Proving the cartesian product of the union of two sets A,B and the set C is equal to the union of two cartesian products $(A \cup B) \times C = (A \times B) \cup (B \times C)$
So far I have:


*

*$x \in (A \times B)$ or $x ∈ (B \times C)$    [definition U]  

*$(x ∈ A \quad \text{and}\quad x ∈ B)$ or $(x ∈ B \quad \text{and}\quad x ∈ C)$ [definition of X]

*$(x \in A \vee x \in B) ∧ (x \in A \vee x \in C) \wedge (x \in B \vee x \in B) \wedge (x \in B \vee x \in C)$ [definition of distribution]


I don't really understand how the Cartesian product is supposed to be interpreted. Is it an and symbol (∧)?
 A: $\begin{align}\forall x\, \forall y:
 (x,y)\in(A\cup B)\times C 
 \iff & x\in (A\cup B), y\in C &\text{def$^n$ cross product}
\\ \iff & (x\in A\lor x\in B), y\in C & \text{def$^n$ union}
\\ \iff & (x\in A, y\in C)\lor(x\in B, y\in C) & \text{distribution}
\\ \iff & (x,y)\in (A\times C) \lor (x,y)\in (B\times C) & \text{def$^n$ cross product}
\\ \iff & (x,y)\in ((A\times C) \cup (B\times C)) & \text{def$^n$ union}
\\[3ex]
\therefore (A\cup B)\times C \quad\equiv\quad & (A\times C)\cup(B\times C)
\end{align}$
A: No, the cartesian product is not an "and" symbol. $A\times B$ is the set of ordered pairs $(a,b)$, with $a\in A$, $b\in B$. For example,
$$
\{J,Q,K\}\times\{\spadesuit,\heartsuit\}=\{(J,\spadesuit),(Q,\spadesuit),(K,\spadesuit),(J,\heartsuit),(Q,\heartsuit),(K,\heartsuit)\}
$$
So, step 2) of your proof should instead say:
2) Either $x=(a,c)$, with $a\in A, c\in C$,or $x=(b,c)$, with $b\in B, c\in C$.
So $x$ is one of two possible ordered pairs. See how you can show this is equivalent to being and ordered pair $(d,c)$, with $d\in A\cup B$, $c\in C$.
A: I think you need this for the right hand side:


*

*$(x,y)\in(A\times B)$ or $(x,y)\in B\times C$.

*($(x\in A)$ and $(y\in B)$) or ($(x\in B)$ and $(y\in C)$).
etc.


However, your formula is wrong anyway.
You need $(A\times C)\cup(B\times C)$ on the right hand side.
So you'll have to fix the question before it can be answered.
