Schema R = (A, B, C, D, E, F)
FD F = {ABC -> D, CD -> B, BCF -> D, CDF -> BE, BCDF -> E}
Find Fc, the minimal cover (aka. canonical cover) of F.
This is the method using in my book:
Example: abc -> xyz
a is redundant (extraneous) if (bc)+ ⊇ a; x is redundant if (abc)+ ⊇ x.
NOTE: Here, the closures are computed using F, with a or x being deleted from abc -> xyz respectively.
I don't understand the last bold sentence.
one solution is:
Consider CDF -> BE
B is redundant: (CDF)+ = (CDFBE) ⊇ (B)
F becomes { ABC -> D, CD -> B, BCF -> D, CDF -> E}
but I don't understand.
according to this logic:
E can be redundant too,
coz:
Consider CDF -> BE
E is redundant: (CDF)+ = (CDFBE) ⊇ (E)
F becomes { ABC -> D, CD -> B, BCF -> D, CDF -> B}
I know I must overlook some important criteria. Can anyone tell me what is that?