Revision of prioritized $\mathcal {E}\mathcal {L}$ ontologies

Abstract

In this paper, we investigated the evolution of prioritized $\varvec{\mathcal {E}}\varvec{\mathcal {L}}$ ontologies in the presence of new information that can be certain or uncertain. We propose an extension of $\varvec{\mathcal {E}}\varvec{\mathcal {L}}$ description logic, named $\varvec{\mathcal {E}}\varvec{\mathcal {L}}_{\bot }^{+}$, within possibility theory to encode such knowledge. This extension provides a natural way to deal with the ordinal scale and represent knowledge in a way that can handle incomplete information and conflicting data. We provided a polynomial algorithm for computing the possibilistic entailment. Then, we defined the evolution process at the semantic and syntactic levels. Interestingly enough, we show that the syntactical algorithm is done in polynomial time.

Appendix A

Proposition 1

Let $\mathcal {O}$ be an assertion free $\mathcal {E}\mathcal {L}_{\bot }^{+}$ ontology and let $A,B \in N_c$ be two concepts of $\mathcal {O}$. Then:

• 1. $\mathcal {O} \models A\sqsubseteq B$ iff $A\sqsubseteq B \in cl(\mathcal {O})$ or $A\sqsubseteq \bot \in cl(\mathcal {O})$.

• 2. The complexity of computing $cl(\mathcal {O})$ is polynomial.

Proof

Suppose that $\mathcal {O} \models A\sqsubseteq B$. This means that in every model of $\mathcal {O}$, every instance of the concept A is also an instance of the concept B. By the definition of the closure operator $cl(\mathcal {O})$, we have $A\sqsubseteq B \in cl(\mathcal {O})$. Now, Considering that $A\sqsubseteq B \in cl(\mathcal {O})$ or $A\sqsubseteq \bot \in cl(\mathcal {O})$. If $A\sqsubseteq B \in cl(\mathcal {O})$, it means that $A\sqsubseteq B$ is a logical consequence of $\mathcal {O}$. Thus, in every model of $\mathcal {O}$, every instance of concept A is also an instance of concept B, which implies $\mathcal {O} \models A\sqsubseteq B$. Similarly, if $A\sqsubseteq \bot \in cl(\mathcal {O})$, it means that $A\sqsubseteq \bot$ is a logical consequence of $\mathcal {O}$. Since $\bot$ represents the bottom concept (the unsatisfiable concept), this implies that there are no instances of concept A in any model of $\mathcal {O}$. Hence, $\mathcal {O} \models A\sqsubseteq B$ holds trivially.

To prove that the complexity of computing $cl(\mathcal {O})$ is polynomial, we need to show that there exists a polynomial-time algorithm that can compute the closure $cl(\mathcal {O})$. Since $\mathcal {O}$ is an assertion-free $\mathcal {E}\mathcal {L}_{\bot }^{+}$ ontology, it consists of concept inclusions of the form $A \sqsubseteq B$ and role inclusions of the form $r \sqsubseteq s$. We can construct an initial closure set $cl(\mathcal {O})$ that contains all the concept inclusions of $\mathcal {O}$. Then, for each concept inclusion $A \sqsubseteq B$ in $cl(\mathcal {O})$, we check if $A \sqsubseteq \bot$ is a logical consequence of $\mathcal {O}$. If it is, we add $A \sqsubseteq \bot$ to $cl(\mathcal {O})$.

This process of checking and adding concept inclusions can be done in polynomial time since the size of $cl(\mathcal {O})$ is bounded by the size of $\mathcal {O}$. Hence, the complexity of computing $cl(\mathcal {O})$ is polynomial. $\square$

Lemma 1

Considering that $\mathcal {O}_\pi$ is the prioritized $\mathcal {E}\mathcal {L}_{\bot }^{+}$ ontology that includes the two axioms $(\phi ,\alpha _1)$ and $(\phi ,\alpha _2)$ then $\mathcal {O}_\pi$ and $\mathcal {O}'_\pi =\{\mathcal {O}_\pi \backslash \{(\phi ,\alpha _1),(\phi ,\alpha _2)\}\}\cup \{(\phi ,\max (\alpha _1,\alpha _2))\}$ are equivalent in the sense that they induce the same possibility distribution, i.e., $\forall \mathcal {I}\in \Omega , \pi _\mathcal {O}(\mathcal {I})=\pi _{\mathcal {O}'}(\mathcal {I})$.

Proof

The proof is given immediately from the definition of the possibility distribution in Definition 3$\square$

Proposition 2

Let $\mathcal {O}_\pi$ be the prioritized $\mathcal {E}\mathcal {L}_{\bot }^{+}$ ontology and $\mathcal {O}'_\pi$ be the ontology obtained by applying the normalization rules in Table 5 into $\mathcal {O}_\pi$. Then $\mathcal {O}_\pi$ and $\mathcal {O}'_\pi$ induce the same possibility distribution.

Sketch of proof

We give the proof for some rules. The proof of the other follows similarly.

