal{I }$ is a model of the TBox, while Properties 17 and 18 that $\mathcal{I }$ is a model of $\mathcal{A }$. Additionally to ${\text{ f }_{KD}\text{- }\mathcal{SHIN }}$ [50], Property 14 ensures that nominals are interpreted as singleton sets, Property 15 that the membership degree of elements in nominals is in accordance with their semantics, and Property 19 that nominals corresponding to ABox individuals are also interpreted correctly. Then, by induction on the structure of concepts, we can show that for all $s\in \mathbf S , \mathcal{L } (s,C)\ge n$ implies $C^\mathcal{I } (s)\ge n$—similarly for $\mathcal{L } (s,C)>n$. We only show the case of nominals and qualified cardinality restrictions, which have not been presented before [50]:

• If $\mathcal{L } (s,\{o\})\ge n$, then by Property 15 and construction of $\mathcal{I }, \{o\}^\mathcal{I } (s)=\mathcal{L } (s,\{o\})\ge 1$.

• If $\mathcal{L } (s,\ge pR.C)\ge n$, then by Property 11 and definition of $R^T$, there are $p$ elements $t_i$, s.t. $\mathcal{E } (R,{\langle } s,t_i{\rangle })\ge n$, and $\mathcal{L } (t_i,C), 1\le i\le p$. By construction, $R^\mathcal{I } (s,t_i)\ge n$ and by induction hypothesis $C^\mathcal{I } (t_i)\ge n$, thus

  $$\begin{aligned} n\le \sup _{t_i\in {\varDelta {^\mathcal I }}}\{\ldots ,\min ^p_{i=1}\{\min (R^\mathcal{I } (s,t_i),C(t_i))\},\ldots \}=(\ge pR.C)^\mathcal{I } (s). \end{aligned}$$

• If $\mathcal{L } (s,\le pR.C)\ge n$, then by Property 12 $\sharp R^T(s,>,1-n,C)\le p$, i.e., there are at most $p$ elements $t_i$ such that, $\mathcal{E } (R, {\langle } s,t_i{\rangle })>1-n$, and $\mathcal{L } (t_i,C)>1-n\, 1\le i\le p$. Nevertheless, we need to show that this is the case also in $\mathcal{I }$, i.e., that for the set $R^\mathcal{I } (s,>,1-n,C)=\{x\in {\varDelta {^\mathcal I }} \mid R^\mathcal{I } (s,x)>1-n \, \text{ and } \, C^\mathcal{I } (x)>1-n\}$ it holds $\sharp R^\mathcal{I } (s,>,1-n,C)\le p$. Assume otherwise, i.e., that there is $t_{p+1}$ different from all other $t_i$ and such that $R^\mathcal{I } (s,t_{p+1})>1-n$ and $C^\mathcal{I } (t_{p +1})>1-n$. By construction, and since $R$ must be a simple role, then $R^\mathcal{I } (s,t_{p+1})=\mathcal{E } (R,{\langle } s,t_{p+1}{\rangle })>1-n$. Consequently, in order to have $\sharp R^T(s,>,1-n,C)\le p$ it must be the case that $\mathcal{L } (t_{p+1},C)\le 1-n$. But then by Property 13, $\mathcal{L } (t_{p+1},\lnot C)\ge n$ must hold and by induction hypothesis $(\lnot C)^\mathcal{I } (t_{p+1})\ge n$, i.e., $C^\mathcal{I } (t_{p+1})\le 1-n$, which contradicts the original assumption that $C^\mathcal{I } (t_{p+1})>1-n$.

For the converse, let $\mathcal{I } =(\Delta ^\mathcal{I },\cdot ^\mathcal{I })$ be a (witnessed) model for ${\Sigma }$ ; then, a fuzzy tableau $T=(\mathbf S ,\mathcal{L },\mathcal{E },\mathcal{V })$ for ${\Sigma }$ can be defined by setting $\mathbf S =\Delta ^\mathcal{I }, \mathcal{E } (R,{\langle } s,t{\rangle })=R^\mathcal{I } (s,t), \mathcal{L } (s,C)=C^\mathcal{I } (s)$ and $\mathcal{V } (a)=a^\mathcal{I }$. It is easy to show that all properties in Definition 2 are satisfied as a direct consequence of the semantics of ${\text{ f }_{KD}\text{- }\mathcal{SHOIQ }}$-concepts and since $\mathcal{I }$ is a witnessed model of ${\Sigma }$. $\square$

Lemma 2

Let ${\Sigma }$ be an ${\text{ f }_{KD}\text{- }\mathcal{SHOIQ }}$ knowledge base. Then,

• 1. Termination: when started for ${\Sigma }$ the tableaux algorithm terminates.

• 2. Correctness: ${\Sigma }$ has a fuzzy tableau if and only if the tableaux algorithm for ${\text{ f }_{KD}\text{- }\mathcal{SHOIQ }}$ can be applied to ${\Sigma }$ such that it yields a complete and clash-free completion-graph.

