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If you have $(a+b)(c+d)(e+f)$ how can you expand this? Someone was mentioning that you get different combinations so that you get $adf+ade+acf+ace+bdf+bde+bce+bcf$? Is that the full expansion?

As an example, how can it be applied to the boolean expression: $F' = [x'w'+z]*[x'+y'z']*[x'+z']$ which should reduce to the sum of products?

user1766888
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You will indeed get $2^3 = 8$ terms when expanding a triple binomial: so yes, the first expansion is indeed the full expansion, barring any information about the relation of the variables $a, b, c, d, e, f$.

It can be applied to your boolean expression $F'$. Based on the given "standard expansion, we obtan:

$$\begin{align} F' & = x'w'y'z' + x'w'z' + x'w'x' + x'w'x' + zy'z' + zz' + zx' + zx'\\ \\ & = x'w'y'z' + zx'\tag{1}\end{align}$$

Can you see how we can reduce the original expansion to two terms, as in $(1),\,$ from the $\,8$-term "formulaic" expansion? Here we have repetition of some of the variables, and we have the appearance of the variables $x, x'$ and $z, z'$:

  • Note that there is reiteration (and hence we can simplify): $x'w'x' + x'w'x' = x'w'x' = x'w'$, and if the first term $x'w'y'z'$ holds, then so does $x'w'z'$ and so does $x'w'$. Similarly, there is reiteration/simplification: $zx' + zx' = zx'$

  • The term $zy'z' = zz'y = F\cdot y = F$, and obviously, $zz' = F$.

So the reduced Sum of Products (which is essentially, disjunctive normal form) of $F'$ is expressed in $(1)$: $$F' = x'w'y'z' + zx'$$

amWhy
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  • Just one question and I've got it. How is it that if x'w'y'z' holds then x'w'z'+x'w' are simplified out? – user1766888 Jun 28 '13 at 03:35
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    Remember that the product of variables is a conjunction ("and-ed"). So if we keep $x'w'y'z'$ then we have that $x'$ and $y'$ and $w'$ and $z'$ hold, so it follows that $x'$ and $w'$ and $z'$ = x'w'z' holds, as does $x'$ and $w'$ = $x'w'$. The $x'w'z'$ and $x'w'$ are redundant if we have that $x'w'y'z'$: they carry no additional information than what's conveyed by $x'w'y'z'$ – amWhy Jun 28 '13 at 03:38
  • Oh! I see it now! Wow, you did a great job explaining that :) Thank you so much. – user1766888 Jun 28 '13 at 03:39
  • You're welcome! – amWhy Jun 28 '13 at 03:41
  • I am Welcome! $+^{+^{+_{+}}}$ – Mikasa Jun 28 '13 at 05:39