Can You Prove (A ∩ B) - C Equals (A - C) ∩ (B - C) in Set Theory?

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Homework Help Overview

The discussion revolves around proving the equality of two set expressions: (A ∩ B) - C and (A - C) ∩ (B - C). The subject area is set theory, focusing on set operations and properties.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the process of proving set equality by examining arbitrary elements and their membership in the respective sets. There is mention of using identities and laws related to set operations, as well as the need to demonstrate both inclusions (A ⊆ B and B ⊆ A).

Discussion Status

The discussion is active, with participants offering various approaches to the proof. Some have outlined steps for proving one direction of the equality, while others seek clarification on the definitions and implications of the expressions involved. There is no explicit consensus yet, as participants are exploring different aspects of the proof.

Contextual Notes

Participants are navigating the requirements of set theory proofs, including the need to establish arbitrary elements and the implications of set membership. There may be assumptions about the familiarity with set operations and notation that are not explicitly stated.

geforce
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Prove that (A n B) - C = (A - C) n (B - C).

n = intersect
≠ε = not a member

I got the first one by doing:
(xεA ^xεB) ^X≠εC ( by identity law and compliment law)


where would I go on from now?
 
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so basically you have to prove the equality of sets. forward way is to let x\in(A\cap B)\setminus C be arbitrary. Then as you have shown

x\in A\;x\in B\;\; x\notin C

which means that x is A and not in C AND x is in B and not in C. So just combine that to arrive at the right side. Then proceed to the reverse direction.
 
What do you mean as in "(A∩B)∖C"to be arbitrary and I have to show how it goes from (A∩B) -C to (A-C) ∩ (B-C)
 
?...
 
when you have to prove that two sets are equal you have to prove that

A\subseteq B\mbox{ and }B\subseteq A

So to prove A\subseteq B you take arbitrary member of A and then prove that
its also member of B. And similar proof for proving B\subseteq A
 

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