# Thread: Logic Check with Set proof

1. ## Logic Check with Set proof

Prove or Give counter example:

$\displaystyle (AXB) \cap (CXD) = (A \cap C) X (B \cap D)$

kind of a long proof. But I was wondering if anyone could help check this for me.

I proved that $\displaystyle x \in A,C$ and $\displaystyle y \in B,D$. So $\displaystyle (AXB) \cap (CXD)$ was a subset of $\displaystyle (A \cap C) X (B \cap D)$.

When trying to prove that $\displaystyle (A \cap C) X (B \cap D)$ was a subset of $\displaystyle (AXB) \cap (CXD)$ I reached a contradiction, well what i believe was a contradiction. I found that
$\displaystyle x \in A$ and $\displaystyle x \in B$. Im just not sure if this is a contradiction or not. If its a contradiction, well im done, if it is not a contradiction I guess I would just continue on to prove it. Im just not sure if the last part of the proof is a contradiction.

Any help with this proof is greatly appreciated, and im sorry if this is somewhat unclear.

2. Originally Posted by p00ndawg
Prove or Give counter example:

$\displaystyle (AXB) \cap (CXD) = (A \cap C) X (B \cap D)$

kind of a long proof. But I was wondering if anyone could help check this for me.

I proved that $\displaystyle x \in A,C$ and $\displaystyle y \in B,D$. So $\displaystyle (AXB) \cap (CXD)$ was a subset of $\displaystyle (A \cap C) X (B \cap D)$.

When trying to prove that $\displaystyle (A \cap C) X (B \cap D)$ was a subset of $\displaystyle (AXB) \cap (CXD)$ I reached a contradiction, well what i believe was a contradiction. I found that
$\displaystyle x \in A$ and $\displaystyle x \in B$. Im just not sure if this is a contradiction or not. If its a contradiction, well im done, if it is not a contradiction I guess I would just continue on to prove it. Im just not sure if the last part of the proof is a contradiction.

Any help with this proof is greatly appreciated, and im sorry if this is somewhat unclear.

proof:

$\displaystyle (x,y)\in [(A\times B)\cap(C\times D)]\Longleftrightarrow$$\displaystyle (x,y)\in(A\times B)\wedge (x, y)\in (C\times D)\Longleftrightarrow$$\displaystyle x\in A\wedge y\in B\wedge x\in C\wedge y\in D \Longleftrightarrow$$\displaystyle (x\in A\wedge x\in C)\wedge (y\in B\wedge y\in D)\Longleftrightarrow$$\displaystyle x\in (A\cap C)\wedge y\in (B\cap D)\Longleftrightarrow$$\displaystyle (x,y)\in[(A\cap C)\times (B\cap D)] 3. Originally Posted by xalk proof: \displaystyle (x,y)\in [(A\times B)\cap(C\times D)]\Longleftrightarrow$$\displaystyle (x,y)\in(A\times B)\wedge (x, y)\in (C\times D)\Longleftrightarrow$$\displaystyle x\in A\wedge y\in B\wedge x\in C\wedge y\in D \Longleftrightarrow$$\displaystyle (x\in A\wedge x\in C)\wedge (y\in B\wedge y\in D)\Longleftrightarrow$$\displaystyle x\in (A\cap C)\wedge y\in (B\cap D)\Longleftrightarrow$$\displaystyle (x,y)\in[(A\cap C)\times (B\cap D)]$

Thanks, hmm I dont think this will fly with my professor tho.
He wants us to prove one is the subset of the other and then prove the other is subset of the other as well. I dont think just saying $\displaystyle \Longleftrightarrow$ will work.

Thank you however, this helps me since I guess there is no contradiction.

4. Originally Posted by p00ndawg
Thanks, hmm I dont think this will fly with my professor tho.
He wants us to prove one is the subset of the other and then prove the other is subset of the other as well. I dont think just saying $\displaystyle \Longleftrightarrow$ will work.

Thank you however, this helps me since I guess there is no contradiction.
That is what is done at the same time . Notice the double implication.

The forward implication proves that:

$\displaystyle (A\times B)\cap (C\times D)$ is a subset of $\displaystyle (A\cap C)\times (B\cap D)$ and

the backwards implication proves that: $\displaystyle (A\cap C)\times (B\cap D)$ is a subset of $\displaystyle (A\times B)\cap (C\times D)$ .

You can check that.