Dr. Alexandra I. Cristea

http://www.dcs.warwick.ac.uk/~acristea/

CS 319: Theory of Databases: C3

(provisionary) Content1. Generalities DB2. Integrity constraints (FD revisited)3. Relational Algebra (revisited)4. Query optimisation5. Tuple calculus6. Domain calculus7. Query equivalence8. LLJ, DP and applications9. Temporal Data10. 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 5

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 iff any two tuples that agree on X value also agree on Y value

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)

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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

• Exampleif F = { A B, B C } then F A C

if F = { S A, SI P } thenF S I AP, F SP SAP etc.

A B C

S API

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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 covercover of F:

• Informal Definitions– F* is the set of all FD’s logically implied by F

(entailed)

• ... usually F* is much too large even to enumerate!

if F is a set of FD’s, then F* { X Y F ├ X Y }

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F*• Usually F* is much too large even to

enumerate!• Example (3 attributes, 2 FD’s and 43 entailed

dependencies)

Attr(R)=ABC and F ={ A B, B C } then F* isA S for all [subset of ABC] 8 FDs

B BC, B B, B C, B 4 FDs

C C, C , 3 FDs

AB S for all subsets S of ABC 8 FDs

AC S for all subsets S of ABC 8 FDs

BC BC, BC B, BC C, BC 4 FDs

ABC S for all subsets S of ABC 8 FDs

A B C

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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

• Formal Definition

F + (the closureclosure of F) is the set of dependencies which can

be deduced from F by applying Armstrong’s axioms

if F is a set of FD’s, then F+ { X Y F |= X Y }

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Armstrong’s AxiomsTheorem: 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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Armstrong’s Theorems• Additional rules derived from axioms

– Union

if A B and A C, then A BC

– Decomposition

if A BC, then A B and A C

A B

C

AB

C

For the proof, we can use:

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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+

Attr(R)=LMNO X=L

F={L M , M N, O N}

then X+ = L+ = LMN

L M N

O

Example:

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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

• Completeness:

(R, X,Y AttrR), 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*)

(In other words) there is a relational instance r in R (rR) in which all the

dependencies in F are true, but X Y does not hold

X X Y Y F*F* enough: enough:

Counter example!!Counter example!! 19

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

Counter example:

R=LMNO X=L

F={L M , M N, O N}

then X+ = LMN

L M N

O

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 (rR)

– in which all the dependencies in F are true

(F true in R ),

– but X Y does not hold 21

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

(assuming Boolean attributes, or more generally, that the two tuples agree on X+ but disagree elsewhere)

Attributes of X+ Other Attributes

1 1 … 1 1 1 … 1

1 1 …. 1 0 0 … 0

Relational instance r0 for R with 2 tuples

L+ = LMNL M N O

1 1 1 1

1 1 1 0

R=LMNO X=L

then X+ = L+

L O cannot be deduced

Example:

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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

Attributes of X+ Other Attributes

1 1 … 1 1 1 … 1

1 1 …. 1 0 0 … 0

Relational instance r0 for R with 2 tuples

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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 r0R=LMNO X=L

F={L M , M N, O N}

then X+ = LMN

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Completeness of the Armstrong’s Axioms

• 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

Attributes of X+ Other Attributes

1 1 … 1 1 1 … 1

1 1 …. 1 0 0 … 0

Relational instance r0 for R with 2 tuples

(R, X,Y Attr(R), F true in R : : X -> Y F+ => X -> Y F*)

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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)

XX so by F1

YX if then by F2

XX

YZXZ so by F1 YYZ YZY

so by F3 YXZ YYZYZXZ , YXZ

YX

ZYYX ,if then by F3 ZX so by F2 ZWXW ZWXW

ZYYX

,

S1

S2

S3

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Solution (completeness)

XY so {S1} Y Y, {S2}YZ Y

if XYZZ ::then hence X Y

YX {S1} X X, X Y, so{S3} XZ YZ

if then henceYZXZ

YX

A1

A2

A3ZYYX ,ifthen {S3}

henceZX

ZYYX

,

00 ZX

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Completeness of the Armstrong’s Axioms

• Exercises– Are the following set of rules a complete set of inference rules?

1111

111

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Ø A B AB Ø

A B

AB

}0{},,{)( AFBARAtt Suppose

Apply above rules exhaustively

ABB

Hence: answer is NO

obtained from F1-3but not from R1-3

R1: X XR2: X Y then XZ YR3: X Y, Y Z then X Z

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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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