Re: ?? Functional Dependency Question ??

On Oct 18, 11:18 am, Bob Badour <bbad..._at_pei.sympatico.ca> wrote:
> tlbaxte..._at_yahoo.com wrote:
>
> > "Although X->A and X->B implies X->AB by the union rule stated above,
> > X->A, and Y->B does *not* imply that XY->AB."
>
> > I'm not seeing this. It seems to me that X->A, and Y->B *DOES* imply
> > that XY->AB.
>
> > I'm sure I'm wrong but I'm not seeing it. Can someone explain?
>
> > Thanks
>
> I'm not seeing it either. By these truth tables, it seems to:
>
> XY  AB X->A  Y->B  (X->A)(Y->B)  XY->AB  (X->A)(Y->B)->(XY->AB)
> 00  00  1     1          1         1                 1
> 00  01  1     1          1         1                 1
> 00  11  1     1          1         1                 1
> 00  10  1     1          1         1                 1
> 01  10  1     0          0         1                 1
> 01  11  1     1          1         1                 1
> 01  01  1     1          1         1                 1
> 01  00  1     0          0         1                 1
> 11  00  0     0          0         0                 1
> 11  01  0     1          0         0                 1
> 11  11  1     1          1         1                 1
> 11  10  1     0          0         0                 1
> 10  10  1     1          1         1                 1
> 10  11  1     1          1         1                 1
> 10  01  0     1          0         1                 1
> 10  00  0     1          0         1                 1
>
> (View with a fixed width font)
>
> Can anyone find a mistake in the above truth tables? Is there a
> difference between functional dependency and implication that I need to
> learn?

Your truth table is correct. You can also prove this with Boolean algebra (below ~ = not, + = or, * = and):

given :

(1) 1 = ~X + A                       :  X implies A
(2) 1 = ~Y + B                       :  Y implies B

prove :

(3) 1 = ~(XY) + AB                   :  XY implies AB

proof :

(4) 1 = (~X + A)(~Y + B)              :  conjuction of (1) and (2)
(5) 1 = ~X~Y + ~XB + ~YA + AB         :  distributive and commutative
(6) 1 = ~X~Y + ~X~Y + ~XB + ~YA + AB  :  idempotent
(7) 1 = ~X(~Y + B) + ~Y(~X + A) + AB  :  distributive and commutative
(8) 1 = ~X + ~Y + AB                  :  substitute (1) and (2)
(9) 1 = ~(XY) + AB                    :  De Morgan

QED KHD Received on Sun Oct 19 2008 - 23:07:23 CEST