Help with some set theory questions

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Discussion Overview

The discussion revolves around set theory proofs, specifically addressing four statements related to set operations and properties. Participants seek assistance in proving these statements and share their attempts and challenges in the process.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents four set theory statements for proof, including operations involving intersection and symmetric difference.
  • Another participant encourages the original poster to share their attempts and specific points of confusion to facilitate better assistance.
  • A participant shares their attempt at proving the first statement, expressing difficulty in progressing further.
  • A later reply suggests a method of "reverse engineering" the proof for the first statement, indicating a potential strategy to approach the problem.

Areas of Agreement / Disagreement

The discussion does not appear to reach a consensus, as participants are still exploring proofs and expressing uncertainty about their approaches.

Contextual Notes

Participants have not provided complete proofs or resolved all steps, indicating potential gaps in understanding or assumptions that may affect the proofs.

Beno
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Greetings all. I was wondering if someone could give me some asistance on a few simple set theory proofs.

1) An(B\C)=(AnB)\(AnC)
2) A+(B+C)=(A+B)+C where + = symmetric difference
3) If A+B = A+C then B=C
and
4) An(B+C) = (AnB)+(AnC)
 
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Welcome to PF!

Hi Beno! Greetings and welcome to PF! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
With number 1
An(B\C)=(AnB)\(AnC)
I tried the following
An(B\C) = An(BnC') =(AnB)n(AnC') but this was as far as it seemed to go.

As for the others I think I've worked them out since posting
 
Beno said:
With number 1
An(B\C)=(AnB)\(AnC)
I tried the following
An(B\C) = An(BnC') =(AnB)n(AnC') but this was as far as it seemed to go.

ok, let's do some reverse engineering :wink:

the last step but one is going to be (AnB)n(AnC)' …

work backwards from there. :smile:
 

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