Proof: Just observe the following sequence of equalities. \begin{align*} A \times (B - C)&=\{(x,y):(x\in A)\land(y\in(B-C))\} \textbf{(def. of $\times$)} \\ &=\{(x,y):(x\in A)\land (y\in B)\land(y\notin C)\} \textbf{(def. of -)} \\ &=\{(x,y):(x\in A)\land(x\in A)\land (y\in B)\land(y\notin C)\} \textbf{(P=$P\land P$)} \\ &=\{(x,y):((x\in A)\land (y\in B))\land((x\in A)\land(y\notin C))\}\textbf{(rearrange)} \\ &=\{(x,y):(x\in A)\land (y\in B)\}\cap\{(x,y):(x\in A)\land(y\notin C)\}\textbf{(def. of $\cap$)} \\ &=\{(x,y):(x\in A)\land (y\in B)\}-\{(x,y):(x\in A)\land(y\notin C)\}\textbf{(Basic properties of $\cap$)} \\ &=(A \times B) - (A \times C)\textbf{(def. of $\times$)} \end{align*}
The last two lines is wrong but I don't know how to arrive at that part.
