# Prove $(A \cup B) \cap C$ = $(A \cap C) \cup (B \cap C)$

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

Starting from the left side,

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

By distributive law, ( distributing the $\cap C$), we have

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

Therefore,

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

If I start from the right, I have

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

By Distributive Law

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

Therefore, $(A \cup B) \cap C$ = $(A \cup B ) \cap C$

Did I do this correctly or do I need to include the set union definition and the set intersection definitions?

Assuming that I need to include the set union definition of $A \cup B$ for the left side $[x: x \in A \lor x \in B]$ so that means that x belongs in A or x belongs in B

For the right side I would have set intersection

$[x: x \in A \land x \in C]$ so x belongs in A and x belongs in C

$[x: x \in B \land x \in C]$ so x belongs in B and x belongs in C

so maybe it's like this?

$[x: x \in A \land x \in C] \lor [x: x \in B \land x \in C]$

and then by distributive law I would have gotten

$[x: x \in A \lor x \in B] \land C$

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

My question is how do I write a better proof than this jumbled mess?

• I'm pretty sure this question is asking you to PROVE the distributive law - I would be surprised if you were allowed to use it. – user98602 Mar 3 '14 at 6:38
• o-o ok so how do I prove the distributive law? uh oh ... I just read something and this is clearly not how it's supposed to be :S – usukidoll Mar 3 '14 at 6:39
• If you invoke the distributive law then it is just one step, since the distributive law establishes an equivalence. – copper.hat Mar 3 '14 at 6:46
• that's what I wanted to do @copper.hat since I saw it then if I just take one side of the equation and use distributive law I can easily have achieved the right side. I don't know if I did this right. I'm really sure that I did since the question was straight forward. – usukidoll Mar 3 '14 at 6:47

\begin{align*} & & x &\in (A \cup B) \cap C \\ &\iff & x &\in (A \cup B) \wedge x \in C \\ &\iff & (x &\in A \vee x \in B) \wedge x \in C \\ &\iff & (x &\in A \wedge x \in C) \vee (x \in B \wedge x \in C) \\ &\iff & (x &\in A \cap C) \vee (x \in B \cap C) \\ &\iff & x &\in (A \cap C) \cup (B \cap C) \\ \end{align*}

We've only used the definitions of union and intersection and the distributive law of logic. Since every line is an equivalence, the first and last lines are equivalent.

• thanks....this is exactly what I was looking for since when I assumed that using set union and set intersection is the right thing to do, I was able to prove it but got some leftover junk while I was using the definitions. – usukidoll Mar 3 '14 at 7:14

I've done some googling and after some hard work ( :3 ) I came up with this http://php.scripts.psu.edu/djh300/cmpsc360/ex-set-equality2.htm

• great for checking, not for cheating :P – usukidoll Mar 3 '14 at 7:13
\begin{align*} & & x &\in (A \cup B) \cap C \ &\iff & x &\in (A \cap C) \cup (B \cap C) \\  \end{align*}