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*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.

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    $\begingroup$ Isn't the identities $FF' = 0; F+F' = 1$ are true by definition of negation? $F' = \overline F$ in Boolean algebra. $\endgroup$ – Kaster Mar 17 '13 at 8:47
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$$xx' = 0\tag1$$?

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 $(2)$

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

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