How to prove this equality?

Question

I would like to prove the following using boolean algebra and not karnaugh maps but I'm stuck:

CD' + CDAB' + C'D'AB' = CD' + CAB' + D'AB'

Answer 1

Here is one way:

If $C \bar{D}$ then clearly both sides are equal, so suppose $\overline{C \bar{D}} = \bar{C}+D $ is true.

If $\bar{C}+D$ is true then $\bar{C} \bar{D} + CD = (\bar{C}+D) (\bar{C} \bar{D} + CD )= \bar{C} \bar{D} + CD$ and $C+ \bar{D} = (C+ \bar{D})(\bar{C}+D) = \bar{C} \bar{D} + CD$, hence $A\bar{B} (\bar{C} \bar{D} + CD) = A\bar{B} (C+ \bar{D})$