4.E: Description Logics (Exercises)
- Page ID
- 6418
Exercise \(\PageIndex{1}\)
How are DLs typically different from full FOL?
Exercise \(\PageIndex{2}\)
What are the components of a DL knowledge base?
Exercise \(\PageIndex{3}\)
What are (in the context of DLs) the concept and role constructors? You may list them for either \(\mathcal{ALC}\) or \(\mathcal{SROIQ}\).
Exercise \(\PageIndex{4}\)
What distinguishes one DL from another? That is, e.g., \(\mathcal{ALC}\) is different from \(\mathcal{SROIQ}\) and from \(\mathcal{EL}\); what is the commonality of those differences?
Exercise \(\PageIndex{5}\)
Explain in your own words what the following \(\mathcal{ALC}\) reasoning tasks involve and why they are important for reasoning with ontologies:
a. Instance checking.
b. Subsumption checking.
c. Checking for concept satisfiability.
Exercise \(\PageIndex{6}\)
Consider the following TBox \(\mathcal{T}\):
\(\texttt{Vegan}\equiv\texttt{Person}\sqcap\forall\texttt{eats.Plant}\)
\(\texttt{Vegetarian}\equiv\texttt{Person}\sqcap\forall\texttt{eats.(Plant}\sqcup\texttt{Dairy)}\)
We want to know if \(\mathcal{T}\vdash Vegan\sqsubseteq Vegetarian\).
This we convert to a constraint system \(S=\{ (Vegan\sqcap\neg Vegetarian)(a)\}\), which is unfolded (here: complex concepts on the left-hand side are replaced with their properties declared on the right-hand side) into:
\[S=\{ Person\sqcap\forall eats.Plant\sqcap\neg (Person\sqcap\forall eats.(Plant\sqcup Dairy))(a)\}\]
Tasks:
a. Rewrite Eq. 3.4.1 into negation normal form
b. Enter the tableau by applying the rules until either you find a completion or only clashes.
c. \(\mathcal{T}\vdash Vegan\sqsubseteq Vegetarian\)?
- Answer
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(a) Rewrite (Eq. 3.4.1) into negation normal form:
\(Person\sqcap\forall eats.Plant\sqcap (\neg Person\sqcup\neg\forall eats.(Plant\sqcup Dairy))\)
\(Person\sqcap\forall eats.Plant\sqcap (\neg Person\sqcup\exists\neg eats.(Plant\sqcup Dairy))\)
\(Person\sqcap\forall eats.Plant\sqcap (\neg Person\sqcup\exists eats.(\neg Plant\sqcap\neg Dairy))\)
So our initial ABox is:
\(S=\{ (Person\sqcap\forall eats.Plant\sqcap (\neg Person\sqcup\exists eats.(\neg Plant\sqcap\neg Dairy)))(a)\}\)
(b) Enter the tableau by applying the rules until either you find a completion or only clashes.
(\(\sqcap\)-rule): \(\{ Person(a),\forall eats.Plant(a),(\neg Person\sqcup\exists eats.(\neg Plant\sqcap\neg Dairy))(a)\}\)
(\(\sqcup\)-rule): (i.e., it generates two branches)
(1) \(\{ Person(a),\forall eats.Plant(a),(\neg Person\sqcup\exists eats.(\neg Plant\sqcap\neg Dairy))(a),\neg Person(a)\}\) \(¡\)clash \(!\)
(2) \(\{ Person(a),\forall eats.Plant(a),(\neg Person\sqcup\exists eats.(\neg Plant\sqcap\neg Dairy))(a),\exists eats.(\neg Plant\sqcap\neg Dairy)(a)\}\)
\((\exists\)-rule): \(\{ Person(a),\forall eats.Plant(a),(\neg Person\sqcup\exists eats.(\neg Plant\sqcap\neg Dairy))(a),\exists eats.(\neg Plant\sqcap\neg Dairy)(a), eats(a, b),(\neg Plant\sqcap\neg Dairy)(b)\}\)
(\(\sqcap\)-rule): \(\{ Person(a),\forall eats.Plant(a),(\neg Person\sqcup\exists eats.(\neg Plant\sqcap\neg Dairy))(a),\exists eats.(\neg Plant\sqcap\neg Dairy)(a), eats(a, b),(\neg Plant\sqcap\neg Dairy)(b),\neg Plant(b),\neg Dairy(b)\}\)
(\(\forall\)-rule): \(\{ Person(a),\forall eats.Plant(a),(\neg Person\sqcup\exists eats.(\neg Plant\sqcap\neg Dairy))(a),\exists eats.(\neg Plant\sqcap\neg Dairy)(a), eats(a, b),(\neg Plant\sqcap\neg Dairy)(b),\neg Plant(b),\neg Dairy(b), Plant(b)\}\)
\(¡\)clash\(!\)
(c) \(\mathcal{T}\vdash Vegan\sqsubseteq Vegetarian\)? yes