Hello shinhidora Originally Posted by

**shinhidora** Find the DNF for:

$\displaystyle f(x,y,z) = y\overline{z} + (y+z)(\overline{z+\overline{x}})$

so by using the axioms and stuff I found:

$\displaystyle f(x,y,z)= xy\overline{z}+ \overline{x}y\overline{z}$

but by putting everything into a table and filling in 1 for the 3 $\displaystyle x,y,z,\overline{x},\overline{y},\overline{z}$ that give $\displaystyle f = 1$ I found:

$\displaystyle f(x,y,z) = \overline{x}y\overline{z} + x\overline{y}z + xy\overline{z} + xyz$

So what did I do wrong or did I overlook something?

If the original expression was:$\displaystyle f(x,y,z) = \bar{x}y\overline{z} + (y+z)(\overline{z+\overline{x}})$?

then I agree with your first answer:$\displaystyle f(x,y,z)= xy\overline{z}+ \overline{x}y\overline{z}$

except that you can then simplify it further:$\displaystyle f(x,y,z)= xy\overline{z}+ \overline{x}y\overline{z}$$\displaystyle = (x+\bar{x})y\bar{z}$

$\displaystyle =y\bar{z}$

But if the original expression was correct, then I should write:$\displaystyle f(x,y,z) = y\overline{z} + (y+z)(\overline{z+\overline{x}})$$\displaystyle = y\overline{z} + (y+z)(\overline{z}x)$

$\displaystyle =y\bar{z}+xy\bar{z} +xz\bar{z}$

$\displaystyle =y\bar{z}(1+x)$

$\displaystyle =y\bar{z}$

which turns out to be the same anyway!

Grandad