   Chapter 6.3, Problem 28ES ### 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
1 views

# In 27 and 28 supply a reason fro each step in the derivation.Forall sets A, B, and C, ( A ∪ B ) − ( C − A ) = A ∪ ( B − C ) . Proof: Suppose A, B, and C, are any sets. Then   ( A ∪ B ) ∩ C = ( A ∪ B )   ∩ ( C − A )   e                                                                             by(a) = ( A ∪ B )       ∩ ( C ∩ A ) e                                                                           by(b) = ( A ∪ B )   ∩ ( ( A e ) ∪ C e )                                                             by(c)

To determine

(AB)(CA)=(AB)(CA)cby (a)=(AB)(C A c)cby (b)=(AB)( A cC)cby (c)=(AB)( ( A c )cCc)by(d)=(AB)(ACc)by (e)=A(BCc)by(f)=A(BC)by(g)

Explanation

Given information

(AB)(CA)=(AB)(CA)cby (a)=(AB)(C A c)cby (b)=(AB)( A cC)cby (c)=(AB)( ( A c )cCc)by(d)=(AB)(ACc)by (e)=A(BCc)by(f)=A(BC)by(g)

Concept used:

A,B,andCare sets.(AB)(CA)=A(BC)

Proof:

(AB)(CA)=(AB)(CA)c by the set difference law=(AB)(C A c)cby the set deference law=(AB)( A cC)cby the commutative law=(AB)( ( A c )cCc)by the De Morgan law=(AB)(ACc)by the double complementary law=A(BCc)bythe distributive law=A(BC)bythe set difference law

Calculation:

A,B,andCare sets

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 