Here is my question:
Give a formal set theoretic argument to prove that, for all sets $X, Y$ and $Z$, $X\setminus(Y\setminus Z)=(X\setminus Y)\cup(X\cap Z)\;.$
I know this is true, using the identity found here, so I'm unsure how to answer the question as it appears to be an identity in itself?!
How do you think I should answer it?
EDIT: This is my final answer (thank you for your help). Is it correct?
//My interpretation of the half of your answer $\begin{align*} &\text{Suppose }X\setminus(Y\setminus Z)\\ &\implies a\in X\setminus(Y\setminus Z)\\ &\implies a\in X\land a\notin (Y\setminus Z) &\quad\text{[Definition of complement]}\\ &\implies (a\in X\land a\notin Y)\lor (a\in X\land (a\in Y\land a\in Z)) &\quad\text{[distributive law]}\\ &\implies X\setminus Y\cup (X\cap Y\cap Z) &\quad\text{[associative law]}\\ &\implies X\setminus Y\cup (X\cap Y\cap Z)\subseteq X\setminus Y\cup (Y\cap Z) &\quad\text{[}A\cap B\subseteq A\text{]} \end{align*}$
//My half of the answer
$\begin{align*} &\text{Suppose }X\setminus Y\cup (X\cap Z)\\ &\implies a\in (X\setminus Y\cup (X\cap Z)) \\ &\implies a\in (X\setminus Y)\lor a\in(X\cap Z) &\text{[definition of union]}\\ &\implies (a\in X\land a\notin Y)\lor (a\in X\land a\in Z)&\text{[def of complement and intersection]}\\ &\implies a\in X\land (a\notin Y\lor a\in Z) &\text{[Distributive Law]}\\ &\implies a\in X\land (T\land (A\notin Y\lor a\in Z))&\text{[Identity]}\\ &\implies a\in X\land ((a\notin Y\lor a\in Y)\land (a\notin Y\lor a\in Z)) &\text{[Excluded middle]}\\ &\implies a\in X\land (a\notin Y\lor(a\in Y\land a\in Z)) &\text{[Distributive law]}\\ &\implies a\in X\land a\notin(Y\setminus Z) &\text{[Definition of complement]}\\ &\implies a\in X\setminus(Y\setminus Z)&\text{[Definition of complement]}\\ &\implies (X\setminus Y)\cup (X\cap Z) \subseteq X\setminus (Y\setminus Z) \end{align*}$
Since A ⊆ B, and B ⊆ A, sets A and B are equal.