CS 319: Theory of Databases: C3
Dr. Alexandra I. Cristea
http://www.dcs.warwick.ac.uk/~acristea/
(provisionary) Content
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Generalities DB
Integrity constraints (FD revisited)
Relational Algebra (revisited)
Query optimisation
Tuple calculus
Domain calculus
Query equivalence
LLJ, DP and applications
Temporal Data
The Askew Wall
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… previous
Proving with Armstrong axioms
(non)Redundancy of FDs
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FD Part 3
• Soundness and Completeness of
Armstrong’s axioms
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Armstrong’s Axioms:
Sound & Complete
Ingredients:
• Functional Dependency (reminder)
– Definition
– Inference rules
– Closure of F : F+
F+: Set of all FDs obtained by applying inferences rules on a basic set
of FDs
– Issues and resolutions
• Armstrong’s Axioms (reminder)
– 3 inferences rules for obtaining the closure of F
– Properties of the Armstrong’s Axioms
They are a sound an complete set of inference rules
• Proof of the completeness
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Functional Dependency
- A functional dependency (FD) has the form X  Y
where X and Y are sets of attributes in a relation R
X  Y if and only if:
for any instance r of R
for any tuples t1 and t2 of r
t1(X) = t2(X)  t1(Y) = t2(Y)
X  Y iff any two tuples that agree on X value also agree on Y value
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Inference of Functional
Dependencies
• Suppose R is a relation scheme, and F is a set of functional
dependencies for R
• If X, Y are subsets of attributes of Attr(R) and if all instances r of
R which satisfy the FDs in F also satisfy X  Y, then we say
that F entails (logically implies) X  Y, written F |- X  Y
• In other words:
there is no instance r of R that does not satisfy X  Y
• Example
if F = { A  B, B  C } then F |- A  C
if F = { S  A, SI  P } then
F |- S I  AP, F |- SP  SAP etc.
A
B
I
S
P
A
C
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Functional Dependency
• Issue:
– How to represent set of ALL FD’s for a relation R?
• Solution
– Find a basic set of FD’s ((canonical) cover)
– Use axioms for inferring
– Represent the set of all FD’s as the set of FD’s that can be
inferred from a basic set of FD’s
• Axioms
– they must be a sound and complete set of inference rules
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Set of Functional Dependencies F*
• Formal Definition of F*, the cover of F:
if F is a set of FD’s, then F* { X  Y  F ├ X  Y }
• Informal Definitions
– F* is the set of all FD’s logically implied by F
(entailed)
• ... usually F* is much too large even to
enumerate!
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F*
• Usually F* is much too large even to
enumerate!
• Example (3 attributes, 2 FD’s and 43 entailed
dependencies)
A
Attr(R)=ABC and F ={ A  B, B  C } then F* is
A  S for all [subset of ABC]
B  BC, B  B, B  C, B  
C  C, C  ,   
AB  S for all subsets S of ABC
AC  S for all subsets S of ABC
BC  BC, BC  B, BC  C, BC  
ABC  S for all subsets S of ABC
B
C
8 FDs
4 FDs
3 FDs
8 FDs
8 FDs
4 FDs
8 FDs
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Armstrong’s Axioms
• Axioms for reasoning about FD’s
F1: reflexivity
if Y  X then X  Y
F2: augmentation
if X  Y then XZ  YZ
F3: transitivity
if X  Y and Y  Z then X  Z
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+
F
• Informal Definition
F + (the closure of F) is the set of dependencies which can
be deduced from F by applying Armstrong’s axioms
• Formal Definition
if F is a set of FD’s, then F+ { X  Y F |= X  Y }
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Armstrong’s Axioms
Theorem:
Armstrong’s axioms are a sound and complete
set of inference rules
– Sound: all Armstrong axioms are valid (correct / hold)
– Complete: all fds that are entailed can be deduced with
the help of the Armstrong axioms
• How to:
– Prove the soundness?
