   Chapter 6.3, Problem 41ES ### 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 41-13 simple the given expression. Cite a property from Theorem 6.2.2 for every step.   A ∩ ( ( B ∪ A e ) ∩ B e )

To determine

To prove:

A((BAc)Bc)

Explanation

Given information

(AB)C=(AC)(BC)

Concept used:

(AB)C=(AC)(BC)

Cite a property from Theorem 6.2.2 for every step of the proof.

Solution let A,B and C be any sets. Then

(AB)C=(AB)Cc               By the set difference law=Cc(AB)               By the commutative law for =(CcA)(CcB)    By the distributive law=(ACc)(BCc)    By the commutative law for =(AC)(BC)         By the set difference law

Calculation:

Consider the expression for the sets A and B written below.

A((BAc)Bc)

Objective is to simply the above expression. For this consider,

A((BAc)Bc)

By the distributive law

A((BAc)Bc)=A{(BBc)(AcBc)}

Now use the complementary law

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