Proving $A \setminus (A \Delta B) = A \cap B$ I am trying to understand a proof that $A \setminus (A \Delta B)$, where we define $X \Delta Y = (X - Y) = (Y - X)$. I have already proved, and am taking for granted, the following two facts:
\begin{align*}
A \setminus (B \cup C) & = (A \setminus B) \setminus C \\ 
(A \cup B) \setminus C & = (A \setminus C) \cup (B \setminus C). 
\end{align*}
The proof I am trying to follow is:

\begin{align*}
A \setminus (A \Delta B) & = (A \setminus (B \setminus A)) \setminus (A \setminus B) \\
& = A \setminus (A \setminus B) \\
& = A \cap B
\end{align*}

The first line is fine. It uses the definition, commutativity of union, and the first fact above:
$$
A \setminus (A \Delta B) = A \setminus ((A - B) \cup (B - A)) = A \setminus ((B - A) \cup (A - B)) = (A - (B - A)) - (A - B).
$$
I don't understand where the second line comes from. It seems to assert that $A \setminus (B \setminus A) = A$. However, I've proved that $A \setminus (B \setminus A) = A \setminus B$:
\begin{align*}
x \in A \setminus (B \setminus A) & \iff x \in A \wedge x \not \in (B \setminus A) \\
& \iff x \in A \wedge \left(x \not \in B \vee x \in A \right) \\
& \iff (x \in A \wedge x \not \in B) \vee (x \in A \wedge x \in A) \\
& \iff x \in A \setminus B 
\end{align*}
This can't be true, because if I substitute this, I get $(A \setminus B) \setminus (A \setminus B)$, which is $\emptyset$.
What am I doing wrong?
 A: Here it is another way to approach it for the sake of curiosity:
\begin{align*}
A - (A\Delta B) & = A - ((A - B)\cup(B - A))\\\\
& = A - ((A\cap B^{c})\cup (B\cap A^{c})\\\\
& = A \cap (A\cap B^{c})^{c}\cap(B\cap A^{c})^{c}\\\\
& = A \cap (A^{c}\cup B)\cap(B^{c}\cup A)\\\\
& = ((A\cap A^{c})\cup(A\cap B))\cap(B^{c}\cup A)\\\\
& = (A\cap B)\cap(B^{c}\cup A)\\\\
& = (A\cap B\cap B^{c})\cup(A\cap B\cap A)\\\\
& = \varnothing\cup(A\cap B)\\\\
& = A\cap B
\end{align*}
Hopefully this helps!
A: For your specific question about line 2- i.e, that $A\setminus (B \setminus A) = A$, intuitively $B \setminus A$ is all the stuff in $B$ not in $A$. So in other words $B \setminus A$, call it $C$, contains nothing in $A$. So $A \setminus C$ removes stuff from $A$ that was not in $A$ to begin with, i.e, it removes nothing from $A$.
Or more formally,
$A \setminus (B \setminus A) = A \cap (B \setminus A)^c = A \cap (B \cap A^c)^c = A \cap (B^c \cup A) = A$
A: The very last line of your derivation $$(x \in A \wedge x \not \in B) \vee (x \in A \wedge x \in A) \iff x \in A \setminus B $$ is wrong. The condition $x\in A\setminus B$ is equivalent to the first clause $(x \in A \wedge x \not \in B)$ of the disjunction on the left, but you seem to have forgotten about the second clause $(x \in A \wedge x \in A)$.  That second clause is equivalent to $x\in A$ and so the left side is true whenever $x\in A$, regardless of whether $x\in B$, so the entire disjunction is equivalent to $x\in A$ rather than $x\in A\setminus B$.