Proof

The termination of the algorithm (claim 1) is a consequence of the same properties that ensure termination in the case of the crisp $\mathcal{SHOIQ }$ language [30]. In brief we have the following observations:

• All rules apart from the shrinking rules ⁷ strictly extend the completion-graph by adding new nodes and edges or extending their labels while neither remove nodes, edges or pairs from them.

• New nodes are added only by the generating rules,⁸ and each of these rule is applied at most once for a given concept in the label of a given node $x$. Even if a shrinking rule is applied and merges an $R$-neighbour $y$ of $x$ into another node $z$, then $\mathcal{L } (y)$ is added into $\mathcal{L } (z), z$ ‘inherits’ all the inequalities from $y$, and either $z$ is an $R$-neighbour of $x$ (if $x$ is a nominal node or $y$ a successor of $x$) or $x$ is removed from the graph by an application of $\mathsf{Prune }$(x) (if $x$ is a blockable node and $x$ is a successor of $y$).

• Since nodes are labelled with non-empty subsets of $cl({\Sigma })$ and edges with subsets of $\mathbf R _{\Sigma }$, obviously there is a finite number of possible labellings for a pair of nodes and an edge, while also the membership degrees that appear in nodes is also finite as in the case of ${\text{ f }_{KD}\text{- }\mathcal{SI }}$. More precisely, for a pair of nodes and an edge, there are at most $2^{8mlk}$ possible labellings, where $k=\vert \mathbf R _{\Sigma } \vert , m=\vert cl({\Sigma })\vert$ and $l=\vert N^\mathcal{A } \vert$. Since a path on which nodes are blocked cannot become longer, paths are of length at most $2^{8mlk}$.

• As it is shown in [30] for $\mathcal{SHOIQ }$, the number of nominal nodes is bounded. This is a consequence of the following facts. The $NN$ can only be applied after a nominal has been added to the label of a blockable node $x$ in a branch of one of the blockable trees rooted in a root node. Otherwise, it is not possible that a blockable node has a nominal node as a successor which is required by the first condition of the rule. Now since $x$ contains one of the initial nominals that exist in $\mathcal{T }$ or $\mathcal{A }$, and the $\{o\}_1$-rule is applied with highest priority, $x$ is merged with an existing nominal node which contains some of the initials nominals. As a consequence of this merging, it is possible that the predecessor of $x$ is merged into a nominal node $n_1$ (created by an application of the $NN$ to $x$) by the shrinking rules (due to the pruning, this cannot happen to a successor of $x$). The merge of the predecessor of $x$ occurs because the $NN$-rule adds $\le mR$ to $x$ together with $m$ already conjugated successors. Hence, $x$ has $m+1$ successors ($m$ created from the rule and one predecessor), and the $\le$-rule will be executed. Now $n_1$ either contains one nominal from the initial ones or only nominals created by the $NN$. Repeating this argument, it is possible that all ancestors of $x$ are merged into nominal nodes. But, since the length of a path of blockable nodes is bounded this repeated merging is bounded. Finally, when the $NN$ has been applied to a concept $(\le pR)$, it can never be applied to $(\le pR)$ again.

The proof of the second claim extends the proof of ${\text{ f }_{KD}\text{- }\mathcal{SHIN }}$ [50], by using the techniques of $\mathcal{SHOIQ }$ [30].

For the ‘if’ direction, we construct a fuzzy tableau $T=(\mathbf S ,\mathcal{L },\mathcal{E },\mathcal{V })$ from a complete and clash-free completion-graph $\mathbf{G }$ by unravelling blockable ‘tree’ parts of the graph due to the lack the finite-model property. Formally, the construction is the following.

An individual in S corresponds to a path in $\mathbf{G }$. Moving down to blocked nodes and up to blocking ones, we can define infinite such paths. More precisely, a path is a sequence of pairs of nodes of $\mathbf{G }$ of the form $p=[\frac{x_{a_0}}{x^{\prime }_{a_0}},\ldots ,\frac{x_{a_n}}{x^{\prime }_{a_n}}]$. For such a path, we define ${\mathop {\mathsf{Tail }}}(p):=x_{a_n}$ and ${\mathop {\mathsf{Tail }}}^{\prime }(p):=x^{\prime }_{a_n}$. With $[p\mid \frac{x_{a_{n+1}}}{x^{\prime }_{a_{n+1}}}]$, we denote the path $[\frac{x_{a_0}}{x^{\prime }_{a_0}},\ldots ,\frac{x_{a_n}}{x^{\prime }_{a_n}},\frac{x_{a_{n+1}}}{x^{\prime }_{a_{n+1}}}]$. The set $\mathsf{Paths } (\mathbf{G })$ is defined inductively as follows:

• For each blockable node $x$ of $\mathbf{G }$ that is a successor of a nominal node or a root node, $[\frac{x}{x}]\in \mathsf{Paths } (\mathbf{G })$, and

• For a path $p\in \mathsf{Paths } (\mathbf{G })$ and a node $z$ in $\mathbf{G }$:

  • if $z$ is a successor of ${\mathop {\mathsf{Tail }}}(p)$ and $z$ is not blocked, then $[p\mid \frac{z}{z}]\in \mathsf{Paths } (\mathbf{G })$, or

