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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?

user1766888
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1 Answers1

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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 $F'$: Truth table

Reading down the list, we note that rows 0,1,2,3,6, and 7 have $1$s. So, $$ F'=m_0+m_1+m_2+m_3+m_6+m_7 $$

Ben Grossmann
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  • I'm confused on your removed redundant terms. at that point I have (x+w+x)(x+w+y+z+z)(z'+x)(x'+y+z+z). So I'm removing the x in the first term, z in the second, and z in the third to get (z+w)(x+w+y+z)(z'+x)(z'+y+z) right? What did I miss there? – user1766888 Jun 27 '13 at 21:54
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    Made a mistake there: $(w+x)(w+x+y+z)=(w+x)$, I did the reverse. The other reduction I made is that $$(z+y+z')=(y+(z+z'))=y+1=1$$ – Ben Grossmann Jun 27 '13 at 22:01
  • Is it possible you could supply the answer by finding the inverse first then the sum of products for that inverse? I know its the same as you have above but I'm having more trouble that way. :( – user1766888 Jun 27 '13 at 22:07
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    All right, I'll write it up that way. Apparently, I was having trouble that way too :P. Are you supposed to do it algebraically, or can you just do it with a truth table? – Ben Grossmann Jun 27 '13 at 22:10
  • Algebraically. With a truth table it might be huge since there are four variables. – user1766888 Jun 27 '13 at 22:11
  • It is surprisingly effective for this stuff. Remember that each 1 on the truth table corresponds to a term in the SOP form – Ben Grossmann Jun 27 '13 at 22:23
  • let us continue this discussion in chat – Ben Grossmann Jun 27 '13 at 22:24