Boolean simplification of $AB'(B' + C)$

Question

Simplifying $AB'(B'+ C)$, then using the distributive property I know I would get

$AB'B' + AB'C$

I am just confused as to how to simplify $B'B'$

Answer 1

$AB'B' + AB'C = AB' + AB'C = AB'(1 + C) = AB'1 = AB'$

$B'B'$ states "$B'$ and $B'$" which by idempotence is equivalent to $B'$. And note that when $AB'$ is satisfied, then $AB'$ OR $ AB'C $ is satisfied, regardless of the truth value of $C$.

Answer 2

Use idempotence: $B'B'=B'.$ Then factor out the $AB'$ again and simplify.

Alternately, consider the induced partial order given by $a\le b$ iff $ab=a$ iff $a+b=b$. Under this partial order, $ab$ is the greatest lower bound of $a,b$ and $a+b$ is the least upper bound of $a,b.$ Then $AB'\le B'\le B'+C$, and so $AB'(B'+C)=AB'$.