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Is the following TBox cyclic or acyclic? If it is a cyclic TBox, how could it be converted to an acyclic one?

A ⊑ ¬E
E ⊑ ¬A

Joshua Taylor
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Welcome789
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  • Do you know what it means for a TBox to be cyclic or acyclic? – Joshua Taylor Feb 23 '15 at 22:02
  • An acyclic TBox is a finite set of concept definitions that it: (i) does not contain multiple definitions; (ii) does not contain cyclic definitions (directly or indirectly). The above ones are equal, so I can use only one of them. Thanks. – Welcome789 Feb 23 '15 at 22:07
  • Neither of these is a concept *definition*, though. They are provide necessary conditions of the classes A and E, but not sufficient conditions. That said, I've provided an answer that explains how you might rewrite their meaning in a more direct way. – Joshua Taylor Feb 23 '15 at 22:09

3 Answers3

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A ⊑ ¬E
E ⊑ ¬A

This TBox doesn't really say anything except that the classes A and E are disjoint. The subclass relations could be read as implications:

  • If something is an A, then it is not an E.
  • If something is an E, then it is not an A.

To express disjointness in description logics, you'd typically say that the intersection of disjoint classes is equivalent, or a subclass, of the bottom concept, ⊥, which by definition has no instances. &bot is also the complement of the top concept, ⊤, which contains everything. Thus you could say any of the following:

A ⊓ E ⊑ ⊥

A ⊓ E ≡ ⊥

A ⊓ E ⊑ ¬⊤

A ⊓ E ≡ ¬⊤

Joshua Taylor
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To add what Joshua said, disjointedness representation depends upon the language you use. Example: EL doesnt support bottom and negation.

The axioms you have written is not cyclic.

Cycle: antecedent and consequent of an axiom should have at least one common predicate (Concept or role).

If an axiom contains a cycle, you have to adopt fixpoint semantics to make it unequivocal.

To the best of my knowledge, axioms are meant to get induced knowledge. Converting a cyclic axiom to an acylic axiom: It is difficult to produce similar semantics.

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How to convert the following TBox axioms into an acyclic Tbox:

 A \sqsubseteq \lnot E
 \exists R.A \sqcap \lnot B \sqsubseteq C
 C \sqsubseteq B \sqcup A
 C = A \sqcup D
 A \sqcap \exists R.E \sqsubseteq D
Welcome789
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