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Armstrong’s Axioms
• Theorem: Armstrong’s axioms are a sound
and complete set of inference rules
– Sound: the Armstrong’s rules generate only FDs in
F*
F+  F*
– Complete: the Armstrong’s rules generate all FDs
in F*
F*  F +
– If complete and sound then F+ = F*
• Here
– Proof of the completeness
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For the proof, we can use:
Armstrong’s Theorems
• Additional rules derived from axioms
– Union
if A  B and A  C, then A  BC
A
– Decomposition
if A BC, then A  B and A  C
B
C
B
A
C
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Completeness of the
Armstrong’s Axioms
• Proving that Armstrong’s axioms are a
complete set of inference rules
Armstrong’s axioms generate all FDs in F*
F*  F +
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Completeness of the
Armstrong’s Axioms
• First define X+, the closure of X with respect to F:
X+ is the set of attributes A such that X  A can be deduced
from F with Armstrong’s axioms
• Note that we can deduce that X  Y for some set Y by
applying Armstrong’s axioms if and only if Y  X+
Example:
Attr(R)=LMNO
X=L
F={L  M , M  N, O  N}
then X+ = L+ = LMN
L
M
N
O
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XY
+
F <=> Y  X+
• Proof: We can deduce that X  Y for some set Y by applying
Armstrong’s axioms if and only if Y  X+
Y  X+  X  Y  F+
• Y  X+ and suppose that A Y then X  A  F+ (definition of X+)
•  A Y: X  A  F+
• X  Y  F+ (union rule)
X  Y  F+  Y  X+
• X  Y  F+ and suppose that A Y then X  A  F+
(decomposition rule)
• A  X+ (definition of X+)
•  A Y: A  X+
• Y  X+
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Completeness (F*  F +) of the
Armstrong’s Axioms
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Completeness:
(R,  X,Y Attr(R), F true in R : : X  Y  F* => X  Y  F +)
•
Idea: (A => B) ≡ (A v B) ≡ (B v A) ≡ (B =>A)
To establish completeness, it is sufficient to show:
if X  Y cannot be deduced from F using Armstrong’s axioms then also X  Y is not
logically implied by F:
(R,  X,Y Attr(R), F true in R : : X  Y  F+ => X  Y  F*)
X  Y  F* enough:
(In other words) there is a relational instance r in R (rR) in which all the
dependencies in F are true, but X  Y does not hold
Counter example!!
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Completeness of the
Armstrong’s Axioms
• Example for the proof idea for a given R, F, X:
If X  Y cannot be deduced using Armstrong’s axioms: then there
is a relational instance for R in which all the dependencies in F are
true, but X  Y does not hold
R=LMNO
X=L
F={L  M , M  N, O  N}
then X+ = LMN
Counter example:
L
M
O
N
L  O cannot be
deduced (so not in F+)
but also does not hold
(not in F*)
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What we want to prove thus:
(R,  X,Y Attr(R), F true in R : : X  Y  F+ => X  Y  F*)
(In other words) Counterexample
– by construction:
– for any R, any X, Y, any F true in R,
– with X->Y not inferable from the axioms
– there is a relational instance r in R (rR)
– in which all the dependencies in F are true
(F true in R ),
– but X  Y does not hold
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Completeness of the
Armstrong’s Axioms
• Suppose one can not deduce X  Y from Armstrong’s
axioms for an arbitrary R, F, X,Y; construct counter-ex.
