# Functional Dependencies PPT Notes

Question asked by ISHAN TOPRE in #Coffee Room on Mar 19, 2011

ISHAN TOPRE 路 Mar 19, 2011

Rank A2 - PRO

I friends I found these notes when I was studying Normalization.I hope you find them useful. 😀
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#Coffee Room

ISHAN TOPRE 路 Mar 19, 2011

Rank A2 - PRO

**Re: Functional Dependencies**

Also see this! 馃槢

kevin88 路 Mar 19, 2011

Rank E1 - BEGINNER

Hello

I am having a problem understanding dependency theory using Armstrong's Axioms, I have researched this online and I understand the axioms however I am struggling to understand a particular example. Can anyone provide me with any advice or help?

Can I post the example here?

Thank you

I am having a problem understanding dependency theory using Armstrong's Axioms, I have researched this online and I understand the axioms however I am struggling to understand a particular example. Can anyone provide me with any advice or help?

Can I post the example here?

Thank you

ISHAN TOPRE 路 Mar 19, 2011

Rank A2 - PRO

Yes of course you can post your query here. Our CS/IT people will try to explain you the theory so that you will be able to solve your problem. 馃榾

kevin88 路 Mar 19, 2011

Rank E1 - BEGINNER

Thank you ishutopre

I have two examples with answers but I do not understand how the answer has been derived:

Given

R < A= {V,W,X,Y,Z},

Gamma = { 1. V -> Y, 2. YW -> Z

3. Y -> W}>

RTP(required to prove) VWY -> Z and VXY ->Z

Answers:

VWY -> WY -> Z

VXY -> Y -> YW -> Z

augment 3 by Y Y -> W

Given the following relation

R < A = {M,N,O,P,Q,R,S},

Gamma = {Q->MO, S->R, M->SP}>

prove that PM->R

Answer:

PM -> SP -> S -> R

M -> SP augment by P

I have two examples with answers but I do not understand how the answer has been derived:

**Example 1:**Given

R < A= {V,W,X,Y,Z},

Gamma = { 1. V -> Y, 2. YW -> Z

3. Y -> W}>

RTP(required to prove) VWY -> Z and VXY ->Z

Answers:

VWY -> WY -> Z

VXY -> Y -> YW -> Z

augment 3 by Y Y -> W

**Example 2:**Given the following relation

R < A = {M,N,O,P,Q,R,S},

Gamma = {Q->MO, S->R, M->SP}>

prove that PM->R

Answer:

PM -> SP -> S -> R

M -> SP augment by P