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Solution:
First solution:
- crewmember(cnr, cname)
- qualification_1(cnr, fname) qualification_2(cnr, pltype)
- function(fname, pltype, min, max)
- plane(pltype, nbseats)
- crew_1(cnr, fname) crew_2(cnr, flnr)
- flight(flnr, pltype, from, to)
Question: can you show that the relations crew_1 and
qualification_2 are redundant and may be removed?
Notes:
In order to show that crew_1 is not needed we can use the constraint that crew members are only allowed to serve in a function for which they are qualified.
In order to show that qualification_2 is not needed we need to apply some common sense (not explicitly given in the description) on the original scheme:
A crew member can only be qualified for a function on a certain plain type if that function exists on that plain type. We also assume that when the function exists in the relation function the “max” number cannot be 0 which would mean that the function in practice does not exist after all. We also need to use the new mvd in order to prove that we can remove qualification_2.
The complete argumentation is given in the answers to the exam of June 30, 2003.