  • if $y$ is a successor of ${\mathop {\mathsf{Tail }}}(p)$ and $z$ blocks $y$, then $[p\mid \frac{z}{y}]\in \mathsf{Paths } (\mathbf{G })$

By construction, it follows that all nodes occurring in a path are blockable nodes. Moreover, if $p\in \mathsf{Paths } (\mathbf{G })$, then ${\mathop {\mathsf{Tail }}}(p)$ is not blocked; ${\mathop {\mathsf{Tail }}}(p)={\mathop {\mathsf{Tail }}}^{\prime }(p)$ iff ${\mathop {\mathsf{Tail }}}^{\prime }(p)$ is not blocked and at last $\mathcal{L } ({\mathop {\mathsf{Tail }}}(p))=\mathcal{L } ({\mathop {\mathsf{Tail }}}^{\prime }(p))$.

Secondly, we make use of the technique introduced in [51] in order to compute the degree that (pairs of) elements of the fuzzy tableau will belong to a concept (role). The function that calculates this degree is called $glb$. Roughly speaking for a specific path $p$ this function is defined as the maximum of the set $\{n\mid {\langle } A,n{\rangle } \in \mathcal{L } (x)\}\cup \{0\}$. Due to normalization there may exist values of the form $n+\epsilon$. It is important to note that an adequately small and carefully selected value $\epsilon$ must be chosen. For example, if ${\langle } \lnot A,0.19+\epsilon {\rangle } \in \mathcal{L } (x)$ and ${\langle } A,0.8+\epsilon {\rangle } \in \mathcal{L } (x)$, then it should hold that $\epsilon \le 1-(0.8+0.19)=0.01$. A similarly important case is if ${\langle } \forall R.C,0.8{\rangle } \in \mathcal{L } (x)$ and ${\langle } R,0.19+\epsilon {\rangle } \in \mathcal{L } ({\langle } x,y{\rangle })$. The existence of such a value is ensured by the clash-freeness of $\mathbf{G }$.

Next we use ${\mathop {\mathsf{Nom }}} (\mathbf{G })$ for the set of nominal and root nodes in $\mathbf{G }$ and define a tableau $T$ as in Table 8.

Table 8 Fuzzy tableaux construction from complete and clash-free completion-graph

It can be shown that $T$ is a fuzzy tableau for ${\Sigma }$ :

• Properties 1–3 of Definition 2 are satisfied because G is clash-free and due to the construction of $T$. Let $\mathcal{L } (p,\lnot A) =n_1\ge n$ and $\mathcal{L } (p,A)=n_2$. The definition of $T$ and since we only consider nonzero assertions this implies that $\{{\langle } \lnot A,n_1{\rangle },{\langle } A,n_2{\rangle } \}\subseteq \mathcal{L } ({\mathop {\mathsf{Tail }}}(p))$. Since $\mathbf{G }$ is clash-free $n_1+n_2\le 1\Rightarrow n_2\le 1-n_1$. Consequently, $\mathcal{L } (p,A)\le 1-n_1\le 1-n$.

• Properties 4 and 5 of Definition 2 are satisfied because none of the $\sqcup , \sqcap$ rules apply to any node in $\mathbf{G }$, and ${\mathop {\mathsf{Tail }}}(p)$ is not blocked. For example, let $\mathcal{L } (p,C\sqcap D)=n_1\ge n$. The definition of $T$ implies that, either ${\langle } C\sqcap D,n_1{\rangle } \in \mathcal{L } ({\mathop {\mathsf{Tail }}}(p))$ or ${\langle } C\sqcap D,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } ({\mathop {\mathsf{Tail }}}(p))$, with $n_1=n^{\prime }+\epsilon$. Completeness of $\mathbf{G }$ implies that either ${\langle } C,n_1{\rangle } \in \mathcal{L } ({\mathop {\mathsf{Tail }}}(p))$ and ${\langle } D,n_1{\rangle } \in \mathcal{L } ({\mathop {\mathsf{Tail }}}(p))$ or ${\langle } C,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } ({\mathop {\mathsf{Tail }}}(p))$ and ${\langle } D,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } ({\mathop {\mathsf{Tail }}}(p))$. Hence, $\mathcal{L } (s,C)\ge \mathcal{L } (s,C\sqcap D)\ge n, \mathcal{L } (s,D)\ge \mathcal{L } (s,C\sqcap D)\ge n$ Property 5 follows for similar reasons.

