The key point about armstrongs axioms is that they

The key point about Armstrong’s axioms is that they are both sound and
complete . That is, if we start with a set F of FDs then:
Repeated application of Armstrong’s axioms to F generates only FDs
that are consequences of F
Any FD that is a consequence of F be obtained by repeated application
of Armstrong’s axioms to F .
Jianxin Li (UWA) Redundancy 28 / 41

Example
Consider a relation with attributes ABC and let
F = { A → B , B → C }
Then from transitivity we get A → C , by augmentation we get AC → BC and
by union we get A → BC .
FDs that arise from reflexivity such as
AB → B
are known as trivial dependencies.
Jianxin Li (UWA) Redundancy 29 / 41

Closure
Given a set X of attributes from some relation R , the closure X
+
is the set of
all attributes that are determined by X . In symbols
X
+
= { A : X → A }
The following properties hold:
X ⊆ X
+
X
+
= R if and only if X is a superkey
X → X
+
is a sort of “maximal” FD
Jianxin Li (UWA) Redundancy 30 / 41

Example
Suppose R ( A , B , C , D , E ) has the following FDs
D → C , CE → A , D → A , AE → D
What is BDE
+
? (We use BDE as shorthand for { B , D , E } .)
31 / 41

Example
Suppose R ( A , B , C , D , E ) has the following FDs
D → C , CE → A , D → A , AE → D
What is BDE
+
? (We use BDE as shorthand for { B , D , E } .)
31 / 41

Example
Suppose R ( A , B , C , D , E ) has the following FDs
D → C , CE → A , D → A , AE → D
What is BDE
+
? (We use BDE as shorthand for { B , D , E } .)
31 / 41

Example
Suppose R ( A , B , C , D , E ) has the following FDs
D → C , CE → A , D → A , AE → D
What is BDE
+
? (We use BDE as shorthand for { B , D , E } .)

Example
Suppose R ( A , B , C , D , E ) has the following FDs
D → C , CE → A , D → A , AE → D
What is BDE
+
? (We use BDE as shorthand for { B , D , E } .)

Example 2
Suppose R ( A , B , C , D , E ) has the following FDs
D → C , CE → A , D → A , AE → D
What is BD
+
?
Jianxin Li (UWA) Redundancy 32 / 41

Example 2
Suppose R ( A , B , C , D , E ) has the following FDs
D → C , CE → A , D → A , AE → D
What is BD
+
?
Jianxin Li (UWA) Redundancy 32 / 41

Example 2
Suppose R ( A , B , C , D , E ) has the following FDs
D → C , CE → A , D → A , AE → D
What is BD
+
?
So far we know that BD
+
contains BD
Jianxin Li (UWA) Redundancy 32 / 41

Example 2
Suppose R ( A , B , C , D , E ) has the following FDs
D → C , CE → A , D → A , AE → D
What is BD
+
?
So far we know that BD
+
contains BD
From the FD D → C , BD
+
contains BCD
Jianxin Li (UWA) Redundancy 32 / 41

Example 2
Suppose R ( A , B , C , D , E ) has the following FDs
D → C , CE → A , D → A , AE → D
What is BD
+
?
So far we know that BD
+
contains BD
From the FD D → C , BD
+
contains BCD
From the FD D → A , BD
+
contains ABCD
Therefore BD
+
is not a superkey because E is not determined by it.