Chapter 5 - SCSI1013 Discrete Structures CHAPTER 5 Boolean...

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15/11/2015 1 CHAPTER 5 BOOLEAN ALGEBRA AND COMBINATORIAL CIRCUITS SCSI1013: Discrete Structures 2014/2015 –Sem. 1 Boolean Algebra Set B = {1, 0} Operation +, , Example 1 + 1 = 1 1 + 0 = 1 0 + 1 = 1 0 + 0 = 0 1 1 = 1 1 0 = 0 0 1 = 0 0 0 = 0 1 = 0 0 = 1 Boolean Variables Any literal symbol such as x , y , z , x 1 , x 2 , ….. , x n used to represent an element of B = {0, 1} is called a Boolean variable

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15/11/2015 2 Boolean Expressions Let x 1 , x 2 , ….. , x n be Boolean expression 0 and 1 are Boolean expressions x 1 , x 2 , ….. , x n are Boolean expressions If α is a Boolean expression, then α′ is a Boolean expression If α 1 and α 2 are Boolean expression, then α 1 + α 2 and α 1 α 2 are also Boolean expression Boolean Expressions In a Boolean expression in which parentheses are not used to specify the order of operations, we assume that the product , “ • ” is evaluated before the sum, “ + ” Example x 1 x 2 + x 3 = ( x 1 x 2 ) + x 3 x 1 + x 2 x 3 = x 1 + ( x 2 x 3 ) Example Construct the truth table for
15/11/2015 3 Example Boolean Expressions Two Boolean expressions α ( x 1 , x 2 , ….. , x n ) and β ( x 1 , x 2 , ….. , x n ) are said to be equal if they assume the same value for every assignment of values 0, 1 to the variables x 1 , x 2 , ….. , x n If two Boolean expressions α and β are equal, then we write α = β Example α ( x 1 , x 2 ) = ( x 1 + x 2 ) β ( x 1 , x 2 ) = x 1 x 2 α = β ? Construct the truth table for α and β Example

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15/11/2015 4 Exercise Construct the truth tables for Boolean expressions 1. x (y + x ) 2. x y + y (x + z) Exercise - Solution x • (y + x′) x y x y + x x•(y + x ′) 1 1 0 1 1 1 0 0 0 0 0 1 1 1 0 0 0 1 1 0 Exercise - Solution x•y′ + y•(x′ + z) x y z x y x•y x ′+z y•(x′ + z) x•y +y•( x′+z ) 1 1 1 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 1 0 1 0 1 1 1 0 1 1 0 0 0 1 1 0 0 1 0 1 1 1 0 0 1 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 0 1 0 0 0 0 0 1 1 0 1 0 0 Properties of Boolean Variables Commutative Laws x 1 + x 2 = x 2 + x 1 x 1 x 2 = x 2 x1 Associative Laws (x 1 + x 2 ) + x 3 = x 1 + ( x 2 + x 3 ) (x 1 x 2 ) • x 3 = x 1 • ( x 2 x 3 ) Distributive Laws x 1 • ( x 2 + x 3 ) = ( x 1 x 2 ) + ( x 1 x 3 )
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