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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.

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    This seems to be the same question as [this one](http://math.stackexchange.com/questions/67861/show-that-a-setminus-b-setminus-c-equiv-a-setminus-b-cup-a-cap-c).2012-07-18

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A formal set-theoretic argument would be to show that for all $a$, $a\in X\setminus(Y\setminus Z)\iff a\in (X\setminus Y)\cup (X\cap Z).$

A common way of doing this is by "double inclusion": show that if $a$ is in the left hand side, then it is in the right hand side (proving that the left hand side is contained in the right hand side), and then proving that if $a$ is in the right hand side, then it is in the left hand side (proving that the right hand side is contained in the left hand side) (hence proving two inclusions which together give equality).

I'll show you one, you do the other:

Suppose $a\in X\setminus(Y\setminus Z)$. Then $a\in X$ and $a\notin (Y\setminus Z)$. Since $a\notin Y\setminus Z$, then either $a\notin Y$, or $a\in Y$ and $a\in Z$; Thus, either $a\in X$ and $a\notin Y$, in which case $a\in X\setminus Y\subseteq (X\setminus Y)\cup (X\cap Z)$, or else $a\in X$ and $a\in Z$, hence $a\in X\cap Z\subseteq (X\setminus Y)\cup (X\cap Z)$. Thus, if $a\in X\setminus (Y\setminus Z)$, then $a\in (X\setminus Y)\cup (X\cap Z)$. That is, $X\setminus(Y\setminus Z) \subseteq (X\setminus Y)\cup (X\cap Z).$

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    I have started the first line of each separate case with the if and only if operator (==>) and started each case with a suppose statement. is this now correct? thank you for your patience.2012-05-01