True or False $A - C = B - C $ if and only if $A \cup C = B \cup C$ True or False
$A - C = B - C $ if and only if $A \cup C = B \cup C$ 
I am in an introduction to proofs class.  I think this is a true statement. I have began the proof and realize I have to do this both directions.  I am stuck on the second part of the proof.
Assume $A - C = B - C$
Let $x \in A$ then $x \in A$ but $x\notin C$
By the assumption $x \in B - C$ then $x \in B$ and $x\notin C$ 
If $x \in A$ then $x \in A \cup C$.  If $x\in B$ then $x \in B \cup C$. Therefore,  $A \cup C = B\cup C$. 
Assume $A \cup C = B \cup C$.
Let $x \in A \cup C$. Then $x \in A$ or $x\in C$.
If $x \in A$ then $x \in A - C$
If $x\notin A$ then $x \in C$. 
This is where I get stuck... I know I am trying to show $A-C = B-C$ for the second part but I don't know how to get there... Am I on the right track?
 A: Hint:You can use $A\cup B=(A-B)\cup B$ 
A: Assuming $A - C = B - C$, I prove $A \cup C \subset B \cup C$; The converse is similar.  
$x\in  A \cup C$ then $x\in A$ or $ x\in C$. if $x\in C$, we are done. if $x\in A- C$, then $x\in A - C = B - C$ so $x\in B$ 

Assuming $A \cup C = B \cup C$ we can prove $A - C \subset B - C$ (and the converse is similar):
if $x\in A - C$ then $x\in A \cup C = B \cup C$ and $x\notin C$. so $x\in B$ and $x\notin C$. therefore $x\in B - C$
A: Your first proof looks a little tangled; it looks like you're splitting into cases or something, but the cases aren't laid out clearly enough for me to see what's happening or to convince me that it's sufficient. I suspect you have the general idea of the proof, but it's not quite communicated.
To lay it out well, consider that your goal is to, given $A-C=B-C$, prove that $A\cup C=B\cup C$. In particular, this means for any $x$, you need to show it is a member of $A\cup C$ if and only if it is in $B\cup C$. To do this, we can split into two cases:


*

*Case 1: $x\in C$. In this case, $x\in A\cup C$ and $x\in B\cup C$.

*Case 2: $x\not\in C$. In this case $x\in A\cup C$ if and only if $x\in A$. This is further equivalent to $x\in A-C$, as $x$ isn't in $C$. Since $A-C=B-C$ we get $x\in B-C$ which is equivalent to $x\in B$ and $x\in B\cup C$. Thus $x\in A\cup C$ if and only if $x\in B\cup C$.


This suffices to show that $A\cup C=B\cup C$ follows from $A-C=B-C$, as both cases result in $x\in A\cup C$ if and only if $x\in B\cup C$.
To set up the other direction of the proof, you can use the same two cases; if $x\in C$ then clearly $x$ is in neither $A-C$ nor $B-C$. If $x\not\in C$, then you can proceed analogously as the second case above, noting that, for such an $x$, the unions and differences involving $C$ won't affect whether $x$ is in the set.
A: As an alternative way to prove this, one can solve this on the 'logic' or element level, using the fact that $\;\lor\;$ distributes over $\;\equiv\;$.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\followsfrom}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$We calculate as follows:
$$\calc
  A - C \;=\; B - C
\op\equiv\hint{definition of $\;=\;$ (i.e., set extensionality); definition of $\;-\;$, twice}
  \langle \forall x :: x \in A \land x \not\in C \;\equiv\; x \in B \land x \not\in C \rangle
\op\equiv\hint{negate both sides of $\;\equiv\;$; DeMorgan -- to introduce $\;\lor\;$}
  \langle \forall x :: x \not\in A \lor x \in C \;\equiv\; x \not\in B \lor x \in C \rangle
\op\equiv\hint{$\;\lor\;$ distributes over $\;\equiv\;$ $\ref 0$}
  \langle \forall x :: (x \not\in A \;\equiv\; x \not\in B) \lor x \in C \rangle
\op\equiv\hint{negate both sides of $\;\equiv\;$}
  \langle \forall x :: (x \in A \;\equiv\; x \in B) \lor x \in C \rangle
\op\equiv\hint{$\;\lor\;$ distributes over $\;\equiv\;$ $\ref 0$}
  \langle \forall x :: x \in A \lor x \in C \;\equiv\; x \in B \lor x \in C \rangle
\op\equiv\hint{definitions of $\;\cup,=\;$}
  A \cup C \;=\; B \cup C
\endcalc$$
The proof becomes still shorter if you're not only allowed to use
$$
\tag 0
P \lor (Q \equiv R) \;\equiv\; (P \lor Q) \equiv (P \lor R)
$$
but also the dual
$$
\tag 1
P \then (Q \equiv R) \;\equiv\; (P \land Q) \equiv (P \land R)
$$
