Databases Reference

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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