$({\textbf {PNR}}_0)$. Let $\mathcal {I}$ be an interpretation. Considering that $\mathcal {I}\models (C_1 \sqcap \top \sqcap C_2 \sqsubseteq D,\alpha )$ by definition of satisfaction relation (for instance see Table 1), we have:

$$(C_1 \cap \top \cap C_2)^ \mathcal {I}\subseteq (D)^\mathcal {I}$$

Hence

$$((C_1)^\mathcal {I}\cap (\Delta )^\mathcal {I}\cap (C_2)^\mathcal {I})\subseteq (D)^\mathcal {I}$$

Then

$$((C_1)^\mathcal {I}\cap (C_2)^\mathcal {I})\subseteq D^\mathcal {I}$$

which means that $\mathcal {I}\models C_1\sqcap C_2\sqsubseteq D$, and $\pi _{\mathcal {O}_\pi }(\mathcal {I})=\pi _{\mathcal {O}'_\pi }(\mathcal {I})$

If $\mathcal {I}\not \models (C_1\sqcap \top \sqcap C_2\sqsubseteq D, \alpha )$, the proof follows similarly.

$({\textbf {PNR}}_1)$. Let $\mathcal {I}$ be an interpretation. Considering that $\mathcal {I}\models (C_1 \sqcap \bot \sqcap C_2 \sqsubseteq D,\alpha )$ by definition of satisfaction relation in Table 1, we have:

$$(C_1 \cap \bot \cap C_2)^ \mathcal {I}\subseteq (D)^\mathcal {I}$$

Hence

$$((C_1)^\mathcal {I}\cap (\bot )^\mathcal {I}\cap (C_2)^\mathcal {I})\subseteq (D)^\mathcal {I}$$

Then

$$((C_1^\mathcal {I}\cap \emptyset \cap (C_2)^\mathcal {I})\subseteq D^\mathcal {I}$$

As a result

$$(\emptyset )\subseteq D^\mathcal {I}$$

which means that $\mathcal {I}\models \bot \sqsubseteq D$, and $\pi _{\mathcal {O}_\pi }(\mathcal {I})=\pi _{\mathcal {O}'_\pi }(\mathcal {I})$

If $\mathcal {I}\not \models C_1\sqcap \bot \sqcap C_2\sqsubseteq D$, the proof follows similarly.

$({\textbf {PNR}}_2)$: let $\mathcal {O}_\pi$ be a prioritized $\mathcal {E}\mathcal {L}_{\bot }^{+}$ ontology where $\mathcal {O}_\pi =\{(C\sqsubseteq D_1\sqcap D_2, \alpha )\}$. Let $\mathcal {O}'_\pi$ be the prioritized ontology obtained after applying $PNR_2$ to the ontology, i.e., $\mathcal {O}_\pi '=\{(C\sqsubseteq D_1,\alpha )~ \text {and} ~(C\sqsubseteq D_2,\alpha )\}$. Then $\mathcal {O}_\pi$ and $\mathcal {O}_\pi '$ induce the same possibility distribution.

Let $\mathcal {I}$ be the interpretation that satisfies $(C\sqsubseteq D_1 \sqcap D_2,\alpha )$, namely $\mathcal {I}\models (C\sqsubseteq D_1 \sqcap D_2,\alpha )$ by definition of satisfaction relation in Table 1, we have:

$$(C)^\mathcal {I}\subseteq (D_1\cap D_2)^\mathcal {I}$$

Hence

$$(C)^\mathcal {I}\subseteq (D_1)^\mathcal {I}\;\;\text {and}\;\; (C)^\mathcal {I}\subseteq (D_2)^\mathcal {I}$$

This means that

$$\mathcal {I}\models C\sqsubseteq D_1 \;\;\text {and}\;\; \mathcal {I}\models C\sqsubseteq D_2$$

Therefore $\pi _{\mathcal {O}_\pi }(\mathcal {I})=\pi _{\mathcal {O}'_\pi }(\mathcal {I})$.

Conversely, assume that $\mathcal {I}\models (C\sqsubseteq D_1,\alpha )$ and $\mathcal {I}\models (C\sqsubseteq D_2,\alpha )$ by definition of satisfaction relation, we have $C^\mathcal {I}\subseteq D_1^\mathcal {I}$ and $C^\mathcal {I}\subseteq D_2^\mathcal {I}$ then $C^\mathcal {I}\subseteq ( D_1^\mathcal {I}\cap D_2^\mathcal {I})$, therefore $\mathcal {I}\models C\sqsubseteq D_1\sqcap D_2$ and $\pi _{\mathcal {O}_\pi }(\mathcal {I})=\pi _{\mathcal {O}'_\pi }(\mathcal {I})$. The other case when $\mathcal {I}\not \models C\sqsubseteq D_1\sqsubseteq D_2$ follows similarly. $\square$

Proposition 3

Let $\mathcal {O}_\pi$ be the prioritized $\mathcal {E}\mathcal {L}_{\bot }^{+}$ ontology and $cl(\mathcal {O}_\pi )$ be its closed ontology, obtained by applying the rules given in Tables 5 and 6 into $\mathcal {O}_\pi$. Then $\mathcal {O}_\pi$ and $cl(\mathcal {O}_\pi )$ induce the same possibility distribution.

Proof

The proof comes down to deduce that the application of normalization rules and inference rules does not change the possibility distribution, it is enough to keep repeating the application of rules and see the possibility distribution. For the normalization rules (Table 5) the proof is given above. Now for the inference rules, considering that $(C_1\sqsubseteq C_2,\alpha _1), (C_2\sqsubseteq C_3,\alpha _2)\in \mathcal {O}_\pi$. Let us show that by applying the rule ($\text {PIR}_3$) into $\mathcal {O}_\pi$ the closed ontology $cl(\mathcal {O}_\pi )$ contains the axiom $(C_1\sqsubseteq C_3,\min (\alpha _1,\alpha _2)\in cl(\mathcal {O}_\pi )$ and the possibility distribution does not change, namely $cl(\mathcal {O}_\pi )=\mathcal {O}_\pi \sqcup \{C_1\sqsubseteq C_3,\min (\alpha _1,\alpha _2)\}$ are equivalent. Let $\mathcal {I}=(\Delta ^\mathcal {I},.^\mathcal {I})$ a DL interpretation. We have four cases:

• $\mathcal {I}\models C_1\sqsubseteq C_2$ and $\mathcal {I}\models C_2\sqsubseteq C_3$ by definition of satisfaction relation in Table 1, we have $C_1^\mathcal {I}\subseteq C_2^\mathcal {I}$ and $C_2^\mathcal {I}\subseteq C_3^\mathcal {I}$ hence $C_1^\mathcal {I}\subseteq C_3^\mathcal {I}$ which means that $\mathcal {I}\models C_1\sqsubseteq C_3$. Therefore $\pi _{\mathcal {O}_\pi }(\mathcal {I})=\pi _{cl(\mathcal {O}_\pi )}(\mathcal {I})$

• $\mathcal {I}\models C_1\sqsubseteq C_2$ and $\mathcal {I}\not \models C_2\sqsubseteq C_3$ let $\mathcal {O}''_\pi =\mathcal {O}_\pi$\$\{(C_1\sqsubseteq C_2,\alpha _1)\text { and }(C_2\sqsubseteq C_3,\alpha _2)\}$ then $\pi _{\mathcal {O}_\pi }(\mathcal {I})=\min (\pi _{\mathcal {O}''_\pi },1-\alpha _2$) $=\min (\pi _{\mathcal {O}''_\pi },1-\alpha _2,1-\min (\alpha _1,\alpha _2))$ $=\pi _{cl(\mathcal {O})_\pi }(\mathcal {I})$

• $\mathcal {I}\not \models C_1\sqsubseteq C_2$ and $\mathcal {I}\models C_2\sqsubseteq C_3$: follow similarly.

• $\mathcal {I}\not \models C_1\sqsubseteq C_2$ and $\mathcal {I}\not \models C_2\sqsubseteq C_3$ let $\mathcal {O}''_\pi =\mathcal {O}_\pi$\$\{(C_1\sqsubseteq C_2,\alpha _1),(C_2\sqsubseteq C_3,\alpha _2)\}$ we have $\pi _{\mathcal {O}_\pi }(\mathcal {I})=\min (\pi _{\mathcal {O}''_\pi }(\mathcal {I}),1-\alpha _1,1-\alpha _2))$ $=\min (\pi _{\mathcal {O}''_\pi }(I),1-\alpha _1,1-\alpha _2,1-\min (\alpha _1,\alpha _2))$ $=\pi _{cl(\mathcal {O}_\pi )}(\mathcal {I})$ $\square$

Proposition 4

Let $\mathcal {O}_\pi$ be a prioritized $\mathcal {E}\mathcal {L}_{\bot }^{+}$ ontology and $cl(\mathcal {O}_\pi )$ be its closure by applying the normalization rules in Table 5 and inference rules in Table 6. Let A and B be two concepts of $\mathcal {O}_\pi$. Then $\mathcal {O}_\pi \models (A\sqsubseteq B, \alpha )$ if

• $(A\sqsubseteq B,\beta )\in cl(\mathcal {O}_\pi )$ with $\beta \ge \alpha$

• $(A\sqsubseteq \bot ,\beta )\in cl(\mathcal {O}_\pi )$ with $\beta \ge \alpha$

Proof

The proof is immediate. It is given from the definition of the inference rules given in Table 6. $\square$

Proposition 5

Let $\mathcal {O}_\pi$ be a prioritized $\mathcal {E}\mathcal {L}_{\bot }^{+}$ ontology and $\pi _{\mathcal {O}_\pi }$ be its associated possibility distribution. Let $(\phi , u)$ be the new input information. Then $\pi _1(\mathcal {I})= \pi _{\mathcal {O}_\pi }(.\mid (\phi ))$ obtained by Definition 4 satisfies the postulates $\textbf{P1}$- $\textbf{P2}$.

Proof

The proof is as follows:

• P1: $\pi '(\mathcal {I})=1$: To prove this, we need to show that $\pi '(\mathcal {I})=1$ for any interpretation $\mathcal {I}$. If $\pi (\mathcal {I})=\Pi (\phi )$ and $\mathcal {I}\models \phi$, then by the definition of $\pi '_{\mathcal {O}_\pi }$, we have $\pi '(\mathcal {I})=1$. If $\pi (\mathcal {I})<\Pi (\phi )$, then by the definition of $\pi '_{\mathcal {O}_\pi }$, we have $\pi '(\mathcal {I})=\pi (\mathcal {I})>0$. Since $\pi$ is a possibility distribution, we have $\pi (\mathcal {I})\le 1$. Therefore, $\pi '(\mathcal {I})>0$ and $\pi '(\mathcal {I})\le 1$, which implies $\pi '(\mathcal {I})=1$. If $\mathcal {I}\not \models \phi$, then by the definition of $\pi '_{\mathcal {O}_\pi }$, we have $\pi '(\mathcal {I})=0\le 1$. Therefore, $\pi '(\mathcal {I})=1$ in all cases, which satisfies P1.

