Before starting this topic, you must know about De Morgan’s Law. To find a complement of a Boolean function or expression, simply apply De Morgan’s Law.
Outline:
How to find the complement of a Boolean expression or function and prove results with the help of a truth table
How to find the dual of a Boolean expression of function and prove results with the help of a truth table
How to find out the logically equivalent Boolean expressions. Prove your results with the help of truth tables and logic circuits.
How to find out the dual && Complement of a Boolean Function | Expression:
For Dual:
Itinvolves the following steps
The dual has exactly the same results in the truth table but the logic is opposite.
For complement:
It involves three steps. First two steps are similar to the above.
Interchange of AND and OR operators
Interchange of 1’s and 0’s
Complement each literal
Starting with the simplest possible examples. Rest of the explanation will be easy to understand.
Find the dual and complement of the A . B
Let’s suppose, A . B is in the positive logic. It means logic high is 1 and logic low is 0
Dual is A + B. The original cricut/ expression (A.B) is in the positive logic then its dual should be in the negative logic
Complement is A’ + B’
Positive Logic AND Gate
A
B
A .B
Logic
1
0
0
0
L
2
0
1
0
L
3
1
0
0
L
4
1
1
1
H
For dual Interchange every 1 with 0 and 0 with 1 (change of logic, “0” means logic high and “1” means logic low)
Dual of A . B is the negative Logic OR Gate
A
B
A + B
Logic
1
0
0
0
H
2
0
1
0
H
3
1
0
0
H
4
1
1
1
L
For complement, each literal should be complemented
Complement of A . B is the negative input OR Gate
A
B
A’
B’
A’ + B’
Logic
1
0
0
1
1
1
H
2
0
1
1
0
1
H
3
1
0
0
1
1
H
4
1
1
0
0
0
L
Example # 1: Find out the dual and complement of the given boolean expression (AB’ +C)D’ + E
Dual:
= AB’D’ + CD’ + E
= (A + B’ + D’)(C + D’)E
Complement:
= (A’ + B + D)(C’ + D)E’
Practice Problems:
Example # 2: Find out the dual and complement of the given boolean expression XY’ + X’Y
Dual:
Interchange AND and OR operators.
= (X + Y’)(X’ + Y)
Complement:
Complement each literal.
= (X’ + Y)(X + Y’)
X
Y
X’
Y’
XY’ + X’Y
(X + Y’)(X’ + Y)
(X’ + Y)(X + Y’)
0
0
1
1
0 (L)
0 (H)
1 (H)
0
1
1
0
1 (H)
1 (L)
0 (L)
1
0
0
1
1 (H)
1 (L)
0 (L)
1
1
0
0
0 (L)
0 (H)
1 (H)
Example # 3: Find out the dual and complement of the given boolean expression
Example # 4: Find out the dual and complement of the given boolean expression
(X + Y’ + Z)(X’ + Z’)(X + Y)
Dual:
= (XY’Z) + (X’Z’) + (XY)
Complement:
= (X’YZ’) + (XZ) + (X’Y’)
X
Y
Z
(X+Y’+Z) (X’+Z’) (X+Y)
(XY’Z) + (X’Z’) + (XY)
(X’YZ’) + (XZ) + (X’Y’)
0
0
0
0 (L)
0 (H)
1 (H)
0
0
1
0 (L)
0 (H)
1 (H)
0
1
0
0 (L)
0 (H)
1 (H)
0
1
1
1 (H)
1 (L)
0 (L)
1
0
0
1 (H)
1 (L)
0 (L)
1
0
1
0 (L)
0 (H)
1 (H)
1
1
0
1 (H)
1 (L)
0 (L)
1
1
1
0 (L)
0 (H)
1 (H)
Example # 5: Find out the dual and complement of the given boolean expression (ABC)’ + (D’ + E)’
Dual:
First simplify using De Morgan’s Law:
= (A’ + B’ + C’) + (DE’)
= (A’ . B’ . C’) . (D + E’)
Complement:
= (A . B . C) . (D’ + E)
A
B
C
D
E
(A+B+C)’+(D’E)
(A’B’C’)(D + E’)
(ABC)(D’ + E)
0
0
0
0
0
1 (H)
1 (L)
0 (L)
0
0
0
0
1
1 (H)
1 (L)
0 (L)
0
0
0
1
0
1 (H)
1 (L)
0 (L)
0
0
0
1
1
1 (H)
1 (L)
0 (L)
0
0
1
0
0
1 (H)
