   Chapter 6.3, Problem 40ES ### 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 30-40, construct an algebraic proof for the given statement, Cite a property from Theorem 6,2,2 for every step.For all sets A,B, and C, ( A − B ) − ( B − C ) = A − B .

To determine

To prove:

(AB)(BC)=AB

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:

The objective is to provide algebraic proof of (AB)(BC)=AB, for all sets A,B and C.

Use the set difference law AB=ABc to (AB)(BC).

(AB)(BC)=(AB)(BC)c=(ABc)(B C c)c   by the set difference law

Apply De Morgan’s law (AB)c=AcBc to (BCc)c

(AB)(BC)=(ABc)(Bc ( C c )c)   by De Morgan's law

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