Question

Reduce using Axioms and theorems of Boolean algebra. Show step-by-step:

(A'B'C)+(A'BC')+(A'BC)+(AB'C)+(ABC)

Answer 1

So, here is the step-by-step solution:

A’B’C + A’BC’ + A’BC + AB’C + ABC

= A’B’C + A’B ( C + C’ ) + AC ( B + B’ )

= A’B’C + A’B ( 1 ) + AC ( 1 ) ( Complementarity Rule X + X’ = 1)

= A’B’C + A’B + AC ( Identity Rule X . 1 = X )

= A’ ( B’C + B ) + AC

= A’ ( B + B’C ) + AC ( Saying this a step-5 )

= A’ ( B (1) + B’C ) + AC ( Identity Rule used as X = X . 1)

= A’ ( B ( C + C’ ) + B’C ) + AC ( Complementary Rule used as 1 = C + C’ )

= A’ ( BC + BC’ + B’C ) + AC

= A’ ( BC + BC + BC’ + B’C ) + AC ( Idempocency Rule used as X = X + X )

= A’ ( BC + BC’ + BC + B’C ) + AC ( Re – arrangement of terms )

= A’ ( B ( C+ C’ ) + C ( B+ B’ ) ) + AC

= A’ ( B ( 1 ) + C ( 1 ) ) + AC ( Complementarity Rule X + X’ = 1)

= A’ ( B + C ) + AC ( Or alternatively, you can directly write after step – 5 that B + B’C = B + C

if you know this rule : X + X’Y = XY or else this is the step – wise solution )

= A’B + A’C + AC

= A’B + C ( A’ + A)

= A’B + C ( 1 ) ( Complementarity Rule X + X’ = 1)

= A’B + C ( Identity Rule X . 1 = X )

So, the answer is : A'B + C . Do comment if there is any query.