• Consider the instance r0 for R with 2 tuples
Relational instance r0 for R with 2 tuples
Attributes of
X+
Other
Attributes
11…1
1 1 …. 1
Example:
L+ = LMN
L
M
N
O
11…1
1
1
1
1
00…0
1
1
1
0
R=LMNO
X=L
then X+ = L+
L  O cannot
be deduced
(assuming Boolean attributes, or more generally, that the two
tuples agree on X+ but disagree elsewhere)
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Completeness of the
Armstrong’s Axioms
• Check that all the dependencies in F are true in R:
– Suppose that V  W is a dependency in F
• If V is not a subset of X+, the dependency holds in r0
• If V is a subset of X+, then both X  V, and then X  W can be
deduced by Armstrong’s axioms. This means that W is a
subset of X+, and thus V  W holds in r0
Relational instance r0 for R with 2 tuples
Attributes of X+
Other Attributes
11…1
11…1
1 1 …. 1
00…0
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Completeness of the
Armstrong’s Axioms
• Check that all the dependencies in F are true
– Extended Example (more tuples)
• O  N is a dependency in F but O is not a subset of X+,
the dependency holds in r0
• M  N is a dependency in F and M is a subset of X+,
then both L  M, and L  N can be deduced by
Armstrong’s axioms. This means that N is a subset of
X+, and thus M  N holds in r0
R=LMNO
X=L
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F={L  M , M  N, O  N}
then X+ = LMN
Completeness of the
Armstrong’s Axioms
(R,  X,Y Attr(R), F true in R : : X -> Y  F+ => X -> Y  F*)
• Proof that X  Y does not hold in r0:
– Recall that we can deduce that X  Y for some set Y by
applying Armstrong’s axioms if and only if Y  X+
– By assumption, we can’t deduce that X  Y holds in r0
– Hence Y contains (at least) an attribute not in the subset X+,
confirming that X  Y does not hold in r0
Relational instance r0 for R with 2 tuples
Attributes of
X+
Other
Attributes
11…1
11…1
1 1 …. 1
00…0
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Completeness of the
Armstrong’s Axioms
• We have proved the correctness (last module)
and here, the completeness of Armstrong’s
Axioms:
– How can we prove the completeness of another
set of rules?
• Repeat the proof for this set
• Deduce the Armstrong’s Axioms from this set
– How can we disprove the completeness of another
set of rules?
• By showing (via a counterexample) that some
consequence of Armstrong's rules cannot be deduced
from them (see the proof technique for non-redundancy)
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Completeness of the
Armstrong’s Axioms
• Exercise
– Are the following set of rules a sound and complete set of
inference rules? (X, Y, Z, W  R)
S1: X  X
S2: if X  Y then XZ  Y
S3: if X  Y, Y  Z then XW  ZW
(This is a typical exam question )
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Solution (soundness)
XX
if X  Y
Y  YZ
so by F1
then by F2
S1
X Y
XZ  Y
S2
X  Y ,Y  Z
XW  ZW
S3
XZ  YZ
so by F1
YZ  Y
XZ  YZ , YZ  Y so by F3
XZ  Y
if X  Y , Y  Z then by F3
so by F2
XX
X Z
XW  ZW
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Solution (completeness)
if Y  X then
Z :: YZ  X so {S1} Y  Y,
{S2}YZ  Y
if X  Y then {S1} X  X, X  Y, so
{S3} XZ  YZ
if X  Y , Y  Z
  Z 0

then {S3} X  0
hence X  Y
X Y
XZ  YZ
A2
X  Y ,Y  Z
X Z
A3
hence
hence
A1
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Completeness of the
Armstrong’s Axioms
• Exercises
– Are the following set of rules a complete set of inference rules?
R1: X  X
R2: X  Y then XZ  Y
R3: X  Y, Y  Z then X  Z
Ø
A
Ø A
1
1
1
1
B AB
1
1
1
1
1
1
Suppose
Att( R)  {A, B}, F  {0  A}
Apply above rules exhaustively
B  AB obtained from F1-3
but not from R1-3
Hence: answer is NO
B
AB
1
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Summary
• We have learned how to prove that the
Armstrong axiom set is complete (and
we already knew it is sound)
• We can now prove the soundness and
completeness of any other set of
axioms
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• Extra Question:
– Can we have a sound and complete set of
inference rules consisting of only 2 rules?
– What about 1 rule?
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… to follow
Dependency preserving, Lossless join,
Normal forms algorithms
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