Set Theory Proof(Using Identities)

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The discussion revolves around solving a proof in set theory, specifically proving that A U B equals (A ∩ B') U (A' ∩ B) U (A ∩ B). The user seeks assistance after struggling for three hours. Participants confirm the use of standard set axioms like distributivity and DeMorgan's laws. The proof is simplified, ultimately leading to the conclusion that A U B is derived correctly. The user expresses satisfaction upon reaching the solution after extensive work.
ozymandris
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Hi, I've been trying for 3 hours to solve this proof using identities. I can't seem to get it.

Can i get a little help please?

Prove: A U B = (A ∩ B') U (A' ∩ B) U (A ∩ B)

thanks
 
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welcome to pf!

hi ozymandris! welcome to pf! :smile:

do the easy bit first …

what is (A' ∩ B) U (A ∩ B) ? :wink:
 
ozymandris said:
Hi, I've been trying for 3 hours to solve this proof using identities. I can't seem to get it.

Can i get a little help please?

Prove: A U B = (A ∩ B') U (A' ∩ B) U (A ∩ B)

thanks

Hey ozymandris and welcome to the forums.

What are you allowed to take for granted? Can you use normal set axioms like distributivity, DeMorgans laws, and so on?
 
chiro said:
Hey ozymandris and welcome to the forums.

What are you allowed to take for granted? Can you use normal set axioms like distributivity, DeMorgans laws, and so on?

(A' ∩ B) U (A ∩ B) = (A' U A) ∩ B


We're allowed all the normal set axioms, distributivity, deMorgan, etc etc
 
ozymandris said:
(A' ∩ B) U (A ∩ B) = (A' U A) ∩ B

yup! :smile:

and A' U A = … ? :wink:
 
tiny-tim said:
yup! :smile:

and A' U A = … ? :wink:

ok, i think I've gotten it. It took me nearly a page of workout. I'll post it when i have a minute to type it all in.
 
It should be only two lines, so you might want to see if you can shorten your reasoning.
 
tiny-tim said:
yup! :smile:

and A' U A = … ? :wink:

A' U A = \varnothing
 
(A ∩ B') U (A' U A) ∩ B

(A ∩ B') U {Universal} ∩ B

(A ∩ B') U B if i distribute from this

(A U B ) ∩ (B' U B)

A U B
 
  • #10
:biggrin: Woohoo! :biggrin:
 

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