I have question about sets. I need to prove that: $X \cap (Y - Z) = (X \cap Y) - (X \cap Z)$.

now, I tried to prove that from both sides of the equation but had no luck.

for example, I tried to do something like this: $X \cap (Y - Z) = X \cap (Y \cap Z')$ but now i don't know how to continue.

from the other side of the equation I tried to do something like that: $(X \cap Y) - (X \cap Z) = (X \cap Y) \cap (X \cap Z)' = (X \cap Y) \cap (X' \cup Z')$ and from here I don't know what to do again.

I will be glad to hear how should I continue from here and what I did wrong. Thanks in advance.

-
Assume $x$ belongs to the left, show it belongs to the right. And vice versa. The advantage of doing sets as compared to an abstract Boolean algebra. –  GEdgar Mar 28 '12 at 20:46
Hint: Set $Y_1 = Y \cap Z$ and $Y_2 = Y \cap Z'$, so that $Y = Y_1 \cup Y_2$, and then prove $X\cap ((Y_1 \cup Y_2) - Z) = (X \cap (Y_1 \cup Y_2)) - (X \cap Z)$. –  dtldarek Mar 28 '12 at 21:02

You can do it two ways: with manipulations using the properties of unions, intersections, and complements, or through double inclusion.

1. To prove it by double inclusion, we must show that $X\cap(Y-Z)\subseteq (X\cap Y)-(X\cap Z)$, and that $(X\cap Y)-(X\cap Z)\subseteq X\cap(Y-Z)$.

I'll show you one of the inclusions: let $a\in X\cap(Y-Z)$. Then $a\in X$, and $a\in Y-Z$. Hence $a\in X$, $x\in Y$, and $x\notin Z$. Since $a\in X$ and $a\in Y$, then $a\in X\cap Y$. Since $a\in X$ and $a\notin Z$, then $a\notin X\cap Z$. Since $a\in X\cap Y$ and $x\notin X\cap Z$, then $a\in (X\cap Y)-(X\cap Z)$. This proves that if $a\in X\cap(Y-Z)$, then $a\in (X\cap Y)-(X\cap Z)$; that is, $X\cap(Y-Z)\subseteq (X\cap Y)-(X\cap Z)$.

Now show that $(X\cap Y)-(X\cap Z)\subseteq X\cap(Y-Z)$.

2. Using the properties, you should use the fact that intersections distribute over unions and vice-versa. So from \begin{align*} (X\cap Y)-(X\cap Z) &= (X\cap Y)\cap(X\cap Z)'\\ &= (X\cap Y)\cap(X'\cup Z')\\ &= (X\cap Y\cap X') \cup (X\cap Y\cap Z'). \end{align*} Can you take it from there?

-
yeah, thank you (and thanks to all other people who answered). it's more simple than I thought. –  UG7JY Mar 28 '12 at 21:12

Hint

• $A \cap (B \cup C)=(A \cap B )\cup(A \cap C)$

• $A \cap A'= \varnothing$

-

In general, to show equality of sets $A = B$ you show $A \subset B$ and $B \subset A$.

To show $X \cap (Y - Z) \subset (X \cap Y) - (X \cap Z)$ assume $a \in X \cap (Y -Z)$. Then $a \in X$ and $a \in Y - Z \subset Y$. Hence $a \in X \cap Y$. Also, $a \notin Z$ hence $a \notin X \cap Z$ and hence $x \in (X \cap Y) - (X \cap Z)$.

Now try to do the other direction.

-

Here is a third approach, which first translates to the element level, and then uses logic manipulation.

Starting at the most complex side, we have for any $\;a\;$ \begin{align} & a \in (X \cap Y) - (X \cap Z) \\ \equiv & \;\;\;\;\;\text{"expand definitions of $\;-\;$ and of $\;\cap\;$ (twice)"} \\ & a \in X \land a \in Y \;\land\; \lnot (a \in X \land a \in Z) \\ \equiv & \;\;\;\;\;\text{"logic: use $\;a \in X\;$ on other side of $\;\land\;$"} \\ & a \in X \land a \in Y \;\land\; \lnot (\text{true} \land a \in Z) \\ \equiv & \;\;\;\;\;\text{"logic: simplify"} \\ & a \in X \;\land\; a \in Y \land \lnot(a \in Z) \\ \equiv & \;\;\;\;\;\text{"reintroduce $\;-\;$ and $\;\cap\;$ using their definitions"} \\ & a \in X \cap (Y - Z) \\ \end{align} Using set extensionality, this proves the statement in question.

-