# Quick Boolean Agebra Question

## Homework Statement

xy + compliment(xy) = 1

## The Attempt at a Solution

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

maybe its not true...

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CompuChip
Homework Helper
Yes. Take $x + \overline{x} = 1$ and rename x to a: $a + \overline{a} = 1$.
Now set $a = xy$.

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

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

tiny-tim
Homework Helper
Hi ZeroPivot!

(guys, thanks for the compliments, but it's complements! )
what if x=1 and y=0 the xy=0 and compliment(xy)=0 …
No, complement(xy) = 1

Hi ZeroPivot!

(guys, thanks for the compliments, but it's complements! )

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

but thanks.

tiny-tim
Homework Helper
Hi ZeroPivot!
xy + compliment(xy) = 1

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

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

but thanks.
So you meant, is $xy + \bar{x}\bar{y} = 1$ ?

No.

CompuChip
Homework Helper
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 $\bar x + \bar y$.