Need help with set equivalence I have two problems related to sets.
how can I prove the following two equivalences:


*

*$$A \cap (B - C) = (A \cap B) - (A \cap C)$$


*$$(A - B) \cup (B - A) = (A \cup B) - (A \cap B)$$

I know that they are true, I'm just not that experienced with set-theoretic equalities yet  to be able to prove these.
Thank you all very much for the quick help!
 A: $x\in A\cap (B\setminus C)\Leftrightarrow x\in A, x\in B, x\notin C\Leftrightarrow x\in (A\cap B)$ but $x\notin A\cap C\Leftrightarrow x\in (A\cap B)\setminus (A\cap C)$.
The second one can similarly be shown.
A: You can prove $X=Y$ by proving $X\subseteq Y$ and $Y\subseteq X$.
For example, in the first:
Suppose $x\in A\cap(B\setminus C)$. Then $x\in A$ and $x\in B$, and so $x\in A\cap B$. Also, since $x\in B\setminus C$, $x\not\in C$ and thus, $x\not\in A\cap C$. Hence, $x\in (A\cap B) \setminus (A \cap C)$. Since $x$ was arbitrary, this proves $A\cap(B\setminus C)\subseteq (A\cap B) \setminus (A \cap C)$. Similarly, we can prove $(A\cap B) \setminus (A \cap C)\subseteq A\cap(B\setminus C)$ and then we have established $A\cap(B\setminus C)=(A\cap B) \setminus (A \cap C)$.
The second one is proved in a similar way.
A: Hint: Draw Venn diagrams of these equalities. Apply De Morgans laws.
A: For problem 1.
$A \cap A = A$ so
$$A \cap (B - C)$$
becomes
$$(A \cap A) \cap (B - C)$$
and you can expand difference into intersections with negation. That is $(B - C)$ becomes$(B \cap  \sim C)$
$$(A \cap A) \cap (B - C)$$
$$(A \cap A) \cap (B \cap \sim C)$$
now you just have a bunch of $\cap$ operations. You can rearrange the terms as needed.
$$A \cap (B - C) = (A \cap B) \cap (A \cup \sim C)$$
and you can turn the second term back into a difference.
$$A \cap (B - C) = (A \cap B) \cap (A  - C)$$
For Problem 2:
This one is a bit more complicated
$$(A - B) \cup (B - A) = (A \cup B) - (A \cap B)$$
lets work from the right
$$(A \cup B) - (A \cap B)$$
$$(A \cup B) \cap  \sim(A \cap B)$$
you can distribute the $\sim$ inside the preference by applying it to both sets and inverting the operation that is $\cap$ to $\cup$ or $\cup$ to $\cap$.
$$(A \cup B) \cap  \sim(A \cap B)$$
$$(A \cup B) \cap  (\sim A \cup \sim B)$$
Now you basically have to do the set version of FOIL and cancel terms
$$(A \cup B) \cap  (\sim A \cup \sim B)$$
$$(\sim A \cap (A \cup B)) \cup  (\sim B \cap (A \cup B))$$
and distribute
$$((\sim A \cap A) \cup (\sim A \cap B)) \cup  ((\sim B \cap A)\cup(\sim b \cap B))$$
now you can reduce $\sim A \cap A$ to $\varnothing$
$$(\varnothing \cup (\sim A \cap B)) \cup  ((\sim B \cap A)\cup \varnothing)$$
and $A \cup \varnothing = A$
$$(\sim A \cap B) \cup  (\sim B \cap A)$$
rearrange
$$(B \cap \sim A) \cup  (A \cap \sim B)$$
convert back to -
$$(B - A) \cup  (A - B)$$
rearrange once more
$$(A - B) \cup (B - A)$$
A: Number 1 I already answered elsewhere (see here).
For number 2 you can use the same method.  Let me give you the first part of the calculation: for any $\;x\;$,
\begin{align}
& x \in (A - B) \cup (B - A) \\
\equiv & \;\;\;\;\;\text{"expand definitions of $\;\cup\;$ and $\;\setminus\;$"} \\
& (x \in A \land x \not\in B) \lor (x \in B \land x \not\in A) \\
\equiv & \;\;\;\;\;\text{"distribute $\;\lor\;$ over $\;\land\;$, three times"} \\
& (x \in A \lor x \in B) \land (x \in A \lor x \not\in A) \land (x \not\in B \lor x \in B) \land (x \not\in B \lor x \not\in A) \\
\equiv & \;\;\;\;\;\text{"simplify using 'excluded middle'"} \\
& \dots \\
\end{align}
