Ontology Languages (COMP321)
Solution to Exercise 7
1. Solution:
Query
Answer for I Certain A for DNemo Certain A for (T , DNemo )
Clownfish(Karl)
Yes
Yes
Yes
Clownfish(Dory)
No
Don’t know
No
Fish(Nemo)
No
Don’t know
Yes
¬Fish(Nemo)
Yes
Don’t know
No
∃has friend.>(Nemo)
Yes
Yes
Yes
∃has friend.Fish(Nemo)
No
Don’t know
Yes
Clownfish u ¬Surgeonfish(Karl)
Yes
Don’t know
Yes
Fish(Dory)
No
Don’t know
Yes
Surgeonfish u ¬Fish(Dory)
Yes
Don’t know
No
∃has friend.Clownfish(Karl)
No
Don’t know
Don’t know
2. Consider the following non-Boolean queries Fi :
• F1 (x) = Clownfish(x)
• F2 (x) = ¬Surgeonfish(x)
• F3 (x, y) = has friend(x, y)
• F4 (x) = Clownfish(x) ∧ ¬has friend(x, Dory)
For each query Fi , give
• answer(Fi , I);
• certanswer(Fi , DNemo );
• certanswer(Fi , (T , DNemo )).
Solution:
• answer(F1 , I) = {Nemo, Karl}.
• certanswer(F1 , DNemo ) = {Nemo, Karl}.
• certanswer(F1 , (T , DNemo )) = {Nemo, Karl}.
• answer(F2 , I) = {Nemo, Karl}.
• certanswer(F2 , DNemo ) = ∅.
• answer(F2 , (T , DNemo )) = {Nemo, Karl}.
• answer(F3 , I) = {(Nemo, Dory)}.
• certanswer(F3 , DNemo ) = {(Nemo, Dory)}.
• certanswer(F3 , (T , DNemo )) = {(Nemo, Dory)}.
• answer(F4 , I) = {Karl}.
• certanswer(F4 , DNemo ) = ∅.
• certanswer(F4 , (T , DNemo )) = ∅.
3. Consider the EL TBox T0 :
Footballplayer
Basketballplayer
Handballplayer
Team
Manager
Manager
v
v
v
v
v
v
∃plays for.Team
∃plays for.Team
∃plays for.Team
∃managed by.Manager
Employee
∃managed by.Manager
and the ABox A0 :
Footballplayer(bob),
Basketballplayer(john),
Handballplayer(peter),
Team(redsocks)
managed by(redsocks, sue)
Compute the interpretation IT0 ,A0 as described in the Comp321
Lecture Notes.
Solution:
The initial assignment is:
S(dFootballplayer )
S(dBasketballplayer )
S(dHandballplayer )
S(dTeam )
S(dManager )
S(dEmployee )
S(bob)
S(john)
S(peter)
S(redsocks)
S(sue)
R(plays for)
R(managed by)
=
=
=
=
=
=
=
=
=
=
=
=
=
{Footballplayer}
{Basketballplayer}
{Handballplayer}
{Team}
{Manager}
{Employee}
{Footballplayer}
{Basketballplayer}
{Handballplayer}
{Team}
∅
∅
{(redsocks, sue)}
• Update R using (rightR) six times:
R(plays for) = {(dFootballplayer , dteam ), (dBasketballplayer , dTeam ),
(dHandballplayer ), dTeam ), (bob, dTeam ),
(john, dTeam ), (peter, dTeam )}
• Update S using (simpleR):
S(dManager ) = {Manager, Employee}
• Update R using (rightR) three times:
R(managed by) = {(redsocks, sue), (dTeam , dManager ),
(dManager , dManager ), (redsocks, dManager )}
Thus, the interpretation IT0 ,A0 is given by:
• The domain ∆IT0 ,A0 consists of dFootballplayer , dBasketballplayer ,
dHandballplayer , dTeam , dManager , dEmployee , bob, john, peter,
redsocks, sue.
• FootballplayerIT0 ,A0 = {dFootballplayer , bob};
• BasketballplayerIT0 ,A0 = {dBasketballplayer , john};
• HandballplayerIT0 ,A0 = {dHandballplayer , peter};
• TeamIT0 ,A0 = {dTeam , redsocks};
• ManagerIT0 ,A0 = {dManager };
• EmployeeIT0 ,A0 = {dManager , dEmployee };
• plays forIT0 ,A0 consists of (dFootballplayer , dteam ), (dBasketballplayer , dTeam ),
(dHandballplayer ), dTeam ) (bob, dTeam ), (john, dTeam ), (peter, dTeam )}.
• managed byIT0 ,A0 consists of (redsocks, sue), (dTeam , dManager ),
(dManager , dManager ), (redsocks, dManager ).
For EL concept queries, we know that IT0 ,A0 gives the answer “Yes” if, and only if, (T0 , A0 ) gives the answer “Yes”.
Check this for the queries:
• ∃plays for.Team(peter);
• ∃managed by.Manager(peter);
• ∃plays for.∃managed by.Manager(peter).
Solution.
• ∃plays for.Team(peter): in both cases the answer is “Yes”.
• ∃managed by.Manager(peter): in both cases the answer
is not “Yes”. For IT0 ,A0 it is “No”; for (T0 , A0 ) it is
“Don’t know”.
• ∃plays for.∃managed by.Manager(peter): in both cases
the answer is “Yes”.
For more complex queries, IT0 ,A0 can give the answer “Yes”
even if (T0 , A0 ) does not give the answer “Yes”. Check this
for
• F (x, y) = ∃z.(plays for(x, z) ∧ plays for(y, z)).
• F = ∃x.managed by(x, x).
Solution.
• john and peter play for dTeam in IT0 ,A0 . Thus IT0 ,A0 |=
F (john, peter). But (T0 , A0 ) 6|= F (john, peter).
• (dManager , dmanager ) ∈ managed byIT0 ,A0 . Thus IT0 ,A0 |=
F . However, (T0 , A0 ) 6|= F .