Equivalence of Sets of Functional Dependencies Example – 

  1. Let F & G are two functional dependency sets. These two sets F & G are equivalent if E+=F+.
    Equivalence means that every functional dependency in F can be inferred from G, and every functional dependency in G an be inferred from F. →
  2. F and G are equal only if
    1. F covers G- means that all functional dependency of G are logically numbers of functional dependency set F⇒F⊇G.
    2. G covers F-means that all functional dependency of F are logically members of functional dependency set G⇒G⊇F
F covers G True True False False
G covers  F True False True False
Result F=G F⊃G G⊃F No Comparison
Question 1 : 
Consider the two sets F and G with their FDs as below :
F :              G:
A  → C           A → CD
AC → D           E → AH
E  → AD
E  → H
Check whether two sets are equivalent or not.
Solution : 
Step 1 : Take Set F and Check G is covered from F or not.
         (A)+ = {ACD}
         (E)+ = {EADHC}
         Hence, both A → CD and E → AH are covered.
      ⇒ G is derived from F. Hence G is covered by F.
      ⇒ F ⊇ G .                  ....(1)
Step 2 : Take Set G and Check F is covered from G or not.
         (A)+  = {ACD}
         (AC)+ = {ACD}
         (E)+  = {EAHCD}
         Hence F = {A → C, AC → D, E → AD, E → H} is covered.
      ⇒ F is derived from G. Hence F is covered from G. 
      ⇒ G ⊇ F.                   ....(2)
From (1) and (2), F and G are equivalent.
Question 2 : 
Consider the two sets P and Q with their FDs as below :
P :              Q :
A  → B           A → BC
AB → C           D → AE
D  → ACE
Check whether two sets are equivalent or not.
Solution : 
Step 1 : Take Set P and Check Q is covered from P or not.
         (A)+ = {ABC}
         (D)+ = {DACEB}
         Hence, both A → BC and D → AE are covered.
      ⇒ Q is derived from P. Hence Q is covered by P.
      ⇒ P ⊇ Q .                  ....(1)
Step 2 : Take Set Q and Check P is covered from Q or not.
         (A)+  = {ABC}
         (AB)+ = {ABC}
         (D)+  = {DAEBC}
         Hence P = {A → B, AB → C, D → ACE} is covered.
      ⇒ P is derived from Q. Hence P is covered by Q. 
      ⇒ Q ⊇ P.                   ....(2)
From (1) and (2), P and Q are equivalent.

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