1. BOOLEAN ALGEBRA AND LOGIC SIMPLIFICATION
DKT 122 / 3 DIGITAL SYSTEMS 1
CHAPTER 4 :
BOOLEAN ALGEBRA AND LOGIC SIMPLIFICATION

2. Boolean Operations & expression Laws & rules of Boolean algebra
DeMorgan’s Theorems
Boolean analysis of logic circuits
Simplification using Boolean Algebra
Standard forms of Boolean Expressions
Boolean Expressions & truth tables
The Karnaugh Map

3. Karnaugh Map SOP minimization
Karnaugh Map POS minimization
5 Variable K-Map
Programmable Logic

4. Boolean Operations & expression
Variable – a symbol used to represent logical quantities (1 or 0)
ex : A, B,..used as variable
Complement – inverse of variable and is indicated by bar over variable
ex : Ā

5. Operation : Boolean Addition – equivalent to the OR operation
X = A + B
Boolean Multiplication – equivalent to the AND operation
X = A∙B A X B A X B

6. Laws & rules of Boolean algebra

7. Commutative law of addition
A+B = B+A
the order of ORing does not matter.

8. Commutative law of Multiplication
AB = BA
the order of ANDing does not matter.

9. Associative law of addition
A + (B + C) = (A + B) + C
The grouping of ORed variables does not matter

10. Associative law of multiplication
A(BC) = (AB)C
The grouping of ANDed variables does not matter

11. (A+B)(C+D) = AC + AD + BC + BD
Distributive Law
A(B + C) = AB + AC
(A+B)(C+D) = AC + AD + BC + BD

12. Boolean Rules 1) A + 0 = A
In math if you add 0 you have changed nothing
In Boolean Algebra ORing with 0 changes nothing

13. Boolean Rules 2) A + 1 = 1
ORing with 1 must give a 1 since if any input
is 1 an OR gate will give a 1

14. Boolean Rules 3) A • 0 = 0
In math if 0 is multiplied with anything you
get 0. If you AND anything with 0 you get 0

15. Boolean Rules 4) A • 1 = A
ANDing anything with 1 will yield the anything

16. Boolean Rules 5) A + A = A
ORing with itself will give the same result

17. Boolean Rules 6) A + A = 1
Either A or A must be 1 so A + A =1

18. Boolean Rules 7) A • A = A
ANDing with itself will give the same result

19. Boolean Rules 8) A • A = 0
In digital Logic 1 =0 and 0 =1, so AA=0 since one of the inputs must be 0.

20. Boolean Rules 9) A = A
If you not something twice you are back to the beginning

21. Boolean Rules 10) A + AB = A Proof: A + AB = A(1 +B) DISTRIBUTIVE LAW
= A∙ RULE 2: (1+B)=1
= A RULE 4: A∙1 = A

22. Boolean Rules 11) A + AB = A + B
If A is 1 the output is 1 , If A is 0 the output is B
Proof:
A + AB = (A + AB) + AB RULE 10
= (AA +AB) + AB RULE 7
= AA + AB + AA +AB RULE 8
= (A + A)(A + B) FACTORING
= 1∙(A + B) RULE 6
= A + B RULE 4

23. Boolean Rules 12) (A + B)(A + C) = A + BC
PROOF
(A + B)(A +C) = AA + AC +AB +BC DISTRIBUTIVE LAW
= A + AC + AB + BC RULE 7
= A(1 + C) +AB + BC FACTORING
= A.1 + AB + BC RULE 2
= A(1 + B) + BC FACTORING
= A.1 + BC RULE 2
= A + BC RULE 4

24. De Morgan’s Theorem,

25. Theorems of Boolean Algebra
1) A + 0 = A
2) A + 1 = 1
3) A • 0 = 0
4) A • 1 = A
5) A + A = A
6) A + A = 1
7) A • A = A
8) A • A = 0

