Boolean Algebra


#1

Please help me with this question sir,
C2A. Show that the complement of (a’ +b’)(a+c’)(b’+c’) is given by ab + a’c.
Write down the Boolean laws in each step of you simplification.
Show your working clearly.

b)Find out how many distinct 4-letter words can be formed from the word MESSAGE.


#2

[(a' + b')(a + c')(b' + c')]' {taking complement of the given expression}
(a' + b')' + (a + c')' + (b' + c')' {from De Morgan’s law}
(a')'(b')' + (a)'(c')' + (b')'(c')' {from De Morgans law}
ab + a'c + bc {complement of x’ = x}
ab + a'c {from Redundancy Theorem}


To know about Redundancy Theorem follow the link: https://youtu.be/3pbH9IhxwOg


If you don’t know about Redundancy Theorem then you may use the solution below
ab + a'c + bc.1 {since x.1 = x}
ab + a'c + bc(a' + a) {since x’ + x = 1}
ab + a'c + a'bc + abc {from Distributive Law}
ab(1 + c) + a'c(1 + b) {taking ab common from the first and last terms and taking a’c common from the two middle terms}
ab.1 + a'c.1 {since 1 + x = 1}
ab + a'c {since 1.x = x}

The b part of the problem is not based on Boolean Algebra so please ask it in math section.


#3

ab + a’c + bc.1 {since x.1 = x}

Hi may i know where do you get the 1 from?


#4

To understand this you must know about the AND operation. When you perform AND operation between any variable and 1 you will get the same variable, as the final result will only depend on the variable. For example, AND operation of A and 1 will be equal to A, when A is 0 we will have 0.1 which is 0 and when A is 1 we will have 1.1 which is 1. If you still have doubt in this you may watch Introduction to Boolean Algebra (Part 1)


#5

i understand the redundancy theorem and also just to clarify
ab + a’c + bc(a’ + a) {since x’ + x = 1}
must the 1 translate to be the variable a or it can be like (b’+b) or (c’+c) etc


#6

it can be anything, but we select it carefully to reduce the number of literals in our expression