Answers to Exercises - Database Design and Relational Theory - page 238

 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: