If A, B and C are the elements of Boolean algebra, simplify the expression (A' + B')(A + C') + B'(B…

Simplified expression: $A'C' + B'$ $(A'+B')(A+C') + B'(B+C)$ $= A'A + A'C' + B'A + B'C' + B'B + B'C$ (distributive law) $= 0 + A'C' + AB' + B'C' + 0 + B'C$ (complement law: $A'A = 0$, $B'B = 0$) $= A'C' + AB' + B'(C' + C)$ $= A'C' + AB' + B'$ (complement law: $C + C' = 1$) $= A'C' + B'(A + 1) = A'C' + B'$ (absorption law) Hence the simplified expression is $A'C' + B'$. Circuit: NOT gates on $A$ and $C$, an AND gate for $A'C'$, a NOT gate on $B$, and an OR gate joining $A'C'$ and $B'$.