# Thread: Proof by cases to prove distributive law for sets

1. ## Proof by cases to prove distributive law for sets

https://gyazo.com/defe1638bbbf385f343684bd6c024c3c

So I know what proof by cases is... and I know how to prove problems like If n^2 is even, then n is even.

I got lost when the problems involve sets and intersection/unions..

Could someone explain this to me Barney-style?

2. ## Re: Proof by cases to prove distributive law for sets

Originally Posted by MrJank
https://gyazo.com/defe1638bbbf385f343684bd6c024c3c

So I know what proof by cases is... and I know how to prove problems like If n^2 is even, then n is even.

I got lost when the problems involve sets and intersection/unions..

Could someone explain this to me Barney-style?
I think what they are after is how the sets "interact" with each other. For example, start with all of A, B, C disjoint from each other. Then move to the case where A and B intersect but C is disjoint from both. Then where A and C intersect and B is disjoint, etc. etc.

-Dan

3. ## Re: Proof by cases to prove distributive law for sets

Originally Posted by MrJank
https://gyazo.com/defe1638bbbf385f343684bd6c024c3c
So I know what proof by cases is... and I know how to prove problems like If n^2 is even, then n is even.
I got lost when the problems involve sets and intersection/unions..
Could someone explain this to me Barney-style?
Why don't you learn to post symbols? Many here will not click on a link.
It is really easy: \$A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\$ gives $A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$

4. ## Re: Proof by cases to prove distributive law for sets

The problem is to prove that "$\displaystyle A\cap (B\cup C)= (A\cap B)\cup (A\cap C)$

To prove that "set X= set Y" do two "cases"
1) show that "if p is in X then p is in Y".
2) show that "if p is in Y then p is in X".

1) "If p is in $\displaystyle A\cap(B\cup C)$ then p is in $\displaystyle (A\cap B)\cup (A\cap C)$"
If p is in $\displaystyle A\cap (B\cup C)$ then p is in A and in $\displaystyle B\cup C$
Since p is in $\displaystyle B\cup C$ then either p is in B or p is in C so we have two "subcases":
1a) p is in B. Then since p is in A and in B p is in $\displaystyle A\cap B$ and so in $\displaystyle (A\cap B)\cup (A\cap C)$.
1b) p is in C. Then since p is in A and in C p is in $\displaystyle A\cap C$ and so in $\displaystyle (A\cap B)\cup (A\cap C)$.

2) "If p is in $\displaystyle (A\cap B)\cup (A\cap C)$ then p is in $\displaystyle A\cap(B\cup C)$".
Since p is in $\displaystyle (A\cap B)\cup (A\cap C)$ p is either in $\displaystyle A\cap B$ or p is in $\displaystyle A\cap C$
so again we have two "subcases":
2a) if p is in $\displaystyle A\cap B$ then p is in both A and B. Since p is in B, b is in $\displaystyle B\cup C$ so p is in $\displaystyle A\cap(B\cup C)$.
2b) if p is in $\displaystyle A\cap C$ then p is in both A and C. Since p is in C, b is in $\displaystyle B\cup C$ so p is in $\displaystyle A\cap(B\cup C)$.

5. ## Re: Proof by cases to prove distributive law for sets

Originally Posted by MrJank
I got lost when the problems involve sets and intersection/unions..
Could someone explain this to me Barney-style?
In thirty-five years of teaching this course, I have never heard of Barney-style.
If $x\in A\cap(B\cup C)$ means that $x\in A$ AND $x\in B\cup C$
$x\in B\cup C$ means that $x\in B$ OR $x\in C$
Putting those two together we get:
$x\in A~\&~x\in B$ OR $x\in A~\&~x\in C$
Thus $x\in (A\cap B)\cup(A\cap C)$

At least you can show some effort to do the reverse.

6. ## Re: Proof by cases to prove distributive law for sets

Originally Posted by Plato
In thirty-five years of teaching this course, I have never heard of Barney-style.
I hate to say I recognize the reference but he's referring to "Barney the Purple Dinosaur." In other words, in simple terms.

-Dan

7. ## Re: Proof by cases to prove distributive law for sets

I thought maybe "Deputy Barney" of the Andy Griffeth Show (In reruns called "Andy of Mayberry").