Firstly I expanded both LHS and RHS as following: LHS: (a+b')(b+c')(c+a') = ab + ac' + b'b + b'c' = ab + ac' + b'c' = abc + ac'c + b'c'c + a'ab + a'ac' + a'b'c' = abc + 0 + 0 + 0 + 0 + a'b'c' = abc + a'b'c'
RHS: (a'+b)(b'+c)(c'+a) = abc + a'b'c'. I get similar expression on RHS and LHS. This is not algebraic expression. To me it seems that both LHS and RHS are in product-of-sums form. How should this boolean equality be proved ?
Here's a useful equivalence principle:
Consensus
$(P \lor Q) \land (\neg Q \lor R) \land (P \lor R) \Leftrightarrow (P \lor Q) \land (\neg Q \lor R)$
$(P \land Q) \lor (\neg Q \land R) \lor (P \land R) \Leftrightarrow (P \land Q) \lor (\neg Q \land R)$
Applied to your statement:
$$(a + b')(b+c')(c+a')= \text{ (Idempotence)}$$
$$(a + b')(b+c')(c+a')(a + b')(b+c')(c+a')= \text{(Consensus x 3)}$$
$$(a + b')(b+c')(a+c')(c+a')(a+b')(c + b')(b+c')(c+a')(b + a')= \text{(Idempotence x 3)}$$
$$(a + b')(b+c')(a+c')(c+a')(c + b')(b + a')= \text{(Commutation x 6)}$$
$$(b' + a)(c'+b)(c'+a)(a'+c)(b'+c)(a'+b)= \text{(Commutation)}$$
$$(a' + b)(b'+c)(a'+c)(c'+a)(c'+b)(b'+a)= \text{ (Idempotence x 3)}$$
$$(a' + b)(b'+c)(a'+c)(c'+a)(a' + b)(c'+b)(b'+c)(c'+a)(b'+a)= \text{ (Consensus x 3)}$$
$$(a' + b)(b'+c)(c'+a)(a' + b)(b'+c)(c'+a)= \text{ (Idempotence)}$$
$$(a' + b)(b'+c)(c'+a)$$