I have:
U-> PT….. 1
Q-> SU……2
etc...
in using the reflexivity axiom can I then say
Q-> S , Q-> U
Q-> PT
I trying to ask how this axiom works using the example above.
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Should I add more information to the original post, not sure if this can be understood? – kt87 Mar 25 '11 at 21:33
1 Answers
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To derive
Q->S
Q->U
from
Q->SU
I'd use the decomposition rule, not the reflexivity axiom. Then I'd apply the transitivity axiom to Q->U, U->PT to derive Q->PT.
If you're asking what the reflexivity axiom means, it means
If Y is a subset of X, then X->Y.
In your example, it looks like you might be trying to say that
SU is a subset of Q, therefore Q->S and Q->U.
But it's not given that SU is a subset of Q. To make sure you get this point, Q->SU doesn't mean SU is a subset of Q.
For example, if you're in the military, your last name and blood type (among other things) are functionally dependent on your service number. Let's let the service number attribute be represented by "S", last name by "L", and blood type by "B". Then
S->LB
But neither "last name" nor "blood type" are subsets of "service number".
On the other hand, let's imagine that you're given this to start with.
U->PT
Q->SU
Q = {SUV} (New information!)
Since Q={SUV}, {S} is a subset of {SUV}, and {U} is a subset of {SUV}, then you can apply the reflexivity axiom to derive
Q->S (or SUV->S)
Q->U (or SUV->U)
But that axiom only applies in this example because you're given Q={SUV}.
Mike Sherrill 'Cat Recall'
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I dont understand the reflexivity rule at all, I am trying to do some more questions and I keep only using transitivity and augmentation and getting no where – kt87 Mar 25 '11 at 22:50
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so the only way it would of be ok to say that SU was a subset is if another dependency said SU -> Q ? – kt87 Mar 25 '11 at 23:02
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@kb88: I edited my answer, and added an example of what reflexivity is, and what it isn't. – Mike Sherrill 'Cat Recall' Mar 25 '11 at 23:24
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