# Full Adder boolean Algebra simplification

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?

• What is your definition of the exclusive-or ($x \oplus y$)? – Adriano Jul 21 '13 at 10:52
• @Adriano: The standard definition : en.wikipedia.org/wiki/Exclusive_or#Truth_table – Games Brainiac Jul 21 '13 at 10:53
• Then, by definition, we can replace ($x \oplus y$) with $(x'y + xy')$. – Adriano Jul 21 '13 at 10:54
• @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. – Games Brainiac Jul 21 '13 at 10:56

• On line 4, how do you get $xy(1+z+z)$ for the last term, because $xyz != xy'z$ – Games Brainiac Jul 21 '13 at 11:12
• @Games The third and fourth terms ($xy'z$ and $xyz$) were swapped before the $xy$ and $z$ were factored out. – Adriano Jul 21 '13 at 11:15