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Provide either a proof or a counterexample.

I haven't been able to find an example where these two statements are not equivalent, as trying to equate any of A, B or C to the Empty Set doesn't work, and I've also tried setting them to shared variables and separate variables. So I'm inclined to believe that this is true, however, I'm not sure how to definitively prove it. If I say, for example, "Let $x∈A, y∈B, z∈C$," that doesn't cover every possibility, such as $x∈A, x∈B$, etc. If anyone could point out even the first step to creating a proof that definitively covers these options, I'd greatly appreciate it.

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    If $x\in A\times (B\setminus C)$ then there are $a,b$ such that $x=(a,b)$ and $a\in A$, $b\in B\setminus C$ – 2017-02-24
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    Ah, I see, and by the same logic you can say $y = (a,b)$ for the right hand side, regardless of $C$ in both cases. Thank you! If you make this an answer I'll accept it. – 2017-02-24

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If $(x,y)\in A\times(B\setminus C)$ then $x\in A$ and $y\in(B\setminus C)$. Hence, in particular $y\in B$ so $(x,y)\in A\times B$ and $y\not\in A\times C$. Summing-up $(x,y)\in A\times B\setminus A\times C$. We conclude: $A\times(B\setminus C)\subseteq A\times B\setminus A\times C$.

For the opposite inclusion. Assume $(x,y)\in A\times B\setminus A\times C$. Then, of course, $x\in A$ and $y\in B$, but since $(x,y)\not\in A\times C$, then $y\in B\setminus C$. Hence, $(x,y)\in A\times(B\setminus C)$.

Remark that the statement is false if $B=C$ or $B\subset C$. As a counter-example in this case choose $A=\{x\}$ and $B=C=\{y\}$.

Edit: After discussion with @J. Davis Taylor, I get convinced that in fact, there is no counter-example. Plz, look at the discussion below.

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    In your "counterexample", both sets are the empty set, thus equal. – 2017-02-24
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    @J.DavidTaylor I consider $A\times\emptyset$ not empty and $\{(x,y)\}\setminus\{(x,y)\}$ really empty. – 2017-02-24
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    Would you name an element of the set $A\times\varnothing$? Where $A=\{x\}$, as in your example. – 2017-02-24
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    I would say $(x)$ ie. there is no pair... but maybe it is matter of conventions? – 2017-02-24
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    Two constructions of the product I know, Choice functions $\{A,\varnothing\}\rightarrow A\cup\varnothing$ Ordered Pairs, $(a,b)$, where $a\in A,b\in\varnothing$ An alternative Defintion: the product in the category of sets, i.e. set functions $Z\rightarrow A\times\varnothing$ are in natural bijection with pairs of set functions, $Z\rightarrow A$ and $Z\rightarrow\varnothing$. The first construction is empty because the only set function into $\varnothing$ is the empty function, $\varnothing\rightarrow\varnothing$. The second, because $\varnothing$ has no elements, – 2017-02-24
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    and the last, because $\varnothing$ has no elements, the only set function to the product is the unique one from $\varnothing$. If the product had at least one element, $x$, then there would be a map $\{x\}\rightarrow A\times\varnothing$. By the categorical characterization, that would mean there is a function $\{x\}\rightarrow\varnothing$. The problem, is that this function cannot take a value at $x$. – 2017-02-24
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    @Maczinga: All elements of the Cartesian product of two sets are pairs. $A\times\emptyset$ is the Cartesian product of two sets. $(x)$ is not a pair. Therefore $(x)\notin A\times\emptyset$. – 2017-02-24