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how can I reduce:

$co = ci'ab+cia'b+ciab'+ciab$

to: $co = ab + cia + cib$

I don't know how they did that on the book... This is for Adders, working with circuits.

Thank you very much

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  • $\begingroup$ Whats means CDA? Is it Compact Disc Audio? And which book are you referring to? $\endgroup$ – J.-E. Pin Oct 11 '17 at 10:31
  • $\begingroup$ ...........loool $\endgroup$ – Spyky Nov 1 '17 at 16:34
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I am not an expert in circuits, but I have some doubt on your question. If you take $c = i = 0$ and $a = b = 1$, then $ci'ab+cia'b+ciab'+ciab = 0$ but $ab + cia + cib = 1$.

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