1. RULES OF BOOLEAN ALGEBRA

2. BASIC RULES OF BOOLEAN ALGERBA Sr. No.Theorem 1.0’=1  1’=0 2.A+0=A 3.A+1=1 4.A+A=A 5.A+A’=1 6.(A’)’=A 7.A+AB=A 8.A+A’B=A+B 9.(A+B)’=(A)’.(B)’ 10.(A.B)’=(A)’+(B)’ 11.(A+B).(A+C)= A+BC 12.A.1=A  A.0=0 13.A.A=A  A.(A)’=0

3. D E M ORGAN ’ S T HEOREMS It states, that the complement of any expression can be obtained by replacing each variable and element with its complement and changing OR operators (+) with AND operators(.) and AND operator (.) with OR operators (+). These theorems can be expressed as follows: (A+B)’=(A)’.(B)’ First De Morgan’s Theorem. (A.B)’=(A)’+(B)’ Second De Morgan’s Theorem. De Morgan’s theorems can also applicable to expressions in which there are more than two variables.

4. T HREE STEPS OF SIMPLIFICATION ( DEMORGANIZATION ) 1. Complement each of the individual variables. 2. Change all the OR’s (+) to AND (.) and AND’s(.) to OR’s(+). 3. Complement the whole Boolean function.

5. E XAMPLES 1 Demorganize the following expression: AB’ + A’B’C +AC Step No. 1: Complement each of the individual variables Means : AB’ + A’B’C + AC  A’B’’ + A’’B’’C’ + A’C’ Step No. 2: Change all the OR’s (+) to AND (.) and AND’s(.) to OR’s(+)

6. E XAMPLES 1 Means: A’B’’ + A’’B’’C’ + A’C’  A’B + ABC’ + A’C’ A’B + ABC’ + A’C’  (A’+B ).(A+B+C’).(A’+C’) Step No. 3: Complement the whole Boolean function Means: (A’+B ).(A+B+C’).(A’+C’)  ((A’+B ).(A+B+C’).(A’+C’))’

7. T RUTH T ABLE F OR T HIS E XAMPLE 1. Before Demorganization. AB’ + A’B’C +AC ABCAB’A’B’CACR=AB’ + A’B’C + AC 0000000 0010101 0100000 0110000 1001001 1011011 1100000 1110011

8. T RUTH T ABLE F OR T HIS E XAMPLE 2. After Demorganization. ((A’+B ).(A+B+C’).(A’+C’))’ ABCA’+BA+B+C’A’+C’R=(A’+B).(A+B+C’).(A’+C’)R’ 00011110 00110101 01011110 01111110 10001101 10101001 11011110 11111001

9. T RUTH T ABLE F OR T HIS E XAMPLE Now Compare the results of the expression before and after Demorganization The Results are identical in both the tables. R=AB’ + A’B’C + AC 0 1 0 0 1 1 0 1 R=(A’+B).(A+B+C’).(A’+C’))’ 0 1 0 0 1 1 0 1 Before DemorganizationAfter Demorganization

10. E XAMPLES 1 Step No.3 A’.B+A’.B’.C+A’+B’(A’.B+A’.B’.C+A’+B’)’ Step No. 2 (A’+B’’).(A’+B’+C’’).(A’B’)A’.B+A’.B’.C+A’+B’ Step No. 1 (A+B’).(A+B+C’).(AB)(A’+B’’).(A’+B’+C’’).(A’B’)

11. T RUTH T ABLE F OR T HIS E XAMPLE 1. Before Demorganization. (A+B’).(A+B+C’).(AB) ABCA+B’(A+B+C’)(AB)R=(A+B’).(A+B+C’).(AB) 0001100 0011000 0100100 0110100 1001100 1011100 1101111 1111111

12. T RUTH T ABLE F OR T HIS E XAMPLE 2. After Demorganization. (A’.B+A’.B’.C+A’+B’)’ ABCA’.BA’.B’.CA’+B’R=A’.B+A’.B’.C+A’+B’R’ 00000110 00101110 01010110 01110110 10000110 10100110 11000001 11100001

13. T RUTH T ABLE F OR T HIS E XAMPLE Now Compare the results of the expression before and after Demorganization The Results are identical in both the tables. R=(A+B’).(A+B+C’).(AB) 0 0 0 0 0 0 1 1 R=(A’.B+A’.B’.C+A’+B’)’ 0 0 0 0 0 0 1 1 Before DemorganizationAfter Demorganization