Quick Boolean Agebra Question

[ZeroPivot]

Homework Statement

xy + compliment(xy) = 1

Homework Equations

The Attempt at a Solution

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

maybe its not true...

 

[CompuChip]

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

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

 

[ZeroPivot]

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]

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

 

[ZeroPivot]

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]

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]

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 $\bar x + \bar y$.