Prove that $A-(B-C)=(A-B) \cup (A\cap C)$ I am trying to prove that
$A-(B-C)=(A-B) \cup (A\cap C)$
For the first statement, I do:
Let $x \in A-(B-C)$ means that $x \in A$ and $x \notin (B-C)$
$x \notin (B-C)$ means that $x \notin B$ or $x \in C$ $(*)$
Therefore $x \in A-(B-C)$ means that $x \in A$ AND ($x \notin B$ or $x \in C$)
For the second statement, I do: Let $x \in (A-B)$ means that $x \in A$ and $x \notin B$
$x \in (A \cap C)$ means that $x \in A$ and $x \in C$
Therefore, $x \in (A-B) \cup (A\cap C)$ means that $x \in A$ AND ($x \notin B$ or $x \in C$) $(*)$
Those two are the same so these sets are equal. I'm not sure if I did the $(*)$ parts correctly. Can someone give me some guidance?
 A: Everything you said makes sense. You do manage to prove the result, though your proof could be greatly clarified by the usage of some logical notation. We use $\land$ to denote "and" and $\lor$ to denote "or". Also we use $\iff$ to denote "if and only if", i.e. $P \iff Q$ means $P$ if and only if $Q$ is true, or in other words $P$ and $Q$ are both true or both false. Finally we use $\neg$ to denote negation. You can read anywhere about propositional logic to learn about the rigorous manipulation of these symbols.
Then we can realize that from the definitions, we have $x \in A \cup B \iff (x \in A \lor x \in B)$, $x \in A \cap B  \iff (x \in A \land x \in B)$, $x \in A - B \iff (x \in A \land x \not \in B)$. Then your proof is that:
$$ x\in A - (B - C) \iff x \in A \land \neg (x \in B - C) \iff x \in A \land \neg (x \in B \land x \not \in C)\\
\iff x \in A \land (x \not \in B \lor x \in C).$$
That was the first argument. Then second reads:
$$x \in (A - B) \cup (A \cap C) \iff x \in (A - B) \lor x \in A \cap C \iff (x \in A \land x \not \in B) \lor (x \in A \land x \in C)\\
\iff x \in A \land (x \not \in B \lor x \in C).$$
Putting it together gives $x \in A - (B -C) \iff x \in (A -B) \cup (A \cap C)$, which then gives that the sets are equal (by the Axiom of Extensionality).
A: I wouldn't ever try to restart a proof in the middle as you have.  Rather than where you write "For the second statement ...", I would continue...
By distributivity, $x \in A$ and ($x \not\in B$ or $x \in C$) is equivalent to ($x \in A$ and $x \not \in B$) or ($x \in A$ and $x \in C$).  Rewriting the parenthesized subexpressions in set notation, $x \in A \smallsetminus B$ or $x \in A \cap C$.  So $x \in (A \smallsetminus B) \cup (A \cap C)$.
A: You actually proved that
$$
A-(B-C)\subseteq A\cap(B^c\cup C)
$$
and you're a bit too fast in stating to have proved that
$$
(A-B)\cup(A\cap C)\subseteq A\cap(B^c\cup C)
$$
(where $B^c$ means the complement with respect to some set that contains $A$, $B$ and $C$).
So, no, you haven't really proved the statement, because you still need to prove the reverse inclusions.
Your “means” should be “implies”.
Of course the reverse inequalities are easy. Why not using some algebra of sets?
If $U=A\cup B\cup C$ and we set $X^c=U-X$, where $X$ is any subset of $U$, you can see that $X-Y=X\cap Y^c$, for all subsets $X,Y\subseteq U$. Then
\begin{align}
A-(B-C)
&=A\cap(B\cap C^c)^c       &&\text{by $X-Y=X\cap Y^c$} \\
&=A\cap(B^c\cup C)         &&\text{De Morgan} \\
&=(A\cap B^c)\cup(A\cap C) &&\text{distributivity} \\
&=(A-B)\cup(A\cap C)       &&\text{by $X\cap Y^c=X-Y$}
\end{align}
