Axiomatic proof and Boolean algebra?

Question

I'm trying to prove that:

$$(c'd') + (bc') + (a'b'c) + (ab'c) = (b' + c')(b + c + d')$$

using an axiomatic proof (i.e. using only the basic axioms and theorems of Boolean algebra).However, no matter what I do, I can't seem to get things to line up correctly. Any help is appreciated.

Answer 1

Note that the left-hand side can be written: $$(c'd') + (bc') + (a'b'c) + (ab'c)= (c'd') + (bc')+ (a'+a)(b'c) = c'd' + bc' + b'c$$

Note that the right-hand side can be written: $$(b' + c')(b + c + d')= b'b + c'b+ b'c + c'c + b'd' + c'd' \\= c'b+b'c + b'd' + c'd' = c'b + b'c + b'd' + c'd'$$

Now, have to prove only that $$c'd' + bc' + b'c = c'd' + bc' + b'c + b'd'$$

Can you see how we might eliminate the term $b'd'$ on the right, so both sides of the equation are equivalent?