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If X Y and YW Z , then XW Z .
7.6
The first step is to rewrite the given set of FDs such that every FD has a singleton right side:
1. AB C
2. C A
3. BC D
4. ACD B
5. BE C
6. CE A
7. CE F
8. CF B
9. CF D
10. D E
11. D F
Now:
2 implies 6, so we can drop 6.
8 implies CF BC by augmentation, which with 3 implies CF D by transitivity, so we can drop 9.
8 implies ACF AB by augmentation, and 11 implies ACD ACF by augmentation, and so ACD AB by
transitivity, and so ACD B by decomposition, so we can drop 4.
No further reductions are possible, and so we're left with the following irreducible cover:
AB C
C A
BC D
BE C
CE F
CF B
D E
D F
Alternatively:
2 implies 6, so we can drop 6 (as before).
2 implies CD AD by augmentation, which implies CD ACD by augmentation again, which with 4
implies CD B by transitivity, so we can replace 4 by CD B .
2 and 9 imply CF AD by composition, which implies CF ADC by augmentation, which with (the
original) 4 implies CF B by transitivity, so we can drop 8.
No further reductions are possible, and so we're left with the following irreducible cover:

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