Question 13.3: Simplification of Logical Expression Using the rules of Tabl...

Simplification of Logical Expression

Using the rules of Table 13.11,

Table 13.11 Rules of Boolean algebra

1. 1. 0+X=X
   2. 1+X=1
   3. X+X=X
   4. X+\overline{X}=1
   5. 0 \cdot X=0
   6. 1 \cdot X=X
   7. X \cdot X=X
   8. X \cdot \overline{X}=0
   9. \overline{\overline{X}}=X
   10. X+Y=Y+X                Commutative law
   11. X \cdot Y=Y \cdot X             Commutative law
   12. X+(Y+Z)=(X+Y)+Z            Associative law
   13. X \cdot(Y \cdot Z)=(X \cdot Y) \cdot Z          Associative law
   14. X \cdot(Y+Z)=X \cdot Y+X \cdot Z          Distributive law
   15. X+X \cdot Z=X           Absorption law
   16. X \cdot(X+Y)=X
   17. (X+Y) \cdot(X+Z)=X+Y \cdot Z
   18. X+\overline{X} \cdot Y=X+Y
   19. X \cdot Y+Y \cdot Z+\overline{X} \cdot Z=X \cdot Y+\overline{X} \cdot Z

simplify the following function using the rules of Boolean algebra.

f(A, B, C, D)=\bar{A} \cdot \bar{B} \cdot D+\bar{A} \cdot B \cdot D+B \cdot C \cdot D+A \cdot C \cdot D