DIGITAL ELECTRONICS- Boolean Algebra Boolean identities Boolean laws Distributive law Commutative law Associative law Absorption law

1. TRINITY INSTITUTE OF PROFESSIONAL STUDIES
Sector – 9, Dwarka Institutional Area, New Delhi-75
Affiliated Institution of G.G.S.IP.U, Delhi
DIGITAL ELECTRONICS
Paper Code :- BCA 106
Keywords:
Boolean Algebra, logic, Laws
By :-HARI MOHAN JAIN

2. TRINITY INSTITUTE OF PROFESSIONAL STUDIES
Sector – 9, Dwarka Institutional Area, New Delhi-75
2
Boolean
Algebra

3. TRINITY INSTITUTE OF PROFESSIONAL STUDIES
Sector – 9, Dwarka Institutional Area, New Delhi-75
3
Example #1: Boolean Algebra
Simplify the following Boolean expression and note the
Boolean theorem used at each step. Put the answer in SOP
form.
DCCBA AF 1
SOP – Sum Of Product (Sum of MINTERMS)
Y = AB + BC
POS – Product Of Sum (Product of MAXTERMS)
Y = (A+B) (B+C)

4. TRINITY INSTITUTE OF PROFESSIONAL STUDIES
Sector – 9, Dwarka Institutional Area, New Delhi-75
4
Example #1: Boolean Algebra
Simplify the following Boolean expression and note the
Boolean theorem used at each step. Put the answer in SOP
form.
DCCBA AF 1
Solution
BA
BA
0BA
D0BA
DCCBA
DCCBAA
1
1
1
1
1
1






F
F
F
F
F
F
; Theorem #3
; Theorem #4
; Theorem #1
; Theorem #5

5. TRINITY INSTITUTE OF PROFESSIONAL STUDIES
Sector – 9, Dwarka Institutional Area, New Delhi-75
5
Example #2: Boolean Algebra
Simplify the following Boolean expression and note the
Boolean theorem used at each step. Put the answer in SOP
form.
0BA1BACCBCBBF 2

6. TRINITY INSTITUTE OF PROFESSIONAL STUDIES
Sector – 9, Dwarka Institutional Area, New Delhi-75
6
Example #2: Boolean Algebra
Simplify the following Boolean expression and note the
Boolean theorem used at each step. Put the answer in SOP
form.
0BA1BACCBCBBF 2
Solution
BACBF
BACBF
0BACBF
0BABACBF
0BA1BACBF
0BA1BACBCBF
0BA1BACCBCBBF







2
2
2
2
2
2
2
; Theorem #3 (twice)
; Theorem #7
; Theorem #2
; Theorem #1
; Theorem #5

7. TRINITY INSTITUTE OF PROFESSIONAL STUDIES
Sector – 9, Dwarka Institutional Area, New Delhi-75
7
Example #3: Boolean Algebra
Simplify the following Boolean expression and note the
Boolean theorem used at each step. Put the answer in SOP
form.
T)R)(S(RTRF3


8. TRINITY INSTITUTE OF PROFESSIONAL STUDIES
Sector – 9, Dwarka Institutional Area, New Delhi-75
8Example #3: Boolean Algebra
Simplify the following Boolean expression and note the
Boolean theorem used at each step. Put the answer in SOP
form. T)R)(S(RTRF3

Solution   
 
 
 
RSTF
RS1TF
RSS1TF
RSSRRTF
TSRSTRTRF
TSRSTR0TRF
TSRSTRRRTRF
TRSRTRF
3
3
3
3
3
3
3
3







 ; Theorem #12B
; Theorem #4
; Theorem #5
; Theorem #12A
; Theorem #8
; Theorem #6
; Theorem #2

9. TRINITY INSTITUTE OF PROFESSIONAL STUDIES
Sector – 9, Dwarka Institutional Area, New Delhi-75
9
Example #4: Boolean Algebra
Simplify the following Boolean expression and note the
Boolean theorem used at each step. Put the answer in SOP
form.
SQPSQPSPF4


10. TRINITY INSTITUTE OF PROFESSIONAL STUDIES
Sector – 9, Dwarka Institutional Area, New Delhi-75
10
Example #4: Boolean Algebra
Simplify the following Boolean expression and note the
Boolean theorem used at each step. Put the answer in SOP
form. SQPSQPSPF4

 
 
 
 
QPPSF
QP1PSF
QPQ1PSF
SQPQPSPF
SQPQSPF
SQPSQSPF
SQPSQPSPF
4
4
4
4
4
4
4






 ; Theorem #12A
; Theorem #13C
; Theorem #12A
; Theorem #12A
; Theorem #6
; Theorem #2
Solution

11. TRINITY INSTITUTE OF PROFESSIONAL STUDIES
Sector – 9, Dwarka Institutional Area, New Delhi-75
11
• Boolean identities
AND Function OR Function NOT function
00=0 0+0=0
01=0 0+1=1
10=0 1+0=1
11=1 1+1=1
A0=0 A+0=A
0A=0 0+A=A
A1=A A+1=1
1A=A 1+A=1
AA=A A+A=A
0 AA 1 AA
10 
01 
AA 

12. TRINITY INSTITUTE OF PROFESSIONAL STUDIES
Sector – 9, Dwarka Institutional Area, New Delhi-75
12
Commutative law Absorption law
Distributive law De Morgan’s law
Associative law Note also
• Boolean laws
ABBA
BAAB


))((
)(
CABABCA
BCABCBA


CBACBA
CABBCA


)()(
)()(
ABAA
AABA


)(
BABA
BABA


ABBAA
BABAA


)(