1. DE MORGAN’S THEOREM

2. De Morgan’s Theorem De Morgan’s Theorem

3. Theorems of Boolean Algebra(1) 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

4. Theorems of Boolean Algebra(2) 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 AB = BA 14) Associative : A+(B+C) =(A+B) + C A(BC) = (AB)C A(BC) = (AB)C 15) Distributive : A(B+C) = AB +AC (A+B)(C+D)=AC + AD + BC + BD (A+B)(C+D)=AC + AD + BC + BD

5. De Morgan’s Theorems 16) (X+Y) = X. Y 17) (X.Y) = X + Y  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 :

6. Implications of De Morgan’s Theorem (a) Equivalent circuit implied by theorem (16) (b) Negative- AND (c) Truth table that illustrates DeMorgan’s Theorem (a) (b) Input Output X Y X+Y XY 0 0 1 1 0 1 0 0 10 0 0 1 1 0 0 (c)

7. Implications of De Morgan’s Theorem (a) Equivalent circuit implied by theorem (17) (b) Negative-OR (c) Truth table that illustrates DeMorgan’s Theorem (a) (b) Input Output X Y XY X+Y 0 0 1 1 0 1 1 1 10 1 1 1 1 0 0 (c)

8. 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

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

10. Example: Analyze the circuit below Y 1. Y=??? 2. Simplify the Boolean expression found in 1

11.  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.

12. Exercises:  Simplify the following Boolean expressions 1. (AB(C + BD) + AB)C 2. ABC + ABC + ABC + ABC + ABC  Write the Boolean expression of the following circuit.

13. 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

14. Standard Forms of Boolean Expressions Converting SOP to Truth Table  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.

15. Standard Forms of Boolean Expressions Converting POS to Truth Table  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.

16. 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

17. 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

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