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Two FD sets F and G are equivalent if:
Practically, we verify this using closure (F⁺ and G⁺)
| F = {A → B, B → C} G = {A → B, A → C} |
Step-by-Step Method
We compute closure using F:
| A → B B → C ⇒ A → C (Transitivity) |
So G is derived from F
Step 2: Check G ⇒ F
| From G: A → B (already present) A → C (given) |
Covers what F provides
Final Result
| F ≡ G (Equivalent) |
| F = {A → B, B → D} G = {A → B, A → D} |
From F:
| A → B B → D ⇒ A → D |
From G:
| A → D already exists |
Conclusion:
| F ≡ G |
Same destination → Equivalent



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