Prove Equivalence of A’D+BCD’=(D+B)(C+D)(A’+D’)

Question

I need to prove that two expressions, obtained from Karnaugh maps, are equivalent. So far I have

A’D+BCD’=(D+B)(C+D)(A’+D’)
A’D+BCD’=(DC+BC+DD+BD)(A’+D’)
A’D+BCD’=A’DC +A’BC+A’DD+A’BD+D’DC+D’BC+D’DD+D’BD
A’D+BCD’=A’DC +A’BC+A’D+A’BD+BCD’
A’D+BCD’=A’D(C+1)+A’BC+A’BD+BCD’
A’D+BCD’=A’D(B+1)+A’BC+BCD’
A’D+BCD’=A’D+A’BC+BCD’

But I don't know how to get rid of A'BC in the last line using only boolean algebra (No Karnaugh maps). What method of simplification should I use?

Answer 1

Figured it out. If you multiply the center term by (D+D') then it can be reduced.

A’D+BCD’=A’D+A’BC+BCD’
A’D+BCD’=A’D+A’BC(D+D')+BCD’
A’D+BCD’=A’D+A’BCD+A'BCD'+BCD’
A’D+BCD’=A’D(1+BC)+BCD’(A'+1)
A’D+BCD’=A’D+BCD’