Element Argument Proofs - Set theory This is an exercise on my study guide for my discrete applications class.
Prove by element argument: A × (B ∩ C) = (A × B) ∩ (A × C)
Now I know that this is the distributive law, but I'm not sure if this proof would work in the exact same way as a union problem would, because I know how to solve that one. Here is my thinking thus far:

Proof:  Suppose A, B, and C are sets.

    
*
    
*A × (B ∩ C) = (A × B) ∩ (A × C)
    
*Case 1 (a is a member of A): if a belongs to A, then by the definition of the cartesian product, a is also a member of A x B and A x C. By definition of intersection, a belongs to (A × B) ∩ (A × C).
    
*Case 2 (a is a member of B ∩ C): a is a member of both B and C by intersection. a is a member of (A × B) ∩ (A × C) by the definition of intersection.
    
*By definition of a subset, (A × B) ∩ (A × C) is a subset of A × (B ∩ C).
    
*Therefore A × (B ∩ C) = (A × B) ∩ (A × C).
    
  

Is that at least a little right?
Thanks.
 A: No, you're not doing it completely right, the cartesian product produces an element that is a pair of elements from both subsets.
The definition of the cartesian product.
Def. $X\times Y = \{ (x,y) : x \in X\text{ and }y \in Y \}$.
PROOF.
$Z =  A \times (B \cap C) = \{ (a,y) : a \in A\text{ and }y \in B \cap C \}$
$W = (A \times B) \cap (A \times C) = \{ (a,b) : a \in A\text{ and }b \in B \} \cap \{ (a,c) : a \in A\text{ and }c \in C \}$
For all $a \in A$:
Case 1. $b \in C$.
If $b \in C$ then $(a,b) \in Z$. Also $(a,b) \in W$.
Case 2. $b \notin C$.
If $b \notin C$, then $b$ is not in $B \cap C$. Then $(a,b)$ is not in $Z$. $b$ is also not in $A \times C$, so it's not in $W$.
The rest follows by the symmetry of intersection. $C \cap B$ is equivalent to $B \cap C$. Relabel $B$ as $C$, and vice versa. Apply case 1 and case 2.
QED.
A: I'm practicing my set proving skills, so it could be that I'm wrong. Frederik E already has given the answer, but it is very concise. The following proof is a bit longer. 
The proof using element argument for:
${A \times (B\cap C) = (A \times B)}\cap(A\times C)$
Starts with proving that ${A \times (B\cap C) \subseteq (A \times B)\cap(A\times C)}$
${x \in A \times (B\cap C)}$
${x \in \{(m,n)|m \in A \land n \in (B\cap C)}\}$ def Cartesian product
${x \in \{(m,n)|m \in A \land (n \in B \land n \in C)}\}$ def intersection
We are short one ${m \in A}$ but we can introduce that using the idempotency law
${x \in \{(m,n)|(m \in A \land m \in A) \land (n \in B \land n \in C)}\}$ idempotency law
${x \in \{(m,n)|(m \in A \land n \in B) \land (m \in A \land n \in C)}\}$ associative and commutative laws
${x \in \{(m,n)|m \in A \land n \in B\} \cap \{(m,n)|m \in A \land n \in C}\}$ def intersection
${x \in (A \times B)\cap(A\times C)}$ def Cartesian product
Since each step is reversible in this proof, it follows that  ${(A \times B)\cap(A\times C) \subseteq A \times (B\cap C)}$ is also true.
And that concludes the proof.
