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I have an expression here from the Full Adder circuit, used for binary addition. One equation used to make it work, is this one:

$$C = xy + xz + yz \tag{1}$$

Now, the book transforms this equation into this: $$C = z(x'y + xy') + xy \tag{2}$$

In the immediate step, the do this: $$C = z(x \oplus y) + xy \tag{3}$$

Now, my question is how does the second equation come from the first one?

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  • $\begingroup$ What is your definition of the exclusive-or ($x \oplus y$)? $\endgroup$ – Adriano Jul 21 '13 at 10:52
  • $\begingroup$ @Adriano: The standard definition : en.wikipedia.org/wiki/Exclusive_or#Truth_table $\endgroup$ – Games Brainiac Jul 21 '13 at 10:53
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    $\begingroup$ Then, by definition, we can replace ($x \oplus y$) with $(x'y + xy')$. $\endgroup$ – Adriano Jul 21 '13 at 10:54
  • $\begingroup$ @Adriano: Yes, ofcourse, but if you take a look at the question, How do you arrive from the first equation to the third one? This was my fault, I will replace 2nd with the 3rd. My mistake, many apologies. $\endgroup$ – Games Brainiac Jul 21 '13 at 10:56
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\begin{align*} C & = xy + xz + yz\\ & = xy + x(y + y')z + (x + x')yz \\ & = xy + xyz + xy'z + xyz + x'yz \\ & = xy(1 + z + z) + (xy' + x'y)z \\ & = xy + (xy' + x'y)z \\ & = xy + (x \oplus y)z \end{align*}

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  • $\begingroup$ On line 4, how do you get $xy(1+z+z)$ for the last term, because $xyz != xy'z$ $\endgroup$ – Games Brainiac Jul 21 '13 at 11:12
  • $\begingroup$ @Games The third and fourth terms ($xy'z$ and $xyz$) were swapped before the $xy$ and $z$ were factored out. $\endgroup$ – Adriano Jul 21 '13 at 11:15
  • $\begingroup$ @Adriano: Yes, I get it now. Sorry to be so dumb. Thanks to the both of you for your help. Hope one day, I can answer questions on the site, like you guys. $\endgroup$ – Games Brainiac Jul 21 '13 at 11:16

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