   Chapter 6.4, Problem 9ES ### 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 4—10 assume that B is a Boolean algebra with operations + and •. Prove each statement using only the axioms for a Boolean algebra and statements proved in the text or in lower-numbered exercises. 9. De Morgan’s law for +: For all a and b in B, a + b ¯ = a ¯ ⋅ b ¯ .

To determine

To Prove:

a.b¯=a¯+b¯

Explanation

Given information:

Let B be the Boolean algebra, with the operations, addition” + ”and multiplication”

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:

Let B be the Boolean algebra, with the operations, addition "+" and multiplication "".

Suppose a and b are any elements of B.

Then, a¯ is the complement of a and b¯ is the complement or b.

The uniqueness of the complement law tells that for all a and x in B, if a+x=1 and ax=0 then x=a¯.

To show that a+b¯=a¯.b¯ it is enough to show that (a+b)+(a¯b¯)=1 and, (a+b)(a¯.b¯)=0 then use the uniqueness of complement law.

Prove (a+b)(a¯.b¯)=1

(a+b)+(a¯b¯)=((a+b)+a¯)((a+b)+b¯) By the Distributive law for addition "+" over multiplication "".

=((b+a)+a¯).((a+b)+b¯) By the Commutative law for addition "+"

=(b+(a+a¯)).(a+(b+b¯)) By the Associative law for addition "+"

=(b+1).(a+1) By the Complement law for addition "+"

=1.1 By the universal bound law for addition "+"

=1 By the identity law for multiplication ""

Therefore, for all a and b in B,(a+b)+(a¯

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