I'm given the equation
$F = (x+w)z' + x(y+z) + xz$
The inverse I got is
$F' = [(x'+z)*(w'+z)]*[(x'+y')*(x'+z')]*[x'+z']$
To start I would expand
$F'= [x'w' + x'z + zw'+zz] * [x'x' + x'z' + y'x' + y'z'] * [x'+z']$
What laws can I apply here to simplify further then keep distributing out?
I'm having trouble expanding this further. What would the next expression be?
Best Answer
I find it's much easier to invert an expression once it's in POS or SOP form. So, I'll begin by getting $F$ as the product of sums, then invert. We have: $$ \begin{align} F &= (x+w)z'+x(y+z)+\{x(z)\} \text{ ......................factor out the $x$}\\ &= (x+w)z'+x(y+z+\{z\}) \text{ ...........................expand $ab+cd=(a+c)(a+d)(b+c)(b+d)$}\\ &= (x+w+x)(x+w+y+z)(z'+x)(z'+y+z) \text{ ...remove redundant terms}\\ &= (w+x)(x+z') \end{align} $$ To invert, apply deMorgan's law: $$ \begin{align} F' &= w'x'+x'z\\ &= (w'x'y'z'+w'x'y'z+w'x'yz'+w'x'yz)\\ &\,\,+ (w'x'y'z+w'x'yz+wx'y'z+wx'yz)\\ &= \end{align} $$
Inverting first you get this:
$$ \begin{align} F'&=[(x+w)z'+x(y+z)+xz]'\\ &=[(x+w)z']'[x(y+z)]'[xz]'\\ &=[x'w'+z][x'+y'z'][x'+z']\\ \end{align} $$ Now, expand that product of binomials $$ \begin{align} F'&=[w'x'+z][x'+y'z'][x'+z']\\ &=w'x'+(w'x'z')+w'x'y'z'+(x'w'y'z')+(\text{stuff containing $zz'$})+x'z\\ &=w'x'+x'z \end{align} $$ From there, expand each sum to get the SOP form as above
Okay so maybe I'll correct all that, but the quickest way to do SOP is just with a truth table. This is the truth table for $F'$:
Reading down the list, we note that rows 0,1,2,3,6, and 7 have $1$s. So, $$ F'=m_0+m_1+m_2+m_3+m_6+m_7 $$