Set Theory Proof(Using Identities)

[ozymandris]

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

 

[tiny-tim]

welcome to pf!

hi ozymandris! welcome to pf!

do the easy bit first …

what is (A' ∩ B) U (A ∩ B) ? ​

 

[chiro]

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?

 

[ozymandris]

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

 

[tiny-tim]

(A' ∩ B) U (A ∩ B) = (A' U A) ∩ B

yup!

and A' U A = … ? ​

 

[ozymandris]

yup!

and A' U A = … ? ​

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.

 

[Landau]

It should be only two lines, so you might want to see if you can shorten your reasoning.

 

[ozymandris]

yup!

and A' U A = … ? ​

A' U A = \varnothing

 

[ozymandris]

(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

 

[tiny-tim]

Woohoo! ​