1 (L)
0 (L)
0
0
1
0
1
1 (H)
1 (L)
0 (L)
0
0
1
1
0
1 (H)
1 (L)
0 (L)
0
0
1
1
1
1 (H)
1 (L)
0 (L)
0
1
0
0
0
1 (H)
1 (L)
0 (L)
0
1
0
0
1
1 (H)
1 (L)
0 (L)
0
1
0
1
0
1 (H)
1 (L)
0 (L)
0
1
0
1
1
1 (H)
1 (L)
0 (L)
0
1
1
0
0
1 (H)
1 (L)
0 (L)
0
1
1
0
1
1 (H)
1 (L)
0 (L)
0
1
1
1
0
1 (H)
1 (L)
0 (L)
0
1
1
1
1
1 (H)
1 (L)
0 (L)
1
0
0
0
0
1 (H)
1 (L)
0 (L)
1
0
0
0
1
1 (H)
1 (L)
0 (L)
1
0
0
1
0
1 (H)
1 (L)
0 (L)
1
0
0
1
1
1 (H)
1 (L)
0 (L)
1
0
1
0
0
1 (H)
1 (L)
0 (L)
1
0
1
0
1
1 (H)
1 (L)
0 (L)
1
0
1
1
0
1 (H)
1 (L)
0 (L)
1
0
1
1
1
1 (H)
1 (L)
0 (L)
1
1
0
0
0
1 (H)
1 (L)
0 (L)
1
1
0
0
1
1 (H)
1 (L)
0 (L)
1
1
0
1
0
1 (H)
1 (L)
0 (L)
1
1
0
1
1
1 (H)
1 (L)
0 (L)
1
1
1
0
0
0 (L)
0 (H)
1 (H)
1
1
1
0
1
0 (L)
0 (H)
1 (H)
1
1
1
1
0
1 (H)
1 (L)
0 (L)
1
1
1
1
1
0 (L)
0 (H)
1 (H)
Example # 6: Find out the dual and complement of the given boolean expression
[(A + B)C]’
Dual:
= (A + B)’ + C’
= A’B’ + C’
= (A’ + B’)C’
Complement:
= (A + B)C
A
B
C
[(A + B)C]’
(A’ + B’)C’
AB + C
0
0
0
1 (H)
1 (L)
0 (L)
0
0
1
1 (H)
1 (L)
0 (L)
0
1
0
1 (H)
1 (L)
0 (L)
0
1
1
0 (L)
0 (H)
1 (H)
1
0
0
1 (H)
1 (L)
0 (L)
1
0
1
0 (L)
0 (H)
1 (H)
1
1
0
1 (H)
1 (L)
0 (L)
1
1
1
0 (L)
0 (H)
1 (H)
Example # 7: Find out the dual and complement of the given boolean expression
(A + B + C)’ + (D’E)’
Dual:
Apply De Morgan’s Law
= (A’B’C’) + (D + E’)
Interchange OR to AND or vice versa.
= (A’ + B’ + C’) . (D . E’)
Complement:
= (A + B + C). (D’ . E)
A
B
C
D
E
(A’+B’+C’)(D’E)
(A’B’C’)(D + E’)
(ABC)(D’ + E)
0
0
0
0
0
1 (H)
1 (L)
0 (L)
0
0
0
0
1
1 (H)
1 (L)
0 (L)
0
0
0
1
0
1 (H)
1 (L)
0 (L)
0
0
0
1
1
1 (H)
1 (L)
0 (L)
0
0
1
0
0
1 (H)
1 (L)
0 (L)
0
0
1
0
1
0 (L)
0 (H)
1 (H)
0
0
1
1
0
1 (H)
1 (L)
0 (L)
0
0
1
1
1
1 (H)
1 (L)
0 (L)
0
1
0
0
0
1 (H)
1 (L)
0 (L)
0
1
0
0
1
0 (L)
0 (H)
1 (H)
0
1
0
1
0
1 (H)
1 (L)
0 (L)
0
1
0
1
1
1 (H)
1 (L)
0 (L)
0
1
1
0
0
1 (H)
1 (L)
0 (L)
0
1
1
0
1
0 (L)
0 (H)
1 (H)
0
1
1
1
0
1 (H)
1 (L)
0 (L)
0
1
1
1
1
1 (H)
1 (L)
0 (L)
1
0
0
0
0
1 (H)
1 (L)
0 (L)
1
0
0
0
1
0 (L)
0 (H)
1 (H)
1
0
0
1
0
1 (H)
1 (L)
0 (L)
1
0
0
1
1
1 (H)
1 (L)
0 (L)
1
0
1
0
0
1 (H)
1 (L)
0 (L)
1
0
1
0
1
0 (L)
0 (H)
1 (H)
1
0
1
1
0
1 (H)
1 (L)
0 (L)
1
0
1
1
1
1 (H)
1 (L)
0 (L)
1
1
0
0
0
1 (H)
1 (L)
0 (L)
1
1
0
0
1
0 (L)
0 (H)
1 (H)
1
1
0
1
0
1 (H)
1 (L)
0 (L)
1
1
0
1
1
1 (H)
1 (L)
0 (L)
1
1
1
0
0
1 (H)
1 (L)
0 (L)
1
1
1
0
1
0 (L)
0 (H)
1 (H)
1
1
1
1
0
1 (H)
1 (L)
0 (L)
1
1
1
1
1
1 (H)
1 (L)
0 (L)
Example # 8: Find out the dual and complement of the given boolean expression
X’YZ’ + X’Y’Z
Dual:
= (X’ + Y + Z’) . (X’ + Y’ + Z)
Complement:
= (X + Y’ + Z) . (X + Y + Z’)
X
Y
Z
X’
Y’
Z’
X’YZ’ + X’Y’Z
(X’+Y+Z’).(X’+Y’+Z)
(X+Y’+Z).(X+Y+Z’)
0
0
0
1
1
1
0 (L)
0 (H)
1 (H)
0
0
1
1
1
0
1 (H)
1 (L)
0 (L)
0
1
0
1
0
1
1 (H)
1 (L)
0 (L)
0
1
1
1
0
0
0 (L)
0 (H)
1 (H)
1
0
0
0
1
1
0 (L)
0 (H)
1 (H)
1
0
1
0
1
0
0 (L)
0 (H)
1 (H)
1
1
0
0
0
1
0 (L)
0 (H)
1 (H)
1
1
1
0
0
0
0 (L)
0 (H)
1 (H)
End Note:
Though it was a straightforward topic, I prepared it with many effects. It takes a lot of time, and it is really difficult to get the time from my busy schedule. I rechecked the topic several times, but still, there are chances of mistakes. If you find any mistakes please notify me. I hope this topic will be helpful. Thank you!
Anonymous
Syeda Amna Ahmed
Online electrical store