Presentation is loading. Please wait. # 1 Section 10.1 Boolean Functions. 2 Computers & Boolean Algebra Circuits in computers have inputs whose values are either 0 or 1 Mathematician George.

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1 Section 10.1 Boolean Functions

2 Computers & Boolean Algebra Circuits in computers have inputs whose values are either 0 or 1 Mathematician George Boole set forth basic rules of logic, which subsequently were adapted to define basic circuits; these rules form basis of Boolean algebra Operation of a circuit is defined by a Boolean function that specifies the output for each set of inputs

3 Boolean algebra Boolean algebra provides operations and rules for working with set {0,1} Most common operations are: –complement (NOT) –Boolean sum (OR) –Boolean product (AND) Rules of precedence: 1) complement, 2) product, 3) sum

4 Notation for Boolean algebra Complement is denoted by bar:0= 1 and1= 0 Boolean sum is denoted by +: 1 + 1 = 1 1 + 0 = 10 + 1 = 10 + 0 = 0 Boolean product is denoted by. Symbol may be omitted 1. 1 = 1 1. 0 = 00. 1 = 00. 0 = 0

5 Example 1 Find the value of 1. 0 + (0 + 1) = 0 + 1 = 0 + 0 = 0

6 Boolean algebra & logical operations Boolean algebraic operations correspond to logical operations: –complement =  –sum =  –product =  –0 = F, 1 = T Results of Boolean algebra can be directly translated into results about propositions, and vice-versa

7 Boolean functions Let B = {0, 1} –a variable x is a Boolean variable if it assumes values only from B –a function from B n = {(x 1, x 2, … x n ) | x i  B, 1<=i<= n} to B is a Boolean function of degree n Values of a Boolean function are often displayed in tables resembling truth tables

8 Boolean expressions Boolean functions can be represented using expressions made up from the variables and Boolean operations Boolean expressions in the variables x 1, x 2, … x n are defined recursively as: –0, 1, x 1, x 2, ….,x n are Boolean expressions; –if E 1 and E 2 are Boolean expressions, then their complements, (E 1 E 2 ) and (E 1 +E 2 ) are Boolean expressions Each Boolean expression represents a Boolean function; values of the function are obtained by substituting 0 and 1 for variables in the expression

9 Example 2 Find the values of the Boolean function represented by F(x,y,z) = xy + z x y z xy z F(x,y,z) 1 1 1 1 0 1 1 1 0 1 1 1 1 0 1 0 0 0 1 0 0 0 1 1 0 1 1 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 0 0 0 1 1

10 Properties & Operations on Boolean Functions Boolean functions F and G of n variables are equal if and only if F(b 1, b 2, … b n ) = G(b 1, b 2, … b n ) whenever b 1, b 2, … b n  B Boolean expressions that represent the same function are equivalent- e.g. xy, xy+0, xy. 1 Complement of a Boolean function F is the function F where F(x 1,…,x n ) = F(x 1,…,x n )

11 Boolean sums & products of functions Let F and G be Boolean functions of degree n –Boolean sum F+G is defined by (F+G)(x 1,…,x n ) = F(x 1,…,x n ) + G(x 1,…,x n ) –Boolean product FG is defined by (FG)(x 1,…,x n ) = F(x 1,…,x n )G(x 1,…,x n )

12 Boolean functions of degree n A Boolean function of degree 2 is a function from a set of 4 elements (pairs of elements from B={0,1}) to B, a set with 2 elements There are 16 different Boolean functions of degree 2: x y f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13 f14 f15 f16 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

13 Identities of Boolean algebra Identities of Boolean algebra are analogous to logical equivalences These identities are useful in simplifying circuit design Each identity can be proven using a table Identities can be used to prove further identities

14 Boolean identities Law of the double complement: x = x Idempotent laws: x + x = x and x. x = x Identity laws: x + 0 = x and x. 1 = x Dominance laws: x + 1 = 1 and x. 0 = 0 Commutative laws: x + y = y + x and xy = yx Associative laws: x + (y + z) = (x + y) + z and x(yz) = (xy)z Distributive laws: x + yz = (x + y)(x + z) and x(y + z) = xy + xz DeMorgan’s laws: (xy) = x + y and (x + y) = x. y

15 Example 3: Proof of DeMorgan’s first law x y x y xy xy x + y 1 1 0 0 1 0 0 1 0 0 1 0 1 1 0 1 1 0 0 1 1 0 0 1 1 0 1 1

16 Example 4 Prove the absorption law: x(x + y) = x using the identities of Boolean algebra x(x + y) = (x + 0)(x + y)identity law for Boolean sum = x + 0. ydistributive law of Boolean sum over Boolean product = x + y. 0commutative law for Boolean product = x + 0dominance law for Boolean product = x identity law for Boolean sum

17 Duality Note that most identities come in pairs The relationship between the 2 identities in a pair can be explained using the concept of a dual: the dual of a Boolean expression is obtained by interchanging Boolean sums and Boolean products, and 1s and 0s

18 Example 5 Find the dual of x(y + 0) Substitute. for + and + for. : x + (y. 0) Substitute 0 for 1: x + (y. 1)

19 Duality The dual of a Boolean function F represented by a Boolean expression is the function represented by the dual of the expression This dual function F d, does not depend of the particular Boolean expression used to represent F; an identity between functions represented by Boolean expressions remains valid when the duals of both sides of the identity are taken This duality principle is useful for obtaining new identities

20 Example 6 Construct an identity from the absorption law: x(x + y) = x Taking duals of both sides: –x + (x. y) –x–x Result is x + xy = x, also called the absorption law

21 Abstract definition of a Boolean algebra Most common way to define a Boolean algebra is to specify properties that operations must satisfy Next slide illustrates such a definition

22 Definition of a Boolean algebra A Boolean algebra is a set B with two binary operations  and , elements 0 and 1 and a unary operation _ such that the following properties hold for all x, y and z in B: Identity laws: x  0 = x and x  1 = x Domination laws: x  x = 1 and x  x = 0 Associative laws:(x  y)  z = x  (y  z) and (x  y)  z = x  (y  z) Commutative laws: x  y = y  x and x  y = y  x Distributive laws:x  (y  z) = (x  y)  (x  z) and x  (y  z) = (x  y)  (x  z)

23 Section 10.1 Boolean Functions

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