Formal Boolean Proof of A ⊕ B' ⊕ C = (A ⊕ B ⊕ C)

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Discussion Overview

The discussion revolves around proving the equation A ⊕ B' ⊕ C = (A ⊕ B ⊕ C)', focusing on formal Boolean proof techniques. The scope includes homework-related inquiries and mathematical reasoning.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents an initial attempt at a solution using the expression A ⊕ B' ⊕ C = ABC' + A'B'C' + A'BC + AB'C, expressing uncertainty about the next steps.
  • Another participant suggests starting with the right-hand side of the equation and indicates that it can be rewritten using certain laws.
  • A later reply specifies that De Morgan's laws are relevant to the discussion, providing definitions for these laws but not elaborating on their application to the problem.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the approach to the proof, and multiple viewpoints regarding the application of laws exist.

Contextual Notes

There is a lack of clarity on how to apply the proposed laws to advance the proof, and the discussion does not resolve the mathematical steps involved.

nahanksh
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Homework Statement


Prove that
A \oplus B' \oplus C = (A \oplus B \oplus C)'

Homework Equations


The Attempt at a Solution


I tried to use A \oplus B' \oplus C = ABC' + A'B'C' + A'BC + AB'C

But i am not sure how to proceed further from there...

Please could someone give me a little bit of help ?
 
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I would start with the right hand side - it can be rewritten with some laws.
 
I'm sorry that was still vague. De Morgan's laws to be specific.
NOT (P OR Q) = (NOT P) AND (NOT Q)
NOT (P AND Q) = (NOT P) OR (NOT Q)
 
I'm sorry that was still vague. De Morgan's laws to be specific.
NOT (P OR Q) = (NOT P) AND (NOT Q)
NOT (P AND Q) = (NOT P) OR (NOT Q)
 

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