Prove that A’C + A’B + BC equals C (A XOR B)’ + A’B

Question

Prove $$A’C + A’B + BC = C (A \oplus B)’ + A’B$$

Here is what i have tried: $$A’C + A’B + BC = C(A’+B) + A’B$$ Now, need to prove $(A \oplus B)’ = A’+B$

$$(A \oplus B)’ = A'B'+AB = \text{now what??}$$

Answer 1

Please note that $(A \oplus B)’ \not = A’+B$

Your mistake is that just because $$C (A \oplus B)’ + A’B = C(A’+B) + A’B$$

does not mean that $(A \oplus B)’ = A’+B$

That is: just because two larger statements are equivalent does not mean that two components of those larger statements are the same, even if the rest is the same. As another example:

Just because $P + P'Q = P + Q$ does not mean that $P'Q=Q$

So, your approach is not going to work.

Instead, I would start with the right side, so you can first rewrite the $\oplus$ into the more familiar boolean operations (e.g. use $A \oplus B = AB'+A'B$) so you can then use the more familiar laws for those more familiar operators to try and end up with the left hand side.

Answer 2

HINT: I am unsure whether the question is correct or not, but here is the hint: use truth tables.

And $(A \oplus B)’ = A’+B$ is wrong...(by symmetric relation in A and B)