• For Property 6, let $p,q\in \mathbf S$ with $\mathcal{L } (p,\forall R.C)=n_1\ge n$ and $\mathcal{E } (\lnot R,{\langle } p,q{\rangle })\ngeq n$. The definition of $T$ implies that either ${\langle } \forall R.C,n_1{\rangle } \in \mathcal{L } (z_p)$ or ${\langle } \forall R.C,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } (z_p)$ with $n_1=n^{\prime }+\epsilon$ and $z_p={\mathop {\mathsf{Tail }}}(p)$ if $p\in \mathsf{Paths } (\mathbf{G })$, or $z_p=z$ if $z\in {\mathop {\mathsf{Nom }}} (\mathbf{G })$. Now we have the following cases,

  • If $p\in \mathsf{Paths } (\mathbf{G })$, then $z_p={\mathop {\mathsf{Tail }}}(p)$ and

    • If $q=[p\mid \frac{x}{x^{\prime }}]$, then $x^{\prime }$ is an $R_{r}$-successor of ${\mathop {\mathsf{Tail }}}(p)$ and, since $glb$ does not create unnecessary conjugations we have that ${\langle } R,r{\rangle } \in \mathcal{L } ({\langle } {\mathop {\mathsf{Tail }}}(p),x^{\prime }{\rangle })$ is such that it conjugates with ${\langle } \lnot R,n{\rangle }$. Hence, due to completeness of $\mathbf{G }$, we have either ${\langle } C,n_1{\rangle } \in \mathcal{L } (x^{\prime })$ or ${\langle } C,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } (x^{\prime })$, and either $x^{\prime }=x$ or the blocking condition implies $\mathcal{L } (x^{\prime })=\mathcal{L } (x)=\mathcal{L } (q)$.

    • If $p=[q\mid \frac{x}{x^{\prime }}]$, then $x^{\prime }$ is an ${\mathop {\mathsf{Inv }}} (R)_{r}$-successor of ${\mathop {\mathsf{Tail }}}(q)$ and again, the definition of $glb$ implies that ${\langle } {\mathop {\mathsf{Inv }}} (R),r{\rangle } \in \mathcal{L } ({\langle } {\mathop {\mathsf{Tail }}}(q),x^{\prime }{\rangle })$ conjugates with ${\langle } \lnot {\mathop {\mathsf{Inv }}} (R),n{\rangle }$. Thus, due to completeness of $\mathbf{G }$, either ${\langle } C,n_1{\rangle } \in \mathcal{L } ({\mathop {\mathsf{Tail }}}(q))=\mathcal{L } (q)$ or ${\langle } C,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } ({\mathop {\mathsf{Tail }}}(q))=\mathcal{L } (q)$.

    • If $p=[\frac{x}{x}]$ and $p=[\frac{y}{y}]$ for two root nodes $x,y$ then $y$ is an $R_{r}$-neighbour of $x$, and since the $\forall$-rule does not apply we have that either ${\langle } C,n_1{\rangle } \in \mathcal{L } (y)=\mathcal{L } (q)$ or ${\langle } C,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } (y)=\mathcal{L } (q)$.

    • If $q=x\in {\mathop {\mathsf{Nom }}} (\mathbf{G })$, then $x$ is an $R_{n}$-neighbour of ${\mathop {\mathsf{Tail }}}(p)$ and completeness implies that either ${\langle } C,n{\rangle } \in \mathcal{L } (x)$ or ${\langle } C,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } (x)$.

  • If $z_p=z\in {\mathop {\mathsf{Nom }}} (\mathbf{G })$, then either

    • $q\in \mathsf{Paths } (\mathbf{G })$, then ${\mathop {\mathsf{Tail }}}(q)$ is an $R_{r}$-neighbour of $z$ and completeness implies that either ${\langle } C,n{\rangle } \in \mathcal{L } (q)$, or ${\langle } C,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } (x)$ or

    • $q=x\in {\mathop {\mathsf{Nom }}} (\mathbf{G })$, then x is an $R_{r}$-neighbour of $z$ and completeness implies that either ${\langle } C,n{\rangle } \in \mathcal{L } (q)$ or ${\langle } C,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } (x)$.

  The same proof applies for Property 6 with $\mathcal{L } (p,\forall R.C)>n$ and for Property 8.

• For Property 7 consider some $p\in \mathbf S$ with $\mathcal{L } (p,\exists R.C)\ge n$.

  • If $p\in \mathsf{Paths } (\mathbf{G })$, then either ${\langle } \forall R.C,n_1{\rangle } \in \mathcal{L } (p)$ or ${\langle } \forall R.C,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } (p)$ with $n_1=n^{\prime }+\epsilon$. Since ${\mathop {\mathsf{Tail }}}(p)$ is not blocked, completeness of $\mathbf{G }$ implies the existence of an $R$-neighbour $y$ of ${\mathop {\mathsf{Tail }}}(p)$ with either ${\langle } C,n_1{\rangle } \in \mathcal{L } (y)$ or ${\langle } C,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } (y)$.

    • If $y$ is a nominal node, then $y\in \mathbf S , \mathcal{L } (y,C)\ge n$ and $\mathcal{E } (R,{\langle } p,y{\rangle })\ge n$.

    • If $y$ is a blockable node and a successor of ${\mathop {\mathsf{Tail }}}(p)$, then ${\langle } p,[p\mid \frac{y^{\prime }}{y}]{\rangle } \in \mathbf S$, and either $y^{\prime }=y$, or $y^{\prime }$ blocks $y$. In both cases either ${\langle } C,n_1{\rangle }$ or ${\langle } C,n^{\prime }+\epsilon {\rangle }$ are in $\mathcal{L } (y^{\prime })$.

