## Equivalence of Sets of Functional Dependencies Example –

- 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. →
- F and G are equal only if
- F covers G- means that all functional dependency of G are logically numbers of functional dependency set F⇒F⊇G.
- 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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