Proving (A∩B)' = A'∪B' Without De Morgan's Method

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Discussion Overview

The discussion revolves around the proof of the set identity \((A \cap B)' = A' \cup B'\) without employing De Morgan's Theorem. Participants explore various methods and definitions to establish this identity, focusing on the need for an "ordinary proof" as opposed to graphical or theorem-based approaches.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest using Karnaugh maps or Venn diagrams as valid proofs, while others insist on a more traditional proof method.
  • One participant attempts to outline a proof using basic definitions, stating that if \(x \in (A \cap B)'\), then \(x\) must not be in \(A\) or \(B\), leading to the conclusion that \(x\) is in \(A' \cup B'\).
  • Another participant points out that the reasoning provided essentially relies on De Morgan's Theorem, which contradicts the request for a proof without it.
  • There is a suggestion to restate the proof of De Morgan's Theorem with specific conditions to derive the desired identity.

Areas of Agreement / Disagreement

Participants do not reach a consensus on how to prove the identity without invoking De Morgan's Theorem. There are competing views on the validity of using different proof methods, and some participants express confusion regarding the definitions and requirements of the proof.

Contextual Notes

Participants express limitations in finding proofs that do not reference De Morgan's Theorem, indicating a potential gap in available resources or methods for proving the identity as requested.

solakis
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Can we prove :([itex]A\cap B[/itex])' = [itex]A^'\cup B^'[/itex] ,without using De Morgan's??
 
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you can if you accept a Karnaugh map or a Venn diagram as proof
 
phinds said:
you can if you accept a Karnaugh map or a Venn diagram as proof

no ,i mean an ordinary proof.
 
solakis said:
no ,i mean an ordinary proof.

well, try googling "proof of DeMorgan's Theorem" (or Law as it seems to be called these days)
 
Just use the basic definitions:

If [itex]x\in \left(A\cap B\right)'[/itex], then x is NOT in[itex]A\cap B[/itex] when means it is either not in A or not in B. If it is not in A then it is in A' and therefore in [itex]A'\cup B'[/itex]. If it is not in B then it is in B' and therefore in [itex]A'\cup B'[/itex]. Thus, [itex]\left(A\cap B\right)'\subset A'\cup B'[/itex].

The other way: if [itex]x \in A' \cup B'[/itex] it is in either A' or in B'. If it is in A', ...
 
HallsofIvy said:
Just use the basic definitions:

If [itex]x\in \left(A\cap B\right)'[/itex], then x is NOT in[itex]A\cap B[/itex] when means it is either not in A or not in B..


That is using De Morgan's. But i asked for a proof without using De Morgans
 
phinds said:
well, try googling "proof of DeMorgan's Theorem" (or Law as it seems to be called these days)

I am afraid i could not find a proof ,apart from one using truth tables
 
solakis said:
I am afraid i could not find a proof ,apart from one using truth tables

And what does that tell you?
 
how is De Morgan's theorem proven? you can restate the proof that uses De Morgan and where you get to the spot where De Morgan is invoked, then proceed with the steps that prove De Morgan but with your specific conditions (or "input") instead of the general A and B case.
 
  • #10
HallsofIvy said:
Just use the basic definitions:

If [itex]x\in \left(A\cap B\right)'[/itex], then x is NOT in[itex]A\cap B[/itex] when means it is either not in A or not in B. If it is not in A then it is in A' and therefore in [itex]A'\cup B'[/itex]. If it is not in B then it is in B' and therefore in [itex]A'\cup B'[/itex]. Thus, [itex]\left(A\cap B\right)'\subset A'\cup B'[/itex].

The other way: if [itex]x \in A' \cup B'[/itex] it is in either A' or in B'. If it is in A', ...

solakis said:
That is using De Morgan's. But i asked for a proof without using De Morgans

Then I am confused as to what you mean by "using DeMorgan's". What I gave is how one would prove DeMorgan's laws but did not use DeMorgan's laws.
 

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