    • If $y$ is a blockable node and a predecessor of ${\mathop {\mathsf{Tail }}}(p)$, then either $p=[r\mid \frac{y}{y}\mid \frac{{\mathop {\mathsf{Tail }}}(p)}{{\mathop {\mathsf{Tail }}}^{\prime }(p)}]$, or $p=[r\mid \frac{z}{y}\mid \frac{{\mathop {\mathsf{Tail }}}(p)}{{\mathop {\mathsf{Tail }}}^{\prime }(p)}]$ and ${\mathop {\mathsf{Tail }}}(p)$ blocks ${\mathop {\mathsf{Tail }}}^{\prime }(p)$, hence ${\mathop {\mathsf{Tail }}}^{\prime }(p)$ is an $R$-successor of $z$. In the first case either ${\langle } C,n_1{\rangle } \in \mathcal{L } (y)$ or ${\langle } C,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } (y)$, while in the second case due to pair-wise blocking $\mathcal{L } (y)=\mathcal{L } (z)$.

  • If $p\in {\mathop {\mathsf{Nom }}} (\mathbf{G })$, then completeness implies the existence of some $R$-successor $x$ of $p$ with either ${\langle } C,n_1{\rangle } \in \mathcal{L } (x)$ or ${\langle } C,n^{\prime }+\epsilon {\rangle } \in \mathcal{L } (x)$

    • If $x$ is a nominal node, then $\mathcal{E } (R,{\langle } p,x{\rangle })\ge n$ and $\mathcal{L } (x,C)\ge n$.

    • If $x$ is a blockable node, then $x$ is a safe $R$-neighbour of $p$ and thus not blocked. Hence, there is a path $q\in \mathsf{Paths } (\mathbf{G })$ with ${\mathop {\mathsf{Tail }}}(q)=x, \mathcal{E } (R,{\langle } p,q{\rangle })\ge n$ and $\mathcal{L } (q,C)\ge n$.

  In any of these cases, $\mathcal{E } (R,{\langle } p,q{\rangle })\ge n_1\ge n, \mathcal{L } (q,C)\ge n_1\ge n$. Similarly for $\mathcal{L } (p,\exists R.C)>n$.

• Property 9 in Definition 2 is satisfied due to the symmetric definition of $\mathcal{E }$.

• Property 10 in Definition 2 is satisfied due to the definition of the $R$-successor that takes into account the role hierarchy .

• Properties 11–13 in Definition 2 are satisfied due to the construction of $T$ as in the classical case [30].

• Properties 14 and 15 are due to completeness of $\mathbf{G }$, the fact that nominal nodes are not ‘unravelled’, while Property 19 due to initialization of $\mathbf{G }$.

• Property 16 is due to completeness of $\mathbf{G }$.

• Properties 17–18 are satisfied cause of the initialization of the completion-graph and the fact that the algorithm never blocks root nodes. Furthermore, for each root node $x_{a_i}$ whose label and edges are removed by the $\mathsf{Merge }$ method, there is another root node $x^j_0$ with $x_{a_i}=x_{a_j}$ and $\{{\langle } C,n{\rangle } \mid (a_i:C)\ge n\in \mathcal{A } \}\subseteq \mathcal{L } (x_{a_j})$.

For the ‘only-if’ direction, if $T=(\mathbf S ,\mathcal{L },\mathcal{E },\mathcal{V })$ is a fuzzy tableau for ${\Sigma }$ we can use $T$, to guide the application of the expansion rules such that they yield a completion-graph $\mathbf{G }$ that is both complete and clash-free. More precisely, we can define a mapping $\pi$ from nodes in the completion-graph to individuals in $\mathbf S$ of the tableau and use this $\pi$ to modify the non-deterministic rules in such a way that we always choose the ‘correct’ fuzzy pair to be added in the label of some node [50]. This, together with the termination property ensure that at some point blocking will occur and the resulting completion-graph would also be clash-free. $\square$

Proposition 2 Let $\mathcal{R } _h$ be a regular role hierarchy, let $\mathcal{B }_S$ be the automaton constructed for a role $S$ w.r.t. $\mathcal{R } _h$ according to the method in [32], and let $L(\mathcal{B }_S)$ denote the language accepted by $\mathcal{B }_S$. Then, $\mathcal{I }$ is a model of $\mathcal{R } _h$ if and only if for each role $S$ occuring in $\mathcal{R } _h$, each word $w\in L(\mathcal{B }_S)$ and each $w^\mathcal{I } (a,b)\ge n$ we have $S^\mathcal{I } (a,b)\ge n$.

Proof

The ‘if’ direction is similar to the one in [32] (i.e. by showing the contrapositive). More precisely, assume that $\mathcal{I }$ is not a model of $\mathcal{R } _h$. Then, there exists a cRIA $w\sqsubseteq S\in \mathcal{R } _h$ not satisfied by $\mathcal{I }$. Hence, there are ${\langle } x,y{\rangle }$ and $n\in (0,1]$ such that $w^\mathcal{I } (x,y)=n$ and $S^\mathcal{I } (x,y)<n$. But, since $\mathcal{B }_S$ is constructed by the method in [32] then it satisfies Lemma 12.1 in [32], i.e., $w\in L(\mathcal{B }_S)$.

