# 3.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**-
(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