1. De Morgan’s Theorem,

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

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

10. 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.

11. Exercises: Simplify the following Boolean expressions
(AB(C + BD) + AB)C
ABC + ABC + ABC + ABC + ABC
Write the Boolean expression of the following circuit.

12. 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

13. 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.

14. 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.

15. 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

16. 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

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