## 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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