1. Laws & Rules of Boolean Algebra

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

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

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

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

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

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

8. 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

9. 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

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

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

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

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

14. 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.

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

16. 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

17. 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

18. 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

19. END OF BOOLEAN
RULES & LAWS

20. DeMorgan’s Theorem
ENT116

21. Theorems Equation: Equation:
The complement of two or more ANDed variables is equivalent to the OR of the complements of the individual variables
Equation:
The complement of two or more ORed variables is equivalent to the AND of the complements of the individual variables
Equation:

22. Gate equivalencies and the corresponding truth tables that illustrate DeMorgan’s theorems

23. Example:

24. Example:

25. Simplification using Boolean Algebra
Example:
Draw your possible logic gate?
Simplify this expression using Boolean algebra?
Try This:

26. Exercise: Analyze the circuit below
2. Simplify the Boolean expression found in 1

27. Exercise:
1. X=???
2. Simplify the Boolean expression found in 1

28. Karnaugh Map Standard Forms of Boolean Expressions
Sum of Product (SOP)
Product of Sum (POS)

29. The Sum-of-Products (SOP) Form
When two or more product terms are summed by Boolean addition

30. Conversion of a General Expression to SOP Form
Any logic expression can be change into SOP form by applying Boolean Algebra techniques
Example:
Try This:

31. The Standard SOP Form
Multiply:

32. The Products-of-Sum (POS) Form
When two or more sum terms are multiplied.

33. The Standard POS Form
Rule 12!
Add:

34. Boolean Expression and Truth Table

35. 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. 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. Converting from Truth Table to SOP and POS
Inputs
Output A B C X 1
POS:

38. The Karnaugh Map

39. The Karnaugh Map
Provides a systematic method for simplifying Boolean expressions
Produces the simplest SOP or POS expression
Similar to a truth table because it presents all of the possible values of input variables

40. The 3-Variable K-Map

41. The 4-Variable K-Map

42. K-Map SOP Minimization
A 1 is placed on the K-Map for each product term in the expression.
Each 1 is placed in a cell corresponding to the value of a product term

43. Example: Map the following standard SOP expression on a K-Map:
Solution:

44. Example: Map the following standard SOP expression on a K-Map:
Solution:

45. Exercise:
Map the following standard SOP expression on a K-Map:

46. Answer:

47. K-Map Simplification of SOP Expressions
A group must contain either 1, 2, 4, 8 or 16 cells.
Each cell in group must be adjacent to one or more cells in that same group but all cells in the group do not have to be adjacent to each other
Always include the largest possible number 1s in a group in accordance with rule 1
Each 1 on the map must be included in at least one group. The 1s already in a group can be included in another group as long as the overlapping groups include noncommon 1s
To maximize the size of the groups and to minimize the number of groups
Goal:

48. Example: Group the 1s in each K-Maps

49. Determining the minimum SOP Expression from the Map
Groups the cells that have 1s.
Each group of cells containing 1s create one product term composed of all variables that occur in only one form (either uncomplemented or complemented) within the group.
Variable that occurs both uncomplemented and complemented within the group are eliminated. These are called contradictory variables.

50. Example: Determine the product term for the K-Map below and write the resulting minimum SOP expression 1

52. Example: Use a K-Map to minimize the following standard SOP expression

53. Example: Use a K-Map to minimize the following standard SOP expression

54. Mapping Directly from a Truth Table

55. Don’t Care (X) Conditions
A situation arises in which input variable combinations are not allowed
Don’t care terms either a 1 or a 0 may be assigned to the output

56. Don’t Care (X) Conditions
Example of the use of “don’t care” conditions to simplify an expression

57. Exercise: Use K-Map to find the minimum SOP from1 2