Question

If $A,B$ and $C$ are the elements of Boolean algebra, simplify the expression $(A^{\prime }+B^{\prime })(A+C^{\prime })+B^{\prime }(B+C)$.
Draw the simplified circuit.

Answer 1

Simplification:

$(A^{\prime }+B^{\prime })(A+C^{\prime })+B^{\prime }(B+C)$

$=A^{\prime }(A+C^{\prime })+B^{\prime }(A+C^{\prime })+B^{\prime }B+B^{\prime }C$ (Distributive Law)

$=A^{\prime }A+A^{\prime }{C}^{\prime }+B^{\prime }A+B^{\prime }{C}^{\prime }+B^{\prime }B+B^{\prime }C$

$=0+A^{\prime }{C}^{\prime }+B^{\prime }A+B^{\prime }{C}^{\prime }+0+B^{\prime }C$ (since,$A^{\prime }A=0,B^{\prime }B=0$)

$=A^{\prime }{C}^{\prime }+B^{\prime }A+B^{\prime }(C^{\prime }+C)$ (Distributive Law)

$=A^{\prime }{C}^{\prime }+B^{\prime }A+B^{\prime }\cdot 1$ (since, $C^{\prime }+C=1$)

$=A^{\prime }{C}^{\prime }+B^{\prime }(A+1)$ (Distributive Law)

$=A^{\prime }{C}^{\prime }+B^{\prime }\cdot 1$ (since $A+1=1$)

$=A^{\prime }{C}^{\prime }+B^{\prime }$

The circuit for $A^{\prime }{C}^{\prime }+B^{\prime }$ is as shown