26. Theorems of Boolean Algebra
9) A = A
10) A + AB = A
11) A + AB = A + B
12) (A + B)(A + C) = A + BC
13) Commutative : A + B = B + A
AB = BA
14) Associative : A+(B+C) =(A+B) + C
A(BC) = (AB)C
15) Distributive : A(B+C) = AB +AC
(A+B)(C+D)=AC + AD + BC + BD

27. De Morgan’s Theorems
Two most important theorems of Boolean Algebra were contributed by De Morgan.
Extremely useful in simplifying expression in which product or sum of variables is inverted.
The TWO theorems are :
16) (X+Y) = X . Y
17) (X.Y) = X + Y

28. Implications of De Morgan’s Theorem
Input Output
X Y X+Y XY
(b)
(c)
(a) Equivalent circuit implied by theorem (16)
(b) Alternative symbol for the NOR function
(c) Truth table that illustrates DeMorgan’s Theorem

29. Implications of De Morgan’s Theorem
Input Output
X Y XY X+Y
(b)
(c)
(a) Equivalent circuit implied by theorem (17)
(b) Alternative symbol for the NAND function
(c) Truth table that illustrates DeMorgan’s Theorem

30. De Morgan’s Theorem Conversion
Step 1: Change all ORs to ANDs and all ANDs to Ors
Step 2: Complement each individual variable
(short overbar)
Step 3: Complement the entire function (long overbars)
Step 4: Eliminate all groups of double overbars
Example : A . B A .B. C
= A + B = A + B + C
= A + B = A + B + C
= A + B

31. De Morgan’s Theorem Conversion
ABC + ABC (A + B +C)D
= (A+B+C).(A+B+C) = (A.B.C)+D
= (A+B+C).(A+B+C) = (A.B.C)+D

32. Examples: Analyze the circuit below
2. Simplify the Boolean expression found in 1

33. Follow the steps list below (constructing truth table)
List all the input variable combinations of 1 and 0 in binary sequentially
Place the output logic for each combination of input
Base on the result found write out the boolean expression.

34. Standard Forms of Boolean Expressions
Sum of Products (SOP)
Products of Sum (POS)
Notes:
SOP and POS expression cannot have more than
one variable combined in a term with an inversion bar
There’s no parentheses in the expression

35. Standard Forms of Boolean Expressions
Converting SOP to Truth Table
Examine each of the products to determine where
the product is equal to a 1.
Set the remaining row outputs to 0.

36. Standard Forms of Boolean Expressions
Converting POS to Truth Table
Opposite process from the SOP expressions.
Each sum term results in a 0.
Set the remaining row outputs to 1.

37. Standard Forms of Boolean Expressions
The standard SOP Expression
All variables appear in each product term.
Each of the product term in the expression is called
as minterm.
Example:
In compact form, f(A,B,C) may be written as

38. Standard Forms of Boolean Expressions
The standard POS Expression
All variables appear in each product term.
Each of the product term in the expression is called as
maxterm.
Example:
In compact form, f(A,B,C) may be written as

39. Standard Forms of Boolean Expressions
Example:
Convert the following SOP expression to an equivalent POS expression:
Example:
Develop a truth table for the expression:

40. THE K-MAP

41. Karnaugh Map (K-Map)
Karnaugh Mapping is used to minimize the number of logic gates that are required in a digital circuit.
This will replace Boolean reduction when the circuit is large.
Write the Boolean equation in a SOP form first and then place each term on a map.

42. Karnaugh Map (K-Map)
The map is made up of a table of every possible SOP using the number of variables that are being used.
If 2 variables are used then a 2X2 map is used
If 3 variables are used then a 4X2 map is used
If 4 variables are used then a 4X4 map is used
If 5 Variables are used then a 8X4 map is used

43. K-Map SOP Minimization

44. 2 Variables Karnaugh Map
B B A
Notice that the map is going false to true, left to right and top to bottom
B B
The upper right hand cell is A B if X= A B then put an X in that cell A 1
This show the expression true when A = 0 and B = 0

45. 2 Variables Karnaugh Map
B B
If X=AB + AB then put an X in both of these cells A 1 1
From Boolean reduction we know that A B + A B = B
B B
From the Karnaugh map we can circle adjacent cell and find that X = B A 1 1

