# Quick Boolean Agebra Question

• ZeroPivot
In summary, The equation xy + compliment(xy) = 1 is true only if x and y are binary variables, and cannot be generalized to other values. The equation x+compliment(x) = 1 does not hold for all values of x. The equation xy + compliment(xy) = 1 and the equation x+compliment(x) = 1 are not equivalent and cannot be used interchangeably.

## Homework Statement

xy + compliment(xy) = 1

## The Attempt at a Solution

is it true? because x+compliment(x) = 1

maybe its not true...

Yes. Take ##x + \overline{x} = 1## and rename x to a: ##a + \overline{a} = 1##.
Now set ##a = xy##.

Note, however, that ##\overline{xy} \neq \overline{x} \overline{y}##.

• 1 person
CompuChip said:
Yes. Take ##x + \overline{x} = 1## and rename x to a: ##a + \overline{a} = 1##.
Now set ##a = xy##.

Note, however, that ##\overline{xy} \neq \overline{x} \overline{y}##.

what if x=1 and y=0 the xy=0 and compliment(xy)=0 then 0+0 != 1

Hi ZeroPivot! (guys, thanks for the compliments, but it's complements! )
ZeroPivot said:
what if x=1 and y=0 the xy=0 and compliment(xy)=0 …

No, complement(xy) = 1 • 1 person
tiny-tim said:
Hi ZeroPivot! (guys, thanks for the compliments, but it's complements! )

No, complement(xy) = 1 i meant compliment(x)compliment(y) = 0

but thanks.

Hi ZeroPivot! ZeroPivot said:
xy + compliment(xy) = 1

is it true? because x+compliment(x) = 1

maybe its not true...
ZeroPivot said:
i meant compliment(x)compliment(y) = 0

but thanks.

So you meant, is ##xy + \bar{x}\bar{y} = 1## ?

No.

• 1 person
tiny-tim said:
Hi ZeroPivot! So you meant, is ##xy + \bar{x}\bar{y} = 1## ?

No.

Because if it were, ##\overline{xy}## would equal ##\bar x \bar y##. In fact it equals ##\bar x + \bar y##.