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AB C
C A
BC D
CD B
BE C
CE F
CF D
D E
D F
Observe, therefore, that there are (at least) two distinct irreducible covers for the original set of FDs. Note
too that those two covers have different cardinalities.
7.7 Yes, it is. The easiest way to prove this result is to compute the closure ACF + of the set ACF , which turns
out to be the entire set ABCDEFG . Alternatively, we can apply Armstrong's axioms and the other rules discussed in
the body of the chapter, as follows:
1. A B (given)
2. ACF BCF (1, augmentation)
3. BC E (given)
4. BCF EF (3, augmentation)
5. ACF EF (2, 4, transitivity)
6. ACF AEF (5, augmentation)
7. AEF G (given)
8. ACF G (6, 7, transitivity)
9. BC DE (given)
10. BC D (9, decomposition)
11. BCF DF (10, augmentation)
12. BCF D (11, decomposition)
13. ACF D (2, 12, transitivity)
14. ACF DG (7, 13, composition)
7.8
Let's number the FDs of the first set as follows:
1.
A B
2.
AB C
3.
D AC
4.
D E
Now, 3 can be replaced by:
3. D A and D C
Next, 1 and 2 together imply (see the “useful rule” mentioned near the end of the answer to Exercise 7.5) that
2 can be replaced by:
2. A C
But now we have D A and A C , so D C is implied by transitivity and can be dropped, leaving:
3. D A

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