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

  • Thread starter solakis
  • Start date
In summary: That is, I showed how to prove them from the definitions.In summary, the conversation discusses the possibility of proving the equation :(A\cap B)' = A^'\cup B^' without using De Morgan's theorem. The suggestion of using a Karnaugh map or a Venn diagram as proof is mentioned, but the person asking the question clarifies that they want an ordinary proof. The conversation ends with the suggestion to search for a proof of De Morgan's theorem online, and the possibility of restating the proof using specific conditions instead of the general case.
  • #1
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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  • #2
you can if you accept a Karnaugh map or a Venn diagram as proof
 
  • #3
phinds said:
you can if you accept a Karnaugh map or a Venn diagram as proof

no ,i mean an ordinary proof.
 
  • #4
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)
 
  • #5
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', ...
 
  • #6
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
 
  • #7
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
 
  • #8
solakis said:
I am afraid i could not find a proof ,apart from one using truth tables

And what does that tell you?
 
  • #9
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.
 

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

What is De Morgan's method?

De Morgan's method is a mathematical rule that states the negation of a union or intersection of two sets is equal to the intersection or union of the negations of those sets. In other words, (A∪B)' = A'∩B' and (A∩B)' = A'∪B'.

Why is it important to prove (A∩B)' = A'∪B' without using De Morgan's method?

Proving (A∩B)' = A'∪B' without using De Morgan's method can help expand one's understanding of set theory and logical reasoning. It also allows for a deeper understanding of the underlying principles and concepts involved in set operations.

What are the steps to prove (A∩B)' = A'∪B' without using De Morgan's method?

The steps to prove (A∩B)' = A'∪B' without using De Morgan's method are as follows:
1. Start with the left side of the equation: (A∩B)'
2. Use the definition of complement to rewrite the left side as (A∪B)'
3. Use the distributive property to expand (A∪B)' as (A'∩B')
4. Rearrange the terms to get (A'∪B')
5. Use the definition of complement again to rewrite (A'∪B') as (A'∪B')'
6. Since (A'∪B')' is the negation of (A'∪B'), it is equal to (A'∪B')' = (A'∪B')'
7. Therefore, (A'∪B')' = A'∪B', which proves the original equation.

What are some common mistakes to avoid when proving (A∩B)' = A'∪B' without using De Morgan's method?

Some common mistakes to avoid when proving (A∩B)' = A'∪B' without using De Morgan's method include:
- Forgetting to use the distributive property when expanding (A∩B)'
- Forgetting to rearrange the terms to get (A'∪B')
- Forgetting to use the definition of complement to rewrite (A'∪B') as (A'∪B')'
- Making errors in the logic or steps of the proof
It is important to carefully follow the steps and double check all the calculations and logic to avoid these common mistakes.

Can this proof be applied to other set operations?

Yes, this proof can be applied to other set operations. The same steps can be followed to prove the following equations without using De Morgan's method:
- (A∪B)' = A'∩B'
- (A∩B∩C)' = A'∪B'∪C'
- (A∪B∪C)' = A'∩B'∩C'
And so on.

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