Use class algebra to prove the following: If $A\cap B = \emptyset$ and $A\cup B = C$, then $A = C-B$ I'm having a bit of trouble proving the following.
If $A\cap B = \emptyset$ and $A\cup B = C$, then $A = C-B$
My initial attempt is to prove it directly, however, I believe I'm assuming the consequent, namely, $A = C-B$, and I'm unsure how to close the proof.
Attempted proof
Suppose that $A = C-B$. We know that $C = A\cup B$ from our premises, therefore:
$A = (A\cup B)-B$, which is $A = (A\cup B) \cap B'$
Now distribute: $A =(A\cap B') \cup (B\cap B')$, but $B \cap B'= \emptyset$, so we have $A = (A\cap B') \cup \emptyset$
Thus, $A = A\cap B'$
Here is where I am at a loss. I see that if some element x was an element of both $A$ and $B$, we would have a disjoint set according to our premises.
Any tips are appreciated and I apologize in advance for the formatting.
 A: Let $A\cap B = \emptyset,\ A\cup B = C$, we are interested in showing $A=C-B$.
This is equivalent to showing $A\subseteq C - B$ and $C-B\subseteq A$.  
Let $x\in A$. Since $A\cup B = C$, $x\in C$.
Similarly, $A\cap B = \emptyset\implies x\not\in B$, taking these two points together, $$x\in A\implies \{x\in C\}\cap\{x\not\in B\} \implies x\in C-B$$
Let $x\in C-B$. Since $A\cup B = C$, then $\{x\in A\}\cup\{x\in B\}$, but we know that $x\not\in B$, so $x\in A$.
We have shown that these two sets are equivalent.
A: First we prove that $A\subset C-B$. 
If $a\in A$, then $a\in C$ (because $A\subset C$ as $C=A\cup B$). And $a\notin B$ because $a\in A$ and $A\cap B=\emptyset$. So $a\in C-B$. Thus $A\subset C-B$.
Not let $a\in C-B$. $a\in C=A\cup B$, so $a\in A$ or $a\in B$. But $a\notin B$. Hence $a\in A$ and $C-B\subset A$.
A: The others showed how to solve the problem already. I'll just add that in your proof formatting, you are in fact assuming the result at the start when you write: "Suppose that $A = C-B$ ".
What you wrote was all correct for the set $C-B$, and you showed that it is equal to the set $A\cap B^{c}$, but at this point you still need to show that $C-B = A$, or equivalently that $A\cap B^{c} = A$. This can be done the same way as the other answers showed, by showing that $A\subset A\cap B^{c}$ and $A\cap B^{c} \subset A$.
A quick way to argue this would be by showing that $A\cap B = \emptyset$ implies that $A\subset B^{c}$.