The ‘only-if’ direction is again proved by similar techniques as in [32] but considering that the semantics of cRIAs are given by $\sup$-$t$ composition. More precisely, let $\mathcal{I }$ be a model of $\mathcal{R } _h$, let $S$ be a role, let $w\in L(\mathcal{B }_S)$ and let $w^\mathcal{I } (a,b)=n$. We need to show that $S^\mathcal{I } (a,b)\ge n$. Since $\mathcal{R } _h$ is regular there exists a strict partial order $\prec$ such that each cRIA in $\mathcal{R } _h$ is $\prec$-regular. Hence, we can use well-founded induction over $\prec$.

First, note that $w\in L(\mathcal{B }_S)$ induces a decomposition $w=w_1\ldots w_k$ and word $\hat{w}=S_1\ldots S_k$ such that

• $S_i\prec S$ or $S_i=S$ for all $1\le i\le k$

• $\hat{w}\in L(\hat{\mathcal{A }}_S)$,⁹ and

• $w_i\in L(\mathcal{B }_{S_i})$

Moreover, $w^\mathcal{I } (a,b)=n$ implies that there exist $k x_i$ with $a=x_0, b=x_k, w^\mathcal{I } _{i+1}(x_i,x_{i+1})=n_{i+1}$, and $n=\min (n_1,\ldots ,n_k)$ for $0\le i<k$. By induction hypothesis, $S^\mathcal{I } _i(x_i,x_{i+1})\ge n_{i+1}$ and $\hat{w}^\mathcal{I } (a,b)=n$. By case analysis on the form of axioms in $\mathcal{R } _h$ that have $S$ in the right-hand side it can be shown that $S^\mathcal{I } (a,b)\ge n$. We show one case, as the rest follow similarly (cf. also [32]).

If $SS\sqsubseteq S\not \in \mathcal{R } _h$ and $S^-\sqsubseteq S\not \in \mathcal{R } _h$, then, by construction of $\hat{\mathcal{A }}_S, \hat{w}$ is of the form

$$\begin{aligned} \begin{array}{lll} \hat{w}=u_1\ldots u_m x v_1\ldots v_\ell &{} and &{} u_iS\sqsubseteq S\in \mathcal{R } _h, \text{ for } \text{ each } 1\le i\le m\\ &{} &{} x\sqsubseteq S\in \mathcal{R } _h \,\text{ or }\, x=S\\ &{} &{} Sv_j\sqsubseteq S\in \mathcal{R } _h, \text{ for } \text{ each } 1\le j\le \ell \end{array} \end{aligned}$$

Since $\mathcal{I }$ is a model of $\mathcal{R } _h$ it satisfies the above axioms and hence it follows that $S^\mathcal{I } (a,b)\ge \hat{w}^\mathcal{I } (a,b)=n$. $\square$

Lemma 5 Let $\mathcal{R } _h$ be a role hierarchy, let $S$ be a role in $\mathcal{R } _h$, let $\mathcal{B }_S$ be the automaton for $S$ w.r.t. $\mathcal{R } _h$ and let $w\in L(\mathcal{B }_S)$, where $w=R_1\ldots R_m$. If $\mathcal{I }$ satisfies $\mathcal{R } _h$, then $\mathcal{I }$ satisfies $\forall S.C\sqsubseteq \forall R_1.(\forall R_2.(\ldots (\forall R_m.C)))$.

Proof

Let $a_i\in \Delta ^\mathcal{I }$ for $0\le i\le m$, where $a_0=a$ be objects in ${\varDelta {^\mathcal I }}$, and let $(\forall S.C){^\mathcal I } (a)\ge n$. By Proposition 2, we have that $S^\mathcal{I } (a_0,a_m)\ge \min (R^\mathcal{I } _1(a_0,a_1),\ldots R^\mathcal{I } _m(a_{n-1},a_n))$, hence we also have $c(S^\mathcal{I } (a_0,a_m))\le c(\min (R^\mathcal{I } _1(a_0,a_1),\ldots R^\mathcal{I } _m(a_{n-1},a_n)))$. Then, we have the following equivalences:

