Proof of sets operations equivalency having a hypothesis This is the exercice:
Let $A, B$ and $C$ be sets. 
Suppose that $C \subseteq A$. Prove that $A - (B - C) = (A - B) \cup C$.
This is what I do:
$x \in A-(B-C) \Longleftrightarrow x \in A \neg \wedge (x \in B \wedge x \notin C) \Longleftrightarrow x \in A \wedge x \notin B \vee x \in C \Longleftrightarrow (x \in A \wedge x \notin B) \vee x \in C \Longleftrightarrow x \in (A-B) \cup C$
With this I proved that they're the same, but I didn't use the hypothesis that  $C \subseteq A$ so I feel like my proof is missing something. Is my proof wrong? If it's ok, then why did they give me that hypothesis? Can you give me a proof that uses that hypothesis?
Thank you.
 A: $x\in A\setminus(B\setminus C)\leftrightarrow x\in A\wedge x\notin(B\setminus C)$
$\leftrightarrow x\in A\wedge\lnot(x\in B\wedge x\notin C)$
$\leftrightarrow x\in A\wedge(x\notin B\vee x\in C)$
$\leftrightarrow(x\in A\wedge x\notin B)\vee(x\in A\wedge x\in C)$
$\leftrightarrow(x\in A\setminus B)\vee x\in C\leftrightarrow x\in(A\setminus B)\cup C$.
The assumption $C\subseteq A$ comes in last step.
A: Your third expression is written down incorrectly: you need to write $\;x \in A \land (x \not\in B \lor x \in C)\;$, so with parentheses.  This makes the next step incorrect.  So instead, at that point, distribute $\;\land\;$ over $\;\lor\;$.  Then you'll have an opportunity to use the hypothesis.
A: Assume $x\in A-(B-C)$. 
$$\implies x\in A \land \neg x\in(B-C)$$ $$\implies x\in A \land \neg(x\in B \land \neg x\in C)$$ $$ \implies x\in A \land (\neg x\in B \vee x \in C) \qquad De\, Morgan \,Rule$$ $$ \implies (x\in A \land \neg x\in B) \vee (x\in A \land x\in C) \qquad Distribution \, of \, and \, on \, or (1)$$
Let's stop here and examine $C \subseteq A$. This impies to $\forall x, x\in C \implies x\in A  \qquad (2) $
Plug (2) to (1), we have:
$$x\in (A-B) \vee x\in C \qquad$$ $$\implies x \in (A-B) \cup C$$
Remember this only tells you that $(A-(B-C)) \subset (A-B) \cup C $ but you can prove the other direction with the same way.
A: Just the set theory definitions and some logical simplification actually show that we have an equivalence:
\begin{align}
& A - (B - C) = (A - B) \cup C \\
\equiv & \;\;\;\;\;\text{"set extensionality; expand definitions of $\;-\;$ and $\;\cup\;$"} \\
& \langle \forall x :: x \in A \land \lnot (x \in B \land \lnot x \in C) \;\equiv\; (x \in A \land \lnot x \in B) \lor x \in C \rangle \\
\equiv & \;\;\;\;\;\text{"DeMorgan on left hand side; distribute on right hand side"} \\
& \langle \forall x :: x \in A \land (\lnot x \in B \lor x \in C) \;\equiv\; (x \in A \lor x \in C) \land (\lnot x \in B \lor x \in C) \rangle \\
\equiv & \;\;\;\;\;\text{"extract common conjunct"} \\
& \langle \forall x :: \lnot x \in B \lor x \in C \;\Rightarrow\; (x \in A  \equiv x \in A \lor x \in C) \rangle \\
\equiv & \;\;\;\;\;\text{"simplify consequent"} \\
& \langle \forall x :: \lnot x \in B \lor x \in C \;\Rightarrow\; (x \in C \Rightarrow  x \in A) \rangle \\
\equiv & \;\;\;\;\;\text{"take both antecedents together"} \\
& \langle \forall x :: (\lnot x \in B \lor x \in C) \land x \in C \;\Rightarrow\;  x \in A \rangle \\
\equiv & \;\;\;\;\;\text{"simplify antecedent by using $\;x \in C\;$ on other side of $\;\land\;$"} \\
& \langle \forall x :: x \in C \Rightarrow  x \in A \rangle \\
\equiv & \;\;\;\;\;\text{"reintroduce $\;\subseteq\;$ using its definition"} \\
& C \subseteq A \\
\end{align}
