Boolean Algebra: Simplefication, $(AC + BD)(AC)'(BD)' = (AC)(AC)' + (BD)(BD)'$

Question

I want to know, how to prove this,

$(AC + BD)(AC)'(BD)' = (AC)(AC)' + (BD)(BD)'.$

Please, help me.... Thanks. What laws in that line?

Answer 1

So what you are trying(as I understand from your comment) is to prove that $(AC + BD)(AC)'(BD)' = 0$. We have: $$(AC + BD)(AC)'(BD)' = (AC)(AC)'(BD)'+ (BD)(AC)'(BD)' = (AC)(AC)'(BD)' + (BD)(BD)'(AC)'$$

We know that $xx' = 0$. So $(AC)(AC)' = 0$ and $(BD)(BD)' = 0$. Therefore: $$(AC + BD)(AC)'(BD)' = (AC)(AC)'(BD)' + (BD)(BD)'(AC)' = 0 + 0 = 0$$

Another proof(by De Morgan's law): $$(AC + BD)(AC)'(BD)' = (AC + BD)(AC + BD)' = 0$$