Prove the following boolean identity using Consensus theorem.

Question

I have been trying to prove it for last 4 hours but couldn't find a solution. Please help me.

$$(A+B')(B+C')(C+D')(D+A')=(A'+B)(B'+C)(C'+D)(D'+A)$$

I solved and got the following answer.

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

How to remove the last 2 terms?

Answer 1

$(A+B′)(B+C′)(C+D′)(D+A′)$

Let's call your terms 1, 2, 3 & 4.

Use consensus on terms 1 and 2 to create term W. 2 & 3 = term X, 3 & 4 = term Y and 4 & 1 to create term Z.

$(A+B′)(B+C′)(C+D′)(D+A′)(A+C′)(B+D′)(C+A′)(D+B′)$

You will now have 8 terms.

Use consensus to create target terms. A=1&X, B=2&Y, C=3&Z, D=4&W. 12 terms.

$(B+C′)(C+A′)\color {red}{(A'+B)}\ (C+D′)(D+B′)\color {red}{(B'+C)}\ $ $\ \ \ (A+C′)(D+A′)\color {red}{(C'+D)}\ (B+D′)(A+B′)\color {red}{(D'+A)}$

Use consensus to absorb terms 1, 2, 3 & 4. 2=W&A, 3=X&B, 4=Y&C and 1=Z&D. 8 terms.

$(B+C′)(A+C′)\color {red}{(A'+B)}\ (C+D′)(B+D′)\color {red}{(B'+C)}\ $ $\ \ \ (D+A′)(C+A′)\color {red}{(C'+D)}\ (A+B′)(D+B′)\color {red}{(D'+A)}$

$(A+C′)\color {red}{(A'+B)}\ (B+D′)\color {red}{(B'+C)}\ (C+A′)\color {red}{(C'+D)}\ (D+B′)\color {red}{(D'+A)}$

You should be able to finish it from here.