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+26 votes

Consider the following implications relating to functional and multivalued dependencies given below, which may or may not be correct.

- if $A \rightarrow \rightarrow B$ and $A \rightarrow \rightarrow C$ then $A \rightarrow \rightarrow BC$
- if $A \rightarrow B$ and $A \rightarrow C$ then $A \rightarrow \rightarrow BC$
- if $A \rightarrow \rightarrow BC$ and $A \rightarrow B$ then $A \rightarrow C$
- if $A \rightarrow BC$ and $A \rightarrow B$ then $A \rightarrow \rightarrow C$

- If A→→ B and A → → C then A → BC
- If A → B and A → C then A → → BC
- If A → → BC and A → B then A → C
- If A → BC and A → B then A → → C

Exactly how many of the above implications are valid?

- $0$
- $1$
- $2$
- $3$

–6

Here single arrow is functional dependency and doble arrow is multivalued dependency. But in axioms rule we only work with functional dependency . So ans. will be 0

+7

That's a very wrong way to put it :)

You should be able to solve by applying the definition of multi-valued dependencies.

You should be able to solve by applying the definition of multi-valued dependencies.

+2

@Arjun, suggest best resource for learning MVD, 4NF. Also till which normal form one should study ? I studied till BCNF. Due to this questions I'll do 4NF. i!

+8

yes. 4NF is required. I had followed Korth for this. I guess any resource is fine- just understand the meaning of MVD.

0

Please refer this video for better understanding of Multivalued dependencies.

Now according to multivalued dependency definition.

SSN -> -> CollegeName means

If we have the below tuples

SSN |
CollegeName |
Hobby |

100 | Global Academy Of Technology | Music |

100 | SJBIT | Swimming |

then we also need to have the below two tuples.

SSN |
CollegeName |
Hobby |

100 | Global Academy Of Technology | Swimming |

100 | SJBIT | Music |

__Counter example of Option (i)If A→→ B and A → → C then A → BC__

A |
B |
C |

100 | X1 | P1 |

100 | X2 | P2 |

100 | X1 | P2 |

100 | X2 | P1 |

200 | Y | Q |

300 | Z | R |

From this table we can see that A $\rightarrow \rightarrow$ B

and A $\rightarrow \rightarrow$ C.

But here A -> B and also A -> C do not hold true.

__Counter example of Option (iii)If A → → BC and A → B then A → C__

A |
B |
C |
D(Rest) |

100 | X1 | P1 | D1 |

100 | X1 | P1 | D2 |

200 | Y1 | Q1 | D3 |

200 | Y1 | Q2 | D4 |

For the same value of A,B,C the rest of the attributes(D here) can be interchanged.

Here A -> -> BC and A -> B hold true but A -> C is not true.

Please correct me if I'm wrong here.

+25 votes

Best answer

a. If A → → B and A → →C then A → BC . So FALSE

b. If A → B and A → C then A→ BC. So A → →BC TRUE..

c. If A → → BC and A → B here B is Subset of AB and (A intersection BC) is phi so

A → B but not A → C so FALSE (** Coalescence rule** )

d. If A → BC then A → C so A → → C TRUE

if A → B then A → → B holds but reverse not true.

Correct Answer: $C$

0

a. This will be true if A->> B and A->>C then A->>BC .So it is false

b. As functional dependency satisfying , it also supports multilevel dependency. So true

c. proved by coalescence rule

d. It is proved by distributive rule of functional dependency and then multilevel dependency

am I right ?

b. As functional dependency satisfying , it also supports multilevel dependency. So true

c. proved by coalescence rule

d. It is proved by distributive rule of functional dependency and then multilevel dependency

am I right ?

+7 votes

Check answer to following question ->

http://dba.stackexchange.com/questions/123510/how-can-i-prove-disprove-if-a-%E2%86%A0-bc-and-a-%E2%86%92-b-then-a-%E2%86%92-c

http://dba.stackexchange.com/questions/123510/how-can-i-prove-disprove-if-a-%E2%86%A0-bc-and-a-%E2%86%92-b-then-a-%E2%86%92-c

+2 votes

**Every FD is a MVD**.

i.e suppose $x\rightarrow y$ $\Rightarrow$ $x\rightarrow \rightarrow y$

If y can be determined by x on y's single value then we can easily say x multi-determines y. as **single value** $\subseteq$ **multiple value.**

Now,

1. can't even possible.

2. A -> BC which implies A -> -> BC. (true)

3. using given data we can't prove the then part.

4. given FDs are A->B & A->C, so using this we can say A->->C.(true)

here 2 implications are valid. i.e **option C**

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