• P2: if $\pi (\mathcal {I})=0$ then $\pi '(\mathcal {I})=0$: To prove this, we need to show that if $\pi (\mathcal {I})=0$ for an interpretation $\mathcal {I}$, then $\pi '(\mathcal {I})=0$. If $\pi (\mathcal {I})=0$, then by the definition of possibility distribution, $\mathcal {I}$ is completely impossible, i.e., $\pi (\mathcal {I})=0$ implies $\Pi (\lnot \mathcal {I})=1$. Therefore, $\Pi (\phi \wedge \lnot \mathcal {I})=\Pi (\phi )\wedge \Pi (\lnot \mathcal {I})=\Pi (\phi )\wedge 0=0$. This means that $\phi \wedge \lnot \mathcal {I}$ is completely impossible, i.e., $\mathcal {I}\not \models \phi$ implies $\phi \wedge \lnot \mathcal {I}$ is completely impossible. By the definition of $\pi '_{\mathcal {O}_\pi }$, if $\mathcal {I}\not \models \phi$, then $\pi '(\mathcal {I})=0$. Therefore, $\pi '(\mathcal {I})=0$ for any interpretation $\mathcal {I}$ with $\pi (\mathcal {I})=0$, which satisfies P2. $\square$

Proposition 6

Let $\mathcal {O}_\pi$ be a prioritized $\mathcal {E}\mathcal {L}_{\bot }^{+}$ ontology and $\pi _{\mathcal {O}_\pi }$ be its associated possibility distribution. Let $(\phi , u)$ be the new input information. Then $\pi '_{\mathcal {O}_\pi }= \pi _{\mathcal {O}_\pi }(.\mid (\phi ,u))$ obtained by Definition 5 satisfies the postulates $\textbf{P1}$- $\textbf{P3}$.

Proof

The proof is as follows:

• To prove that the above definition satisfies P1, we need to show that for any interpretation $\mathcal {I}$, the possibility distribution $\pi '(\mathcal {I})$ obtained after min-based conditioning satisfies $\pi '(\mathcal {I})=1$. Let $(\phi , u)$ be the uncertain input, and let $\mathcal {I}$ be an interpretation such that $\mathcal {I}\models \phi$. Then, according to the definition of min-based conditioning, we have:

  $$\begin{aligned} \pi '(\mathcal {I})= & {} \pi _{\mathcal {O}\pi }(\mathcal {I}\mid _m(\phi , u))\\= & {} {\left\{ \begin{array}{ll} 1\;\;\text {if}\;\; \pi _{\mathcal {O}_\pi }(\mathcal {I})=\Pi (\phi )\\ 1-u\;\; \text {if}\;\; \Pi (\phi )\le \pi _{\mathcal {O}_\pi }\le 1-u\\ \pi _{\mathcal {O}_\pi }(\mathcal {I})\;\text {Otherwise} \end{array}\right. } \end{aligned}$$

  Since $\mathcal {I}\models \phi$, we have $\pi _{\mathcal {O}_\pi }(\mathcal {I})\ge \Pi (\phi )$, so the first two cases do not apply. Therefore, we have:

  $$\begin{aligned} \pi '(\mathcal {I})=\pi _{\mathcal {O}_\pi }(\mathcal {I})=1 \end{aligned}$$

  On the other hand, if $\mathcal {I}$ is such that $\mathcal {I}\not \models \phi$, then we have:

  $$\begin{aligned} \forall \mathcal {I}\not \models \phi , \pi _{\mathcal {O}_\pi }(.\mid _m(\phi , u))={\left\{ \begin{array}{ll} 1-u\;\; \text {if}\;\; \pi _{\mathcal {O}_\pi }(\mathcal {I})=N_\pi (\phi )\\ 1-u\;\; \text {if}\;\; \pi _{\mathcal {O}_\pi }(\mathcal {I})> 1-u\\ \pi ({\mathcal {I}})\;\text {Otherwise} \end{array}\right. } \end{aligned}$$

  Since $\pi _{\mathcal {O}\pi }(\mathcal {I})\le N_\pi (\phi )$ if $\mathcal {I}\not \models \phi$, the first case does not apply. Since $\pi _{\mathcal {O}\pi }(\mathcal {I})\le 1-u$ by definition of $N_\pi (\phi )$, the second case does not apply either. Therefore, we have:

  $$\begin{aligned} \pi '(\mathcal {I})=\pi _{\mathcal {O}_\pi }(\mathcal {I})=1 \end{aligned}$$

  In both cases, we have $\pi '(\mathcal {I})=1$, which completes the proof that the above definition satisfies P1.

