So, once again, I'm back with a proof that I've been trying to write up. The problem is as follows:
Let $A$ and $B$ be sets. Then, $A-(B-A) = A$, where $A-B$ denotes the relative complement of $B$ in $A$.
Proof:
Let $A$ and $B$ be sets. We also introduce the set $U$ such that $A \subset U$ and $B \subset U$. Let $x$ be an object in $U$. Then:
$x \in A – (B – A) \iff (x \in A) \land \lnot{(x \in B-A)}$
$x \in A – (B – A) \iff (x \in A) \land \lnot{[(x \in B) \land \lnot{(x \in A)}]}$
$x \in A – (B – A) \iff (x \in A) \land [\lnot{(x \in B)} \lor (x \in A)]$
$x \in A – (B – A) \iff [(x \in A) \land \lnot{(x \in B)}] \lor (x \in A)$
$x \in A -(B – A) \iff x \in A$
Now, the claim that I'm making is that there is an equivalence between the last line and the 2nd last line. To prove this, suppose there wasn't. Hence, one can be false and the other can be true.
Let $x \in A$ be false and that massive proposition in the 2nd last line (we'll just call it P) be true. Then, $(x \in A) \land \lnot{(x \in B)}$ must be true. However, that's not possible as that would have to mean that $x \in A$ is true. That's a contradiction, so $x \in A$ cannot be false and P cannot be true.
Let $x \in A$ be true and P be false. However, that's a problem because $x \in A$ being true is enough to allow for P to be true, by the definition of the disjunction. Hence, this is a contradiction so $x \in A$ cannot be false and P cannot be true.
Thus, they both have to be true at the same time or they both have to be false. That means that they are equivalent and by the Axiom of Extension, that last line proves that $A – (B – A) = A$ holds.
Yea, I'd liked to know if I've messed up anywhere in my approach. If you can provide a shorter proof, that would be great but please also let me know if my argument above makes sense.
Best Answer
I think your proof is correct, albeit a bit overcomplicated.
My suggestion of proof: \begin{eqnarray*} x \in (A - (B - A)) &\implies& (x \in A) \land (x \notin (B -A)) \\ &\implies& x \in A. \end{eqnarray*} Since every element in the first set must exist in the second set, we can conclude that $(A - (B - A)) \subseteq A$. On the other hand, if $x \in A$, then $x$ cannot be an element in any set subtracted with $A$. In other words, $x \notin (B - A)$. We therefore have that $$ \begin{eqnarray} x \in A &\implies& (x \in A) \land (x \notin (B - A)) \\ &\implies& x \in (A - (B - A)). \end{eqnarray} $$ By the same argument as earlier, we have that $A \subseteq (A - (B - A))$. Since we now have two sets that are subsets of each other, they have to be equal, i.e. $A = (A - (B - A))$.