How to prove that two sets of functional dependencies are equal or not - Proving equality of two sets of functional dependencies - How to find whether two sets of functional dependencies are equal or not? - Equivalent sets of functional dependencies
How to find whether two sets of functional dependencies are equal or not?
Let F and G be two different sets
of functional dependencies holding on a relation R. These two functional dependencies
are called equivalent if F+ = G+, i.e., if closure of the
set F is equal to the closure of set G.
In other words,
Two sets of functional dependencies
F and G are said to be equal if;
- Every FD in G can be inferred (derived) from the functional dependencies in F, and
- Every FD in F can be inferred (derived) from the functional dependencies in G.
Alternative definition;
Two sets of functional dependencies
F and G are said to be equal if;
- F can cover G, and
- G can cover F.
Example:
Let R (A, B, C, D, E) be a
relation with set of functional dependencies F = { A → BC, A → D, CD → E } and G = { A → BCE, A → ABD, CD → E }. Is F = G?
Does F cover G?
If set of FDs of G can be
inferred from F, then we would say that F covers G.
The FD A → BCE of G can be
inferred from the FDs A →
BC, A →
D, and CD →
E of F. [here, A gives BCD. If you know C and D then E can be derived]
The FD A → ABD of G can be
inferred from the FDs A →
BC, and A →
D of F.
The FD CD → E of G can be
inferred from the FD CD →
E of F.
All the three FDs of G can be
inferred from FDs of F. Hence, F covers G.
Does G cover F?
If set of FDs of F can be
inferred from G, then we would say that G covers F.
The FD A → BC of F can be
inferred from the FD A →
BCE of G.
The FD A → D of F can be
inferred from the FD A →
ABD of G.
The FD CD → E of F can be
inferred from the FD CD →
E of G.
All the three FDs of F can be
inferred from FDs of G. Hence, G covers F.
Hence, we would say F is
equivalent to G.
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