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I have to prove this theorem - Let $A$, $B$, and $C$ be sets. Then, $A \times (B \Delta C) = (A \times B) \Delta (A \times C)$.

I was wondering if this was the right way to start the proof?

Let $m$ be an arbitrary element of $A \times (B \Delta C)$. Then, $m=(x,y)$ s.t. $x \in A$ and $y \in (B \Delta C)$ by the definition of cartesian products. Then, $(m \in A) \land (y \in B) \land (y \not\in C)$ by the definition of set difference. So, $((x,y) \in A \times B) \land ((x,y) \not\in A \times C)$ by the definition of cartesian product. Thus, $(x,y) \in (A \times B) \Delta (A \times C)$.

So basically the LHS = RHS because (x,y) is an element of both sides?

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    First thing: $(x,y)$ is an *element*, not a subset, of these sets. – 2017-02-21
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    Right, fixed that. Thanks – 2017-02-21
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    Second: you should take a closer look at the definition of a *symmetric* difference. – 2017-02-21

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Def. let be $E,F$ sets: $$E \bigtriangleup F := (E \setminus F) \cup (F\setminus E)$$

The. let be $A,B,C$ sets: $$A \times (B \bigtriangleup C)=(A \times B) \bigtriangleup (A \times C)$$ Proof: $$ \begin{align} A \times (B \bigtriangleup C) &= A \times ((B \setminus C) \cup (C \setminus B)) \\&= (A \times (B\setminus C)) \cup (A \times (C \setminus B)) \\&=((A \times B)\setminus (A \times C)) \cup ((A \times C) \setminus (A \times B)) \\&=(A\times B)\bigtriangleup (A \times C)\end{align}$$ I used "Cartesian Product Distributes over Union" and "Cartesian Product Distributes over Set Difference"