1. Dr. Alexandra I. Cristea http://www.dcs.warwick.ac.uk/~acristea/ CS 319: Theory of Databases: C3

2. Exam preparation Perform example problems by yourself, then check results; If different, try to understand why; Search also for alternative solutions.

3. (provisionary) Content 1.Generalities DB 2.Temporal Data 3.Integrity constraints (FD revisited) 4.Relational Algebra (revisited) 5.Query optimisation 6.Tuple calculus 7.Domain calculus 8.Query equivalence 9.LLJ, DP and applications 10.The Askew Wall 11.Datalog

4. … previous ( HD: Temporal Data;) FD revisited; proofs with FD with definition & counter- example

5. FD Part 2: Proving with FDs: Proving with Armstrong axioms (non)Redundancy of FDs

6. Armstrong’s Axioms Axioms for reasoning about FD’s reflexivity F1: reflexivity if Y  X then X  Y augmentation F2: augmentation if X  Y then XZ  YZ transitivity F3: transitivityif X  Y and Y  Z then X  Z

7. Armstrong’s Axioms ++ Additional rules derived from axioms: Union F4. Union if A  B and A  C, then A  BC Decomposition F5. Decomposition if A  BC, then A  B and A  C Prove them ! AB C A B C

8. 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.

9. 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.

10. Rules hold vs redundant? Armstrong Rules hold – but are they all necessary? Can we leave some out? –How do we check this?

11. 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}.

12. F+ F+ = {fd | F |= fd} closure of F F* = {fd | F |- fd} cover of F

13. 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.

14. 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 R – {r} all possible consequences. Show then that there is another consequence to be computed with the help of r.

15. 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?

16. 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.

17. 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.

18. 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+.

19. How do we show something is redundant (superfluous)? Show that it is inferable from the other axioms

20. 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.

21. 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

22. FD Part 3 Soundness and Completeness of Armstrong’s axioms

23. Armstrong’s Axioms: Sound & Complete Ingredients: Functional Dependency (reminder) –Definition –Inference rules –Closure of 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

24. 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 t 1 and t 2 of r t 1 (X) = t 2 (X)  t 1 (Y) = t 2 (Y)

25. 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 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. ABC SA PI

26. 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

27. Set of Functional Dependencies F* coverFormal Definition of F*, the cover 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 }

28. 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 * is A  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 ABC

29. Armstrong’s Axioms Axioms for reasoning about FD’s F1: reflexivityif Y  X then X  Y F2: augmentation if X  Y then XZ  YZ F3: transitivityif X  Y and Y  Z then X  Z

30. F+F+ Informal Definition Formal Definition closure F + (the closure 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 }

31. 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?

32. 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

33. Armstrong’s Axioms ++ 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 AB C A B C For the proof, we can use:

34. Completeness of the Armstrong’s Axioms Proving that Armstrong’s axioms are a complete set of inference rules Armstrong’s rules generate all FDs in F*

35. 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 LMN O Example:

36. 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 +

37. Completeness ( F*  F + ) of the Armstrong’s Axioms 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 * ) (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 X  Y  F* enough: Counter example!!

38. 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 LMN O L  O cannot be deduced (so not in F+) but also does not hold (not in F*)

39. 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: –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

40. 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 r 0 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 …. 10 0 … 0 Relational instance r 0 for R with 2 tuples L + = LMN LMNO 1111 1110 R=LMNO X=L then X + = L + L  O cannot be deduced Example:

41. 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 r 0 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 r 0 Attributes of X + Other Attributes 1 1 … 1 1 1 …. 10 0 … 0 Relational instance r 0 for R with 2 tuples

42. 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 r 0 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 r 0 R=LMNO X=L F={L  M, M  N, O  N} then X + = LMN

43. Completeness of the Armstrong’s Axioms Proof that X  Y does not hold in r 0 : –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 r 0 –Hence Y contains (at least) an attribute not in the subset X +, confirming that X  Y does not hold in r 0 Attributes of X + Other Attributes 1 1 … 1 1 1 …. 10 0 … 0 Relational instance r 0 for R with 2 tuples

44. Completeness of the Armstrong’s Axioms We have proved the correctness (last colstruction) 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)

45. 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 ) Extra Question: –Can we have a sound and complete set of inference rules consisting of only 2 rules? –What about 1 rule?

46. Solution (soundness) so by F1 then by F2 so by F1 if so by F3 ifthen by F3 so by F2 S1 S2 S3

47. Solution (completeness) so {S1} Y  Y, {S2}YZ  Y ifthen hence X  Y {S1} X  X, X  Y, so {S3} XZ  YZ ifthen hence A1 A2 A3 if then {S3} hence

48. Completeness of the Armstrong’s Axioms Exercises –Are the following set of rules a complete set of inference rules? 1111 111 11 11 Ø A B AB Suppose Apply above rules exhaustivelyHence: answer is NO obtained from F1-3 but not from R1-3 R1: X  X R2: X  Y then XZ  Y R3: X  Y, Y  Z then X  Z

49. 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

50. … to follow RA revisited