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)

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

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

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.

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.

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

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.

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

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

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}

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

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

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