## Duality Principle:

Dual of an expression can be obtained by :

- Changing all + with .
- Complement 0’s and 1’s i.e. 1’=0 and 0’=1

- Keep the variable sam (i.e. do not complement the variable)

## Example:

- Find the dual of : A(B+C) = AB + AC

Solution:

- Find the Dual of F(xyz)=(x+y) (x+z)(y+z)

Solution:

(x+y) (x+z)(y+z)

= (xx+xz+xy +yz)(y+z)

= (x+xz+xy+yz)(y+z)

= (x +xy +yz )(y+z)

= (x+yz)(y+z)

= xy +xz + yyz + yz

= xy + xz + yz

Also dual of (x+y) (x+z)(y+z) = xy + xz + yz

Therefore, we find that in this case dual of a function is the function itself

3. For n variables expression haw many total dual functions are possible

Solution:

- Which of the following is TRUE

S1: The dual of NAND function is NOR

S2:The dual of X-OR function is X-NOR

(a). S1 and S2 are true

(b). S1 is true

(c ). S2 is true

(d). None of these

Solution:

NAND function = (xy)’

Dual of NAND = (x + y)’ ; i.e. Dual of NAND is NOR

X-OR function = xy’+x’y =

Dual of X-OR = (x+y’)(x’+y)

= xx’ +xy + x’y’ + yy’

= xy + x’y’ ; Dual of NOR is XNOR

So, Both S1 and S2 are correct

## DeMorgan’s Theorem

### De-Morgan’s 1^{st} Law:

Complement of sums equals product of complements

(A+B)’ = A’ . B’

### De-Morgan’s 2^{nd} Law:

Complement of products equals the complement of sums

(A.B)’ = A’ + B’

## Examples:

1 Simplify the Boolean expression Y =A(A+B’)

Solution:

AA + AB’

A+AB’

A(1+B’)

A

- Simplify A+A’B

A(B+1) + A’B

AB +A + A’B

AB + A’B + A

B(A+A’) + A

A + B

- Simplify the Boolean Expression Y=A(A+B) + B(A’+B)

=A(A+B) + B(A’+B)

= AA + AB + BB + A’B

=A + AB + B + A’B

= A(1+B) + B(1+A’)

= A + B

- Simplify A’B’C’ + A’BC’ + ABC’ + AB’C’

Solution

= A’B’C’ + A’BC’ + ABC’ + AB’C’

= A’C’ ( B’ + B ) + AC’ ( B + B’)

= A’C’ + AC’

=C’( A’ + A)

= C’

5 Simplify the Boolean function Y = (A’BC + A’BC’ + A’B’C)’

Solution

= (A’BC + A’BC’ + A’B’C)’

= (A’BC + A’B’C + A’BC’)’

=(A’C(B + B’) + A’BC’)’

=(A’C + A’B’C’)’

= (A’(C+B’C’)’

=(A’ (C + B))’ ; NOW USE DeMorgan’s Theorem

=A +C’B’