• P2: if $\pi (\mathcal {I})=0$ then $\pi '(\mathcal {I})=0$. Let $\mathcal {I}\not \models \phi$ and $\pi _{\mathcal {O}_\pi }(\mathcal {I})=0$. Then we have:

  $$\begin{aligned} \forall \mathcal {I}\not \models \phi , \pi _{\mathcal {O}_\pi }(.\mid _m(\phi , u))={\left\{ \begin{array}{ll} 1-u\;\; \text {if}\;\; \pi _{\mathcal {O}_\pi }(\mathcal {I})=N_\pi (\phi )\\ 1-u\;\; \text {if}\;\; \pi _{\mathcal {O}_\pi }(\mathcal {I})> 1-u\\ \pi ({\mathcal {I}})\;\text {Otherwise} \end{array}\right. } \end{aligned}$$

  Therefore, the property P2 is satisfied.

• To prove that Definition 5 satisfies P3, we need to show that: $\Pi '(\phi )=1$ and $N'(\phi )\ge u$. To show that $\Pi '(\phi )=1$, we need to show that for all interpretations $\mathcal {I}$ that satisfy $\phi$, $\pi '(\mathcal {I})=1$. There are two cases to consider:

  • 1. $\pi _{\mathcal {O}_\pi }(\mathcal {I})=\Pi (\phi )$ In this case, we have: $\pi _{\mathcal {O}_\pi }(.\mid _m(\phi ,u))=1$ then, $\pi '(\mathcal {I})=1$

  • 2. $\Pi (\phi )\le \pi _{\mathcal {O}_\pi }(\mathcal {I})\le 1-u$ In this case, we have: $\pi _{\mathcal {O}_\pi }(.\mid _m(\phi ,u))=1-u$ then, $\pi '(\mathcal {I})=1-u<1$

  However, this case cannot happen because of the assumption that $\Pi (\phi )>1-u$. Therefore, we have shown that for all interpretations $\mathcal {I}$ that satisfy $\phi$, $\pi '(\mathcal {I})=1$, and hence $\Pi '(\phi )=1$. Now, to show that $N'(\phi )\ge u$, we need to show that for all interpretations $\mathcal {I}$ that do not satisfy $\phi$, $\pi '(\mathcal {I})\le 1-u$. There are two cases to consider:

  • 1. $\pi _{\mathcal {O}\pi }(\mathcal {I})=N\pi (\phi )$ In this case, we have: $\pi _{\mathcal {O}_\pi }(.\mid _m(\phi ,u))=1-u$, then $\pi '(\mathcal {I})=1-u\le 1-u$

  • 2. $\pi _{\mathcal {O}_\pi }(\mathcal {I})> 1-u$ In this case, we have: $\pi _{\mathcal {O}_\pi }(.\mid _m(\phi ,u))=1-u$ then, $\pi '(\mathcal {I})=1-u\le 1-u$

  Therefore, we have shown that for all interpretations $\mathcal {I}$ that do not satisfy $\phi$, $\pi '(\mathcal {I})\le 1-u$, and hence $N'(\phi )\ge u$. $\square$

Proposition 7

Let $\mathcal {O}_\pi$ be a prioritized $\mathcal {E}\mathcal {L}_{\bot }^{+}$ ontology, and $\pi _{\mathcal {O}_\pi }$ be the associated possibility distribution. Let $(\phi , u)$ be the new input information and $\lambda =Inc(\mathcal {O}'_\pi )$, where $\mathcal {O}'_\pi =\mathcal {O}_\pi \cup \{(\phi ,1)\}$. Then, the new prioritized ontology $\mathcal {O}'_\pi$ is defined as follows:

$$\mathcal {O}'_\pi = \{(\phi ,u)\}\cup \{(\Phi ,\alpha ): (\Phi ,\alpha )\in \mathcal {O}_\pi \;\;\text {and}\;\; \alpha > \lambda \} $$

The associated possibility distribution obtained by conditioning defined in Definition 5 is as follows:

$$\forall \mathcal {I}\in \Omega , \pi '_{\mathcal {O}_\pi }(\mathcal {I})=\pi _{\mathcal {O}_\pi }(\mathcal {I}\mid _m (\phi , u))$$

With $\pi _{\mathcal {O}_\pi }(\mathcal {I}\mid _m (\phi ,u))$ represents the possibility distribution obtained by $\pi _{\mathcal {O}_\pi }$ with min-based conditioning in Definition 5.

Proof

First, point out that $\Pi (\phi )=1-\lambda$. Let $\mathcal {I}$ be an interpretation that falsifies the input information $\phi$, then for each interpretation that falsified $\phi$, $\pi _{\mathcal {O}_\pi }= \pi _{\mathcal {O}'_\pi }(\mathcal {I})=0$. Now, considering that $\mathcal {I}$ satisfies $\phi$, then, if $\pi _{\mathcal {O}_\pi }(\mathcal {I})=\Pi (\phi )= 1-\lambda$, this means that the interpretation $\mathcal {I}$ satisfies all the axioms with a weight greater than $1-\lambda$, hence $\mathcal {I}$ satisfies all the axioms of $\mathcal {O}'_\pi$. As a result $\pi _{\mathcal {O}'_\pi }(\mathcal {I})=1$. Considering now that $\pi _{\mathcal {O}_\pi }(\mathcal {I})< 1-\lambda$, then:

