# How to reduce C′A′B + CAB′ to C′B + CB′?

I faced this Boolean expression:

$C'(A'B+A)+C(A'+AB')$

It was solved as follows:

$C'(A'B+A)+C(A'+AB')$
$=C'(A+B)+C(A'+B')$ ...by applying absorption laws $(I)$ $=C'A+C'B+CA'+CB'$
$=(C\oplus A)+(C\oplus B)$

However somehow it did not clicked to me to apply absorption laws. So I did following:

$C'(A'B+A)+C(A'+AB')$
$=C'A'B+C'A+CA'+CAB'$ $...(II)$

Here, I can have $C'A+CA'=C\oplus A$, but I have following doubt:

What Boolean law / identity I can use to reduce $C'A'B+CAB'$ to $C'B+CB'$ (which then can be reduced to $C\oplus B$) ? Is it even possible? or I have to somehow remember to apply absorption law as shown above and their is not other way to arrive at $(C\oplus A)+(C\oplus B)$ by expanding as in step $(II)$?

What Boolean law / identity I can use to reduce $C'A'B+CAB'$ to $C'B+CB'$?
None. You can't reduce $C'A'B+CAB'$ to $C'B+CB'$. Those two Boolean expressions are not equivalent. For example, when $A=\text{True}$, $B=\text{True}$, and $C=\text{False}$, $C'A'B+CAB'$ becomes $\text{False}$, but $C'B+CB'$ becomes $\text{True}$.
• oohhh!!!!!! then where is the mistake in question? If $f=(C\oplus A)+(C\oplus B)$ and if I am getting $f=C'A'B+CAB'+(C\oplus A)$, then shouldn't $C'A'B+CAB'=(C\oplus B)$? Or I made mistake while saying the former itself? Do sound alcoholic? – anir Nov 4 '16 at 19:59