(1) ${\mathop {\mathop {\inf }}\limits }_{a_m\in {\varDelta {^\mathcal I }}} \max (c(S^\mathcal{I } (a_0,a_m)),C{^\mathcal I } (a_m))\ge n$	$\Rightarrow _{\text{ monotonicity }}$
(2) ${\mathop {\mathop {\inf }}\limits }_{a_m\in {\varDelta {^\mathcal I }}}\max (c({\mathop {\mathop {\min }}\limits }^{m}_{i=1}\ R_i{^\mathcal I } (a_{i-1},a_i)),C{^\mathcal I } (a_m))\ge n$	$\Rightarrow _{\text{ De } \text{ Morgan }}$
(3) ${\mathop {\mathop {\inf }}\limits }_{a_m\in {\varDelta {^\mathcal I }}}\max ({\mathop {\mathop {\max }}\limits }^{m}_{i=1}\ c(R_i{^\mathcal I } (a_{i-1},a_i)),C{^\mathcal I } (a_m))\ge n$	$\Rightarrow _{\text{ associativity }}$
(4) ${\mathop {\mathop {\inf }}\limits }_{a_m\in {\varDelta {^\mathcal I }}}\max ({\mathop {\mathop {\max }}\limits }^{m-1}_{i=1}\ c(R_i{^\mathcal I } (a_{i-1},a_i)),\max (c(R_m{^\mathcal I } (a_{m-1},a_m)),C{^\mathcal I } (a_m)))\ge n$	$\Rightarrow _{\text{ Property } (\blacklozenge )}$
(5) $\max ({\mathop {\mathop {\max }}\limits }^{m-1}_{i=1}\ c(R_i{^\mathcal I } (a_{i-1},a_i)),{\mathop {\mathop {\inf }}\limits }_{a_m\in {\varDelta {^\mathcal I }}}\max (c(R_m{^\mathcal I } (a_{m-1},a_m)),C{^\mathcal I } (a_m)))\ge n$	$\Rightarrow$
(6) $\max ({\mathop {\mathop {\max }}\limits }^{m-1}_{i=1}\ c(R_i{^\mathcal I } (a_{i-1},a_i)),(\forall R_m.C){^\mathcal I } (a_{m-1}))\ge n$

Now, note that $a_{m-1}$ is an arbitrary object of ${\varDelta {^\mathcal I }}$ and that fuzzy DLs under the standard fuzzy operators satisfy the witnessed model property; hence, from (6) we have ${\mathop {\mathop {\inf }}\limits }_{a_{m-1}}\max ({\mathop {\mathop {\max }}\limits }^{m-1}_{i=1}\ c(R_i{^\mathcal I } (a_{i-1},a_i)),(\forall R_m.C){^\mathcal I } (a_{m-1}))\ge n$. Working similarly as above we can infer $\max ({\mathop {\mathop {\max }}\limits }^{m-2}_{i=1}\ c(R_i{^\mathcal I } (a_{i-1},a_i)),(\forall R_{m-1}(\forall R_m.C)){^\mathcal I } (a_{m-2}))\ge n$. Consequently, after $m$ steps we can infer $\forall R_1.(\forall R_2.(\ldots (\forall R_m.C)))^\mathcal{I } (a)\ge n$. $\square$

Theorem 1

Let ${\Sigma }$ be an ${\text{ f }_{KD}\text{- }\mathcal{SROIQ }}$ KB. ${\Sigma }$ is satisfiable if and only if $\tau ({\Sigma })$ is satisfiable.

Proof

For the ‘only-if’ direction assume that $\mathcal{I }$ is a model of ${\Sigma }$ but not of $\tau ({\Sigma })$. By construction, $\tau ({\Sigma })$ is almost like ${\Sigma }$ but with some concepts of the form $\forall S.C$ replaced with fresh concepts of the form $i_S$ and some additional axioms of the form $s_i \sqsubseteq \forall R_i.s_{i+1}$. All axioms of ${\Sigma }$ that have not been modified in $\tau ({\Sigma })$ are clearly satisfied by $\mathcal{I }$. Moreover, for those axioms of ${\Sigma }$ where $\forall S.C$ has been re-written using $i_S$, since $i_S$ is a fresh concept the model $\mathcal{I }$ can be trivially extended to $\mathcal{I }^{\prime }$ which satisfies them. More precisely, if $(\forall S.C)^\mathcal{I } (a)=n$, then we can set $i^{\mathcal{I }^{\prime }}_S(a)=n$. Hence, that $\mathcal{I }$ does not satisfy $\tau ({\Sigma })$ it must be due to the newly added axioms $s_i \sqsubseteq \forall R_i.s_{i+1}$. However, again, since each $s_{i}$ is a fresh concept, $\mathcal{I }^{\prime }$ can be further extended to satisfy these axioms. Thus, that $\mathcal{I }^{\prime }$ is not a model it must be due to a set of axioms of the form $i_S \sqsubseteq \forall R_1.s_1,\ldots ,s_{n}\sqsubseteq \forall R_m.f_S, f_S\sqsubseteq C$, i.e., that also involves the axiom with $C$. These axioms can be unfolded into a single axiom of the form $i_S\sqsubseteq \forall R_1.(\forall R_2.(\ldots (\forall R_m.C)))$ and finally to $\forall S.C\sqsubseteq \forall R_1.(\forall R_2.(\ldots (\forall R_m.C)))$. By construction of $\tau ({\Sigma })$, it follows that $R_1\ldots R_m\in L(\mathcal{B }_S)$, hence, by Lemma 5 $\mathcal{I }$ must satisfy the axiom which leads to a contradiction.

