De Morgan’s Theorem,

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

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

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

Implications of De Morgan’s Theorem Input Output X Y X+Y XY 0 0 1 1 0 1 0 0 0 0 0 1 1 0 0 (b) (c) (a) Equivalent circuit implied by theorem (16) (b) Alternative symbol for the NOR function (c) Truth table that illustrates DeMorgan’s Theorem

Implications of De Morgan’s Theorem Input Output X Y XY X+Y 0 0 1 1 0 1 1 1 0 1 1 1 1 0 0 (b) (c) (a) Equivalent circuit implied by theorem (17) (b) Alternative symbol for the NAND function (c) Truth table that illustrates DeMorgan’s Theorem

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

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

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

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.

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

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

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.

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.

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

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

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