$$\begin{aligned} \pi _{\mathcal {O}_\pi }(\mathcal {I}) \quad= & {} \min \{1-\alpha :(\Phi ,\alpha \in \mathcal {O}_\pi \;\text {and}\; \mathcal {I}\not \models \phi \} \\ \nonumber= & {} \min \{\min \{1\!-\!\alpha :(\Phi ,\alpha )\in \mathcal {O}_\pi \;\text {and}\; \mathcal {I}\not \models \Phi \;\text {and}\;\alpha \!>\!\lambda \},\\ \nonumber{} & {} min\{1-\alpha :(\Phi ,\alpha )\in \mathcal {O}_\pi \;\text {and}\; \mathcal {I}\not \models \Phi \;\text {and}\;\alpha \le \lambda \}\}\\ \nonumber= & {} \min \{1-\alpha :(\Phi ,\alpha )\in \mathcal {O}_\pi \;\text {and}\; \mathcal {I}\not \models \Phi \;\text {and}\;\alpha >\lambda \}\\{} & {} \times (\text {since}\; \pi _{\mathcal {O}_\pi }<1-\lambda )\\ \nonumber= & {} \pi _{\mathcal {O}'_\pi } \end{aligned}$$

$\square$

Proposition 8

The computational of the Algorithm 1 is efficient and done in polynomial time.

Proof

The computation of the obtained ontology by the Algorithm 1 is efficient. Its complexity is the same as the one of computing the inconsistency of $\pi$ $\mathcal {E}\mathcal {L}_{\bot }^{+}$ ontology that is given in [29], which is done in polynomial time. $\square$

Proposition 9

Let $\mathcal {O}_\pi$ be the prioritized $\mathcal {E}\mathcal {L}_{\bot }^{+}$ ontology and $(\phi , u)$ be the uncertain input. Considering that $\mathcal {O}'_\pi$ is the augmented ontology by the assumption that $\phi$ is false. Then the degree of inconsistency of $\mathcal {O}'_\pi$ is $\lambda _{inc}= Inc(\mathcal {O}'_\pi )$. The revised prioritized $\mathcal {E}\mathcal {L}_{\bot }^{+}$ ontology is defined as follows:

$$\mathcal {O}'_\pi = \{(\phi ,u)\}\cup \{(\phi , \alpha ): (\phi , \alpha )\in \mathcal {O}_\pi \; \text {and}\; u>\lambda _{inc}\}$$

$$\cup \{(\phi , \alpha ): (\phi , u)\in \mathcal {O}_\pi \; \text {and}\; \alpha <u\}$$

$$\cup \{ (\phi , \alpha ): (\phi , \alpha )\in \mathcal {O}_\pi \; \text {and}\; u\le \alpha \le \lambda _{inc}\}$$

Their associated possibility distributions using the min operator defined in Definition 5 are defined as follows:

$$\forall \mathcal {I}\in \Omega , \pi _{\mathcal {O}'_\pi }(\mathcal {I})=\pi _{\mathcal {O}_\pi }(\mathcal {I}\mid _m(\phi , u))$$

Proof

Considering the case when the ontology $\mathcal {O}_\pi$ is inconsistent with $\lnot \phi$, this means that $\Pi (\phi )=1$ and $\Pi (\lnot \phi )=1-\lambda _{inc}$. Let $\mathcal {I}$ be an interpretation that satisfies $\phi$. Then:

$$\begin{aligned}{} & {} \pi _{\mathcal {O}_\pi }(\mathcal {I}) \quad = Min \{\min \{1-\alpha _i:(\phi ,\alpha _i)\in \mathcal {O}_\pi \;\text {and}\; \mathcal {I}\not \models \phi _i \text {and}\; \alpha _i\\{} & {} \qquad \qquad \qquad \qquad \qquad> \lambda _{inc}\} \\ \nonumber{} & {} \qquad \qquad \quad \min \{1-\alpha _i:(\phi ,\alpha _i)\in \mathcal {O}_\pi \;\text {and}\; \mathcal {I}\not \models \phi _i\;\text {and}\;\alpha _i\le \lambda _{inc}\}\}\\ \nonumber{} & {} \qquad \qquad \quad = Min \{min\{1-\alpha _i:(\phi _i,\alpha _i)\in \mathcal {O}_\pi \;\text {and}\; \mathcal {I}\not \models \phi _i\;\text {and}\;\alpha _i\\{} & {} \qquad \qquad \qquad \qquad \qquad >\lambda _{inc}\},\\ \nonumber{} & {} \qquad \qquad \quad \min \{1-\alpha _i:(\phi _i,\alpha _i)\in \mathcal {O}_\pi \;\text {and}\; \mathcal {I}\not \models \phi _i\;\text {and}\;\alpha _i\le \lambda _{inc}\}\}\\ \nonumber{} & {} \qquad \qquad \quad = \min \{1-\alpha _i:(\phi _i,\alpha _i)\in \mathcal {O}_\pi \;\text {text}\; \mathcal {I}\phi _i\}\\ \nonumber{} & {} \qquad \qquad \quad = \pi _{\mathcal {O}_\pi }(\mathcal {I} \end{aligned}$$

