CS 319: Theory of Databases
Dr. Alexandra I. Cristea
http://www.dcs.warwick.ac.uk/~acristea/
Content
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Generalities DB
Temporal Data
Integrity constraints (FD revisited)
Relational Algebra (revisited)
Query optimisation
Tuple calculus
Domain calculus
Query equivalence
LLJ, DP and applications
The Askew Wall
Datalog
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… previous
FD revisited; proofs with FD with definition & counterexample
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FD Part 2: Proving with FDs:
• Proving with Armstrong axioms
• (non)Redundancy of 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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Theorems
• Additional rules derived from axioms:
F4. Union
if A  B and A  C, then A  BC
A
F5. Decomposition
if A BC, then A  B and A  C
B
C
B
A
• Prove them!
C
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Union Rule
if A  B and A  C, then A  BC
• Let A  B and A  C
• A  B, augument (F2) with A: A  AB
• A  C, augument (F2) with B: AB  BC
• A  AB and AB  BC, apply transitivity
(F3): A  BC q.e.d.
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Decomposition Rule
if A  BC, then A  B and A  C
• Let A  BC
• BBC, apply reflexivity (F1) : BC  B
• A  BC and BC  B, apply
transitivity (F3): A  B
• Idem for A  C q.e.d.
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Rules hold vs redundant?
• Armstrong Rules hold – but are they all
necessary?
• Can we leave some out?
– How do we check this?
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Redundancy
DEF: An inference rule inf in a set of
inference rules Rules for a certain type
of constraint C
is redundant (superfluous)
when for all sets F of constraints of type
C it holds that:
F+{Rules –{inf}} = F+{Rules}
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F+, F*
• F+ = {fd | F |= fd}
• F* = {fd | F |- fd}
closure of F
cover of F
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Exercises
1. Show that Armstrong’s inference rules
for FDs (F1-3) are not redundant.
2. Show that Rules = {F1, F2, F3, F4} is
redundant.
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Hint (Ex. 1)
• Show with the help of an example that, if one
of the three axioms is omitted, the remaining
set of functional dependencies is not
complete.
• Take therefore an appropriate set of
constraints and compute with the help of
Rules – {inf} all possible consequences.
Show then that there is another consequence
to be computed with the help of inf.
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Solution
• We start from a relation scheme R and an arbitrary
legal instance r(R). Let ,  and  be sets of attributes
(headers), so that Attr(R), Attr(R) and Attr(R).
We have the following axioms:
• F1: (Reflexivity) Let  be valid (holds). Then we
also have →.
• F2: (Augmentation) Let → be valid. Then we also
have →.
• F3: (Transitivity) Let → and → be valid. Then
we also have →.
• Now we omit in turn one of the axioms.
– Why in turn?
– Why not just one?
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Case 1: F1 is not superfluous:
• Let Attr(R) = {X} and F = . Because F is
empty, neither F2 nor F3 can be used to
deduce new fds. Therefore, F+ = F = .
• From F1 we could however deduce that X
 X is valid, which is not present in the
above set.
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Case 3: F2 is not superfluous:
• Let R = {X, Y} and F = {X  Y}.
• With the help of F1 and F3 we deduce:
• F+ = {   , X  X, Y  Y, X  ,
Y  , XY  XY, XY  Y, XY  X,
XY  }
• However, with X  Y and with the help
of F2 we can infer that X  XY is valid,
which is not present in the above set.
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Case 3: F3 is not superfluous:
• Let R = {X, Y, Z} and F = {X  Y, Y  Z }.
F+ = {
XYZ  XYZ, XY  XY,
YZ  YZ,
X  Y,
Y  Z,
XYZ  XY, XY  X,
YZ  Y,
X  XY,
Y  YZ,
XYZ  XZ, XY  Y,
YZ  Z,
XY  Y,
XY  XZ,
XYZ  YZ, XY  ,
YZ  ,
XZ  YZ,
YZ  Z,
XYZ  X, XZ  XZ,
X  X, X  , Y Y, Y  , Z  Z, Z  
XYZ  Y,
XYZ  Z, XYZ  ,
XZ  ,   }
• With the help of F3 we can also infer X  Z, which is not in F+.
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How do we show something is
redundant (superfluous)?
• Show that it is inferable from the other
axioms
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F4 is superfluous:
• F4 (union rule) : Let → and  → be valid.
• Then →  is also valid.
• We show now that F = {F1, F2, F3, F4} is redundant is by, e.g., inferring F4
from the other three.
• By using augumentation, from → we deduce that also → is valid
(augmentation with ).
• By using augumentation, from → we deduce that also → is valid
(augmentation with ).
• By using transitivity, from → and →, we deduce that also → is
valid.
• Note that to prove that a set of rules (axioms) is redundant we can use
normal calculus; however, to prove that a set of rules is not redundant, we
need to know the meaning of the rules.
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Summary
• We have learned how to prove fds based
on the Armstrong axioms
– and also why & when it’s ok to do so
• We have learned how to prove that a set
of axioms is redundant or not
• We have learned that the Armstrong
axioms are not redundant
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… to follow
Constrains revisited: Soundness and Completeness of
Armstrong Axioms
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