Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

*Find the complement of $F=x+yz$; then show that $FF’ = 0$ and $F + F’ = 1$

$F(x,y) = x+yz$

$F’(x,y) = (x+yz)’ = x’(yz)’ = x’(y’+z’)$

$FF’ = (x+yz)x’(y’+z’) = (xx’+x’yz)(y’+z’) = x’yz(y’+z’) = x’yy’z+x’yzz’ = 0+0 = 0$

$F+F'= x+yz)+x’(y’+z’)=x+yz+x’y’+x’z’ = x(y’+y)(z’+z)+(x’+x)yz+x’y’(z’+z)+x’(y’+y)z’$

$ = xy’z’+xy’z+xyz’+xyz+x’yz+xyz+x’y’z’+x’y’z+x’y’z’+x’yz’$

$ = x’y’z’+x’y’z+x’yz’+x’yz+xy’z’+xy’z+ xyz’+xyz = Σ(0,1,2,3,4,5,6,7) = 1 $

My problem is with these parts right here:

$x’yy’z+x’yzz’ = 0+0 = 0$

$ Σ(0,1,2,3,4,5,6,7) = 1 $

Why are they true? I've understood how to get up to those points, but I don't understand why $x’yy’z+x’yzz$ are both 0 and why $Σ(0,1,2,3,4,5,6,7) = 1$

PS: 1 is true, 0 is false.

share|improve this question
1  
Isn't the identities $FF' = 0; F+F' = 1$ are true by definition of negation? $F' = \overline F$ in Boolean algebra. –  Kaster Mar 17 '13 at 8:47

1 Answer 1

up vote 1 down vote accepted

$$xx' = 0\tag1$$?

The following truth table gives the reasoning

$$ \begin{array}{c|c|c} x & \text{x`} & \text{x.x`}\\ \hline 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{array} $$

Now your boolean expression

$$x.y.y`z+x`.y.z.z`$$ $$x.(y.y)`z+x`.y.(z.z`)$$

Using $(1)$ $$x.(0)`z+x`.y.(0)$$ $$=0+0$$ $$=0$$

$$x’y’z’+x’y’z+x’yz’+x’yz+xy’z’+xy’z+ xyz’+xyz = Σ(0,1,2,3,4,5,6,7)$$ $$x’y’(z’+z)+x’y(z`+z)+x`y(z’+z)+ xy(z’+z)$$

$$z+z` = 1\tag2$$ The following truth table gives the reasoning

$$ \begin{array}{c|c|c} x & \text{x`} & \text{x+x`}\\ \hline 0 & 1 & 1 \\ 1 & 0 & 1 \\ \end{array} $$

So we have from $(2)$

$$x’y’(1)+x’y(1)+xy1(1)+ xy(1)$$ $$=x’y’+x’y+xy`+ xy$$ $$=x’(y’+y)+x(y`+ y)$$ $$=x’(1)+x(1)$$ $$=x’+x$$ $$=1$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.