When $\mathcal {I}\not \models \phi$, then:

$$\begin{aligned} \pi _{\mathcal {O}_\pi }(\mathcal {I})= & {} Min\{1-a, \min \{1-\alpha _i:(\phi _i,\alpha _i)\in \mathcal {O}_\pi \;\text {and}\\{} & {} \qquad \quad \mathcal {I}\not \models \phi _i\;\text {and}\;\alpha _i>\lambda _{inc}\}\} \end{aligned}$$

We have two scenarios:

• When $\pi _{\mathcal {O}_\pi }(\mathcal {I})=\Pi (\lnot \phi )=1-\lambda _{inc}$ which means that the interpretation $\mathcal {I}$ satisfies all the axioms of the ontology $\mathcal {O}_\pi$ having weight strictly greater than $\lambda _{inc}$, then:

  $$\begin{aligned}{} & {} \min \{1\!-\!\alpha _i:(\phi _i,\alpha _i)\in \mathcal {O}_\pi \;\text {and}\; \mathcal {I}\not \models \phi _i\;\text {and}\; \alpha _i>\lambda _{inc}\}\\{} & {} \quad =1 \end{aligned}$$

  Therefore $\pi _{\mathcal {O}_\pi }(\mathcal {I})=1-u$

• When $\pi _{\mathcal {O}_\pi }(\mathcal {I})<1-\lambda _{inc}$, then:

  $$\begin{aligned}{} & {} \pi _{\mathcal {O}'_\pi }(\mathcal {I}) =Min \{1-u,\min \{1-\alpha _i:(\phi ,\alpha _i)\in \mathcal {O}_\pi \;\text {and}\\{} & {} \qquad \qquad \qquad \qquad \qquad \mathcal {I}\not \models \phi _i \text {and}\; \alpha _i> \lambda _{inc}\}\} \\{} & {} \quad \quad \qquad = Min 1-u,\{min\{1-\alpha _i:(\phi _i,\alpha _i)\in \mathcal {O}_\pi \;\text {and}\\{} & {} \qquad \qquad \qquad \qquad \qquad \mathcal {I}\not \models \phi _i\;\text {and}\;\alpha _i>\lambda _{inc}\},\\{} & {} \quad \quad \qquad \min \{1\!-\!\alpha _i:(\phi _i,\alpha _i)\!\in \!\mathcal {O}_\pi \;\text {and}\; \mathcal {I}\!\not \models \!\phi _i\;\text {and}\;\alpha _i\!\le \!\lambda _{inc}\}\}\\{} & {} \quad \quad \qquad (since \pi _{\mathcal {O}_\pi }(\mathcal {I})<1-\lambda _{inc})\\{} & {} \quad \quad \qquad =Min\{1-u, \pi _{\mathcal {O}_\pi }(\mathcal {I})\} \end{aligned}$$

  $\square$

Proposition 10

Let $\mathcal {O}_\pi$ be a prioritized $\mathcal {E}\mathcal {L}$ ontology and $\pi _{\mathcal {O}_\pi }$ its associated possibility distribution. Let $(\phi , u)$ be the new input information. Therefore: $\pi '_{\mathcal {O}_\pi }=\pi _{\mathcal {O}_\pi }(.\mid _m(\phi , u))$ obtained by the Definition 6 satisfies the logical properties P1, P2 and P3

Proof

To prove that the given definition of min-based conditioning satisfies properties P1, P2, and P3, we need to show that the updated possibility distribution $\pi '$ satisfies these properties.

P1: $\pi '(\mathcal {I})=1$

Let $\mathcal {I}$ be any interpretation. We need to show that $\pi '(\mathcal {I})=1$. There are four cases to consider:

• Case 1: $\mathcal {I}\models \phi \cup \mathcal {S}$ In this case, $\pi '(\mathcal {I})=\pi (\mathcal {I})$ by the first condition in the definition of min-based conditioning. Since $\pi$ is a possibility distribution and $\mathcal {I}\models \phi \cup \mathcal {S}$, we have $\pi (\mathcal {I})\ge \Pi (\phi )>0$, which implies that $\pi '(\mathcal {I})=\pi (\mathcal {I})>0$.

• Case 2: $\mathcal {I}\models \phi \cup (\mathcal {O}_\pi \setminus \mathcal {S})$ and $\mathcal {I}\not \models \mathcal {S}$ In this case, $\pi '(\mathcal {I})=\pi (\mathcal {I})$ by the second condition in the definition of min-based conditioning. Since $\pi$ is a possibility distribution and $\mathcal {I}\models \phi \cup (\mathcal {O}_\pi \setminus \mathcal {S})$, we have $\pi (\mathcal {I})\ge N(\phi )>0$, which implies that $\pi '(\mathcal {I})=\pi (\mathcal {I})>0$.

• Case 3: $\mathcal {I}\models \phi$ and $\mathcal {I}\not \models \mathcal {S}$ and $\mathcal {I}\not \models \mathcal {O}_\pi \setminus \mathcal {S}$ and $\pi (\mathcal {I})=N(\phi )$ and $1-\alpha '\ge 1-u$ In this case, $\pi '(\mathcal {I})=1-u$ by the third condition in the definition of min-based conditioning. Since $1-\alpha '\ge 1-u$, we have $u\ge \alpha '$. Also, since $\pi (\mathcal {I})=N(\phi )>0$ and $\mathcal {I}\models \phi$, we have $\Pi (\phi )>0$. Therefore, $\pi (\mathcal {I})\ge \Pi (\phi )$, which implies that $\pi '(\mathcal {I})=1-u<1$.

