   Chapter 6.4, Problem 2ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# In 1-3 assume that B is a Boolean algebra with operation s = and ‘. Give the reasons needed to fill in the blanks in the proofs using only the axioms for a Boolean algebra. Universal bound law for +: For every a in B. a + 1 = 1 Proof: Let a be any element of B. Then a + 1 = a + ( a + a ¯ )               ( a ) = ( a + a ) + a ¯                   ( b ) = a + a ¯                                           by   Example 6 .4 .2 =1                                                               (c)

To determine

Let a be any element of B. Then

a+1=a+(a+a¯)   (a)_=(a+a)+a¯    (b)_=a+a¯             by Example 6.4.2=1                   (c)_

Explanation

Given information:

Let a be any element of B.

Concept used:

a¯+a=a+a¯     by the commutative law=1            by the complement law for 1

And

a¯a=aa¯     by the commutative law=0            by the complement law for 0

Calculation:

For all elements a in the Boolean algebra B,a+a=a.

Consider B as a Boolean algebra, with the operations, addition + and multiplication.

Objective is to provide the reason to fill in the blanks in the proof.

Suppose a is any element of +. Then a¯ is the complement of a

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