For the ‘if’ direction, assume that $\mathcal{I }$ is a model of $\tau ({\Sigma })$. A model $\mathcal{I }^{\prime }$ for ${\Sigma }$ can be constructed as follows:

• $\Delta ^{\mathcal{I }^{\prime }}={\varDelta {^\mathcal I }}$

• For each individual $a, a^{\mathcal{I }^{\prime }}=a^\mathcal{I }$

• For each atomic concept $A\in cl({\Sigma }), A^{\mathcal{I }^{\prime }}(a)=A^\mathcal{I } (a)$

• If $R$ is minimal w.r.t. $\prec$, then $R^{\mathcal{I }^{\prime }}(a,b)=R^\mathcal{I } (a,b)$

• If $R$ is not minimal w.r.t. $\prec$, then

  $$\begin{aligned} R^{\mathcal{I }^{\prime }}(a,b)=\max (R^\mathcal{I } (a,b),\min (R^\mathcal{I } _1(a,x_1),R^\mathcal{I } _2(x_1,x_2),\ldots ,R^\mathcal{I } _m(x_{m-1},b))) \end{aligned}$$

  where $R_i\prec S$ and there are axioms of the form $s_i\sqsubseteq \forall R_i.s_{i+1}$ in $\tau ({\Sigma })$, with $1\le i\le m$.

Note that the interpretation of complex roles is inductive. However, since $\mathcal{R } _h$ is regular the induction on $\prec$ is well founded. By Proposition 2 and the inductive interpretation of complex roles using $\prec$ and $\mathcal{B }_S$ it follows that $\mathcal{I }^{\prime }$ is a model of the role hierarchy of ${\Sigma }$. Moreover, by induction on the structure of concepts, it also follows that $D^\mathcal{I } (a)=D^{\mathcal{I }^{\prime }}(a)$. More precisely, if $D=C_1\sqcap C_2$ and $(C_1\sqcap C_2)^\mathcal{I } (a)=n$ we have $C_1^\mathcal{I } (a)=n_1, C_2^\mathcal{I } (a)=n_2$, with $n=\min (n_1,n_2)$. Then, by the induction hypothesis, we have $C_1^{\mathcal{I }^{\prime }}(a)=n_1, C_2^{\mathcal{I }^{\prime }}(a)=n_2$ and hence $(C_1\sqcap C_2)^{\mathcal{I }^{\prime }}(a)=\min (n_1,n_2)=n$. All other cases follow in a similar way. The only non-trivial interesting case is if $D=\forall S.C,(\forall S.C)^\mathcal{I } (a)=n,S^{\mathcal{I }^{\prime }}(a,b)=p$, and $S$ is complex. If $1-p=n$ (i.e. $p=1-n$), then $\max (1-p,C^{\mathcal{I }^{\prime }}(b))=n$ regardless of $C^{\mathcal{I }^{\prime }}(b)$ and hence $(\forall S.C)^{\mathcal{I }^{\prime }}(a)=n$. Assume that $p>n$. Since $(\forall S.C)^\mathcal{I } (a)=n$ we have that $i^\mathcal{I } _S(a)=n$. Now there are two cases:

• $S^\mathcal{I } (a,b)=p$. Since $S$ is complex $\tau ({\Sigma })$ contains $i_S\sqsubseteq \forall S.f_S$ and $f_S\sqsubseteq C$. Since $p>n$ and $\mathcal{I }$ is a model of $\tau ({\Sigma })$ we must have $C^\mathcal{I } (b)=n$.

• $S^\mathcal{I } (a,b)\ne p$. Then, there exist $m R_i$ with $R^\mathcal{I } _1(a,x_1)=p_1,\ldots R^\mathcal{I } _m(x_m,b)=p_m$ s.t. $p=\min (p_1,\ldots .p_m)$. By construction of $\mathcal{I }^{\prime }$ there are axioms $s_i\sqsubseteq \forall R_i.s_{i+1}$ and $s_{m+1}\sqsubseteq f_S, f_S\sqsubseteq C$, in $\tau ({\Sigma })$. Moreover, since $p>n$, then also $p_i>n$ for each $1\le i\le m$. But then, since $\mathcal{I }$ is a model of $\tau ({\Sigma })$, for all $s_i$ we must have $s^\mathcal{I } _{i}(x_i)=n$ and hence, also $C^\mathcal{I } (b)=n$.

In both cases, by induction hypothesis, $C^{\mathcal{I }^{\prime }}(a)=n$, hence we finally get $(\forall S.C)^{\mathcal{I }^{\prime }}(a)=n$.

Concluding, we need to show that $\mathcal{I }^{\prime }$ satisfies each axiom $C\sqsubseteq D$. Since $\mathcal{I }$ is a model of $\tau ({\Sigma })$ we have $C^\mathcal{I } (a)\le D^\mathcal{I } (a)$. However, as shown $C^{\mathcal{I }^{\prime }}(a)=C^\mathcal{I } (a)\le D^\mathcal{I } (a)=D^{\mathcal{I }^{\prime }}(a)$, hence also $\mathcal{I }$ satisfies $C\sqsubseteq D$. $\square$