# Thread: sets prove

1. ## sets prove

Prove the following identities , for any sets A , B and C .

(a) (A u B) n ( B u C) N ( C u A )=( A n B ) u ( A n C ) u ( B n C )

(b) A n ( B - C) = ( A n B ) - ( A n C )

(c) ( A-B ) u B = A if and only if $B \subset C$

Is there a way to prove all these besides illustrating them with a venn diagram or use examples .

2. Hint to do Qn b and c. Note A-B = AnB'

3. ## Re :

Originally Posted by mathaddict
Prove the following identities , for any sets A , B and C .

(a) (A u B) n ( B u C) N ( C u A )=( A n B ) u ( A n C ) u ( B n C )

(b) A n ( B - C) = ( A n B ) - ( A n C )

(c) ( A-B ) u B = A if and only if $B \subset C$

Is there a way to prove all these besides illustrating them with a venn diagram or use examples .

(a) I have no idea how to start at all .

(b) i use the distributive law to prove .

(c) ( B' n A ) u B = ( B' u B ) n ( A u B ) = B n ( A u B ) = B
haiz , also wrong .

Can someone pls check my A and C . THanks .

4. ## Proofs using the Laws of Sets

Hello mathaddict

I've posted some notes on the Laws of Sets, with some worked examples on using them on Wikibooks (where I'm known as nigeltn35). You'll find them at Discrete mathematics/Set theory/Page 2 - Wikibooks, collection of open-content textbooks

Have a read through. If you still need some help with these examples, let me know.

Grandad

5. ## Re :

Thanks Grandad , i have read through the notes but i am still having some problems with these questions .

Prove :

(a) (A u B) n ( B u C) n ( C u A )=( A n B ) u ( A n C ) u ( B n C )
which laws can be applied to prove it ?

(b)Prove :
(A' n B n C) u {A' u (B n C')}' = (A n B') u (B n C)

My working :
(A' n B n C) u {A n (B n C')' } ------ De Morgan's law
= (A' n B n C) u {A n (B' u C) } ------ De Morgan's law
From here onwards , I am not sure what to do .

6. ## Set Proofs

Hello mathaddict

Originally Posted by mathaddict
Thanks Grandad , i have read through the notes but i am still having some problems with these questions .

Prove :

(a) (A u B) n ( B u C) n ( C u A )=( A n B ) u ( A n C ) u ( B n C )
which laws can be applied to prove it ?

(b)Prove :
(A' n B n C) u {A' u (B n C')}' = (A n B') u (B n C)

My working :
(A' n B n C) u {A n (B n C')' } ------ De Morgan's law
= (A' n B n C) u {A n (B' u C) } ------ De Morgan's law
From here onwards , I am not sure what to do .
These are really quite tricky, aren't they?

For the first one, begin by proving that $C \cap (A \cup C) = C$, as follows:

$C \cap (A \cup C) = (C \cup \oslash) \cap(C \cup A)$, Identity Law

$= C \cup (\oslash \cap A)$, Distributive Law

$= C \cup \oslash$, Identity Law

$= C$, Identity Law. Call this equation (1).

Then:

$(A \cup B) \cap(B \cup C) \cap (A \cup C)$

$=((B \cup A) \cap (B \cup C)) \cap (A \cup C)$, Commutative Law

$=(B \cup (A \cap C)) \cap (A \cup C)$, Distributive Law

$=(B \cap (A \cup C)) \cup ((A \cap C) \cap (A \cup C))$, Distributive Law

$=(B \cap A) \cup (B \cap C) \cup ((A \cap (C \cap (A \cup C))$, Distributive and Associative Laws

$=(A \cap B) \cup (B \cap C) \cup (A \cap C)$, from equation (1)

For the second one, look at the RHS: $(A \cap B') \cup (B \cap C)$.

You'll see that it doesn't contain a term in $A'$, which the LHS does. So you must 'create' one, by using the fact that $A \cup A' =\text{ U }$

I'll do the first two or three lines for you (but you supply the name of the Laws I'm using) - then you see if you can finish it. Your two uses of De Morgan's Law are correct - it's just that it's easier to finish with them, rather than starting with them. So:

$(A \cap B') \cup (B \cap C)$

$= (A \cap B') \cup (\text{ U } \cap (B \cap C)$

$= (A \cap B') \cup ((A \cup A') \cap (B \cap C)$

$=(A \cap B') \cup ((A \cap B \cap C) \cup (A' \cap B \cap C))$

$=(A' \cap B \cap C) \cup (A \cap B \cap C) \cup (A \cap B')$

You'll see that this gives us the term we want at the beginning, $(A' \cap B \cap C)$, so leave this alone! Now use the Distributive Law on the rest a couple of times, and with care you should be able to complete the proof.

I hope you can do it from here.

Grandad