• Case 4: $\mathcal {I}\not \models \phi$ and $\pi (\mathcal {I})>1-u$ In this case, $\pi '(\mathcal {I})=1-\alpha '$ by the fourth condition in the definition of min-based conditioning. Since $\pi (\mathcal {I})>1-u$, we have $\alpha '<u$, which implies that $1-\alpha '>1-u$. Also, since $\pi (\mathcal {I})>0$, we have $\pi '(\mathcal {I})=1-\alpha '<1$.

P2: if $\pi (\mathcal {I})=0$ then $\pi '(\mathcal {I})=0$. Let’s consider the different cases for min-based conditioning:

• Case 1: $\mathcal {I}\models \phi \cup \mathcal {S}$: In this case, by definition of min-based conditioning, we have $\pi '(\mathcal {I})=\pi (\mathcal {I})$, since $\pi (\mathcal {I})$ is either 1 or greater than 0 (if it were 0, then $\mathcal {I}$ would not be a model of $\phi \cup \mathcal {S}$). Therefore, if $\pi (\mathcal {I})=0$, then we also have $\pi '(\mathcal {I})=0$, which satisfies P2.

• Case 2: $\mathcal {I}\models \phi \cup (\mathcal {O}_\pi \setminus \mathcal {S})$, $\mathcal {I}\not \models \mathcal {S}$, and $\pi (\mathcal {I})=N(\phi )$: In this case, by definition of min-based conditioning, we have $\pi '(\mathcal {I})=1-u$. Therefore, if $\pi (\mathcal {I})=0$, then we also have $\pi '(\mathcal {I})=0$, which satisfies P2.

• Case 3: $\mathcal {I}\models \phi \cup (\mathcal {O}_\pi \setminus \mathcal {S})$, $\mathcal {I}\not \models \mathcal {S}$, and $\pi (\mathcal {I})>N(\phi )$: In this case, by definition of min-based conditioning, we have $\pi '(\mathcal {I})=\pi (\mathcal {I})$, since $\pi (\mathcal {I})$ is either 1 or greater than $N(\phi )$. Therefore, if $\pi (\mathcal {I})=0$, then we also have $\pi '(\mathcal {I})=0$, which satisfies P2.

• Case 4: $\mathcal {I}\models \phi$, $\mathcal {I}\not \models \mathcal {S}$, $\mathcal {I}\not \models \mathcal {O}_\pi \setminus \mathcal {S}$, and $\pi (\mathcal {I})=N(\phi )$: In this case, by definition of min-based conditioning, we have $\pi '(\mathcal {I})=1-u$ if $1-\alpha '\ge 1-u$, and $\pi '(\mathcal {I})=1-\alpha '$ if $1-\alpha '<1-u$. Therefore, if $\pi (\mathcal {I})=0$, then we also have $\pi '(\mathcal {I})=0$, which satisfies P2.

• Case 5: $\mathcal {I}\not \models \phi$ and $\pi (\mathcal {I})>1-u$: In this case, by definition of min-based conditioning, we have $\pi '(\mathcal {I})=\pi (\mathcal {I})$, since $\pi (\mathcal {I})$ is either 1 or greater than $1-u$. Therefore, if $\pi (\mathcal {I})=0$, then we also have $\pi '(\mathcal {I})=0$, which satisfies P2.

P3: $\Pi '(\phi )=1$ and $N'(\phi )\ge u$ Considering the case where $\mathcal {S}=\mathcal {O}_\pi$. Then $\mathcal {O}_\pi \models \phi$, so we have $\Pi (\phi )=1$ and $N(\phi )=0$. Therefore, by definition of min-based conditioning, we have $\Pi '(\phi )=1$ and $N'(\phi )=0\ge u$, which satisfies P3. Now, let’s consider the case where $\mathcal {S}\subsetneq \mathcal {O}_\pi$. We know that $\phi$ is not entailed by $\mathcal {O}_\pi$, so we have $N(\phi )>0$. We need to show that $\Pi '(\phi )=1$ and $N'(\phi )\ge u$. First, consider the case where $\alpha '\ge 1-u$. Then, we have $\Pi (\phi )\ge 1-\alpha '>u$, since $\Pi (\phi )\le 1-\alpha '$ and $\alpha '\ge 1-u$. Therefore, by definition of min-based conditioning, we have $\Pi '(\phi )=1$ and $N'(\phi )\ge u$. Now, let’s consider the case where $\alpha '<1-u$. In this case, we have $\Pi (\phi )\le 1-\alpha '<1-u$, since $\Pi (\phi )\ge 1-\alpha '$ and $\alpha '<1-u$. Therefore, by definition of min-based conditioning, we have $\Pi '(\phi )=1$ and $N'(\phi )\ge u$. Therefore, in both cases, we have $\Pi '(\phi )=1$ and $N'(\phi )\ge u$, which satisfies P3. $\square$