3.E: Description Logics (Exercises)

• • Contributed by Maria Keet
• Associate Professor (Computer Science) at University of Cape Town

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:

$S=\{ Person\sqcap\forall eats.Plant\sqcap\neg (Person\sqcap\forall eats.(Plant\sqcup Dairy))(a)\}$

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. $$\mathcal{T}\vdash Vegan\sqsubseteq Vegetarian$$?

(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