46. 3 Variables Karnaugh Map
C C
Gray Code
00 A B
01 A B
11 A B
10 A B

47. 3 Variables Karnaugh Map (cont’d)
X = A B C + A B C + A B C + A B C
Gray Code
00 A B
01 A B
11 A B
10 A B
C C
One simplification could be
X = A B + A B 1 1 1 1

48. 3 Variables Karnaugh Map (cont’d)
X = A B C + A B C + A B C + A B C
Gray Code
00 A B
01 A B
11 A B
10 A B
C C
Another simplification could be
X = B C + B C
A Karnaugh Map does wrap around 1 1 1 1

49. 3 Variables Karnaugh Map (cont’d)
X = A B C + A B C + A B C + A B C
Gray Code
00 A B
01 A B
11 A B
10 A B
C C
The Best simplification would be
X = B 1 1 1 1

50. On a 3 Variables Karnaugh Map
One cell requires 3 Variables
Two adjacent cells require 2 variables
Four adjacent cells require 1 variable
Eight adjacent cells is a 1

51. 4 Variables Karnaugh Map
Gray Code
00 A B
01 A B
11 A B
10 A B
C D C D C D C D

52. Simplify : X = A B C D + A B C D + A B C D + A B C D + A B C D + A B C D
Gray Code
00 A B
01 A B
11 A B
10 A B
C D C D C D C D
Now try it with Boolean reductions 1 1 1 1 1 1
X = ABD + ABC + CD

53. On a 4 Variables Karnaugh map
One Cell requires 4 variables
Two adjacent cells require 3 variables
Four adjacent cells require 2 variables
Eight adjacent cells require 1 variable
Sixteen adjacent cells give a 1 or true

54. Simplify using Karnaugh map
First, we need to change the circuit to an SOP expression

55. Simplify using Karnaugh map (cont’d) comfirm back (the right answer)
Y= A + B + B C + ( A + B ) ( C + D)
Y = A B + B C + A B ( C + D )
Y = A B + B C + A B C + A B D
Y = A B + B C + A B C A B D
Y = A B + B C + (A + B + C ) ( A + B + D)
Y = A B + B C + A + A B + A D + B + B D + AC + C D
SOP expression

56. Simplify using Karnaugh map (cont’d)
Gray Code
00 A B
01 A B
11 A B
10 A B
C D C D C D C D
Y = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

57. K-Map POS Minimization

58. 3 Variables Karnaugh Map
Gray Code
0 0
0 1
1 1
1 0 C AB

59. 3 Variables Karnaugh Map (cont’d)

60. 4 Variables Karnaugh Map
0 0
0 1
1 1
1 0
C D
A B

61. 4 Variables Karnaugh Map (cont’d)

62. 4 Variables Karnaugh Map (cont’d)

63. Karnaugh Map - Example Mapping a Standard SOP expression Example:
Answer:
Mapping a Standard POS expression
Using K-Map, convert the following standard POS
expression into a minimum SOP expression
Y = AB + AC or standard SOP :

64. K-Map with “Don’t Care” Conditions
Example :
Input Output
3 variables with output “don’t care (X)”

65. K-Map with “Don’t Care” Conditions (cont’d)
4 variables with output “don’t care (X)”

66. K-Map with “Don’t Care” Conditions (cont’d)
Example:
Determine the minimal SOP using K-Map:
Answer:

67. Minimum SOP expression is
Solution : CD AB X
X X X X 00 01 11 10 AD BC CD
Minimum SOP expression is

68. K-map Product of Sums simplification
Example: Simplify the Boolean function F(ABCD)=(0,1,2,5,8,9,10) in
(a) S-of-p (b) P-of-s
Using the maxterms (0’s) and complimenting F Grouping as if they were minterms, then using De Morgen’s theorem to get F.
F’(ABCD)= BD’+CD+AB
F(ABCD)= (B’+D)(C’+D’)(A’+B’)
Using the minterms (1’s)
F(ABCD)= B’D’+B’C’+A’C’D