Prove the following boolean identity using Consensus theorem. 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?
 A: $(A+B′)(B+C′)(C+D′)(D+A′)$
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.  
A: There are many ways to solve it:
First, through your results $(A'+B)(B'+C)(C'+D)(D'+A)(A+C')(A'+C)$
Recall the identity $(x'+y)(x+z)=(x'+y)(x+z)(y+z)$, apply it from the right side to the left side, the last two terms can be eliminated to obtain $(A'+B)(B'+C)(C'+D)(D'+A)$ and it suffices.
Or just try this: Using the identity: $x(x'+y)=xy$, $x=x(y+y')$ and $(x+y)C=xC+yC$ (and law of commutation, but it may be unnecessary to say)
$(A+B')(B+C')(C+D')(D+A')=(A+A')(A+B')(B+C')(C+D')(D+A')=A(A+B')(B+C')(C+D')(D+A')+A'(A+B')(B+C')(C+D')(D+A')=A(A+B')(B+C')(C+D')(D)+A'(B')(B+C')(C+D')(D+A')=A(A+B')(B+C')(C)(D)+A'(B')(C')(C+D')(D+A')=ABCD+A'B'C'D'$ 
Now by symmetry you can obtain the idenetity you want to prove, or continue as follows,
$(A'+B)(B'+C)(C'+D)(D'+A)=(A+A')(A'+B)(B'+C)(C'+D)(D'+A)=(A)(A'+B)(B'+C)(C'+D)(D'+A)+(A')(A'+B)(B'+C)(C'+D)(D'+A)=AB(B'+C)(C'+D)+A'(A'+B)(B'+C)(C'+D)D'=ABCD+A'B'C'D'$
So $(A+B')(B+C')(C+D')(D+A')=(A'+B)(B'+C)(C'+D)(D'+A)$
