1. Problem Axiom of addition:
If X ® Y and W ® V then XW ® YV
Prove this axiom using Armstrong’s Axioms or show that it is not sound
X ® Y (given)
XW ® YW (augmentation of 1)
W ® V (given)
YW ® YV (augmentation of 4)
XW ® YV (transitivity of 2, 4)

2. Problem Axiom of confusion:
If XZ ® Y, YZU®X, W ® V, V ® WU then
XW ® YV
Prove this axiom using Armstrong’s Axioms or show that it is not sound U V W X Y Z u v w x y1 z1 y2 z2

3. Problem R = (A,B,C,D). For each of the following:
find all keys
determine the normal form of R
C → D, C → A, B → C B, not 3NF
ABC → D, D → A ABC, DBC 3NF
A → B, BC → D, A → C A, not 3NF
AB → C, AB → D, C → A, D → B AB, CB, AD, CD, 3NF

4. Problem
Proposition: Let R be a relation containing exactly two attributes. Let F be a set of functional dependencies. Then, R is in BCNF.
Prove or give a counter-example.

5. Proof: Let R(A,B) be a relation with 2 attributes and let F be functional dependencies over R. Every dependency other than AB, B A, A AB and B BA is trivial. In all the dependencies detailed, the left side is a key.

6. Problem
Proposition: Let R be a relation and F be a set of functional dependencies. Suppose that every key in F has a single attribute. Then, R is in BCNF.
Prove or give a counter-example.
Counter Example: R(A,B,C) A B, B C

7. Problem
Proposition: Let R be a relation and F be a set of functional dependencies. Suppose that every key in F has a single attribute. Then, R is in BCNF if and only if R is in 3NF.
Prove or give a counter-example.

8. Correct, by an easy case analysis
Correct, by an easy case analysis. If R is in BCNF, then it is in 3NF by definition. Suppose R is in 3NF. Let X A be a dependency in R. If this dependency is trivial or X is a super key, then this dependency does not contradict the BNCF requirement. If A is a field in a key, then A is a key, and hence, X is a superkey

9. Problem Prove or show a simple counter example.
Proposition: Let R be a relation and F be a set of functional dependencies. Suppose that every dependency in F has one attribute on the left side. Then, R is in BCNF if and only if R is in 3NF
Prove or show a simple counter example.
Counter example: R(A,B,C) A B, B A

10. Problem Find small(est) examples of R and F such that:
R is in BCNF R(A), F is empty
R is in 3NF, but not in BCNF R(A,B,C), F = {A B, B A}
R is not in 3NF R(A,B,C), F={A B}

11. Problem R=(ABCD) F={A->B,C->D,A->D}
What is the normal form of R? Not 3NF
We have a decomposition R1=(ABC), R2=(BCD)
Is this a lossless join decomposition? Yes

12. Problem R=(ABC) F={A->B,BC->A} What is the normal form of R? 3NF
We have a decomposition R1=(AB), R2=(AC)
Is this a lossless join decomposition? Yes

13. Problem Let R=ABCDEG Let F={C->D, E->C, EG->A, G->B}
What are all the keys of R? EG
What is the normal form of R? Not 3NF
Is the following decomposition lossless? Yes
R1=AEG, R2=CE, R3=BG, R4=DEG