1. Logics for Data and Knowledge Representation
Exercises: Description Logics
Fausto Giunchiglia, Rui Zhang and Vincenzo Maltese

2. Normalization of a TBox
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Normalize the TBox below:
MonkeyLow ⊑  GetBanana
GetBanana ≡ Survive
Possible solution:
MonkeyLow ≡  GetBanana ⊓  ClimbBox
Note that, with this theory, the monkey necessarily needs to get the banana to survive. 2

3. TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Expansion of a TBox
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Expand the TBox below:
MonkeyLow ≡  GetBanana ⊓  ClimbBox
GetBanana ≡ Survive
T’, expansion of T (The Venn diagram gives a possible model):
MonkeyLow ≡  Survive ⊓  ClimbBox
Survive
GetBanana
MonkeyLow
ClimbBox
Notice that the fact that a monkey climbs the box does not necessarily mean that it survives. 3

4. Satisfiability with respect to a TBox T (I)
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
RECALL: A concept P is satisfiable w.r.t. a terminology T, if there exists an interpretation I with I ⊨ θ for all θ ∈ T, and such that I ⊨ P, namely I(P) is not empty
ADD EXAMPLE 4

5. Satisfiability with respect to a TBox T (II)
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Suppose we model the Monkey-Banana problem as follows:
“If the monkey is low in position then it cannot get the banana. If the monkey gets the banana it survives”.
TBox T
MonkeyLow ⊑  GetBanana
GetBanana ⊑ Survive
Is T satisfiable?
YES! Look at the
Venn diagram
Survive
GetBanana
MonkeyLow 5

6. Satisfiability with respect to a TBox T (III)
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Suppose we model the Monkey-Banana problem as follows:
TBox T
MonkeyLow ⊑  GetBanana
GetBanana ⊑ Survive
Is it possible for a monkey to survive even if it does not get the banana?
We can restate the problem as follow:
does T ⊨  GetBanana ⊓ Survive ?
YES! Look at the
Venn diagram
Survive
GetBanana
MonkeyLow 6

7. Translate the following statements in DL (I)
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
All users can download files
User ⊑ ∃Download.File
A read access user can read files only
ReadAccesUser ⊑ ∀Read.File
Those who can write can also read
Write ⊑ Read
A limited user can download no more than 3 files
LimitedUser ⊑ ≤3 Download.File

8. Translate the following statements in DL (I)
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Programmers write programs
Programmer ⊑ ∃Write.Programs
Producers can produce music and song files only
Music ⊑ File
Song ⊑ File
Producer ⊑ ∀Produce.(Music ⊔ Song)
A program can be downloaded by developers only
Program ⊑ ∀Download-1.Developer

9. Define a TBox and ABox for the following problem
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Diana wants to give read access to her friends to the files on her PC, and write access to her best friend Gloria only. Bill is just a friend.
Friend ⊑ ∃Read.File
BestFriend ⊑ ∃Write.File ⊓ Friend
BestFriend(Gloria)
Friend(Bill)
Can Gloria read Diana’s files?
Yes
BestFriend(Gloria)  Friend(Gloria)  ∃Read.File(Gloria)

10. Define a TBox and ABox for the following problem
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
In a library, employees have access to books. However, a loaned book is a non precious book and can be read by clients. Mary is a client and reads a book titled “Top stories”.
Employee ⊑ ∃Access.Book
LoanedBook ⊑ Book ⊓ Precious ⊓ ∀Read-1.Client
Client(Mary)
Book(TopStories)
Read(Mary,TopStories)
Is it possible for the book to be precious?
Yes
We can verify that Precious(TopStories) is consistent with the ABox and TBox above.

11. Defining the TBox and ABox: the LDKR Class
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Define a TBox and ABox for the following database:
ABox =
{Italian(Fausto), Italian(Enzo),
Chinese(Rui), Indian(Bisu),
BlackHair(Enzo), BlackHair(Rui), BlackHair(Bisu),
WhiteHair(Fausto)}
TBox =
{Italian ⊑ LDKR,
Indian ⊑ LDKR,
Chinese ⊑ LDKR,
BlackHair ⊑ LDKR,
WhiteHair ⊑ LDKR}
LDKR
Name
Nationality
Hair
Fausto
Italian
White
Enzo
Black
Rui
Chinese
Bisu
Indian
NOTE: ClassL is not expressive enough to represent database constrains such as keys involving two fields. 11

12. Formalization of a semantic network
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
instance-of
Person ⊑ ∃Drives.Car Person ⊑ ∃HasHobby.SportCar Person ⊑ ∃HasHobby.Opera Student ⊑ Person SportCar ⊑ Car Student(Ralf) Opera(DonCarlos)

13. TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Venn diagrams
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Provide the Venn diagram for: A ⊑ B ⊓ ¬C as a way to prove the satisfiability B A C

14. TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING

15. TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING

16. TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING

17. TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING

18. Using DPLL for reasoning tasks
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
DPLL solves the CNFSAT-problem by searching a truth-assignment that satisfies all clauses θi in the input proposition P = θ1  …  θn
DL sentences must to be translated in PL (via TBox and ABox elimination)
Model checking Does ν satisfy P? (ν ⊨ P?)
Check if ν(P) = true
Satisfiability Is there any ν such that ν ⊨ P?
Check that DPLL(P) succeeds and returns a ν
Unsatisfiability Is it true that there are no ν satisfying P?
Check that DPLL(P) fails
Validity Is P a tautology? (true for all ν)
Check that DPLL(P) fails 18

19. TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Reduce to PL reasoning
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Consider the TBox T = {A ⊑ B ⊔ C, C ⊑ D ⊓ E}. Determine if A ⊑ E by elimination of T.
T’ = {A ≡ (B ⊔ C) ⊓ X, C ≡ D ⊓ E ⊓ Y}
A’ = (B ⊔ (D ⊓ E ⊓ Y)) ⊓ X
E’ = E
A’ = (B  (D  E  Y))  X
Call DPLL((B  (D  E  Y))  X → E)
(Note that the formula has to be converted in CNF first)
The steps:
T’ = Normalize(T);
C’ = Expand(C, T’);
D’ = Expand(D, T’);
C’ = RewriteInPL(C’);
D’ = RewriteInPL(D’);
return DPLL(C’ → D’);

20. Satisfiability of a TBox (a set of formulas)
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Say if the following TBox is satisfiable and provide a class valuation in the case:
FederalAgent ⊑ ∃Access.TopSecretDocument
XFile ⊑ TopSecretDocument ⊓ Restricted ⊓ ∀Read-1.Policeman
I(FederalAgent) = {A} We have that I ⊨ T
I(TopSecretDocument) = {D1, D2}
I(Access) = {(A, D1), (A, D2)}
I(XFile) = {D1}
I(Restricted) = {D2}
I(Read) = {(B, D1)}
I(Policeman) = {B}

21. Satisfiability w.r.t. a TBox
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Consider the TBox T:
FederalAgent ⊑ ∃Access.TopSecretDocument
XFile ⊑ TopSecretDocument ⊓ Restricted ⊓ ∀Read-1.Policeman
and the formula P: TopSecretDocument ⊓ Restricted
does T ⊨ P ?
Yes, if fact these is an interpretation I (e.g. the one of the previous exercise) such that I ⊨ T and I ⊨ P.
I(TopSecretDocument) = {D1, D2}
I(Restricted) = {D1}
Notice that T does not affect P at all (i.e. we cannot further expand P w.r.t. T)

22. Subsumption FederalAgent ⊑ ∃Access.TopSecretDocument
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Consider the TBox T:
FederalAgent ⊑ ∃Access.TopSecretDocument
XFile ⊑ TopSecretDocument ⊓ Restricted ⊓ ∀Read-1.Policeman
does T ⊨ TopSecretDocument ⊑ Restricted ?
By definition, it must be I(TopSecretDocument )  I(Restricted ) for every model I of T. Even if this is true for the I of the previous exercise, this is not true in general. It is enough to provide a counterexample:
I(FederalAgent) = {A}
I(TopSecretDocument) = {D2}
I(Access) = {(A, D1), (A, D2)}
I(XFile) = {D2}
I(Restricted) = {D1}
I(Read) = {(B, D2)}
I(Policeman) = {B}

23. Satisfiability of a TBox using tableaux
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Say if the following TBox is satisfiable using the tableaux:
FederalAgent ⊑ ∃Access.TopSecretDocument
XFile ⊑ TopSecretDocument ⊓ Restricted ⊓ ∀Read-1.Policeman
Translate into a formula:
( FederalAgent ⊔ ∃Access.TopSecretDocument) ⊓
( XFile ⊔ (TopSecretDocument ⊓ Restricted ⊓ ∀Read-1.Policeman))
Apply the tableaux rules:
(1)  FederalAgent(x) or ∃Access.TopSecretDocument(x)
and
(2)  Xfile(x) or {TopSecretDocument(x), Restricted(x), ∀Read1.Policeman(x)} …

24. TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Expansion of an ABox
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Provide the expansion of A w.r.t. T (without normalization), where:
TBox T = {Student ⊑ Faculty, Professor ⊑ Faculty ⊓ Teach}
ABox A = {Professor(Bob), Faculty(Rui)}
Professor(Bob), Faculty(Bob), Teach(Bob)
Faculty(Rui)

25. TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
ABox Consistency
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
An ABox A is consistent with respect to a TBox T if there is an interpretation I which is a model of both A and T.
T = {Parent⊑≤1hasChild}
A = {hasChild(mary, bob), hasChild(mary, cate), Parent(mary)}
A is consistent ALONE but is not consistent with respect T.
In fact, from A mary has two children while T imposes maximum one 25

26. ABox: Instance checking (I)
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Given the TBox and ABox below T:
Female⊑Human
Male⊑Human
Mother⊑Female
Father⊑Male
Child≡∃has.Mother⊓ ∃has.Father
Male⊓Female⊑⊥ A:
Mother(Anna)
Father(Bob)
has(Cate,Anna)
has(Cate,Bob)
Prove:
Human(Anna)
¬Female(Bob)
Child(Cate) 26

27. ABox: Instance checking (II)
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Expand A w.r.t. T: A:
Mother(Anna)  Female(Anna)  Human(Anna)
Father(Bob)  Male(Bob)  Human(Bob) , ¬Female(Bob)
has(Cate,Anna) , has(Cate,Bob)  Child(Cate) 27

28. ABox: Instance checking (III)
TBOX :: MODELING IN DL :: PROOFS :: TBOX REASONING :: ABOX REASONING
Given the ABox A = {R(a,b), R(a, c), A(b)} is a an instance of ∃R.A?
Take ∃R.A(a). Given its semantics, there should be {R(a, x), A(x)}  A for some x.
If we take x = b we see that this actually holds. 28