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:

It involves 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’

### For dual Interchange every 1 with 0 and 0 with 1 (change of logic, “0” means logic high and “1” means logic low)

### For complement, each literal should be complemented

### 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’)

### Example # 3: Find out the dual and complement of the given boolean expression

## AB(C’D + CD’) + A’B'(C’ + D)(C + D’)

### Dual:

= [(A + B)+(C’ + D)(C + D’)] [(A’ + B’) + C’D + CD’]

### Complement:

= [(A’ + B’)+(C + D’)(C’ + D)] [(A + B) + CD’ + C’D]

### 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’)

### 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)

### 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

### 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)

### 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’)

## 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

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